Plus Two

Students can Download Chapter 1 The Solid State Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 1 The Solid State

Plus Two Chemistry The Solid State One Mark Questions and Answers

Question 1.
The 14 possible three dimensional crystal lattices are called ____________.
Answer:
Bravais Lattices

Question 2.
Which of the following type of cubic lattices has maximum number of atoms per unit cell?
(a) simple cubic
(b) body centred cubic
(c) face centred cubic
(d) all have same
Answer:
(c) face centred cubic (4 atoms per unit cell)

Question 3.
F- centres in an ionic crystal are
(a) lattice sites containing electrones
(b) interstitial sites containing electrons
(c) lattice cites that are vacant
(d) interstitial sites containing electrones
Answer:
(a) lattice sites containing electrons

Question 4.
The colour imparted by excess potassium in KCl crystal is ________.
Answer:
Violet (or lilac)

Question 5.
Which of the following substances show anti-ferromagnetism?
(a) ZrO₂
(b) CdO
(c) CrO₂
(d) Mn₂O₃
Answer:
(d) Mn₂O₃

Question 6.
The number of tetrahedral and octahedral voids in a ccp array of 100 atoms are respectively.
Answer:
200 and 100

Question 7.
Potassium dichromate belongs to which crystal system.
Answer:
Triclinic

Question 8.
A solid compound contains XYZ atoms in a cubic lattice with X atoms occupying the corners, Y atoms in the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound Answer:
XYZ₃

Question 9.
The empty space in body centred cubic lattice is
Answer:
32%

Question 10.
Which solid has the weakest intermolecular force
Answer:
ice

Question 11.
In crystalline solid, atoms, ions or molecules are held in an orderly array. But some point defect is observed in a crystal, when a vacancy is created by an atom or ion dislocated from its normal position to an interstitial site. What is the defect called?
Answer:
Frenkel defect

Plus Two Chemistry The Solid State Two Mark Questions and Answers

Question 1.

1. Write the names of A and B?
2. Identify and write the name of the shaded parts of A and B?

Answer:
1. Names of A and B are:

• Hexagonal close packing.
• Square close packing,

2. shaded parts of A and B are:

• Is octahedral void; in
• Is tetrahedral void.

Question 2.
Teacher said that Frenkel defect will not happen in alkali metal halides. Ramu asked the reason for this. Can you explain?
Answer:
Frenkel defect is favoured by the small size of cations. In alkali metal halides both cations and anions are of almost same size.

Question 3.
“Dielectric substances are related to conductors.”
“Dielectric substances do not conduct electricity at normal condition”
These are two arguments of a class discussion.

1. Do you agree with these arguments?
2. If yes, justify both statements?

Answer:

1. Yes.
2. Dielectric substances do not conduct electricity through them. But they can be made conductors either by heating the substances or by applying mechanical stress.

Question 4.

1. Identify A and B.
2. Explain them.

Answer:
1. A and B.

• A – Square close packed (scp) arrangement in two dimensions.
• B- Hexagonal close packed (hep) arrangement in two dimensions.

2. Explanation:

(a) A – Square close packed:
(scp) arrangement in two dimensions – particles of the second row are arranged just below the first row. Similarly, particles of third row are arranged just below the particles of the second row and so on. Here vertical as well as horizontal alignment is possible. In this arrangement each particle is in direct contact with four other particles and the coordination number is 4.

(b) B- Hexagonal close packed:
(hep) arrangement in two dimensions – In this arrangement, particles of second row are arranged in the depressions of the particles of the first row. Similarly the particles of the third row are arranged in the depressions of the particles of the second row. In this arrangement, each particle is in direct contact with six other particles and the coordination number is 6.

Question 5.

1. Name the unit of magnetic moment.
2. Match the following.
   • Paramagnetic – Fe₃O₄
   • Ferromagnetic – O₂
   • Antiferromagnetic – CrO₂
   • Ferrimagnetic – MnO

Answer:

1. Bohr magneton (μ_B)
2. Match
   • Paramagnetic – O₂
   • Ferromagnetic – CrO₂
   • Antiferromagnetic – MnO
   • Ferrimagnetic – Fe_3O4

Question 6.
All crystals exhibit imperfections.

1. Which law is related to this statement?
2. Draw the picture showing Frenkel defect.

Answer:
1. Third law of thermodynamics.

2.

Question 7.
What do you mean by Anisotropy how is it differ from isotropy
Answer:

Anisotropic	Isotropic
Physical properties have different values along different directions	Physical property would be same in any direction
Because they have different arrangement in different directions	Because arrangement is irregular in any direction

Question 8.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer:
The atom at the corner makes \(\frac{1}{8}\) contribution while atom at body centre makes 1 contribution to the unit cell.
No. of atoms of Q per unit cell = 8 × \(\frac{1}{8}\) = 1
No. of atoms of P per unit cell = 1 × 1 = 1 Therefore, formula of the compound is PQ.
The atom at the body centre would be in contact with all the atoms at the corners. Hence, the coordination number of P is 8.
Similarly, coordination number of Q is 8 because it is shared by 8 other atoms.

Question 9.
Packing efficiency differ for B.C.C, F.C.C and Simple cube what is packing efficiency?
Answer:
Packing efficiency is the % of total space filled by the particle.

OR

Packing efficiency = \(\frac{\text { Volume occupied by spheres in the unit cell }}{\text { Total volume of the unit cell }} \times 100[/latex ]

Question 10.
If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies?
Answer:
Every Sr²⁺ ion causes one cation vacancy (because two Na⁺ ions are replaced by one Sr²⁺ and it occupies the site of one Na⁺ ion and the other site remains vacant.) Therefore, introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10^-3 moles cation vacancies in 100 moles of NaCl.
No. of vacancies per mole of NaCl = [latex]\frac{10^{-3}}{100}\) × 6.02 × 10²³ = 6.02 × 10¹⁸

Question 11.
Even crystal, the substance which we consider as the most perfect solid, shows some defects or imperfections.

1. Which law in thermodynamics deals with this topic?
2. Explain the law.

Answer:

1. Third law of thermodynamics
2. Entropy of a perfectly crystalline substance is zero at absolute zero temperature.

Question 12.
Excess of Li makes LiCl crystal pink and excess of K makes KCl crystals lilac. Is this true? How will you account for the above processes?
Answer:
Yes. This is an example for metal excess defect due to anion vacancies. The colour is due to the presence of F – centre in the crystals.

Question 13.
What are the consequences of Schottky and Frenkel defects?
Answer:
Frenkel defect does not change the density of the crystal but Schottky defect decreases the density of the substance.

Question 14.
Calculate the number of atoms per unit cell of silver which crystallizes in fee lattice.
Answer:

• Contribution of particles from one corner = 1/8
• Contribution from 8 corners = 1/8 × 8 = 1
• Contribution from 6 face centres = 1/2 × 6 = 3
• Total number of particles present in the unit cell of the crystal = 1 + 3 = 4
• Number of atoms present in one unit cell of Ag = 4.

Question 15.
Explain the type of voids found in three-dimensional close packing in crystals.
Answer:
Two voids

Plus Two Chemistry The Solid State Three Mark Questions and Answers

Question 1.
In an answer paper a student wrote as “carborundum crystals are very soft.”

1. Do you agree with this?
2. What is your opinion?
3. In which crystal type carborundum is included?
4. Substantiate your view.

Answer:

1. No.
2. Carborundum crystals are very hard.
3. Covalent crystal.
4. In the case of covalent crystals the constituent particles are atoms and they are held together by strong covalent bond network. In carborundum covalent bond network is constituted by Si and C atom which are held very strongly at their positions. Hence, carborundum is a covalent crystal and is very hard.

Question 2.
During a seminar a student asked another student, “Can NaCl give flame test?”

1. Write your answer.
2. Write the colour of sodium during flame test.
3. Write the name of the point which is responsible for the colour of alkali metal halides having excess metal ions.

Answer:

1. Yes
2. Golden yellow
3. F-centre electron trapped anion vaccancy

Question 3.
Solids are classified into two types.

1. What are they?
2. Give two examples.
3. Give 3 features of them.

Answer:
1. Crystalline and amorphous.

2. Crystalline – NaCl, KCl. Amorphous-Plastic, rubber.

3.

Crystalline Solids	Amorphous Solids
1. Sharp melting point	1. Melting point is in a range of temperature
2. Long range order	2. Short range order
3. Anisotropic	3. Isotropic
4. Definite shape	4. Irregular shape

Question 4.
Teacher explained that due to a stoichiometric defect, the density of a crystal changes.

1. Name the defect.
2. What change can we observe?
3. Give an example.

Answer:

1. Schottky defect
2. Density decrease
3. NaCl

Question 5.
1. Packing efficiency is the percentage of total space filled by the particles. Which of the following lattices has the highest packing efficiency? Simple cubic lattice, body centered cubic lattice, hexagonal close packed lattice.

2. An element has a body centred cubic structure with a cell edge of 288 pm. The density of the element is 7.2 g/cm³. How many atoms are present in 208 g of the element?
Answer:
1. Hexagonal close packed lattice.

2. a = 288 pm = 288 × 10^-10 cm
Volume of the unit cell = (288 × 10^-10cm)³
= 23.9 × 10^-24cm³
Volume of 208 g of the element
\(\frac{208 \mathrm{g}}{7.2 \mathrm{g} \mathrm{cm}^{-3}}\) = 28.88 cm³
∴ Number of units cells in 208 g =
\(\frac{28.88 \mathrm{cm}^{3}}{23.9 \times 10^{-24} \mathrm{cm}^{3}}\) = 1.208 × 10²⁴
For bcc structure, number of atoms in one unit cell = 2
∴ Number of atoms in 208 g = 2 × 1.208 × 10²⁴
= 2.416 × 10²⁴

Question 6.
Derive packing efficiency of

1. ccp and hep structure
2. Body centered cubic
3. Simple cubic

Answer:
1. Packing Efficiency in ccp and hep Structures:
In the case of ccp and hep, the edge length

We know that each unit cell in ccp structure has 4 spheres.
Volume of sphere = \(\frac{4 \pi r^{3}}{3}\)
Volume of the cube = a³

2. Packing Efficiency of Body Centred Cubic Structures:
In this case radius of a sphere.

We know that bcc has 2 spheres in the unit cell.
∴ Packing efficiency = \(\frac{2 \times \frac{4}{3} \pi r^{3}}{\left[\left(\frac{4}{\sqrt{3}} r\right)\right]^{3}}\) × 100% = 68%

3. Packing Efficiency is Simple Cubic Lattice:
In simple cubic lattice edge length ‘a’ and radius of the sphere ‘r’ are related as,

We know that a simple cubic unit cell contains only one sphere.
∴ Packing efficiency = \(\frac{1 \times \frac{4}{3} \pi r^{3}}{(2 r)^{3}}\) × 100%
= 52.36%
= 52.4%

Question 7.
Iron (II) oxide crystallise in cubic structure with unit cell edge of 5.0 Å If the density of the oxide is 3.8 g cm^-3. Calculate the no.of Fe²⁺ and O^2- present in each unit cell, [atomic mass of Fe = 56, O = 16] (3)
Answer:
Cell edge a = 5.0Å = 5.0 × 10^-1 m = 5 × 10^-8 cm
Density = 3.8 g/cm³ Molecular mass of FeO = 56 + 16 = 72 u
N_A = 6.022 × 10²³
d = \(\frac{Z M}{N_{A} a^{3}}\)
Z = \(\frac{d N_{A} a^{3}}{M}=\frac{3.8 \times 6.022 \times 10^{23}}{72} \times\left(5 \times 10^{-8}\right)^{3}\)
= 3.97 = 4
Each cell contain 4 FeO molecule every FeO molecule contain one Fe²⁺ and one O^2- ion. So no. of Fe²⁺ ion =4 no.of O^2- ion = 4

Question 8.
Which are the two types of close packing in two dimension. What are its differences?
Answer:
1. Square close packing or AAA type.

• Coordination no.4. One sphere is in touch with 4 other spheres
• 2^nd raw placed exactly under the first raw

2. Hexagonal close packing or AB AB Type.

• Coordination no.6 ie. one spheres is in touch with other six spheres.
• 2^nd raw is placed in the depressions of first raw.

Question 9.
Classify the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

• Tetraphosphorus decoxide (P₄O₁₀)
• Ammonium phosphate (NH₄PO₄)
• SiC
• I₂
• P₄
• Plastic
• Graphite
• Brass
• Rb
• LiBr
• Si

Answer:

• Tetraphosphorus decoxide (P₄O₁₀) – Molecularsolid
• Ammonium phosphate (NH₄PO₄) – Ionic solid
• SiC – Network(covalent) solid
• I₂ – Molecular solid
• P₄ – Molecular solid
• Plastic-Amorphous solid
• Graphite – Network(covalent) solid
• Brass – Metallic solid
• Rb- Metallic solid
• LiBr – Ionic solid
• Si – Network(covalent) solid

Question 10.
A substance A’ crystallizes in fee lattice.

1. Calculate the number of atoms present per unit cell of ‘A’.
2. In a crystalline solid AB, some vacancy is produced by missing of equal number of oppositely charged ions. What is the defect called?

Answer:

1. The number of atoms present per unit cell of ‘A’.
   • Contribution from 8 corners = 1/8 × 8 = 1
   • Contribution from centres of 6 faces = 1/2 × 6 = 3
   • Total number of atoms of A’ per unit cell = 1 +3 = 4
2. Schottky defect

Question 11.

1. What is meant by the term coordination number?
2. What is the coordination number of atoms:
   • In a cubic close-packed lattice?
   • In a body-centred cubic structure?

Answer:

1. The number of nearest neighbours in a packing is called coordination number.
2. The coordination number of atoms
   • In cubic-close packed structure each atom is in direct contact with 12 other atoms. Hence, its coordination number is 12.
   • In a body centred cubic structure coordination number is 8

Question 12.
Identify the crystal.

1. Write the name of the crystal.
2. How many particles are present the unit cell of this crystal?
3. Write the relation connecting edge length and radius of the particle.

Answer:

1. Face centred cubic crystal.
2. Particles are present the unit cell of this crystal.
   • Contribution from 8 corners = 1/8 × 8 = 1
   • Contribution from 6 face centers = 1/2 × 6 = 3
   • The total number of particles = 4
3. r = \(\frac{\sqrt{2} a}{4}=\frac{a}{2 \sqrt{2}}\).

Question 13.
1. The following diagram shows the alignment of magnetic moments for some magnetic properties.

• ↑↑↑↑↑↑
• ↑↓↑↓↑↓
• ↑↑↓↑↑↓

Identify the magnetic properties denoted by (a), (b), (c).

2. Examine the substances H₂O, NaCl, C₆H₆ and name the magnetic property common to them.
Answer:
1. The alignment of magnetic moments for some magnetic properties.

• Ferromagnetic
• Antiferromagnetic
• Ferrimagnetic

2. Diamagnetism

Question 14.
How many lattice points are there in one unit cell of the following lattices?

1. Face centred cubic
2. Body centred cubic
3. Simple cubic

Answer:

1. Face centred cubic – 4
2. Body centred cubic – 2
3. Simple cubic – 1

Plus Two Chemistry The Solid State Four Mark Questions and Answers

Question 1.

• Classify the following into crystalline and amorphous solids.
  1. NaCl
  2. Graphite
  3. Plastic
  4. Diamond
  5. Rubber
  6. KCl
  7. Wood
  8. CaCO₃
  9. Iodine
• Write any two properties of graphite.

Answer:

Crystalline solids	Amorphous solids
NaCl, Graphite	Plastic
Diamond, CsCl	Rubber
CaCO3, Iodine	Wood

• Two properties of graphite
  1. It is a good conductor of electricity.
  2. Graphite is a good solid lubricant.

Question 2.
Fill in the blanks.

Answer:

Question 3.

1. NaCl shows mainly Schottky defect and AgCl shows Frenkel defect. Do you agree with this statement? Justify.
2. Classify the following solids into isotropic and anisotropic.
   Polyvinylchloride, Rubber, Glucose, Glass

Answer:

1. Yes. In NaCl, both Na⁺ and Cl^– are of almost same size and hence it shows Schottky defect. But in AgCl, Ag+ is smaller than Cf. Hence, it shows Frenkel defect.
2. Anisotropic-Glucose
   Isotropic – Glass, Rubber, Polyvinylchloride

Question 4.
The common salt, sodium chloride is an example for a crystal system with edge length a = b = c and axial angles α = β = γ = 90°.

1. Identify the crystal system.
2. What happens when sodium chloride crystal is heated in presence of sodium?

Answer:

1. Cubic crystal
2. Gives yellow colour to the crystal due to F centre, ie. electron trapped anion vaccancy. It is a metal excess defect.

Question 5.
Stoichiometric defects are of two types such as vacancy defects and interstitial defects.

1. Which defect is basically a vacancy defect in ionic solids?
2. Which stoichiometric defect causes the decrease in density of solid?
3. On heating white ZnO it turns yellow. Which is the crystal involved here? Explain.

Answer:

1. Schottky defect
2. Schottky defect
3. It is due to metal excess defect caused by the presence of extra cations at interstitial sites. ZnO is white in colour at room temperature. On heating it loses oxygen and turns yellow.

The excess Zn²⁺ ions move to interstitial sites and the electrons to neighbouring interstitial sites. The yellow colour of non-stoichiometric ZnO (Zn_1+xO) is due to these trapped electrons.

Question 6.

1. Why is Frenkel defect not found in alkali metal halides?
2. A crystalline solid has simple cubic structure in which P atoms are present at the corners, Q atoms are present at the edge centres and R atoms are present at the centre of the unit cell. What is the formula of the compound?

Answer:
1. Frenkel defect is favoured by the small size of cations. In alkali metal halides both cations and anions are of almost same size.

2. Number of atoms of P = \(\frac{1}{8}\) × 8 = 1
Number of atoms of Q = \(\frac{1}{4}\) × 12 = 3 4
Number of atoms of R = 1
∴ Formula of the compound is PQ₃R

Question 7.

1. Classify each of the following solids as ionic, metallic, molecular, network or amorphous.
   • I₂
   • Plastic
   • LiBr
   • SiC
2. In terms of band theory differentiate Conductors, insulators & semi conductors

Answer:

1. Classification of solids as ionic, metallic, molecular, network or amorphous.
   • I₂ – Molecular
   • Plastic – Amorphous
   • LiBr – Ionic
   • SiC – Covalent
2. Differences between Conductors, insulators & semi conductors.
   • Conductors:
     The valence band overlaps with the conduction band or no energy gap exists between them.
   • Insulators:
     The energy gap between valance band and conduction band is very large. Hence electrons from valence band cannot move into the conduction band. Semi conductors have small energy gap between valence band and conduction band.

Question 8.
On the basis of nature of constituent particles crystals are classified into four types.

1. Which are they?
2. In which type does diamond belongs to? Why?
3. Can you say whether Iodine can be written as an example of ionic crystal? Why?

Answer:

1. Ionic crystals, Molecular crystals, Covalent crystals and Metallic crystals.
2. Diamond belongs to covalent crystals, because in diamond, the constituent particles are carbon atoms connected by strong covalent bonds.
3. No. In ionic crystal the constituent particles are ions. Since the constituent particles of iodine are molecules, it is a molecular crystal.

Question 9.

1. Distinguish between A and B?
2. Explain the defect in figure B.
3. Give two examples of crystals showing this defect.

Answer:

1. Fig. A – Ideal crystal. Fig. B – Frenkel defect.
2. Frenkel defect is due to the migration of cation from its original position to the void. This type of defect is favoured by the small size of cation. As a result of this defect, the neutrality, density and stoichiometry remain the same. But the conductivity increases.
3. AgCl, AgBr

Question 10.
The teacher explained crystal defects in classroom.

1. What are the different types of crystal defects?
2. Explain with the help of diagram the important difference between Schottky and Frenkel defects?

Answer:
1. Stoichiometric defects and Non-Stoichiometric defects.

2.

Question 11.
NaCl is an example for diamagnetic substance.

1. Write an example for paramagnetic substance.
2. What is the difference between ferromagnetic and anti ferromagnetic substances?
3. In case of ferri magnetic substances net magnetic moment is not zero. Justify.

Answer:
1. CuO
2. & 3.

Question 12.
The diagram of a cubic crystal whose molecular mass = M, edge length = a, density = ρ,  is given below. N is the Avogadro number.

1. From the above given details find the mass of this cube and also the mass of N particles?
2. By equating these two equations, try to find out a suitable equation for the density of this cube.

Answer:
1. Mass of cube = volume × density = a³ ρ ……………. (1)
Mass of N particles = N × Mass of one particle
= N × \(\frac{M}{N_{A}}\) (M = Gram Atomic Mass) …………. (2)

2. The equations (1) & (2) represent the mass of the cube.
i.e., a³p = N × \(\frac{M}{N_{A}}\)
∴ \(\rho=\frac{N \times M}{a^{3} \times N_{A}}\)

Question 13.

1. Classify each of the following as being either a p- type or a n-type semiconductor.
   • Ge dopped with In
   • Si dopped with B
2. A compound is formed by two elements P and Q. The element Q forms ccp and atoms of P occupy 1/3^rd of the tetrahedral voids. What is the formula of the compound?

Answer:
1. Either a p- type or a n-type semiconductor

• Ge dopped with In → p-type
• Si dopped with B → p-type

2. The formula of the compound
Q = 4
No. of tetrahedral voids = 2N = 2 × 4 = 8
P = 1/3 × 8 = 8/3
Q_8/3P₄ = Q₈P₁₂ = Q₂P₃

Plus Two Chemistry The Solid State NCERT Questions and Answers

Question 1.
Classify the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

• Tetraphosphorus decoxide (P₄O₁₀)
• Ammonium phosphate (NH₄PO₄)
• SiC
• I₂
• P₄
• Plastic
• Graphite
• Brass
• Rb
• LiBr
• Si

Answer:

• Tetraphosphorus decoxide (P₄O₁₀) – Molecularsolid
• Ammonium phosphate (NH₄PO₄) – Ionic solid
• SiC – Network(covalent) solid
• I₂ – Molecular solid
• P₄ – Molecular solid
• Plastic-Amorphous solid
• Graphite – Network(covalent) solid
• Brass – Metallic solid
• Rb- Metallic solid
• LiBr – Ionic solid
• Si – Network(covalent) solid

Question 2.

1. What is meant by the term coordination number?
2. What is the coordination number of atoms:
   • In a cubic close-packed lattice?
   • In a body-centred cubic structure?

Answer:

1. The number of nearest neighbours in a packing is called coordination number.
2. The coordination number of atoms
   • In cubic-close packed structure each atom is in direct contact with 12 other atoms. Hence, its coordination number is 12.
   • In a body centred cubic structure coordination number is 8

Question 3.
How many lattice points are there in one unit cell of each of the following lattice?

1. Face-centred cubic
2. End-centred monoclinic
3. Body-centred

Answer:
1. A face-centred cubic unit cell has lattice points at the corners of the cube and at face centres. There are eight comers and six faces of the cube. Each atom at corner makes a contribution of \(\frac{1}{8}\) while each atom at face centre makes a contribution of \(\frac{1}{2}\) to the unit cell.

Therefore, the number of atoms per unit cell
= 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 1 + 3 = 4

2. A end-centred monoclinic unit cell has lattice points at the face centres of only one set (two) of faces, in addition to the lattice points at the comers of the unit cell.

Therefore, the number of atoms per unit cell =
8 × \(\frac{1}{8}\) + 2 × \(\frac{1}{2}\) = 1 + 1 = 2

3. A body centred cubic unit cell has lattice points at the comers of the cube and at the body centre.

Therefore, the number of atoms per unit cell
= 8 × \(\frac{1}{8}\) + 1 × 1 = 1 + 1 = 2

Question 4.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?
Answer:
The atom at the corner makes \(\frac{1}{8}\) contribution while atom at body centre makes 1 contribution to the unit cell.
No. of atoms of Q per unit cell = 8 × \(\frac{1}{8}\) = 1
No. of atoms of P per unit cell = 1 × 1 = 1 Therefore, formula of the compound is PQ.
The atom at the body centre would be in contact with all the atoms at the corners. Hence, the coordination number of P is 8.
Similarly, the coordination number of Q is 8 because it is shared by 8 other atoms.

Question 5.
If NaCl is doped with 10^-3 mol % of SrCl₂, what is the concentration of cation vacancies?
Answer:
Every Sr²⁺ ion causes one cation vacancy (because two Na⁺ ions are replaced by one Sr²⁺ and it occupies the site of one Na⁺ ion and the other site remains vacant.) Therefore, introduction of 10^-3 moles of SrCl₂ per 100 moles of NaCl would introduce 10^-3 moles cation vacancies in 100 moles of NaCl.
No. of vacancies per mole of NaCl = \(\frac{10^{-3}}{100}\) × 6.02 × 10²³ = 6.02 × 10¹⁸

Students can Download Chapter 7 Web Hosting Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 7 Web Hosting

Plus Two Computer Science Web Hosting One Mark Questions and Answers

Question 1.
The companies that provide web hosting services are called ______.
Answer:
Web Hosts

Question 2.
The service of providing storage space in a web server to make a website available on Internet is called
Answer:
web hosting

Question 3.
Which of the following is true in the case of dedicated hosting?
(a) It shares server with other websites.
(b) It is usually inexpensive.
(c) It does not guarantee performance.
(d) It offers freedom for the clients to choose the hardware and the software.
Answer:
(d) only

Question 4.
Choose the odd one out, and justify your answer.
(a) Shared hosting
(b) Dedicated hosting
(c) DNS
(d) Virtual Private Server
Answer:
(c) DNS others are types of web hosting.

Question 5.
______ is the service of providing storage space in a Webserver.

OR

Storing the web pages of a website in a server is popularly known as _____.
Answer:
Web hosting

Question 6.
The companies that provide web hosting services are called _______.
Answer:
Web hosts

Question 7.
Odd one out.
(a) Shared
(b) Dedicated
(c) VPS
(d) DNS
Answer:
(d) DNS means Domain Name System. The others are types of web hosting

Question 8.
VPS stands for _____ .
Answer:
Virtual Private Server.

Question 9.
From the following select the most commonly used web hosting.
(a) Shared
(b) Dedicated
(c) VPS
(d) DNS
Answer:
(a) Shared hosting

Question 10.
DNS stands for ____.
Answer:
Domain Name System

Question 11.
‘A record’ means ______ .
Answer:
Address record

Question 12.
FTP stands for _____ .
Answer:
File Transfer Protocol.

Question 13.
Odd one out.
(a) Mozilla Firefox
(b) FileZilla
(c) CuteFTP
(d) SmartFTP
Answer:
(a) Mozilla Firefox, it is a web browser, the others are popular FTP client software

Question 14.
______ hosting provides web hosting services free of charge.
Answer:
Free hosting

Question 15.
Give an example for Free web hosting services with sub-domain website address.
Answer:
www.bvmhsskalparamba.facebook.com

Question 16.
Give an example for Free web hosting services with directory service website address.
Answer:
www.facebook.com/bvmhsSkalparamba

Question 17.
CMS stands for ____.
Answer:
Content Management System

Question 18.
The term ‘responsive web designing’ was introduced by _____.
Answer:
Ethan Marcotte

Question 19.
The method of Flexible designing of web pages to suit the various types of screens is called _____.
Answer:
Responsive web design method

Question 20.
In dedicated hosting, if the client is allowed to place his own purchased web server in the service provider’s facility, then it is called _______.
Answer:
Co-location

Question 21.
What are the information contained in a ICANN database?
Answer:
Registered domain names/name, address, telephone number and e-mail address of the registrants.

Question 22.
What is‘A record’?
Answer:
‘A record’ is used to store the IP address and the corresponding domain name

Question 23.
Joomla is an example for ______ .
(a) CMS
(b) ISP
(c) DNS
(d) None of the above
Answer:
(a) CMS

Question 24.
The responsive web design feature that converts horizontal menu to a drop-down menu in mobile phones is called ______.
Answer:
Media queries

Question 25.
The organization that maintains the WHOIS database of domain names is ______.
Answer:
ICANN

Plus Two Computer Science Web Hosting Two Mark Questions and Answers

Question 1.
Consider that your school is planning to host a website. What are the factors that you will consider while choosing the type of web hosting?
Answer:
Following factors to be considered

1. Buying sufficient amount of memory space for storing our website files
2. If the web pages contain programming contents supporting technology must be considered
3. Based upon the programs select Windows hosting or Linux hosting

Question 2.
Emmanuel wishes to buy a suitable domain for his company. Unfortunately, the domain name he chose is already registered by someone else. Name the feature that will help him to find the current owner. List the details will he get.
Answer:
WHOIS
Name, address, telephone number and e-mail address of the registrant

Question 3.
What is the use of FTP client software? Give an example.
Answer:
FTP (File Transfer Protocol) client software:
When a client requests a website by entering website address. Then FTP client software helps to establish a connection between client computer and remote server computer. Unauthorised access is denied by using user name and password hence secure our website files for that SSH(Secure Shell) FTP simply SFTP is used.

Instead of http://, it uses ftp://. By using FTP client s/w we can transfer(upload) the files from our computer to the web server by using the ‘drag and drop’ method. The popular FTP client software are FileZilla, CuteFTP, SmartFTP, etc.

Question 4.
Haseena has decided to host her new website using free hosting facility; her friend Rinisha is against this move. Can you guess her argument against the utilization of free hosting facility?
Answer:
In free web hosting service, the expense is meet by the advertisements. Some service providers allow limited facilities such as limited storage space, do not allow multimedia (audio and video) files.

Plus Two Computer Science Web Hosting Three Mark Questions and Answers

Question 1.
Priya has developed a website for her shop. She has purchased a domain name and hosting space.

1. Name the software that will help her to transfer her files from her computer to the webserver.
2. List the requirements in that software that are necessary to connect to the webserver.

Answer:

1. FTP software such as FileZilla, Cute FTP, Smart FTP
2. Following are the requirements to connect to the Webserver
   • Domain name /IP address
   • User name.
   • Password

Question 2.
Explain the advantages of using SFTP protocol in FTP client software.
Answer:
Unauthorized access is denied by using username and password hence secure our website files, for this SSH (Secure Shell) FTP simply SFTP is used. It encrypts and sends usernames, passwords, and data to the webserver.

Question 3.
Merin plans to create a website for their family without spending money.

1. List some of the limitations that Merin will face regarding the hosting space for website.
2. How will she provide a domain name for the website?

Answer:
1. In free web hosting service, the expense is meet by the advertisements. Some service providers allow limited facilities such as limited storage space, do not allow multimedia (audio and video) files.

2. Usually, two types of free web hosting services as follows.

• As a directory service: service provider’s website address/ our website address
  eg: www.facebook.com/bvmhsskalparamba
• As a subdomain: Our website address. service providers website address
  eg: www.bvmhsskalparamba.facebook.com

Question 4.
Recently more and more people are using Content Management Systems (CMS) for developing professional websites. What can be the reasons for this?
Answer:
CMS means Content Management System. Do you heard about Data Base Management System (DBMS). DBMS is a software (collection of programs) used to create, alter, modify, delete and retrieve records of a DataBase. Similarly CMS is a collection of programs that is used to create, modify, update and publish website contents.

CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc. are the examples of CMS.

Question 5.
Suggest a hosting type for the following websites given below. Justify.

1. Website for a medical shop in your city.
2. Website for Public Service Commission (PSC) of Kerala.
3. Website for an online shopping facility.

Answer:

1. Shared hosting
2. Dedicated/VPS
3. Dedicated/VPS

Question 6.
Consider that a college in your locality plans to shift its website from shared type of hosting to VPS hosting. List the advantages that the website will gain from this change.
Answer:
Virtual Private Server (VPS):
A VPS is a virtual machine sold as a service by an Internet hosting Service. A VPS runs its own copy of an OS(Operating System) and customers have super level access to that OS instance, so they can install almost any s/ w that runs on that OS.

This type is suitable for websites that require more features than shared hosting but less features than dedicated hosting.
Eg: It is similar to owning a Condo

Question 7.
Mr. Mohan wants to host a personal website with minimal cost. Which type of web hosting would you advise for him? Justify your answer.
Answer:
Shared Hosting:
This type of hosting sharing re-sources, like memory, disk space, and CPU hence the name shared. Several websites share the same server. This is suitable for small websites that have less traffic and it is not suitable for large websites that have large bandwidth, large storage space and have large volume of traffic.

Eg: Shared hosting is very similar to living in an Apartment(Villas) complex. All residents are in the same location and must share the available resources(Car parking area, Swimming pool, Gymnasium, playground, etc) with everyone.

Question 8.
A supermarket in a city wishes to take its business online. It plans to accept orders for its products through a website and receive payments online.

1. Which type of hosting is suitable for this website?
2. Explain the reason for your choice.

Answer:
1. Dedicated Hosting

2. A web server and its resources are exclusively for one website that have large volume of traffic means large volume of requests by the visitors. Some Govt, departments or large organizations require uninterrupted services for that round the clock power supply is needed. It is too expensive but it is more reliable and provides good service to the public.

Eg: It is similar to living in an Our own house. All the resources in your house is only for you. No one else’s account resides on the computer and would not be capable of tapping into your resourses

Question 9.
Emil wishes to purchase the web hosting space required to host a website for his medical shop. List the features to be taken into consideration while buying hosting space on a web server.
Answer:
Following factors to be considered:

1. Buying sufficient amount of memory space for storing our website files
2. If the web pages contain programming contents supporting technology must be considered
3. Based upon the programs select Windows hosting or Linux hosting

Question 10.
How can we connect a website hosted in a Web server to a domain name?
Answer:
Millions of websites are available over Internet so that our website must be registered with a suitable name. Domain Name registration is used to identify a website over Internet. A domain name must be unique(i.e. no two websites with same name is available).

So you have to check the availability of domain name before you register it, for this www.whois.net website will help. If the domain name entered is available then we can register it by paying the Annual registration fees online. Consider a Post Office, it has two addresses one string address (Irinjalakuda) and one numeric(pin) code (680121).

Just like this the website has also two addresses a string address, for example, www.agker.cag.gov.in and a numeric address (http:/ /210.212.239.70/). We are following string address, hence this domain name has to be connected to the corresponding IP address of the webserver. This is done by using ‘A record’(Address record) of the domain. ‘A record’ is used to store the IP address and the corresponding domain name.

Question 11.
What is the advantage of using SFTP protocol in FTP software?
Answer:
FTP(File Transfer Protocol) client software:
When a client requests a website by entering website address. Then FTP client software helps to establish a connection between client computer and remote server computer. Unauthorised access is denied by using username and password hence secure our website files for that SSH(Secure Shell) FTP simply SFTP is used.

Instead of http://, it uses ftp://. By using FTP client s/w we can transfer(upload) the files from our computer to the web server by using the ‘drag and drop’ method. The popular FTP client software are FileZilla, CuteFTP, SmartFTP, etc.

Question 12.
Raju wishes to host a website for his family. What are the advantages that free web hosting companies provide?
Answer:
The name implies it is free of cost service and the expense is meet by the advertisements. Some service providers allow limited facilities such as limited storage space, do not allow multimedia(audio and video) files. A paid service website’s address is as follows:
Eg: www.bvmhsskalparamba.com

Usually two types of free web hosting services as follows:
1. As a directory service:
Service provider’s website address/ our website address.
eg: www.facebook.com / bvm hss kalparambu.

2. As a Sub domain:
Our website address.service providers website address.
eg: www.bvmhsskalparamba.facebook.com Earlier web hosting services are expensive but nowadays it is cheaper hence reduced the need for free web hosting.

Question 13.
What is CMS? What are the features of CMS? Give Examples.
Answer:
CMS means Content Management System. Do you heard about Data Base Management System (DBMS)? DBMS is a software(collection of programs) used to create, alter, modify, delete and retrieve records of a DataBase. Similarly, CMS is a collection of programs that is used to create, modify, update and publish website contents.

CMS can be downloaded freely and is useful to design and manage attractive and interactive websites with the help of templates that are available in CMS. WordPress, Joomla, etc. are examples of CMS.

Plus Two Computer Science Web Hosting Five Mark Questions and Answers

Question 1.
Explain different types of web hosting?
Answer:
Types of web hosting:
Various types of web hosting services are available. We can choose the web hosting services according to our needs depends upon the storage space needed for hosting, the number of visitors expected to visit, etc.

1. Shared Hosting:
This type of hosting sharing resources, like memory, disk space, and CPU hence the name shared. Several websites share the same server. This is suitable for small websites that have less traffic and it is not suitable for large websites that have large bandwidth, large storage space and have large volume of traffic.

Eg: Shared hosting is very similar to living in an Apartment(Villas) complex. All residents are in the same location and must share the available resources(Car parking area, Swimming pool, Gymnasium, playground, etc) with everyone.

2. Dedicated Hosting:
A web server and its resources are exclusively for one website that have large volume of traffic means large volume of requests by the visitors. Some Govt, departments or large organizations require uninterrupted services for that round the clock power supply is needed. It is too expensive but it is more reliable and provides good service to the public.

Eg: It is similar to living in an Our own house. All the resources in your house is only for you. No one else’s account resides on the computer and would not be capable, of tapping into your resources.

3. Virtual Private Server (VPS):
A VPS is a virtual machine sold as a service by an Internet hosting Service. A VPS runs its own copy of an OS (Operating System) and customers have super level access to that OS instance, so they can install almost any s/what runs on that OS. This type is suitable for websites that require more features than shared hosting but less features than dedicated hosting.
Eg: It is similar to owning a Condo

Question 2.

1. What is responsive web design?
2. Why is it gaining importance recently?

Answer:
1. The home page is displayed differently according to the screen size of the browser window (different screen sized devices mobile phone, palmtop, tablet, laptop, and desktop) we used. The website is designed dynamically (flexibly) that suit the screen size of different device introduced by Ethan Marcotte.

Before this, companies have to design different websites for different screen sized devices. By responsive web design, companies have to design only one website that suitably displayed according to the screen size of the devices. It is implemented by using flexible grid layout, images, and media queries.

• Flexible grid layouts: It helps to set the size of the web page to fit the screen size of the device.
• Flexible image and video: It helps to set the image or video dimension to fit the screen size of the device.
• Media queries: There is an option (settings) to select the size of the web page to match our device, this can be done by using media queries inside the CSS file.

A well known Malayalam daily Malaysia Manorama launched their responsive website.

2. Instead of using desktops or laptops many people nowadays visit websites using tablets and mobile phones. Portability is the main reason for this.

Question 3.
Today, we visit websites using tablets and mobile phones also. You might have noticed that the same website is displayed in a different layout in different devices.

1. Name the concept used for this.
2. List and explain the technologies used for implementing this concept

Answer:
1. Responsive web design.

2. It is implemented by using flexible grid layout, images, and media queries:

• Flexible grid layouts: It helps to set the size of the web page to fit the screen size of the device.
• Flexible image and video: It helps to set the image or video dimension to fit the screen size of the device.
• Media queries: There is an option (settings) to select the size of the web page to match our device, this can be done by using media queries inside the CSS file.

A well known Malayalam daily Malayala Manorama launched their responsive website.

Question 4.

1. What is responsive web design?
2. How is responsive web design implemented?

Answer:
1. The home page is displayed differently according to the screen size of the browser window (different screen sized devices mobile phone, palm top, tablet, laptop and desktop) we used. The website is designed dynamically (flexibly) that suit the screen size of different device introduced by Ethan Marcotte.

Before this, companies have to design different websites for different screen sized devices. By responsive web design, companies have to design only one website that suitably displayed according to the screen size of the devices.

2. It is implemented by using flexible grid layout, images, and media queries:

• Flexible grid layouts: It helps to set the size of the web page to fit the screen size of the device.
• Flexible image and video: It helps to set the image or video dimension to fit the screen size of the device.
• Media queries: There is an option (settings) to select the size of the web page to match our device, this can be done by using media queries inside the CSS file.

A well known Malayalam daily Malayala Manorama launched their responsive website.

Students can Download Chapter 2 Spread Sheet Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Accountancy Chapter Wise Questions and Answers Chapter 2 Spread Sheet

Plus Two Accountancy Spread Sheet One Mark Questions and Answers

Question 1.
The best way to get started in Libre Office calc is _____
Answer:
Application → Office → LibreOffice Calc

Question 2.
___________ is a configuration of rows and columns
Answer:
Spread sheet

Question 3.
A spread sheet is also known as ____________
(a) Work book
(b) Work area
(c) Work sheet
(d) Spread book
Answer:
(c) Work sheet

Question 4.
____________ in spread sheet are horizontal vectors while ______ are vertical vectors
Answer:
Rows, columns

Question 5.
Spread sheet is used to
(a) Record data
(b) Calculate data
(c) Compare data
(d) All the above
Answer:
(d) All of the above

Question 6.
Each cell value can either be an independent (basic) value or it may be derived on the basis of ____________.
Answer:
Arithmetic expression or a function.

Question 7.
A file in LibreOffice Calc is known as a ______
(a) Work sheet
(b) Page
(c) Work book
(d) All the above
Answer:
(c) Work book

Question 8.
A work book is a collection of a number of ______
Answer:
Work Sheets

Question 9.
Where is the address of the active cell displayed?
(a) Row heading
(b) Status Bar
(c) Name Box
(d) Formula Bar
Answer:
(c) Name Box

Question 10.
Which command reverses the last action performed in the worksheet?
(a) Cut
(b) Undo
(c) Redo
(d) Paste
Answer:
(b) Undo

Question 11.
__________ is a text or special character or descriptive information for rows or columns.
Answer:
Label

Question 12.
A formula must starts with a ___________ sign
(a) =
(b) >
(c) *
(d) {}
Answer:
(a) =

Question 13.
_______ Function is commonly used to get the addition of various numbers orthe contents of various cells.
Answer:
Autosum (Σ)

Question 14.
The cell A5 indicate Column _______ and Row _____
Answer:
Column A & Row 5

Question 15.
The dark box which distinguishes the active cell is called _______
Answer:
Cell Pointer

Question 16.
One or more cells selected is called ______
Answer:
a range

Question 17.
Without the equal sign, the entry in a cell is treated as _________
(a) Text
(b) Label
(c) TextorLabel
(d) None of the above
Answer:
(c) Text or Label

Question 18.
Libre Office calc has three type of cell entries, they are _______ , ______and _______
Answer:
Value, Label and Formula.

Question 19.
Which command allows you to reverse an undo command?
(a) Redo
(b) Repeat
(c) Reset
(d) Reverse
Answer:
(a) Redo

Question 20.
Which cell alignment is assigned to most values by default?
(a) Right
(b) Left
(c) Centre
(d) None of the above
Answer:
(b) Left

Question 21.
Which function automatically totals a column or row of Values?
(a) TOTAL
(b) ADD
(c) SUM
(d) AVG
Answer:
(c) SUM

Question 22.
Which Mathematical Operator is represented by an asterisk (*)
(a) Exponentiation (square)
(b) Addition
(c) Subtraction
(d) Multiplication
Answer:
(d) Multiplication

Question 23.
How many blank work sheets are shown when a new workbook is created?
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(c) Three

Question 24.
The cell co-ordinate in the formula are known as ………….
Answer:
cell references

Question 25.
IF function is …………….
Answer:
Logical function

Question 26.
The cell references for cell range of G2to M12 is …………….
(a) G2.M12
(b) G2: M12
(c) G2; M12
(d) G2 – M12
Answer:
(c) G2: M12

Question 27.
If 4/6 entered in a cell without applying any format, LibreOffice Calc will treat this as
(a) Fraction
(b) Number
(c) Text
(d) Date
Answer:
(d) Date

Question 28.
The cell labelled F5 refers to …………….
(a) Row F column 5
(b) Column F row 5
(c) Function available in cells
(d) Function key F5
Answer:
(b) Column F row 5

Question 29.
LibreOffice Calc is a FOSS. What is FOSS?
Answer:
Free and Open Source Software.

Question 30.
Which among the following is not a spreadsheet software.
(a) MS Office Excel
(b) Open Office Spredsheet
(c) Libre Office Calc
(d) MS Office Word
Answer:
(d) M S Office Word

Question 31.
How many worksheets can be made as active worksheet at a time?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 32.
The name of the worksheet will be shown in the ______ at the bottom left of the windows.
Answer:
Sheet Tab

Question 33.
To add column in a worksheet, click __________, there we get an option to add column.
Answer:
Column Header

Question 34.
A ………… is identified by a combination of a column header (letter) and a row header (number)
Answer:
cell

Question 35.
_______ is a group of adjacent cells that forms a rectangular area.
Answer:
Range

Question 36.
Which among the following is used as the range operator
(a) ;
(b) :
(c) /
(d) #
Answer:
(b) :

Question 37.
One ENTER key stroke means
(a) One cell down
(b) One cell up
(c) One cell right
(d) One cell left
Answer:
(a) One cell down

Question 38.
………………. Function counts the number of cells which contain any value.
(COUNT, COUNTA, COUNT BLANK, COUNTIF)
Answer:
COUNTA

Plus Two Accountancy Spread Sheet Two Mark Questions and Answers

Question 1.
Match the following

A	B
1. Rows	1. Intersection of a row & a column
2. Columns	2. Numerical numbers from top to bottom
3. Cell	3. Unique identification code of a cell
4. Cell address	4. Alpha characters from left to right

Answer:

A	B
1. Rows	1. Numerical numbers from top to bottom
2. Columns	2. Alpha characters from left to right
3. Cell	3. Intersection of a row & a column
4. Cell address	4. Unique identification code of a cell

Question 2.
What do you mean by spreadsheet?
Answer:
Spreadsheet:
Spreadsheet application is a computer program that allows to record, calculate and compare numerical or financial data. Using a spreadsheet program. We can store a lot of data in the worksheet and also arrange and analyse the data by using different functions and formulae for the meaningful object.

It is used to establish relationship between two or more sets of data. Libre Office Calc, MS Office Excel, Open Office Spreadsheet etc. are examples of spreadsheet software.

Question 3.
Give a short note on

1. Work book
2. Work sheet

Answer:
1. Workbook:
A file in spread sheet is known as a workbook. A work book is a collection of a number of work sheets.

2. Work sheets:
The work area which consists of rows and columns in a spreadsheet is called a worksheet. By default three work sheets-sheet 1, sheet -2, sheet -3 are available in work book.

Question 4.
Explain about rows and columns in Libre Office Calc?
Answer:
1. Rows:
Rows are the horizontal vectors in the worksheet. These are numbered numerically from Top to Bottom.

2. Columns:
Columns are vertical vectors in the worksheet. These are referred by alpha characters from left to right such as A, B, C, …, AA, AB, AC …etc.

Question 5.
What do you mean be Relative cell Reference?
Answer:
1. Relative Cell references:
By default cell reference is relative; which means that as a formula or function is copied and pasted to other cells, the cell references in the formula or function change to reflect the new location.

Question 6.
Define the following

1. Label
2. Formula

Answer:
1. Labels:
Descriptive information for rows or columns in the form of a text or a special character is called Label.
Eg: Name, Roll No, Address.

2. Formula:
The formula means a mathematical calculation on a set of cells. The formula must start with an = (equal to) sign. When a cell contains a formula, it often contains reference to other cells. Eg:= Basic pay + DA + HRA

Question 7.
List down any two features fo Spreadsheet.
Answer:

• A spreadsheet is a configuration of rows and columns.
• A spreadsheet is also known as worksheet.

Question 8.
Name the different spreadsheet software available.
Answer:
LibreOffice Calc, MS Office Excel, Open Office Spreadsheet.

Question 9.
Briefly explain any two Date and Time functions availabe in LibreOffice Calc
Answer:
1. TODAY():
It is the function for today’s date in the worksheet. This helps to update the date value when we reopen the spreadsheet or modify the values of the document
Syntax: =TODAY()

2. NOW ():
It is the function for today’s date and present time. This helps to update the date and the time value when the cell value is modified.
Syntax: =NOW()

Question 10.
Give the Syntax for

1. YEAR()
2. DATE()

Answer:
1. YEAR()
Syntax: = YEAR (Date value) OR = YEAR (“Date”)

2. DATE()
Syntax: = DATE (Year, Month, Day)

Question 11.
Complete the table by using ROUNDUP()

Answer:

1. 1880
2. 1900

Question 12.
What are the Text Manipulation Function in LibreOffice Calc
Answer:

• TEXT
• CONCATENATE

Question 13.
Give the explanation of the following errors
Error Message

1. # DIV/o!
2. VALUE!

Answer:

1. # DIV/0! → When a number is divided by zero
2. VALUE! → When a wrong argument is given in a fromula

Question 14.
Spot the correct pairs by observing the nature of software, and give justification
(a) Microsoft Excel and LibreOffice Calc
(b) Linux and Tata Ex.
(c) Windows and MS Office
(d) GINUGhata and Microsoft Access
Answer:
(a) Microsoft Excel and LibreOffice Calc. Both are spread sheet packages.

Plus Two Accountancy Spread Sheet Three Mark Questions and Answers

Question 1.
What are the advantage of Libre Office Calc?
Answer:

1. It is both free software and open source software.
2. It can be used to calculate, analyse and manage data.
3. Libre Office Calc is available for a variety of platforms including Linux, OSX, Microsoft windows and Free BSD.

Question 2.
Give the cell address or range reference in the following situations.

1. Cell at 10^th column and 6^th row.
2. Cell at 27th column and 15^th row.
3. Range starting from 5^th column, 9^throw and spread till 12^th column and 15^th row.

Answer:

1. Cell at 10^th column and 6^th row = J6.
2. Cell at 27^th column and 15^th row = AA15.
3. Range starting from 5^th column, 9^th row and spread till 12^th column and 15^th row = E9: L15.

Question 3.

1. What is FOSS?
2. What are the benefits of using FOSS?

Answer:

1. FOSS means Free and Open Source Software
2. The benefits of using FOSS:
   • Decreased software costs.
   • Increased Security and stability.
   • Any one can use, copy, study and change the software in any manner.
   • Source code is openly shared.

Question 4.
How can we save Libre Office Calc file?
Answer:
Step 1: Go to the File menu
File menu

Step 2: Click on Save
Save

OR

Step 1: & Step 2 – Press Ctrl + S
Ctrl + S

Step 3: Type the file in the name field instead of default name Untitled 1
Name field

Step 4: Choose the place where we want to save the new file.

Step 5: Click on Save

Question 5.
Write the steps to be followed to

1. Rename a worksheet
2. Delete a worksheet
3. Copy a worksheet

Answer:
1. Rename a worksheet:
Step – 1 Select the work sheet in the Sheet Tab which we want to Rename.

Step – 2 Right click and select Rename sheet from the drop up menu

Step – 3 Type new name in the Name field and press OK button

2. Delete a worksheet:
Step – 1 Select the worksheet in the Sheet Tab which we want to delete

Step – 2 Right click and select Delete sheet from the drop up menu

Step – 3 Click on Yes to the conformation question.

3. Copy a worksheet:
Step – 1 Click on blank rectangle Top left corner of the worksheet (Range Adress A1: AM J1048576)

Step – 2 Move the curser inside the worksheet and Right click the mouse.

Step – 3 Click on Copy from the dropdown menu.

Step – 4 Open the worksheet where we want to copy the sheet

Step – 5 Right click on cell A1. Click on Paste.

Question 6.
Give the cell address or range reference in the following situations.

1. Cell of 10th column and 9^th row
2. Range starting from 2^nd column 4^th row and spread till 8^th column 12^th row
3. Range starting from 4^th column 5^th row and spread till 10^thcolumn 15^th row

Answer:

1. J9
2. B4: H12
3. D5: J15

Question 7.
What is the purpose of the COUNTIF function?
Answer:
COUNTIF():
This function counts the number of cells within a given that meet the criteria or condition.
Syntax: = COUNTIF (Range, Criteria)

Question 8.
Write the command to calculate the State Life Insurance Premium (SLI) of an employee using IF Function. The condition is SLI Premium Rs. 500/-below Basic Pay (BP) of Rs. 15000/- and for others Rs. 800/- (BP is given in cell C3)
Answer:
= IF(C3 < 15000, 500, 800)

Question 9.
4 numbers are entered in Libre Office Calc spreadsheet starting from cell A1 to A4. Write any two formulae for getting total of those numbers in the cell A5

Answer:

1. A5 = A1 + A2 + A3 + A4
2. A5 = Σ(A1 : A4)

Question 10.
For preparation of Payslip of an employee, Govind entered Basic Pay in cell B2 of a worksheet in LibreOffice calc. The D.A. is 76% of Basic Pay and Gross Salary is the sum of Basic Pay and DA. In order to calculate the amount of DA in cell C2 and Gross salary in cell D2, what entries are to be made?
Answer:
Basic Pay → B2
DA → C2
Gross salary → D2
C2 = B2 * 76%
D2 = B2 + C2

Question 11.
Consider the following features of a software tool. Identify the software.

1. It can be used as a text editor
2. List out the important uses of this software

Answer:

1. LibreOffice calc has the above mentioned features.
2. Importance of LibreOffice calc.
   • It can be used for the preparation of statement of depreciation.
   • It can be used for preparation of Payroll of employees.
   • It can be used for the preparation of the loan re-payment schedule.

Question 12.
Complete the following.

1. One cell down → arrow key
2. One cell up → …………………………
3. One cell left → …………………………
4. One cell right → ………………………

Answer:

2. Up arrow key
3. Left arrow key
4. Right arrow key

Question 13.
Can you give some key navigations and short cut in LibreOffice calc?
Answer:
Spreadsheet navigation:
We can move around a worksheet through four arrow keys

1. Left-arrow key
2. Right arrow key
3. Up arrow key
4. Down arrow key

The mouse can also be used for navigation in a spreadsheet except for data entry. Some common operations/navigations are listed below:

Movement	Keystroke (Press Key)
One cell up	Up arrow key/ Shift + Enter key
One cell down	Down arrow key/ Enter key
One cell right	Right arrow key/ Tab key
One cell left	Left arrow key / Shift + Tab key
Top of sheet (Cell A1)	Ctrl + Home Key
Move to last cell containing data	Ctrl + End Key
Move to beginning of the Row	Home Key or Ctrl + Left arrow key
Move to last filled cell in column	End key

Question 14.
What commands are used to insert a column and delete a column in LibreOffice Calc?
Answer:
1. To insert a column:
To insert a column click a cell in the column immediately to the right of where you want to insert the new column. Then on the Insert menu, click Columns.

2. To delete a column:
Select the column you want to delete. On the Edit menu, click Delete.

Question 15.
Give the procedure to insert a new worksheet and to delete a worksheet.
Answer:
To insert a new worksheet, right click the worksheet, select ‘insert’. To delete a worksheet, select the sheet you want to delete and ‘click delete sheet’ on the edit menu.

Question 16.
Give the cell address or range reference in the following situations.

1. Cell at 12^th Column and 8^th row.
2. Range starting from 6^th Column 10^th row and spread till 12^th Column and 16^th row.

Answer:

1. L8.
2. F10: L16

Question 17.
Name the appropriate Statistical functions.

1. Number of cells contain numbers.
2. Number of cells contain any value.
3. Number of empty cell.
4. Number of cells that meet the given criteria.

Answer:

1. COUNT
2. COUNTA
3. COUNTBLANK
4. COUNTIF

Question 18.
What is the use of financial function ACCRINT in LibreOffice Calc
Answer:
ACCRINT ():
This function returns the accrued interest for a security that pays periodic interest.
Syntax: = ACCRINT(Issue, First_Interest, settlement, rate, Par, frequency, basis, calc_method)

Question 19.
Explain the difference between relative cell reference and absolute cell reference?
Answer:
1. Relative Cell references:
By default cell reference is relative; which means that as a formula or function is copied and pasted to other cells, the cell references in the formula or function change to reflect the new location.

2. Absolute Cell reference:
The absolute cell reference consists of the column letter and row number surrounded by dollar ($) signs. Eg $A$5. An absolute cell reference is used when we want a cell reference to stay fixed on a specific cell.

Question 20.
What is the use of PIVOT TABLE?
Answer:
Preparation of reports using Pivot Tables:
A Pivot Table is a way to present information in a report format. A Pivot Table report provides enhanced layout, attractive and formatted report with improved readability. There are two types of data table.

1. One-Variable Data Table (One – Variable )
2. Two-Variable Data Table (Two-Variable)
   • The one variable Data Table allows us to identify a single decision variable in our model and see how changing the values for that variable affects the values calculated by one or more formulas in our model.
   • The two variable Data table allows us to specify two decision variables and a variety of inputs and only a single formula.

Plus Two Accountancy Spread Sheet Five Mark Questions and Answers

Question 1.
Define the following

1. Cell
2. Range
3. Worksheet
4. Workbook

Answer:
1. Cell:
The intersection of a row and a column is called a cell. A cell is identified by a combination of an alpha – numeric character eg: A1, B6, C10, etc. This alpha numeric character is called cell address. Hence each cell has a unique address.

2. Ranges:
Range is a group of adjacent cells that forms a rectangular area. A range is specified by giving the address for first cell in range and the last cell in the rage, eg: range starting from A10 to A20 is written as A10: A20 where colon (:) is the range operator.

3. Work sheets:
The work area which consists of rows and columns in a spreadsheet is called worksheet. By default three worksheets-sheet 1, sheet -2, sheet -3 are available in work book.

4. Workbook:
A file in spread sheet is known as a workbook. A work book is a collection of a number of work sheets.

Question 2.
What are the different spreadsheet reference functions in Libre Office Calc?
Answer:
Spreadsheet Reference Functions:
The important spreadsheet reference functions are

1. LOOKUP () functions:
The LOOKUP function returns a value either from a one-row or one-column range or from an array. The lookup function has two syntax forms: Vector form and Array form.

The vector form of LOOUP looks in a one-row or one column range (known as a vector) for a value, and then returns a value from the same position in a second one-row or one-column range.

The array form of LOOKUP looks in the first row or column of an array for the specified value, and then returns a value from the same position in the last row or column of the array.

LOOKUP (Vector from)
Syntax: = LOOKUP (search criterion, Search vector, Result vector)
LOOKUP (Array form)
Syntax: =(LOOKUP (lookup_value, array)

2. VLOOK UP ():
VLOOK UP is the vertical LOOKUP function. Use VLOOK UP to search the first column (columns are vertical) of a block of data and return the value from another column in the same row.
Syntax: = VLOOKUP (Search criterion; Array; Index; Sort Order)

3. HLOOKUP ( ):
It is the Horizontal LOOKUP function, searches for a value in the first row of a table array and returns the corresponding value in the same column from another row of the same table array.
Syntax: HLOOKUP (search criteria; index; sorted)

Question 3.
Give the cell address or range reference in the following situations.

1. Cell at 10^th Column and 9^th row.
2. Range starting from 2^nd Column 4^th row and spread till 8^th Column and 12^th row.

Answer:

1. J9.
2. B4: H12

Question 4.
What are the logical functions in LibreOffice Calc
Answer:
Logical Functions:
Logical functions are used for comparison and checking a test condition. The major logical functions are IF, AND and OR.

Question 5.
Explain the steps involved to insert a new work sheet in LibreOffice Calc
Answer:
Steps involved to insert a new work sheet in LibreOffice Calc.
Step 1 – Click on the [+] button in the sheet Tab

OR

Step 1 – Click on the work sheet in the sheet Tab Sheet Tab

Step 2 – Right click the mouse

Step – 3 Select insert sheet from the Drop up menu box
Drop up

Step 4- Add number of sheets and give name, press
OK

Question 6.
You are required to prepare a rank list for admission to the B.Com course. The Index mark is calculated as follows.
Sum of mark secured in Plus 2, Mark obtained in Accountancy and -25% of Mark obtained in Business studies. Give the entries in Libre Office Calc to get the index mark.
Answer:

• A1 → Total Mark
• B1 → Mark in Accountancy
• C1 → Mark in Business Studies
• D1 → Index Mark
• A2 → Enter the Marks
• B2 → Enter the Marks
• C2 → Enter the Marks
• D2 → =A2 + B2 + (C2 * 25%)

Question 7.
Match the following:

A	B
1. One cell to the left	1. Right arrow key
2. To cell A1	2. Ctrl + End
3. One cell to the right	3. Ctrl + Home
4. To the last cell in the worksheet that contains data	4. Shift + Tab

Answer:

A	B
1. One cell to the left	1. Shift + Tab
2. To cell A1	2. Ctrl + Home
3. One cell to the right	3. Right arrow key
4. To the last cell in the work sheet that contains data	4. Ctrl + End

Question 8.
The monthly sales of a company for the first six months are given below:

A	B
1. January	25000
2. February	15000
3. March	28000
4. April	32000
5. May	20000
6. June	30000

7. Find the total sales for the six months
8. Find the average sales of the six months
9. Find the lowest sales of the six months
10. Find the highest sales of the six months

Answer:

A	B
7. Total sales	= SUM (B1:B6)
8. Average sales	= Average (B1:B6)
9. Lowest sales	= MIN (B1:B6)
10. Highest sales	= MAX(B1:B6)

Question 9.
Give the range reference in the following cases.

1. Range beginning from 1^st column, 1^st row and ending 16^th column, 8^th row
2. Range beginning from 5^th column, 7^th Row and ending 27^th column, 37^th row

Answer:

1. A1: P8
2. E7: AA37

Question 10.
Match the following:

A	B
a) Today	a) Today’s date & Time
b) Now	b) Convert date into corresponding value
c) Day	c) Today’s date
d) Date value	d) Day of the data referred in the formula

Answer:

A	B
a) Today	a) Today’s date
b) Now	b) Today’s date & time
c) Day	c) Day of the date referred in the formula
d) Date value	d) Convert date into corresponding value

Question 11.
Give the cell address or range reference in the following situations

1. Cell at 8^th column and 10^th row
2. Cell at 27^th column and 6^th row
3. Range starting from 5^th column, 9^th row and spread till 12^th column and 15^th row

Answer:

1. H10
2. AA6
3. E9: L15

Question 12.
What are the steps to be followed for naming cells and ranges.
Answer:
Step 1 – Select the cells or ranges that are to be named.

Step 2 – Select Define Range from the Data menu Data.

Step 3 – This will display a dialogue box us “Define Database Range”. It will provide a place to enter “Name”.

Step 4 – Click OK on the dialogue box.

Plus Two Accountancy Spread Sheet Practical Lab Work Questions and Answers

Question 1.
The following marks are obtained by 8 students in an examination.

Ascertain the grade obtained by students based on the following criteria

Procedure:
Step 1 – Open Libre Office Cal work sheet
Applications → Office → Libre Office calc

Step 2 – Enter the data in the given cells

Step 3 – Enter the following formula in cell D2
= IF(C2 < 30, “FAIL”, IF(C2 < 40, “D+”, IF(C2 < 50, “C”, IF(C2<60, “C+”, IF(C2 < 70, “B”, IF (C2 < 80, “B+”, IF(C2 < 90, “A”, IF(C2 < 100, “A+”))))))))

Step 4 – Drag the formula to D9
Output:

Question 2.
The sales made by 6 salesmen during three months are given below. You are required to prepare a statement showing the total sales of each salesman and total sales of the firm.

Procedure:
Step 1 – Open LibreOffice calc worksheet
Applications → Office → LibreOffice Calc

Step 2 – Enter the following data in the given cells

Step 3 – Enter the given formula in cell E2 = B2 + C2 + D2

Step 4 – Drag the formula to E7 to get the total sales of each salesman

Step 5 – Enter the given formula in cell E8 to get the total sales of the firm = SUM (E2: E7)
Output:

Question 3.
The monthly production of a company are given below

Month	Production (Units)
January	25000
February	20000
March	22000
April	18000
May	19000
June	24000

 

1. Find the total production for the six months
2. Find the average production of the six months
3. Find the number of months during the period
4. Find the lowest production of the six months
5. Find the highest production of the six months

Procedure:
Step 1 – Open LibreOffice Calc worksheet
Applications → Office → LibreOffice Calc

Step 2 – Enter the following data in the given cells

Step 3 – Enter the following formula in the given cells

B8	= SUM (B2:B7)
B9	= AVERAGE(B2: B7)
B10	= COUNT (B2:B7)
B11	= MIN (B2:B7)
B 12	= MAX (B2:B7)

Output:

Question 4.
For the recruitment of managers in different departments of a company, Applicant’s age should be greater than 35 and less than 45 as on 31/03/2018. Write the Spreadsheet statement to test when a candidate is eligible for recruitment or not, when his/ her date of birth is entered in a cell as input
Procedure:
Step 1 – Open LibreOffice Calc Spread sheet.
Applications → Office → LibreOffice Calc

Step 2 – Enter the given data in the following cells

Step 3 – Enter the date of birth in cell B2

Step 4 – Enter the following formula in cell B3 = Round ((B1-B2)/365,0). Then the age will display in the cell B3.

Step 5 – Enter the following formula in cell B4 to test the eligibility of the candidate = IF(B3<35, “Not eligible”, IF(B3>45, “Not eligible”, “Eligible”)

OR

= IF (AND (B>34, B3<46), “Eligible”, “Not eligible”)
Note: The data value entered in cell B2 and B3 are to be formatted as “Date”

Question 5.
The given table shows name of employees, Designation and monthly salary paid for different employees in Jose Traders.

Find out the following:

1. The total monthly salary by naming range (TOTAL SALARY)
2. The total monthly salary paid to marketing managers (MM) in the firm
3. The name of employees with a monthly salary of Rs.30000 by using LOOK UP function.

Procedure:
Step 1 – Open LibreOffice Calc worksheet
Applications → Office → LibreOffice Calc

Step 2 – Enter the data in the appropriate cells

Step 3 – Naming a range
Select the range which shows monthly salary ie C2: C7 click on “Data” from the menu bar. Select “Define Range” Type the name “TOTAL_SALARY” in the name box and press OK or Enter key

Select the Range (C2: C7) → Data → Define Ranges Type TOTAL_SALARY. Press OK/Enter Key

Step 4 – Enter the following formulas in respective cells .

B8	= SUM (TOTAL_SALARY)
B9	= SUMIF(B2:B7, “MM”, C2:C7)
B10	= LOOKUP(30000, C2:C7, A2:A7)

Output:

Question 6.
The following Data is given in the form of a Table.

• How many cells contain numbers only?
• How many cells contain any value?
• Count the number of cells containing the value exceeding 500.
• How many blank cells are there in the table?

Procedure:
Step 1 – Open LibreOffice Calc
Applications → Office → LibreOffice Calc

Step 2- Enter the given details in appropriate cells.

Step 3 – Enter the given details and formula in the Following cells.

Step 3 – Enter the given details and formula in the

Output:

Question 7.
The following details are given

Find out

• Find the name of student whose admission number is 8267
• Look up value 8136 and locate the fee paid status by using VLOOKUP
• Look up the name of student against admission number 8124
• Ad. No. of student who paid fee Rs. 580

Procedure:
Step 1 – Open LibreOffice Calc worksheet
Application → Office → LibreOffice Calc.

Step 2 – Enter the details in the given cells

Step 3 – Enter the given details and formula in the following cells

	A
8	Name of student Ad.No.8267
9	Fee paid status of Ad No.8136
10	Name of student Ad. No.8124
11	Ad.No. of student, paid Rs.580

	B
8	= LOOKUP(8267,A2:A7,B2:B7)
9	= VLOOKUP(8136,A2:C7,3,0)
10	= LOOKUP(8124,A2:A7,B2:B7)
11	= VLOOKUP (8370,A2:C7,3,0)

Output:

Question 8.
The marks in Accountancy of some students are given below.

Name	Mark
Priya	89
Indira	-ab-
Sindhu	56
Reny	64
Beena	49
Bindhu	50
Resmi	-ab-

Calculate:

1. Number of students in the class
2. Number of students appeared in the Accountancy examination
3. Total marks in Accountancy examination
4. Average Marks in Accountancy examination
5. Lowest mark in Accountancy
6. Highest mark in Accountancy

Procedure:
Step 1 – Open LibreOffice Calc Work sheet
Application → Office → LibreOffice Calc.

Step 2 – Enter the given data in appropriate cells.

Step 3 – Enter the following details and formula in the given cells

Output:

Question 9.
Mark summary of some students are given below.

Calculate

1. Number of cells containing 90 marks
2. Count the number of paper scored less than 40
3. Find the result of each student. (Pass or Fail) minimum mark required to pass is 30

Procedure:
Step 1 – Open LibreOffice Calc worksheet
Application → Office → LibreOffice calc

Step 2 – Enter the given data in appropriate cells

Step3 – Enter the following details and formula in appropriate cells

A8	Number of cells containing 90 mark
B8	= COUNTIF(B2: G7, “90”)
A9	Count the number of paper scored less than 40 Mark
B9	= COUNTIF(B2: G7, “40”)

Step4 – Find the result of the student

1^ststep – Count the number of marks less than 30 scored by each student For this,
Enter the formula in H2 =COUNT IF (B2: G2, “<30”)
Drag the equation to H7

2^nd step – Based on the result in H2: H7 range, we can find out the result in I2 cell by giving the following formula
=IF(H2 = 0, “PASS”, “FAIL”)
Drag (Copy) the formula to I7
Output:

Question 10.
List of debtors and creditors, and the amount due from them are given below

Calculate the total amount of receivables and payables using SUMIF functions.
Procedure:
Step -1 Open Libre Office Calc worksheet
Applications → Office → LibreOffice calc

Step 2 – Enter the details in the given cells.

Output:

Question 11.
Calculate Income Tax of following employees based on the following criteria.

Conditions

1. Tax rate is 40% of Total taxable Income.
2. For male, standard deduction is 150,000 3 For female, if taxable income is less than or equal to Rs.4,00,000, then the standard deduction is 2,00,000, otherwise 1,50,000

Procedure:
Step 1 – Open LibreOffice Calc worksheet
Application → Office → LibreOffice calc

Step 2 – Enter the given details in the given range

Step 3 – Calculation of Tax
Enter the formula in D2 cell and drag with the fill handle up to D8
= IF(B2 = “Male”, (C2 – 150000)*40%, IF (AND(C2 = “Female”, C2 <= 400000), (C2 – 200000)*40%, (C2 – 150000)*40%))
Output:

Name of Employee	Tax
Manoj	60,000
Praveena	1,20,000
Naveen	2,20,000
Riya	2,08,000
Rohit	1,40,000
Latha	80,000
Arya	2,84,000

Question 12.
For SI selection in Kerala polices there is a physical test, which consists of three items. A candidate has to qualify ANY ONE of the three tests to qualify for the final. The standard for the physical test is given below.

• Shot put: 5 meters or above.
• Ball throws 50 meters.
• 500-meter race within 5 minutes.

The following data is furnished:

Check whether the candidates quality or not.
Procedure:
Step 1 – Open LibreOffice Calc worksheet
Application → Office → LibreOffice calc

Step 2 – Enter the details in the form of a Table

Step 3 – Enter the following formula in F2 and drag with fill up to F6 = IF(OR(C2>=5, D2>=50, E2<=5), ‘‘Qualified”, “Not Qualified”)
Output:

Chest No.	Result
203	Qualified
216	Qualified
275	Qualified
304	Not Qualified
361	Qualified

Question 13.
Calculate DAY, MONTH and YEAR of 36525 Procedure:
Step 1 – Open LibreOffice Calc worksheet.

Step 2 – Enter the details in appropriate cells.

Output:

Question 14. Calculate the Date value of 28/10/1978/
Procedure:
Step 1 – Open LibreOffice calc work sheet
Application → Office → LibreOffice Calc

Step-2 – Enter the given details in Cell A1 = DATEVALUE (“28/10/1978”)
Output:

Question 15.
Find the age of Resi Jos based on her date of birth and today’s date Date of birth 06-05-1981.
Procedure:
Step 1 – Open Libre office Calc work sheet Application officer LibreOffice calc

Step 2 – Enter the following details in appropriate cells.

Output:

Question 16.
Below is given the table showing the name, department, and salary paid for different employees

• Find number of employees in the firm.
• Find number of employees in Production Department.
• Find the total monthly salary paid in Purchase Department.
• Find the total monthly salary paid in Finance Department.

Procedure:
Step 1 – Open a new blank work sheet in LibreOffice Calc.

Step 2 – Enter the following details as given below.

Step 3 – Enter the following text in different cells as given below.

Cell	Text
A9	No. of employees in the firm
A10	No. of employees in the production department
A11	Total monthly salary paid in purchase depart­ment
A12	Total monthly salary paid in finance depart­ment

Step 4 – Enter the following formula in different cells as given below.

Cell	Formula
B9	= COUNTA(A2: A8)
B10	= COUNTIF(B2: B8, “Production”)
B11	= SUMIF (B2: B8, “Purchase”, C2: C8)
B12	= SUMIF (B2: B8, “Finance”, C2: C8)

Output:

Question 17.
Below is given the table showing the Name, Department, and Salary paid for different employees.

• Name the employee name column as “Emp Name”, Department column as “Dept” and monthly salary column as “Salary”.
• Find the total monthly salary.

procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2 – Enter the following details in respective cells.

Step 3 – Select the range A2: A6, From the Data tool menu, select “Define Range” and – give name as “ Emp Name” and click (OK) button.

Step 4 – Select the range B2: B6, From the Data tool menu, select “Define Range” and give name as “Dept” and click (OK) button.

Step 5 – Select the range C2: C6, From the Data tool menu, select “Define Range” and give name as “Salary” and click (OK) button.

Step 6 – Enter the following formula to find out the total monthly salary.

Cell	Text / Formula
A 7	Total Monthly Salary
B 7	= SUM (Salary)

Output:

Total Monthly Salary

12500

Question 18.
Below is given the table showing the Name, Class, and Fees due to different students.

• Find out the total fees due from the students.
• Find the average amount of fees
• Find the highest amount of fees
• Find the lowest amount of fees

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following details in the worksheet as follows.

Step 3 – Enter the following text in respective cells.

Cell	Text
A9	Total Fees due
A10	Average Fees
A11	Highest amount of fees
A12	Lowest amount of fees

Step 4 – Enter the following formula in respective cells

Cell	Formula
B9	= SUM (D2: D8)
B10	= Average (D2: D8)
B11	= Max (D2;D8)
B12	= Min (D2: D8)

Output:

Total Fees Due	2700
Average Fees	386
Highest Amount of Fees	625
Lowest Amount of Fees	60

Question 19.
From the given values, calculate the following:

• Find the number of values
• Find the total sum of the values
• Find the average
• Find the largest value
• Find the smallest value

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the values in the work sheet as follows.

Step 3 – Enter the following text in the respective cells.

Cell	Text
A8	Number of values
A9	Sum of the values
A10	Average
AH	Largest value
A12	Smallest value

Step 4 – Enter the following formula in the respective cells.

Cell	Formula
B8	= COUNT (A1: A7)
B9	= SUM(A1: A7)
B10	= AVERAGE/A1: A7)
B11	= MAX/A1: A7)
B12	= MIN(A1: A7)

Output:

Number of values	7
Sum of the values	3000
Average	429
Largest value	700
Smallest value	200

Question 20.
Prepare a statement showing advance tax collected form the employees @ 10% of the yearly salary from those who receive Rs. 500000 or more.

Employees	Yearly salary
Fijo	855000
Joshy	650000
Roby	720000
Bose	425000
Prince	570000
Binoy	380000

Procedure:
Step 1 – Open a new blank worksheet in LiberOffice Calc.

Step 2 – Enter the following details in respective cells.

Step 3 – Select the range A2: A7, From Data Tools menu, select ‘Name a Range’ and give name as “Employees” and click (OK) button.

Step 4 – Enter the text /formula in the following cells.

	C
1	Advance Tax
2	= IF (B2>=500000, B2*10%, 0)
3
4
5
6
7

Step 5 – Copy formula to C3: C7 to the last employee.

Step 6 – Enter “Total Advance Tax” in cell B8 and enter the formula ‘‘SUM C2: C7” to get total amount.
Output:

Fijo	85500
Joshy	65000
Roby	72000
Prince	57000

Question 21.
Mrs. Leela, the class teacher is analysing the performance of her students in a class test.

Find out

• Total number of students
• Number of students appeared in the class test
• Number of students with no grade
• Number of A+ holders
• Number of B grade holders.

Procedure:
Step 1 – Open a new blank worksheet in LiberOfficeCalc

Step 2 – Enter the following details in the respective cells.

Step 3 – Enter the text /formula in the corresponding

Output:

Total number of students	8
Students appeared in-class test	6
Students with no grade	2
Number of A+ holders	2
Number of B grade holders	1

Question 22.
The monthly sales effected by 6 salesmen are given below.

Calculate the commission earned by each salesman on the basis of the following rules.

Total sales	Commission
Less than 8000	Nil
8000- 10000	5%
10000- 12000	8%
More than 12000	10%

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc.

Step 2 – Enter the following details in the respective cells.

Step 3 – Enter the formula = SUM (B2: D2) in cell E2 to get the total sales. Copy the formula to the last employee.

Step 4 – Enter the formula = IF (E2 >= 12000, E2 * 10%, IF (E2 >= 10000, E2*8%, IF(E2 >= 8000, E2 * 5%,0)))
Copy the formula to the last employee.
Output:

Question 23.
ABC Ltd categorises their salesmen into four on the basis of sales targets achieved in each quarter. The criteria and sales are given below.

Performance criteria
Total sales – Grade
More than 100000 – Excellent
50000-100000 – Good
30000-50000 – Average
Less than 30000 – Bad
Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following details in the respective cells.

Step 3 – Enter the formula in E2 to get total sales = SUM (B2: D2)
Copy the formula to the last employee.

Step 4 – Enter the formula in F2 to get the commission = IF(E2 >= 100000, “EXCELLENT”, IF (E2 >= 50000, “GOOD”, IF (E2 >= 30000, “AVERAGE”, “BAD”)))
Copy the formula to the last employee.
Output:

Ramesh	BAD
Suresh	GOOD
Mahesh	AVERAGE
Rajesh	GOOD
Sukesh	GOOD

Question 24.
Below is given the name and address of some students. Combine and show details in an address format using CONCATENATE function.

Procedure:
Step 1 – Open a new blank worksheet in LiberOffice Calc
Application → Office → LibreOffice Calc

Step 2 – Enter the following details in appropriate cells

Step 3 – Enter the following formula in F2 cell = CONCATENATE (A2, “ ”, B2, “ ”, C2, “ ”, D2, “ ”, E2)

Step – 4 Copy down the formula up to the row of the last employee.
Output:

Address

Sanjan Kollannur Kechery 680579.

Shaji Amala Bhavan Ollur 680514 Thrissur.

Nithin MRA/258 Mannuthy 680007 Ernakulam.

Question 25.
Mr Venugoapl is planning to invest Rs. 10000 in the beginning of each year in an annual investment scheme. The interest rate is 8% and the term of the scheme is 10 years. Using FV function, find out how much amount he will get at the expiry of the scheme.
Procedure:
Step 1 – Open a new blank worksheet in LiberOffice Calc

Step 2 – Enter the given detials and formula in diffemt cells

Output:

Future value

156454.87

Question 26.
Anakha Ltd. wants to select one machinery, out of the two alternatives available on the basis of net present value. The cost and inflows of these machineries are given below.

Assuming annual interest rate of 10%, find out net present values of these two machineries.
Procedure:
Step 1 – Open a blank work sheet in LiberOffice Calc

Step 2 – Enter the following details in respective cells.

Step 3 – Enter the formula = NPV (10%, C2:F2) – B2 in G2 to get the net present value of semi automatic machinery.
Copy the formula to G3
Output:

Machinery	NPV
Semi-Automatic	4683.42
Fully Automatic	52891.2

Question 27.
Calculate Net Present Value (NPV) from the following data

Cost of Machinery	2000000
Cash inflows -1 year	60000
Cash inflows – II year	80000
Cash inflows – III year	82000
Cost of capital	12%

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following details in the respective cells.

Step 3 – Enter the formula in B6 = NPV (12%, B2:B4)-B1 to get NPV
Output:

Net Present value

24287.08

Question 28.
Write the Libre Office Calc formula to multiply a given number entered in a cell with the following conditions

• If the cell value is less than 20, then multiply by 1
• If the cell value is greater than or equal to 20 but less than 40, then multiply by 2
• If the cell value is greater than or equal to 40 but less than 80, then multiply by 3
• If the cell value is greater than or equal to 80 but less than 100, then multiply by 4.
• If the cell value is greater than or equal to 100, then display “Enter a value less than 100” Use nested IF function.

Procedure:
Step 1 – Open a blank work sheet in LibreOffice Calc

Step 2 – Enter the following details in respective cells.

Step 3 – Enter any value in cell A2; then the result will be in cell B2.
Output:

Cell value	Result
18	18
37	74
115	Enter value less than 100
85	340

Question 29.
Salary detail of 8 employees are given below.

Develop a formula to compute tax under the following conditions.
1. Tax rate-20%

2. For males, Rs. 2,00,000 is allowed as standard deduction. For females, if taxable income is less than or equal to Rs. 5,00,000, then Rs. 3,00,000 is allowed as standard deduction, otherwise Rs. 2, 50,000. Use suitable logical function.
Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following details in respective cells.

Step 3 – Enter the formula in cell E2 and copy it to the last employee.
=IF (C2 = “MALE”, (D2-20Q000)*20%, IF (AND(C2 = “FEMALE”, D2<=500000), (D2 – 300000) * 20%, (D2-250000)*20%))
Output:

Name	Tax
Elsy	40,000
Jose	90,000
Rosily	20,000
Rejina	40,000
Antony	1,00,000
Jessy	30,000
George	80,0000
Baby	60,000

Question 30.
The recruitment process of language teachers in a school consists of three items like interview, group discussion and paper presentation. A candidate has to qualify any one of the three tests to qualify for the written test. The prescribed standard for the items are given below.

• Interview – 16 score out of 20.
• Group Discussion – 25 score out of 30.
• Paper Presentation – 40 score out of 50

Other informations are also available.

Write a formula using OR function to check whether a candidate qualify or not.
Procedure:
Step 1 – Open a blank work sheet in LiberOffice Calc.

Step 2 – Enter the available data in work sheet

Step 3 – Enter the formula in F2 and copy it to the last candidate
= IF(OR (C2>=16, D2>=25, E2>=40) “Qualified for the written Test”, “Not Qualified”)
Output:

Question 31.
Business studies, Accountancy and Economics are commerce subjects. Write a formula to check whether text is a commerce subject. If yes, print **** is a commerce subject, otherwise print **** is not a commerce subject. Write procedure based on the available hints
Procedure:
Step 1 – Open a blank work sheet in LiberOffice Calc

Step 2 – Enter the text in the following cells.

Step 3 – Enter the name of subject in A2 and type the formula in B2 and drag it
= IF (OR (A2= “Business studies”, A2 = “Accountancy”, A2= “Economics”), A2 & “is a” & “Commerce Subject”, A2 & “is” & “not a commerce subject”)
Output:

Subject	Remarks
Malayalam	Malayalam is not a commerce subject
Business studies	Business studies is a commerce subject
Chemistry	Chemistry is not a commerce subject
Mathematics	Mathematics is not a commerce subject
Accountancy	Accountancy is a commerce subject.

Question 32.
ABC company issued a security with par value Rs. 50000 on 1/1/2013. The first interest date is 1-4-2013, the settlement date is 31-12-2015 an the annual coupon rate is 6%. The security’s payments are made quarterly, and a US (NASD) 30/360 day count basis is used. Use ACCRINT() function to calculated the accrued interest of a security that pays periodic interest.
Procedure:
Step 1 – Open a blank new worksheet in LibreOffice Calc.

Step 2 – Enter the data in the following cells.

Step 3 – Enter the formula in B9 = ACCRINT (B2, B3, B4, B5, B6, B7, B8)
Output:

Accrued Interest

9000

Question 33.
Prince took a loan of Rs. 500000 from banks @ 12% interest p.a. repayable after 15 years. Compute interest payable at the end of (a) First year (b) Second year (c) Fifth year and the last year.
procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the given details in respective cells

Step 3 – Enter the given formula in respective cells.

• B8 = CUMIPMT(B3/12, B2*12, B4, B5, B6, B7)
• C8 = CUMIPMT(C3/12, C2*12, C4, C5, C6, C7)
• D8 = CUMIPMT(D3/12, D2*12, D4, D5, D6, D7)
• E8 = CUMIPMT(E3/12, E2*12, E4, E5, E6, E7)

Output:

Year	Interest
First year	59316.92
Second year	57707.11
Fifth year	51545.84
Last year	4470.16

Question 34.
Calculate the present value of an annuity that pays Rs. 3000 per month for a period of 5 years. The interest is 11.5% per annum and each payment is made at the end of the month.
Assume the payments are made.

• At the beginning of each month
• At the end of each month

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following data/ Formula in respective cells.

Step 3 – Enter the formula as follows
B6 = PV (B2/12, B3*12, B4, B5)
C6 = PV(C2/12, C3*12, C4, C5)
Output:

Present value – at the beginning	136409.4
Present value – at the end	136410.04

Question 35.
The table given below shows the First name, Middle name and Last name of some employees. Show their Full name in the next column using CONCATENATE Function.

Procedure:
Step 1 – Open a new blank worksheet in Liber Office Calc
Application → Office → Libre Office Calc

Step 2 – Enter the following details in appropriate cells

Step 3 – Enter the following formula in D2 cell = CONCATENATE (A2, “ ”, B2, “ ”, C2)

Step 4 – Copy down the formula up to the row of last employee.
Output:

FULL NAME

Roby Antony Alappatt

Hema Gangadharan Menon

Cili Jose Vazhappilly

Santhosh Jacob Kannanaikal

Students can Download Chapter 4 Determinants Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 4 Determinants

Plus Two Maths Determinants Three Mark Questions and Answers

Question 1.
Using properties of determinants prove \(\left|\begin{array}{ccc}{x} & {y} & {x+y} \\{y} & {x+y} & {x} \\{x+y} & {x} & {y}\end{array}\right|\) = -2(x³ + y³).
Answer:

= 2(x + y)(-x² + xy – y²) = -2(x³ + y³).

Question 2.
If a, b, c are real numbers and \(\left|\begin{array}{lll}{b+c} & {c+a} & {a+b} \\{c+a} & {a+b} & {b+c} \\{a+b} & {b+c} & {c+a}\end{array}\right|\) = 0, Show that a = b = c.
Answer:

2(a + b + c) [(b – c) (c – b) – (b – a) (c – a)] =0 (a+b+c) = 0
(a + b + c) = 0 or (b – c) (c – b) = (b – a) (c – a)
(a + b + c) = 0 or a = b = c.

Question 3.
Solve using properties of determinants.
\(\left|\begin{array}{ccc}{2 x-1} & {x+7} & {x+4} \\{x} & {6} & {2} \\{x-1} & {x+1} & {3}
\end{array}\right|\) = 0
Answer:

⇒ (x – 1) (x² + x – 6x + 6) = 0
⇒ (x – 1)(x² – 5x + 6) = 0
⇒ (x – 1) (x – 3) (x – 2) = 0
⇒ x = 1, x = 3, x = 2.

Question 4.
If \(\left|\begin{array}{cc}{3} & {x} \\{x} & {x}\end{array}\right|=\left|\begin{array}{cc}{-2} & {2} \\{4} & {1}\end{array}\right|\), find the value of x.
Answer:
\(\left|\begin{array}{cc}{3} & {x} \\{x} & {x}\end{array}\right|=\left|\begin{array}{cc}{-2} & {2} \\{4} & {1}\end{array}\right|\)
⇒ 3x – x² = – 2 – 8
⇒ x² – 3x – 10 = 0
⇒ x = 5, -2.

Question 5.
A = \(\left[\begin{array}{ccc}{1} & {-3} & {1} \\{2} & {0} & {4} \\{1} & {2} & {-2}
\end{array}\right]\)

1. Calculate |A| (1)
2. Find |adjA| {Hint: using the property A × adjA = |A|I} (1)
3. Find |3A| (1)

Answer:
1. |A| = \(\left[\begin{array}{ccc}{1} & {-3} & {1} \\{2} & {0} & {4} \\{1} & {2} & {-2}
\end{array}\right]\) = – 28.

2. A × adjA = |A|I

3. |3A| = 27 × |A| = 27 × -28 = -756.

Question 6.
Using properties of determinants prove the following.

Answer:

= 2{-{-c){b{a – c)) – b(-c(c + a))}
= 2{c(ab – cb) + b(c² + ac)}
= 2{abc – c²b + bc² + abc)} = 4abc.

Plus Two Maths Determinants Four Mark Questions and Answers

Question 1.
(i) If \(\left|\begin{array}{rrr}{1} & {-3} & {2} \\{4} & {-1} & {2} \\{3} & {5} & {2}\end{array}\right|\) = 40, then \(\left|\begin{array}{ccc}{1} & {4} & {3} \\{-3} & {-1} & {5} \\{2} & {2} & {2}\end{array}\right|\) = ?
(a)   0
(b)  – 40
(c)  40
(d)  2 (1)
(ii) \(\left|\begin{array}{rrr}{3} & {-3} & {2} \\{12} & {-1} & {2} \\{9} & {5} & {2}\end{array}\right|\) = ?
(a) 120
(b)  40
(c)  – 40
(d)  0 (1)
(iii) Show that ∆ = \(\left|\begin{array}{ccc}{-a^{2}} & {a b} & {a c} \\{b a} & {-b^{2}} & {b c} \\{a c} & {b c} & {-c^{2}}\end{array}\right|\) = 4a²b²c² (2)
Answer:
(i) (c) 40

(ii) (a)120

(iii) ∆ = abc\(\left|\begin{array}{ccc}{-a} & {a} & {a} \\{b} & {-b} & {b} \\{c} & {c} & {-c}\end{array}\right|\) take a, b, c from C₁, C₂, C₃

Question 2.

Answer:
(i) \(\left|\begin{array}{ll}{2} & {4} \\{5} & {1}\end{array}\right|=\left|\begin{array}{ll}{2 x} & {4} \\{6} & {x}\end{array}\right|\) ⇒ -18 = 2x² – 24.
⇒ 2x² = 6 ⇒ x² = 3 ⇒ x = \(\pm \sqrt{3}\).

Question 3.
Prove that \(\left|\begin{array}{ccc}{(b+c)^{2}} & {a^{2}} & {a^{2}} \\{b^{2}} & {(c+a)^{2}} & {b^{2}} \\{c^{2}} & {c^{2}} & {(a+b)^{2}}\end{array}\right|\) = 2abc(a + b + c)³.
Answer:


= (a + b + c)² × 2ab [(b + c) (c + a) – ab]
= (a + b + c)² × 2ab [bc + ab + c² + ac – ab)
= (a + b + c)² × 2abc [a + b + c]
= 2abc (a + b + c)³.

Question 4.
(i) Let the value of a determinant is ∆. Then the value of a determinant obtained by interchanging two rows is
(a) ∆
(b) -∆
(c) 0
(d) 1 (1)
(ii) Show that \(\left|\begin{array}{ccc}{a+b} & {b+c} & {c+a} \\{b+c} & {c+a} & {a+b} \\{c+a} & {a+b} & {b+c}\end{array}\right|=2\left|\begin{array}{lll}{a} & {b} & {c} \\{b} & {c} & {a} \\{c} & {a} & {b}\end{array}\right|\) (3)
Answer:
(i) (b) -∆

(ii) Operating C₁ → C₁ + C₂ + C₃, we have

Question 5.
Test the consistency 3x – y – 2z = 2, 2y – z = -1, 3x – 5y = 3.
Answer:
The given system of equations can be put in the matrix form, AX = B, where

|A| = 3(0 – 5) + 1(0 + 3) – 2(0 – 6) = 0
C₁₁ = -5, C₁₂ = -3, C₂₁ = -6, C₂₂ = 10, C₂₃ = 6, C₃₁ = 12, C₃₂ = 5, C₃₃ = 6


Therefore the system is inconsistent and has no solutions.

Question 6.
Consider the system of equations 2x – 3y = 7 and 3x + 4y = 5

1. Express the system in AX = B form. (1)
2. Find adj A (2)
3. Solve the system of equations. (1)

Answer:
1. |A| = \(\left|\begin{array}{cc}{2} & {-3} \\{3} & {4}\end{array}\right|\) = 8 + 9 = 17.

2. c₁₁ = 4, c₁₂ = -3, c₂₁ = 3, c₂₂ = 2,

3. The given equations can be expressed in the form AX = B,

Question 7.
(i) If A and B are matrices of order 3 such that|A| = -1; |B| = 3, then |3AB| is
(a) -9
(b) -27
(c) -81
(d) 9 (1)
(ii) If A = \(\left[\begin{array}{cc}{1} & {\tan x} \\{-\tan x} & {1}\end{array}\right]\), Show that A^T A^-1 = \(\left[\begin{array}{cc}{\cos 2 x} & {-\sin 2 x} \\{\sin 2 x} & {\cos 2 x}\end{array}\right]\) (3)
Answer:
(i) (c) -81 (since |3AB| = 27|A||B|).

(ii) |A| = \(\left[\begin{array}{cc}{1} & {\tan x} \\{-\tan x} & {1}\end{array}\right]\) = sec²x ≠ 0, therefore A is invertible.

Question 8.
Consider the determinant ∆ = \(\left|\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\{y} & {y^{2}} & {1+y^{3}} \\{z} & {z^{2}} & {1+z^{3}}\end{array}\right|\), Where x, y, z, are different.
(i) Express the above determinant as sum of two determinants. (1)
(ii) Show that if ∆ = 0, then 1 + xyz = 0. (3)
Answer:
(i) Given,
∆ = \(\left|\begin{array}{ccc}{x} & {x^{2}} & {1+x^{3}} \\{y} & {y^{2}} & {1+y^{3}} \\{z} & {z^{2}} & {1+z^{3}}\end{array}\right|=\left|\begin{array}{ccc}{x} & {x^{2}} & {1} \\{y} & {y^{2}} & {1} \\{z} & {z^{2}} & {1}\end{array}\right|+\left|\begin{array}{ccc}{x} & {x^{2}} & {x^{3}} \\{y} & {y^{2}} & {y^{3}} \\{z} & {z^{2}} & {z^{3}}\end{array}\right|\)

Given, ∆ = 0 ⇒ (1 + xyz)(y – x)(z – x)(z – y) = 0 ⇒ 1 + xyz = 0
∵ x ≠ y ≠ z.

Question 9.
(i) The value of the determinant \(\left|\begin{array}{cc}{\sin 10} & {-\cos 10} \\{\sin 80} & {\cos 80}\end{array}\right|\) is
(a) – 1
(b) 1
(c) 0
(d) – 2 (1)
(ii) Using properties of determinants, show that (3)
\(\left|\begin{array}{lll}{a} & {a^{2}} & {b+c} \\{b} & {b^{2}} & {c+a} \\{c} & {c^{2}} & {a+b}\end{array}\right| = (b – c) (c – a) (a – b) (a + b + c)\)
Answer:
(i) (b) Since,
sin 10 cos 80 + cos 10 sin 80 = sin (10 + 80) =sin 90 = 1.

(ii) Let C₃ → C₃ + C₁

= (a + b + c)(b – a)(c – a)(c + a – b – a)
= (a + b + c)(b – a)(c – a)(c – b)
= (b – c)(c – a)(a – b)(a + b + c).

Question 10.
(i) Choose the correct answer from the bracket. Consider a square matrix of order 3. Let C₁₁, C₁₂, C₁₃ are cofactors of the elements a₁₁, a₁₂, a₁₃ respectively, then a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ is (1)
(a) 0
(b) |A|
(c) 1
(d) none of these.
(ii) Verify A(adjA) = (adjA)A = |A|I for the matrix A = \(\left[\begin{array}{ll}{5} & {-2} \\{3} & {-2}\end{array}\right]\) that, where I = \(\left[\begin{array}{ll}{1} & {0} \\{0} & {1}\end{array}\right]\) (3)
Answer:
(i) (b) |A|

(ii) |A| = \(\left|\begin{array}{cc}{5} & {-2} \\{3} & {-2}\end{array}\right|\) = – 4
C₁₁ = – 2, C₁₂ = – 3, C₂₁ = 2, C₂₂ = 5

Hence A(adjA) = (adjA)A = |A|I.

Question 11.
Consider the following system of equations x + 2y = 4,2x + 5y = 9

1. If A = \(\left[\begin{array}{ll}{1} & {2} \\{2} & {5}\end{array}\right]\), find |A| (1)
2. Express the above system of equations in the form AX = B (1)
3. Find adj A, A^-1 (1)
4. Solve the system of equations. (1)

Answer:
1. |A| = \(\left[\begin{array}{ll}{1} & {2} \\{2} & {5}\end{array}\right]\) = 5 – 4 = 1

2. The given system of equation can be expressed in the form AX = B.

3. Cofactor matrix of A = \(\left[\begin{array}{cc}{5} & {-2} \\{-2} & {1}\end{array}\right]\)

4. We have,

x = 2, y = 1.

Question 12.
Consider the point X(-2, -3), B(3, 2), C(-1, -8)

1. Find the area of ∆ABC (2)
2. Find third vertex of any other triangle with same area and base AB. (2)

Answer:
1. \(\frac{1}{2}\left|\begin{array}{ccc}{-2} & {-3} & {1} \\{3} & {2} & {1} \\{-1} & {-8} & {1}\end{array}\right|\)
\(\frac{1}{2}\) (- 2(2 + 8) + 3(3 + 1) + 1(- 24 + 2)) = – 15
Area of ∆ ABC = 15.

2. The base AB is fixed and the third point is variable. Therefore we can choose any x coordinate and find y coordinate or vice versa.

⇒ – 2(2 – y) + 3(3 – 1) + 1(3y – 2) = 30
⇒ – 4 + 2y + 6 + 3y – 2 = 30
⇒ 5y = 30 ⇒ y – 6
Therefore point is(1, 6).

Question 13.
Find the inverse of the following

Answer:
(i) Let |A| = \(\left|\begin{array}{lll}{1} & {2} & {3} \\{0} & {2} & {4} \\{0} & {0} & {5}\end{array}\right|\) = 10
C₁₁ = 10, C₁₂ = 0, C₁₃ = 0, C₂₁ = – 10, C₂₂ = 5, C₂₃ = 0, C₃₁ = – 2, C₃₂ = – 4, C₃₃ = 2

(ii) Let |A| = \(\left|\begin{array}{ccc}{1} & {0} & {0} \\{3} & {3} & {0} \\{5} & {2} & {-1}\end{array}\right|\) = -3
C₁₁ = -3, C₁₂ = 3, C₁₃ = -9, C₂₁ = 0, C₂₂ = -1, C₂₃ = -2, C₃₁ = 0, C₃₂ = 0, C₃₃ = 3

(iii) Let |A| = \(\left|\begin{array}{ccc}{2} & {1} & {3} \\{4} & {-1} & {0} \\{-7} & {2} & {1}\end{array}\right|\)
= 2(-1 – 0) -1(4 – 0) + 3(8 – 7) = -3
C₁₁ = -1, C₁₂ = -4, C₁₃ = 1, C₂₁ = 5, C₂₂ = 23, C₂₃ = -11, C₃₁ = 3, C₃₂ = 12, C₃₃ = -6

(iv) Let |A| = \(\left|\begin{array}{ccc}{1} & {-1} & {2} \\{0} & {2} & {-3} \\{3} & {-2} & {4}\end{array}\right|\)
= 1(8 – 6) + 1(0 + 9) + 2(0 – 6) = -1
C₁₁ = 2, C₁₂ = -9, C₁₃ = -6, C₂₁ = 0, C₂₂ = -2, C₂₃ = -1, C₃₁ = 3, C₃₂ = 3, C₃₃ = 2

Question 14.
Consider the system of equations 5x + 2y = 4, 7x + 3y = 5. If A = \(\left[\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right]\), X = \(\left[\begin{array}{l}{\mathrm{r}} \\{y}\end{array}\right]\) and B = \(\left[\begin{array}{l}{4} \\{5}\end{array}\right]\)

1. Find |A| (1)
2. Find A^-1 (2)
3. Solve the above system of equations. (1)

Answer:
1. |A| = \(\left|\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right|\) = 15 – 14 = 1.

2. Given, A = \(\left[\begin{array}{ll}{5} & {2} \\{7} & {3}\end{array}\right]\)

3. X = A^-1B

⇒ x = 2, y = -3.

Plus Two Maths Determinants Six Mark Questions and Answers

Question 1.
(i) Let A be a square matrix of order ‘n’ then |KA| = …….. (1)
(ii) Find x if \(\left|\begin{array}{cc}{x} & {2} \\{18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\{18} & {6}\end{array}\right|\) (2)
(iii) Choose the correct answer from the bracket. The value of the determinant \(\left|\begin{array}{ccc}{0} & {p-q} & {p-r} \\{q-p} & {0} & {q-r} \\{r-p} & {r-q} & {0}
\end{array}\right|\) is ….. (1)
(iv) Consider \(\left|\begin{array}{ccc}{a} & {a+b} & {a+b+c} \\{2 a} & {3 a+2 b} & {4 a+3 b+2 c} \\{3 a} & {6 a+3 b} & {10 a+6 b+3 c}\end{array}\right|\) (2)
Answer:
(i) If A be a square matrix of order n, then |KA| = Kⁿ|A|

(ii) \(\left|\begin{array}{cc}{x} & {2} \\{18} & {x}\end{array}\right|=\left|\begin{array}{cc}{6} & {2} \\{18} & {6}\end{array}\right|\) ⇒ x² – 36 = 0
⇒ x² = 36 ⇒ x = ±6.

(iii) (c) 0 (since the given determinant is the determinant of a third order skew symmetric matrix)

= a [7a² + 3ab – 6a² – 3ab] = a(a²) = a³

Question 2.
(i) Let \(\left|\begin{array}{lll}{1} & {3} & {2} \\{2} & {0} & {1} \\{3} & {4} & {3}
\end{array}\right|\) = 3, then what is the value of \(\left|\begin{array}{lll}{1} & {3} & {2} \\{4} & {0} & {2} \\{3} & {4} & {3}\end{array}\right|\) = ? and\(\left|\begin{array}{lll}{6} & {7} & {6} \\{2} & {0} & {1} \\{3} & {4} & {3}\end{array}\right| \) = ? (2)
(Hint: Use the properties of determinants)
(ii) Using properties of determinants show that (4)
\(\left|\begin{array}{ccc}{1+a} & {1} & {1} \\{1} & {1+b} & {1} \\{1} & {1} & {1+c}\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Answer:

(ii) Taking ‘a’ from R₁, ‘b‘ from R₂,’C’ from R₃

Question 3.
If A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)

1. Find |A| (1)
2. Find adj.A. (2)
3. Solve 2x – 3y + 5z = 11, 3x + 2y – 4z = -5, x + y – 2z = -3 (3)

Answer:
1. A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)
|A| = 2 × 0 + 3x – 2 + 5 = -1.

2. Co.factor A

3. Given

i.e; AX = B ⇒ X = A^-1 B

Question 4.
Let A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {1} & {-3} \\{1} & {1} & {1}\end{array}\right]\)

1. Is A singular? (1)
2. Find adj A. (2)
3. Obtain A^-1 (1)
4. Using A^-1 solve the system of equations x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2 (2)

Answer:
1. A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {1} & {-3} \\{1} & {1} & {1}\end{array}\right]\)
⇒ |A| = 4 + 5 + 1 = 10 ≠ 0
A is non singular matrix.

2. Cofactor A

3. A^-1 = \(\frac{1}{10}\) \(\left[\begin{array}{ccc}{4} & {2} & {2} \\{-5} & {0} & {5} \\{1} & {-2} {3}\end{array}\right]\)

4. Given, AX = B ⇒ X = A^-1 B

⇒ x = 2, y = -1, z = 1.

Question 5.
Solve the following system of linear equations.

1. x + y + z = 3, y – z = 0, 2x – y = 1 (6)
2. 5x – 6y + 4z = 15 , 7x + 4y – 3z = 19, 2x + y + 6z = 46 (6)
3. x + 2y + 5z = 10, x – y – z = -2, 2x + 3_y-2 = -11 (6)

Answer:
1. Let AX = B
Where A = \(\left[\begin{array}{ccc}{1} & {1} & {1} \\{0} & {1} & {-1} \\{2} & {-1} & {0}\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right], B=\left[\begin{array}{l}{3} \\{0} \\{1}\end{array}\right]\)
|A| = 1(0 – 1) – 1(0 + 2) + 1(0 – 2) = -5
C₁₁ = -1, C₁₂ = -2, C₁₃ = -2, C₂₁ = -1, C₂₂ = 3, C₂₃ = 3, C₃₁ = -2, C₃₂ = 1, C₃₃ = 1

2. Let AX = B,
Where A = \(\left[\begin{array}{ccc}{5} & {-6} & {4} \\{7} & {4} & {-3} \\{2} & {1} & {6}\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right],B=\left[\begin{array}{c}{15} \\{19} \\{46}\end{array}\right]\)
|A| = 5(24 + 3) + 6(42 + 6) + 4(7 – 8) = 419
C₁₁ = 27, C₁₂ = -48, C₁₃ = -1, C₂₁ = -1, C₂₂ = 22, C₂₃ = -17, C₃₁ = 2, C₃₂ = 43, C₃₃ = 62

3. Let AX = B
\(\text { Where } A=\left[\begin{array}{ccc}{1} & {2} & {5} \\{1} & {-1} & {-1} \\{2} & {3} & {-1}
\end{array}\right], X=\left[\begin{array}{c}{x} \\{y} \\{z}\end{array}\right], B=\left[\begin{array}{c}{10} \\{-2} \\{-11}\end{array}\right]\)
|A| = 1(4) – 2(1) + 5(5) = 27
C₁₁ = 4, C₁₂ = -1, C₁₃ = 5, C₂₁ = 17, C₂₂ = -11, C₂₃ = 1, C₃₁ = 3, C₃₂ = 6, C₃₃ = -3

⇒ x = -1, y = -2, z = 3.

Question 6.
If f(x) = \(\left[\begin{array}{ccc}{\cos x} & {-\sin x} & {0} \\{\sin x} & {\cos x} & {0} \\{0} & {0} & {1}\end{array}\right]\)
(i) Find f(-x) (2)
(ii) Find (f(x)]^-1 (2)
(iii) Is |f(x)]^-1 = f(-x)? (2)
Answer:

(ii) |f(x)| = \(\left[\begin{array}{ccc}{\cos x} & {-\sin x} & {0} \\{\sin x} & {\cos x} & {0} \\{0} & {0} & {1}\end{array}\right]\) = cos x (cos x) + sin x (sin x) = 1 ≠ 0
Therefore , [f(x)]^-1 exists.
The cofactors are as follows.
C₁₁ = cos x, C₁₂ = -sin x, C₁₃ = 0, C₂₁ = sin x, C₂₂ = cos x, C₂₃ = 0, C₃₁ = 0, C₃₂ = 0, C₃₃ = 1

Since, |f(x)|= 1

(iii) Yes. From (1) and (2) we have,
[f(x)]^-1 =f(-x).

Question 7.
(i) Choose the correct answer from the bracket. If A = \(\left[\begin{array}{cc}{2} & {3} \\{1} & {-2}\end{array}\right]\) and A^-1 = kA, then the value of ‘k’ is
(a) 7
(b) -7
(c) \(\frac{1}{7}\)
(d)\(-\frac{1}{7}\) (1)
(ii) If A = \(\left[\begin{array}{ccc}{1} & {-1} & {1} \\{2} & {-1} & {0} \\{1} & {0} & {0}
\end{array}\right]\),
(a) A² (2)
(b) Show that A² = A^-1 (3)
Answer:


C₁₁ = 0, C₁₂ = 0, C₁₃ = 1, C₂₁ = 0, C₂₂ = -1, C₂₃ = -1, C₃₁ = 1, C₃₂ = 2, C₃₃ = 1

Question 8.
‘Arjun’ purchased 3 pens, 2 purses, and 1 instrument box and pays Rs. 410. From the same Shop ‘Deeraj’ purchases 2 pens, 1 purse, and 2 instrument boxes and pays Rs.290, while ‘Sindhu’ purchases 2pens, 2 purses, 2 instrument boxes and pays Rs. 440.

1. Translate the equation into system of linear equations. (2)
2. The cost of one pen, one purse and one instrument box using matrix method. (4)

Answer:
1. Let The price of one pen is Rs.x, one purse is Rs.y and one instrument box be Rs.z
3x + 2y + z = 410; 2x + y + 2z =290; 2x + 2y + 2z = 440(1) 2 mts.

2. The system can be represented by the matrix equation AX = B


C₁₁ = -2, C₁₂ = 0, C₁₃ = 2, C₂₁ = -2, C₂₂ = 4, C₂₃ = -2, C₃₁ = 3, C₃₂ = -4, C₃₃ = -1

Hence the cost one pen is Rs.20, one purse is Rs. 150 and one instrument box is Rs. 50.

Question 9.
If A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)

1. Find A^-1 (3)
2. Using it solve the system of equations 2x – 3y + 5z = 16, 3x + 2y – 4z = -4, x + y – 2z = -3 (3)

Answer:
1. A = \(\left[\begin{array}{ccc}{2} & {-3} & {5} \\{3} & {2} & {-4} \\{1} & {1} & {-2}\end{array}\right]\)
⇒ |A| = 0 + 3x – 2 + 5 = -1

2. Given AX = B
⇒ X = A^-1B

⇒ x = 2, y = 1, z = 3.

Question 10.
Consider the following system of equations x + y + 3z = 5, x + 3y – 3z = 1, -2x – 4y – 4z = -10
(i) Convert the given system in the form AX = B (1)
(ii) Find A^-1 (3)
(iii) Hence solve the system of equations. (2)
Answer:

(ii) i.e; AX = B, ⇒ X = A^-1 B ⇒ |A| = -24 + 10 + 6 = -8

(iii) X = A^-1B

= \(-\frac{1}{8}\) \(\left[\begin{array}{l}{-8} \\{-8} \\{-8}\end{array}\right]\)
⇒ x = 1, y = 1, z = 1.

Question 11.
Solve the following system by equations by matrix method x + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = -11.
Answer:
x + 2y + 5z = 10; x – y – z = -2; 2x + 3y – z = 11

⇒ x = -1, y = -2, z = 3.

Question 12.
If A = \(\left[\begin{array}{ccc}{3} & {-2} & {3} \\{2} & {1} & {-1} \\{4} & {-3} & {2}\end{array}\right]\)

1. Find |A| (1)
2. Find A^-1 (3)
3. Solve the linear equations 3x – 2y + 3z = 8; 2x + y – z = 1; 4x – 3y + 2z = 4 (2)

Answer:
1. |A| = \(\left[\begin{array}{ccc}{3} & {-2} & {3} \\{2} & {1} & {-1} \\{4} & {-3} & {2}\end{array}\right]\)
= 3(2 – 3) + 2(4 + 4) + 3(- 6 – 4) = -17.

2. |A| ≠ 0, hence its inverse exists.
A^-1 = \(\frac{1}{|A|}\)adj A
C₁₁ = -1, C₁₂ = -8, C₁₃ = -10, C₂₁ = -5, C₂₂ = -6, C₂₃ = 1, C₃₁ = -1, C₃₂ = 9, C₃₃ = 7

3. The given system of linear equations is of the form

∴ We have, x = 1, y = 2, z = 3.

Question 13.
if \(\left[\begin{array}{cc}{2} & {5} \\{-3} & {7}\end{array}\right] \times A=\left[\begin{array}{cc}{17} & {-1} \\{47} & {-13}\end{array}\right]\) then
(i) Find the 2 × 2 matrix A. (3)
(ii) Find A². (1)
(iii) Show that A² + 5A – 6I = 0, where I is the identity matrix of order 2. (2)
Answer:

Students can Download Chapter 7 Integrals Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals

Plus Two Maths Integrals Three Mark Questions and Answers

Question 1.
Integrate the following. (3 Score each)

1. ∫sin x sin 2x sin 3 xdx
2. ∫sec²x cos²2x dx

Answer:
1. We have sinxsin2xsin3x
= 1/2 (2sinxsin3x) sin2x
= 1/2 (cos2x – cos4x) sin2x
= 1/4 (2sin2xcos2x – 2cos4xsi n2x)
= 1/4 [sin4x – (sin6x – sin2x)]
= 1/4(sin4x + sin2x – sin6x)
∫sin x sin 2x sin 3 xdx
= \(\frac{1}{4}\) ∫(sin 4x + sin 2x – sin 6x)dx
= –\(\frac{1}{16}\) cos4x – \(\frac{1}{8}\) cos2x + \(\frac{1}{24}\) cos6x + c.

2. sec²x cos²2x = \(\frac{\left(2 \cos ^{2} x-1\right)^{2}}{\cos ^{2} x}\)
= \(\left(\frac{2 \cos ^{2} x}{\cos x}-\frac{1}{\cos x}\right)^{2}\) = (2cosx – secx)²
= 4cos²x + sec²x – 4
= 2(1 + cos2x) + sec²x – 4
= 2cos2x + sec²x – 2
∫sec² x cos² 2x dx = ∫(2 cos 2x + sec² x – 2)dx
= sin 2x + tan x – 2x + c.

Question 2.
Find \(\int \frac{2+\sin 2 x}{1+\cos 2 x} e^{x} d x\)?
Answer:

= ∫e^x [sec² x + tan x]dx
= ∫e^x[tanx + sec²x]dx = e^x tanx + c.

Question 3.
Evaluate \(\int \frac{\sec ^{2} x d x}{\sqrt{\tan ^{2} x+4}}\)?
Answer:
Put tanx = u, sec²xdx = dy

Question 4.
Find the following integrals.

Answer:
(i) I = \(\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x\)
Put cosx = t ⇒ -sin xdx = dt
When x = 0 ⇒ t = cos0 = 1,

(ii) I = \(\int_{0}^{1} x e^{x^{2}} d x\)
Put x² = t ⇒ 2xdx = dt
When x = 0 ⇒ t = 0,
x = 1 ⇒ t = 1
I = \(\frac{1}{2} \int_{0}^{1} e^{t} d t\) =

= [e¹ – e⁰] = e – 1.

Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,

(iv) I = \(\int_{0}^{2} x \sqrt{x+2} d x\)
Put x + 2 = t² ⇒ dx = 2tdt
When x = 0 ⇒ t = \(\sqrt{2}\), x = 2 ⇒ t = 2

(v) I = \(\int_{0}^{\frac{\pi}{2}} \sqrt{\sin x} \cos x d x\)
Put sin x = t ⇒ cos xdx = dt
When x = 0 ⇒ t = sin0 = 0,


Put tan x = t ⇒ sec² xdx = dt
When x = 0 ⇒ t = tan 0 = 0,

Question 5.
(i) If f (x) is an odd function, then \(\int_{-a}^{a} f(x)\) = ?
(a) 0
(b) 1
(c) 2\(\int_{0}^{a} f(x)\) dx
(d) 2a
Evaluate
(ii) \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x\)
(iii) \(\int_{-1}^{1} e^{|x|} d x\)
Answer:
(i) (a) 0.

(ii) Here, f(x) = sin⁹⁹x.cos¹⁰⁰x .then,
f(-x) = sin⁹⁹(- x).cos¹⁰⁰(- x) = – sin⁹⁹ x. cos¹⁰⁰ x = -f(x)
∴ odd function ⇒ \(\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x=0\).

(iii) Here, f(x) = e^|x|, f(-x) = e^|-x| = e^|x| = f(x)
∴ even function.

we have |x| = x, 0 ≤ x ≤ 1

Question 6.

1. Show that cos² x is an even function. (1)
2. Evaluate \(\int_{-\pi / 4}^{\pi / 4} \cos ^{2} x d x\) (2)

Answer:
1. Let f(x) = cos²x ⇒ f(-x) = cos² (-x) = cos² x = f(x) even.

2.

Question 7.
Find the following integrals.

Answer:

Question 8.
Find the following integrals.

Answer:

Add (1) and (2)

Question 9.
Find the following integrals.

1. \(\int \frac{1}{3+\cos x} d x\)
2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\)

Answer:
1. \(\int \frac{1}{3+\cos x} d x\)
Put t = tanx/2 ⇒ dt = 1/2 sec² x/2 dx

2. \(\int \frac{2 x}{x^{2}+3 x+2} d x\) = \(\int \frac{2 x}{(x+2)(x+1)} d x\)

2x = A(x + 1) + B (x + 2)
when x = -1, -2 = B ; B = -2
when x = -2, -4 = -A ; A = 4

= 4log(x + 2) – 2log (x + 1) + C.

Plus Two Maths Integrals Four Mark Questions and Answers

Question 1.
Find the following integrals.

Answer:

x² + x +1 = A(x² + 1) + (Bx + C)(x + 2)
Put x = -2 ⇒ 4 – 2 + 1 = 5A ⇒ A = \(\frac{3}{5}\)
Equating the coefficients of x²
⇒ 1 = A + B ⇒ B = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
Equating the constants
⇒ 1 = A + 2C ⇒ 2C = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\) ⇒ C = \(\frac{1}{5}\)


⇒ 1 = A(x – 1) + B(x + 3)
Put x = 1 ⇒ 1 = 2A ⇒ A = \(\frac{1}{2}\)
Put x = -3 ⇒ 1 = -4B ⇒ B = – \(\frac{1}{4}\)


Equating the constants; ⇒ 1 = A
Equating the coefficients if t;
⇒ 0 = A + B ⇒ B = -1

Question 2.
Find the following integrals.

1. ∫ e^2x sin3xdx
2. ∫ x sin^-1xdx

Answer:
1. I = ∫e^2x sin3xdx = ∫ sin 3x × e^2xdx

2. ∫ x sin^-1xdx = ∫ sin^-1x × xdx

Question 3.
(i) Which of the following is the value of \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}\)? (1)

(ii) Evaluate \(\int \frac{2 x}{x^{2}+3 x+2} d x\) (3)
Answer:
(i) [sin^-1\(\frac{x}{a}\) + c]

(ii)

⇒ 2x = A(x + 1) + B(x + 2) ⇒
Put x = -2 and x = -1, we get A = 4, B = -2

Question 4.

1. Choose the correct answer from the bracket.
   ∫e^x dx = — (e^2x + c, e^-x + c, e^2x + c) (1)
2. Evaluate: ∫ e^x sinxdx

Answer:
1. e^x + c

2. I = ∫e^x sinxdx = sinx.e^x – ∫cos x.e^xdx
= sin x.e^x – (cos x.e^x – ∫(- sin x).e^x dx)
= sinx.e^x – cosxe^x – ∫sinx.e^xdx
= sin x.e^x – cos xe^x – I
2I = sin x.e^x – cos xe^x
I = \(\frac{1}{2}\)e^x(sinx – cosx) + c.

Question 5.
(i) f(x)∫g(x) dx – ∫(f'(x)∫g(x) dx)dx (1)
(a) ∫f'(x)g{x)dx
(b) ∫f(x)g'(x)dx
(c) ∫\(\frac{f(x)}{g(x)}\)dx
(d) ∫f(x)g(x)dx
(ii) Integrate sin^-1\(\sqrt{\frac{x}{a+x}}\)dx w.r.to x. (3)
Answer:
(i) (d) ∫f(x)g(x)dx

(ii) ∫sin^-1\(\sqrt{\frac{x}{a+x}}\)dx,
Put x = a tan²θ, θ = tan^-1\(\sqrt{\frac{x}{a}}\)
⇒ dx = 2a tanθ sec²θ dθ
I = ∫sin^-1\(\left(\frac{\tan \theta}{\sec \theta}\right)\) 2a tanθ sec²θ dθ
= ∫sin^-1(sinθ)2a tanθ sec²θ dθ
= 2a∫θ tanθ sec²θ dθ
Put tanθ = t, θ = tan^-1 t ⇒ sec²θ dθ = dt
= 2a ∫ tan^-1 t (t) dθ

= a[tan²θ.θ – tanθ + θ] + c
= a[θ(1 + tan²θ) – tanθ] + c

Question 6.
Match the following. (4)

Answer:

Question 7.
Evaluate \(\int \frac{x}{\sqrt{x+a}+\sqrt{x+b}} d x\)?
Answer:

Question 8.
Match the following.

Answer:
1.

2. ∫sec x(sec x + tan x)dx = ∫(sec² x + sec x. tan x)dx
= tanx + secx + c.

3. ∫e^3xdx = \(\frac{e^{3 x}}{3}\) + c.

4. ∫(sin x + cos x)dx = sin x – cosx + c.

Question 9.
Consider the integral I = \(\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x\)?

1. What substitution can be given for simplifying the above integral? (1)
2. Express I in terms of the above substitution. (1)
3. Evaluate I. (2)

Answer:
1. Substitute sin^-1 x = t.

2. We have, sin^-1 x = t ⇒ x = sint
Differentiating w.r.t. x; we get,
\(\frac{1}{\sqrt{1-x^{2}}}\)dx = dt
∴ I = ∫t sin t dt.

3. I = ∫t sin t dt = t.(-cost) -∫(-cost)dt = -t cost + sint + c
= -sin^-1 x. cos (sin^-1 x) + sin(sin^-1 x) + c
x – sin^-1 x.cos(sin^-1 x) + c.

Question 10.
Evaluate \(\int_{0}^{\pi / 4} \log (\tan x) d x\).
Answer:

Question 11.
Find the following integrals.

1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) (2)
2. \(\int \frac{1}{x^{2}-6 x+13} d x\) (2)

Answer:
1. \(\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x\) = \(\int \frac{\sin ^{2} x}{\cos ^{2} x} d x\) = ∫tan² xdx
= ∫(sec²x – 1)dx = tanx – x + c.

2. \(\int \frac{1}{x^{2}-6 x+13} d x\)

Question 12.
Match the following. Justify your answer.

Answer:

Question 13.
(i) ∫sin2x dx = ? (1)
(a) 2 cos x + c
(b) -2 sin x + c
(c) \(\frac{\cos 2 x}{2}\) + c
(d) \(-\frac{\cos 2 x}{2}\) + c
(ii) Evaluate ∫e^x sin 2x dx (3)
Answer:
(i) (d) \(-\frac{\cos 2 x}{2}\) + c.

(ii) Consider I = ∫e^x sin 2x dx
= ∫sin 2x. e^xdx = sinx.e^x – 2∫cos 2x. e^xdx
= sin 2x.e^x – 2 (cos 2x.e^x + 2∫sin 2x. e^xdx)
= sin 2x. e^x – 2 cos 2x e^x – 4 ∫sin 2x. e^xdx
= sin 2x. e^x – 2 cos 2x e^x – 4I
5 I = sin 2x. e^x – 2 cos 2x e^x
I = \(\frac{e^{x}}{5}\) (sin 2x – 2 cos 2x).

Question 14.

1. Resolve \(\frac{x^{2}+1}{x^{2}-5 x+6}\) into partial fractions. (2)
2. Hence evaluate ∫\(\frac{x^{2}+1}{x^{2}-5 x+6}\). (2)

Answer:
1.

2.

5x – 5 = A(x – 2) + B(x – 3)
x = 2, 5 = -B, B = -5
x = 3, 10 = A, A = 10
(1) ⇒ I = ∫ 1dx + ∫\(\frac{10}{x-3}\) dx – ∫\(\frac{5}{x-2}\) dx
= x + 10log(x – 3) – 5log(x – 2) + c.

Question 15.
Evaluate \(\int_{0}^{4}\) xdx as a limit of sum.
Answer:
By definition,
\(\int_{a}^{b}\) f(x) dx =
(b – a)\(\lim _{n \rightarrow \infty} \frac{1}{n}\){f(a) + f(a + h) +…….+f(a + {n – 1)h)}
Here, a = 0, b = 4, f(x) = x, h = \(\frac{4-0}{n}=\frac{4}{n}\) ⇒ nh = 4

Question 16.

1. Define the real valued function f(x) = |x² + 2x – 3| (2)
2. Evaluate \(\int_{0}^{2}\)|x² + 2x – 3|dx. (2)

Answer:
1. f(x) = |x² + 2x – 3| = |(x – 1) (x + 3)|
We have;

2. I = \(\int_{0}^{2}\)|x² + 2x – 3|dx

Question 17.
Consider the function f(x) = |x|+|x + 1|

1. Define the function f (x) in the interval [-2, 1]. (2)
2. Find the integral \(\int_{-2}^{1}\) f(x) dx (2)

Answer:
1. Given, f(x) = |x|+|x + 1|.
We have,

Combining these two functions, we get the function f(x).

2.

Question 18.
Evaluate \(\int_{\sqrt{6}}^{\sqrt{3}} \frac{d x}{1+\sqrt{\tan x}} d x\). (4)
Answer:

Plus Two Maths Integrals Six Mark Questions and Answers

Question 1.
(i) Fill in the blanks. (3)
(a) ∫ tan xdx = —
(b) ∫ cos xdx = —
(c) ∫\(\frac{1}{x}\)dx = —
(ii) Evaluate ∫sin³ xcos² xdx (3)
Answer:
(i) (a) log|secx| + c
(b) sinx + c
(c) log|x| + c.

(ii) ∫sin³ xcos² xdx = ∫sin² xcos² x sin xdx
= ∫(1 – cos² x)cos² x sin xdx
Put cos x = t ⇒ – sin xdx = dt
∴ ∫(1 – cos² x)cos² xsin xdx = -∫(1 – t² )t²dt
= ∫(t⁴ – t²)dt = \(\frac{t^{5}}{5}-\frac{t^{3}}{3}\) + c
= \(\frac{\cos ^{5} x}{5}-\frac{\cos ^{3} x}{3}\) + c.

Question 2.
Find the following integrals.

Answer:
(i) I = ∫(3x – 2)\(\sqrt{x^{2}+x+1} d x\)
Let 3x – 2 = A(2x + 1) + B
⇒ 3 = 2 A ⇒ A = \(\frac{3}{2}\)
⇒ -2 = A + B ⇒ -2 = \(\frac{3}{2}\) + B
⇒ B = -2 – \(\frac{3}{2}\) = – \(\frac{7}{2}\)


Using (2) and (3) in (1) we have;

(ii) I = \(\int \frac{2 x-3}{x^{2}+3 x-18} d x\)
Let 2x – 3 = A(2x + 3) + B
⇒ 2 = 2A ⇒ A = 1
⇒ -3 = 3A + B ⇒ -3 = 3 + B ⇒ B = -6

(iii) I = \(\int \frac{5 x+2}{1+2 x+3 x^{2}} d x\)
Let 5x + 2 = A{6x + 2) + B
⇒ 5 = 6 A ⇒ A = \(\frac{5}{6}\)
⇒ 2 = 2A + B ⇒ 2 = \(\frac{5}{3}\) + B ⇒ 2 – \(\frac{5}{3}\) = \(\frac{1}{3}\)

(iv) I = \(\int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x\)
Let 5x + 3 = A(2x + 4) + B
⇒ 5 = 2A ⇒ A = \(\frac{5}{2}\)
⇒ 3 = 4A + B ⇒ 3 = 10 + B ⇒ B = -7


Using (2) and (3) in (1) we have;

Question 3.
Consider the expression \(\frac{1}{x^{3}-1}\)

1. Split it into partial fraction. (2)
2. Evaluate ∫ \(\frac{1}{x^{3}-1}\) dx (4)

Answer:
1.

1 = A (x² + x + 1) + (Bx + c)(x + 1),
Put x = -1 ⇒ 1 = A(1 + 1 + 1) ⇒ A= \(\frac{1}{3}\)
Equating like terms.
0 = A + B ⇒ B = – \(\frac{1}{3}\), 1 = A + C ⇒ C = \(\frac{2}{3}\)

2.

Put, x – 2 = D (2x – 1) + E ,
1 = 2 D ⇒ D = \(\frac{1}{2}\),
-2 = -D + E ⇒ E = –\(\frac{3}{2}\)

Question 4.
(i) Match the following (4)

(ii) Consider the function f(x) = \(\frac{x^{4}}{x+1}\) Evaluate ∫f(x)dx (2)
Answer:
(i)

(ii) Here the numerator is of degree 4 and denominator of degree 1. So to make it a proper fraction we have to divide Nr by Dr.

Question 5.

1. Evaluate the as \(\int_{0}^{2}\)x²dx the limit of a sum. (3)
2. Hence evaluate \(\int_{-2}^{2}\)x²dx (1)
3. If \(\int_{0}^{2}\) f(x)dx = 5 and \(\int_{-2}^{2}\) f(x)dx = 0, then \(\int_{-2}^{0}\) f(x)dx = …….. (2)

Answer:
1. Here the function is f(x) = x², a = 0, b = 2 and h = \(\frac{b-a}{n}=\frac{2}{n}\)
\(\int_{0}^{2}\)x²dx =

2. \(\int_{-2}^{2}\) x²dx = 2 \(\int_{0}^{2}\)x²dx = \(\frac{16}{3}\)

3.

Question 6.
Find ∫\(\sqrt{\tan x}\)xdx.
Answer:
Given;
I = ∫\(\sqrt{\tan x}\)xdx,
Put tanx = t² ⇒ sec²xdx = 2tdt ⇒ dx = \(\frac{2 t d t}{1+t^{4}}\)

Question 7.
(i) Match the following. (2)

(ii) Integrate \(\frac{\sec ^{2} x}{5 \tan ^{2} x-12 \tan x+14}\) w.r.to x. (4)
Answer:
(i)

Question 8.

1. Evaluate \(\int_{0}^{1} \sqrt{x} d x\) (1)
2. If \(\int_{0}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of a. (3)
3. Hence find \(\int_{a}^{a+1}\)x dx. (2)

Answer:
1.

2. Given;

3. When a = 0

When, a = 4

Question 9.
(i) Let f (x) be a function, then \(\int_{0}^{a}\) f(x) dx = ? (1)
(a) 2 \(\int_{0}^{a}\) f(x – a) dx
(b) \(\int_{0}^{a}\) f(a – x) dx
(c) f(a)
(d) 2\(\int_{0}^{a}\) f(a – x) dx
Evaluate

Answer:
(i) (b) \(\int_{0}^{a}\) f(a – x) dx

(ii)

(1) + (2)

⇒ I = 1.

(iii)

Question 10.
Find the following integrals.

1. ∫\(\frac{2 e^{x}}{e^{3 x}-6 e^{2 x}+11 e^{x}-6} d x\)
2. ∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)

Answer:
1.

⇒ 1 = A(t – 2)(t – 3) + B(t – 1)(t – 3) + C(t – 1)(t – 2)
Put t = 1 ⇒ 1 = A(-1)(-2) ⇒ A = \(\frac{1}{2}\)
Put t = 2 ⇒ 1 = B(1)(-1) ⇒ B = -1
Put t = 3 ⇒ 1 = B(2)(1) ⇒ B = \(\frac{1}{2}\)

2. I = ∫\(\frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x\)dx
Put sin x = t ⇒ cosxdx = dt

⇒ 3t – 2 = A(t – 2) + B
Equating the coefficients if t; ⇒ 3 = A
Equating the constants
⇒ -2 = -2A + B ⇒ -2 = -6 + B ⇒ B = 4

Question 11.

1. Find ∫\(\frac{1}{x^{2}+a^{2}}\)dx (1)
2. Show that 3x + 1 = \(\frac{3}{4}\)(4x – 2) + \(\frac{5}{2}\) (2)
3. Evaluate \(\int \frac{3 x+1}{2 x^{2}-2 x+3} d x\) (3)

Answer:
1. ∫\(\frac{1}{x^{2}+a^{2}}\)dx = 1/a tan^-1 x/a + c.

2. 3x + 1 = A \(\frac{d}{d x}\)(2x² – 2x + 3) + B
= A(4x – 2) + B
3 = 4A; A = 3/4
1 = -2A + B
1 = -3/2 + B, B = 1 + 3/2 = 5/2
∴ 3x + 1 = 3/4(4x – 2) + 5/2

3.

Students can Download Chapter 12 ICT and Society Questions and Answers, Plus Two Computer Science Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Computer Science Chapter Wise Questions and Answers Chapter 12 ICT and Society

Plus Two Computer Science ICT and Society One Mark Questions and Answers

Question 1.
IPR stands for ______.
Answer:
Intellectual Property Right.

Question 2.
WIPO stands for _____.
Answer:
World Intellectual Property Organisation

Question 3.
______ is the exclusive rights to prevent unauthorized copying of inventions by a Creator from the Unauthorised person or company.
Answer:
Patent

Question 4.
_____ is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company.
Answer:
Trademark

Question 5.
A product or article is designed so beautifully to attract customers. This type of design is called
Answer:
Industrial Design.

Question 6.
Aranmula Kannadi, Palakkadan Matta, Marayoor Sarkkara, etc are example of _______.
Answer:
Geographical indications.

Question 7.
_____ is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator.
Answer:
Copyright

Question 8.
From the following which is the symbol for copyright.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(b) ©

Question 9.
From the following which is the symbol for Unregistered trademark.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(d) ™

Question 10.
From the following which is the symbol for Registered trademark.
(a) $
(b) ©
(c) ®
(d) ™
Answer:
(c) ®

Question 11.
Unauthorized copying or use of Intellectual property rights such as Patents, Copyrights and Trademarks are called ____.
Answer:
Intellectual Property Infringement.

Question 12.
_____ prevents others from the unauthorized or intentional copying or use of Patent without the permission of the creator.
Answer:
Patent Infringement.

Question 13.
______ is the illegal copying, distribution, or use of software.
Answer:
Piracy.

Question 14.
______ prevents others from the unauthorized or intentional copying or use of Trademark without the permission of the creator.
Answer:
Trademark Infringement

Question 15.
_____ prevents others from the unauthorized or intentional copying or use of Copy right without the permission of the creator.
Answer:
Copy right Infringement

Question 16.
______ is a virtual environment created by computer systems connected to the internet
Answer:
Cyberspace

Question 17.
A person committing crimes and illegal activities with the use of computers over Internet. This crime is included as _____ crime.
Answer:
Cybercrime

Question 18.
State True or False.
Cybercrimes can be classified into three categories such as against individual, property, and Government.
Answer:
True

Question 19.
Phishing, hacking, denial of service attacks, etc are ____ crimes.
Answer:
Cyber

Question 20.
Odd one out
(а) Identity theft
(b) Harassment
(c) violation of privacy
(d) credit card fraud
Answer:
(d) credit card fraud, it is a cybercrime against individual others are cyber crimes against property.

Question 21.
Odd one out
(a) Credit card theft
(b) Intellectual property theft
(c) Internet time theft
(d) Dissemination of obscene material
Answer:
(d) Dissemination of obscene material, It is cyber , crime against individual, the others are cyber against property.

Question 22.
Odd one out
(a) cyberterrorism
(b) Attacks against e-Governance websites
(c) Impersonation and cheating
(d) Website defacement
Answer:
(c) Impersonation and cheating, it is cybercrime against individual others are cyber crimes against Government.

Question 23.
IT Act amended in _____.
(a) 2015
(b) 2008
(c) 1900
(d) 1998
Answer:
(b) 2008

Question 24.
IT Act passed in Indian parliament is ____.
Answer:
2000.

Question 25.
The laws to prevent cyber crimes is termed as ____.
Answer:
Cyberlaw

Question 26.
_____ excessive enthusiasm for acquiring knowledge.
Answer:
Infomania

Question 27.
Phishing is an example of ______.
Answer:
Cybercrime.

Question 28.
ICT stands for _______.
(a) Internet and Communication Technology
(b) Information and Computer Technology
(c) Information and Communication Technology
(d) Integrated Communication Technology
Answer:
(c) Information and Communication Technology

Question 29.
Which of the following e-Governance helps citizens for interacting with the Government?
(a) G2E
(b) G2B
(c) G2C
(d) G2G
Answer:
(c) G2C

Question 30.
What are the different types of interactions in e-Governance?
Answer:
G2G, G2E.G2B, G2C.

Question 31.
The unauthorized use of intellectual property rights is termed as
Answer:
Infringement

Question 32.
Expand the term WIPO in connection with IPR.
Answer:
World Intellectual Property Organization.

Question 33.
The exclusive right granted to an invention is called
(a) Trademark
(b) Copy right
(c) Patent
(d) Design
Answer:
(c) Patent

Question 34.
The exclusive right given to a person over the creation of his/her mind for a period of time is called
Answer:
Patent / Intellectual Property Right

Question 35.
What is the name given to the process of using scientific knowledge for analyzing and presenting evidence of cyber related crimes before court?
Answer:
Cyber forensics

Question 36.
Which among the following are considered as violation to privacy?
1. Keeping hidden cameras in private places
2. Publishing private photos of individual in social media without their permission
3. Use of unauthorized software
4. Using simple password
(A) All the above are correct
(B) 1,2 and 3 only
(C) 1 and 4 only
(D) 1 and 2
Answer:
(D) 1 and 2

Plus Two Computer Science ICT and Society Two Mark Questions and Answers

Question 1.
“IPR (Intellectual Property Right) encourages innovation” Justify.
Some people spend lots of money,time body and mental power to create some products such as a classical movie, album, artistic work, discoveries, invention, software, etc. These type of Intellectual properties must be protected from unauthorized access by law. This is called Intellectual Property right(IPR). It enables to earn recognition, financial benefit, can sell the innovation, etc. It motivates further innovation.

Question 2.
Define the following terms.

1. Cyber space
2. Cyber crime

Answer:
1. CyberSpace:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap.

2. Cyber crime:
Just like normal crimes (theft, trespassing private area, destroy, etc,) Cyber crimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc) also increased significantly. Due to cyber crime, the victims lose money, reputation, etc and some of them commit suicide.

Question 3.
Write a short note on

1. Trademark
2. Industrial design

Answer:
1. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

2. Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 4.
Compare patent and Trademark.
1. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

2. Trademark:
This is a unique, simple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Question 5.
Write any one website for the following services.

1. e-Governance
2. e-Business
3. e-Banking
4. e-Learning

Answer:

1. e-Governance(any One) www.dhsekerala.gov.in, www.incometaxindia.gov.in, www.spark.gov.in,www.ceo.kerala.gov. in
2. e-Business www.indane.co.in, www.amazon.com,www.ebay.in
3. e-Banking www.onlinesbi.co.in
4. e-Learning www.ignouonline.ac.in,www.nptel.iitm.ac.in

Question 6.
Write a short note about EPS.
Answer:
Electronic Payment System(EPS): It is also called plastic money that is electronically exchange money between two individuals or firms(buyers and sellers) in an online environment.

Question 7.
What is cyberspace?
Answer:
Earlier Traditional communication services such as postal service(Snail mail) are used for communication. It is a low speed and not reliable service. In order to increase the speed Telegram Services were used. Its speed was high but it has lot of limitations and expensive too.

Later telephones were used for voice communication. Nowadays telephone system and computer system are integrated and create a virtual(unreal) environment. This is called cyberspace. The result for this integration is that tremendous speed and it is very cheap.

Question 8.
Why is cyberspace called a virtual world?
Answer:
The telephone system and computer system are integrated and create a virtual(un real) environment. This is called cyber space. The result for this integration is that tremendous speed and it is very cheap. This is an imaginary world. We can see persons with different behaviour. Because of good and bad people we can’t believe blindly. If we search a solution for a problem thousands of answers will get instantly and may confused us.

Question 9.
What is copyright? How does it differ from patent?
Answer:
1. Copyright:
The trademark is ©, copyright is the property right that arises automatically when a person creates a new work by his own and by Law it prevents the others from the unauthorized or intentional copying of this without the permission of the creator for 60 years after the death of the author.

2. Patents:
A person or organization invented a product or a creation can be protected from unauthorized copying or creation without the permission of the creator by law. This right is called Patent. In India the validity of the right is up to 20 years. After this anybody can use freely.

Question 10.
Explain the exclusive right given to the owner by IPR?
Answer:
The exclusive right given to the owner by I PR is owner can disclose their creations for money.

Question 11.
it is the unauthorized copying, distribution, and use of a creation without the permission of the creator. It is against the copyright act and hence the person committed deserve the punishment.

Question 12.
Match the following

Answer:
a – 2
b – 3
c – 4
d – 1

Question 13.
What do you meant by infringement?
Answer:
Unauthorized copying or use of Intellectual property rights such as Patents, Copy rights and Trademarks are called intellectual property lnfringement(violation). It is a punishable offence.

Plus Two Computer Science ICT and Society Three Mark Questions and Answers

Question 1.
Write a short note on the importance of IT Act 2000.
Answer:
Information Technology Act 2000(amended in 2008):
IT Act 2000 controls the use of Computer(client), Server, Computer Networks, data and Information in Electronic format and provide legal infrastructure for E-commerce, in India. This is developed to promote IT industry, control e-commerce also ensures the smooth functioning of E-Governance and it prevents cyber crimes.

The person those who violate this will be prosecuted. In India, IT bill introduced in the May 2000 Parliament Session and it is known as Information Technology Act 2000. Some exclusions and inclusions are introduced in December 2008.

Question 2.
“Infomania affects peoples’ lives and their loved ones.”
Comment on this statement.
Answer:
Info mania is excessive desire(infatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media. Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet than others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Question 3.
Define the term e-Business. What are the advantages and challenges of e-Business? Write any two e-Business websites.
Answer:
E-business(electronic Business): Providing ser¬vices or running business through internet is called E-business.
Advantages of e-business:

• It overcomes geographical limitations
• It reduces the operational cost
• It minimizes the time and cost
• It remains open all the time
• We can locate the product faster from a wider range of choices
  Challenges to E business
• Peoples are unaware of IT applications and its uses
• Most peoples don’t have plastic money(credit / debit card) and net banking
• It requires high security measurements otherwise you may lose money
• We can’t touch or smell products through online
• Some companies may not have proper Goods delivery service
  Useful e-Business websites

Question 4.
How do trademark and industrial design differ?
Answer:
Trademark:
This is a unique, sirhple and memorable sign to promote a brand and hence increase the business and goodwill of a company. It must be registered. The period of registration is for 10 years and can be renewed. The registered trademark under Controller General of Patents Design and Trademarks cannot use or copy by anybody else.

Industrial designs:
A product or article is designed so beautifully to attract the customers. This type of designs is called industrial design. This is a prototype and used as a model for large scale production.

Question 5.
Why is Cyberlaw important?
Answer:
Just like normal crimes (theft, trespassing private area, destroy, etc.) Cybercrimes (Virus, Trojan Horse, Phishing, Denial of Service, Pornography, etc.) also increased significantly. Due to cybercrime, the victims lose money, reputation, etc. and some of them commit suicide.

Cyberlaw ensures the use of computers and Internet by the people safely and legally. It consists of rules and regulations like Indian Penal Code (IPC) to stop crimes and for the smooth functions of Cyberworld. Two Acts are IT Act 2000 and IT Act Amended in 2008.

Question 6.
“Infomania has became a psychological problem”. Write your opinion.
Answer:
Info mania is the excessive desire(lnfatuation) for acquiring knowledge from various modern sources like Internet, Email, Social media, Instant Message Application(WhatsApp) and Smart Phones. Due to this the person may neglect daily routine such as family, friends, food, sleep, etc. hence they get tired.

They give first preference to Internet others. They create their own Cyber World and no interaction to the surroundings and the family. They are more anxious and afraid that they will be out from the cyber world unless they updated.

Plus Two Computer Science ICT and Society Five Mark Questions and Answers

Question 1.
“Due to anonymous nature of Internet it is possible for the people to engage in variety of criminal activities.” Justify the statement with special reference to cyber crimes taking place against individual.
Answer:
Cyber crimes against individuals
i. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc) , Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

ii. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

iii. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original ICT and Society one for the purpose of cheating or misleading others. Eg: Fake accounts in Social Medias (Facebook, Twitter, etc), fake SMS, fake emails etc.

iv. Violation of privacy:
Trespassing into another person’s life and try to spoil life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

v. Dissemination of obscene material: With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

Question 2.
Explain different categories of cyber crimes in detail.
Answer:
Just like normal crimes( theft, trespassing private area, destroy, etc,) Cyber crimes(Virus, Trojan Horse, Phishing, Denial of Service, Pornography etc) also increased significantly . Due to cyber crime, the victims lose money, reputation,etc and some of them commit suicide.
Answer:
A. Cyber crimes against individuals:
1. Identity theft:
The various information such as personal details(name, Date of Birth, Address, Phone number etc.), Credit / Debit Card details(Card number, PIN, Expiry Date, CW, etc), Bank details, etc. are the identity of a person. Stealing these information by acting as the authorized person without the permission of a person is called Identity theft. The misuse of this information is a punishable offence.

2. Harassment:
Commenting badly about a particular person’s gender, colour, race, religion, nationality, in Social Media is considered as harassment. This is done with the help of Internet is called Cyber stalking (Nuisance). This is a kind of torturing and it may lead to spoil friend ship, career, self image and confidence. Sometimes may lead to a big tragedy of a whole family or a group of persons.

3. Impersonation and cheating:
Fake accounts are created in Social Medias and act as the original one for the purpose of cheating or misleading others. Eg: F.ake accounts in Social Medias (Facebook, Twitter,etc), fake sms, fake emails, etc.

4. Violation of privacy:
Trespassing into another person’s life and try to spoil the life. It is a punishable offence. Hidden camera is used to capture the video or picture and black mailing them.

5. Dissemination of obscene material:
With the help of hidden camera capture unwanted video or picture. Distribute or publish this obscene clips on Internet without the consent of the victims may mislead the people specifically the younger ones.

B. Cyber crimes against property:
Stealing credit card details, hacking passwords of social media accounts or mail account or Net banking, uploading latest movies etc, are considered as cyber crimes against property.
1. Credit card fraud:
Stealing the details such as credit card number, company name, expiry date, cw number,password etc. and use these details to make payment for purchasing goods or transfer funds also.

2. Intellectual property theft:
The violation of Intellectual Property Right of Copy right, Trademark, Patent, etc. In film industry crores of investment is needed to create a movie. Intellectual Property thieves upload the movies on the Releasing day itself. Hence the revenue from the theatres are less significantly and undergoes huge loss.(Eg: Premam, Bahubali, etc) Copying a person’s creation and present as a new creation is called plagiarism. This can be identified some tools(programs) available in the Internet

3. Internet time theft:
This is deals with the misuse of WiFi Internet facility. If it is not protected by good password there is a chance of misuse our devices(Modem/Router) to access Internet without our consent by unauthorized persons. Hence our money and volume of data(Package) will lose and we may face the consequences if others make any crimes.

C. Cyber crimes against government:
The cyber crimes against Govt, websites is increased significantly. For example in 2015 the website of Registration Department of Kerala is hacked and destroys data from 2012 onwards.

1. Cyber terrorism:
It is deals with the attacks against very sensitive computer networks like computer controlled atomic energy power plants, air traffic controls, Gas line controls, telecom, Metro rail controls, Satellites, etc. This is a very serious matter and may lead to huge loss (money and life of citizens). So Govt, is very conscious and give tight security mechanism for their services.

2. Website defacement:
It means spoil or hacking websites and posting bad comments about the Govt.

3. Attacks against e-governance websites :
Its main target is a Web server. Due to this attack the Web server/ computer forced to restart and this results refusal of service to the genuine users. If we want to access a website first you have to type the web site address in the URL and press Enter key, the browser requests that page from the web server. Dos attacks send huge number of requests to the web server until it collapses due to the load and stops functioning.

Question 3.
“For the implementation of e-Learning different tools.
Answer:
e Learning tools
1. Electronic books reader(eBooks): With the help of a tablet or portable computer or any other device we can read digital files by using an s/w is called electronic books reader.

2. e-text: The electronic format of textual data is called e-Text.

3. Online chat: Realtime exchange of text or audio or video messages between two or more person over the Internet.

4. e-Content: The data or information such as text, audio, video, presentations, images, animations, etc, are stored in electronic format.

5. Educational TV channels: TV channels dedicated only for the e-Learning purpose.
Eg. VICTERS (Virtual Classroom Technology on Edusat for Rural Schools OR Versatile ICT Enabled Resources for Students)

Students can Download Chapter 6 Application of Derivatives Questions and Answers, Plus Two Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 6 Application of Derivatives

Plus Two Maths Application of Derivatives Three Mark Questions and Answers

Question 1.
Find the equation of tangents and normals to the given curves x = cost, y = sin t at t = \(\frac{π}{4}\).
Answer:
Given; x = cost, y = sin t

Equation of tangent at t = \(\frac{π}{4}\) is;

Equation of normal at t = \(\frac{π}{4}\) is;

⇒ \(\sqrt{2}\)y + \(\sqrt{2}\)x = 0 ⇒ y + x = 0.

Question 2.
A ladder Sm long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the decreasing when the foot of the ladder is 4m away from the wall?
Answer:
From the figure we have;
x² + y² = 25 ____(1)

Differentiating w.r.t t;

From (1) when x = 4 ⇒ 16 + y² = 25 ⇒ y = 3
Given; \(\frac{d x}{d t}\) = 2cm/s = 0.02 m/s
(2) ⇒ 4(0.02) + 3 \(\frac{d x}{d t}\) = 0

Question 3.
Find the points on the curve y = x³, the tangents at which are inclined at an angle of 60° to x-axis?
Answer:
\(\frac{d y}{d x}\) = 3x²
Slope of the tangent = tan60°
i.e. 3x² = \(\sqrt{3}\)

Question 4.
Find the equation of the tangent to the parabola y² = 4x + 5 which is parallel to y = 2x + 7.
Answer:
y² = 4x + 5 _____(1)
2y \(\frac{d y}{d x}\) = 4
\(\frac{d y}{d x}\) = \(\frac{4}{2y}\) = \(\frac{2}{y}\)
Given tangent is parallel to y = 2x + 7
ie. Slope of the tangent is 2 ⇒ \(\frac{2}{y}\) = 2 ⇒ y = 1
∴ from (1) ⇒ 1 = 4x+ 5 ⇒ 4x = -4 ⇒ x = -1
So the point of contact is (-1, 1).
∴ Equation of tangent is
y -1 = 2(x + 1) ⇒ y = 2x + 3.

Question 5.
Find the intervals in which the function f given f(x) = 2x² – 3x is

1. Strictly increasing.
2. Strictly decreasing.

Answer:
Given; f(x) = 2x² – 3x ⇒ f'(x) = 4x – 3
For turning points; f'(x) = 0
⇒ 4x – 3 = 0 ⇒ x = \(\frac{3}{4}\)
The intervals are \(\left(-\infty, \frac{3}{4}\right),\left(\frac{3}{4}, \infty\right)\)
f'(0) = – 3 < 0
∴ Strictly decreasing in \(\left(-\infty, \frac{3}{4}\right)\)
f'(1) = 1 > 0
∴ Strictly increasing in \(\left(\frac{3}{4}, \infty\right)\).

Question 6.
Find the intervals in which the function f(x) = (x + 1)³ (x – 3)³ strictly increasing or decreasing.
Answer:
Given; f(x) = (x + 1)³ (x – 3)³
⇒ f'(x) = (x + 1)³ 3(x – 3)² + (x – 3)³3(x + 1)²
= 3(x + 1)²(x – 3)²(x + 1 + x – 3)
= 3(x + 1)²(x – 3)²(2x – 2)
= 6(x +1)² (x – 3)² (x -1)
⇒ 6(x +1)² (x – 3)² (x – 1) = 0
⇒ x = -1, 1, 3
The intervals are
(-∞, -1), (-1, 1), (1, 3), (3, ∞)
f'(-2) = (-2 – 1) < 0
∴ Strictly decreasing in (-∞, -1)
f'(0) = (0 – 1) < 0
∴ Strictly decreasing in (-1, 1)
f'(2) = (2 – 1) > 0
∴ Strictly increasing in (1, 3)
f'(4) = (4 – 1) > 0
∴ Strictly increasing in (3, ∞).

Question 7.
Find the intervals in which the function f(x) = x + \(\frac{1}{x}\) strictly increasing or decreasing.
Answer:

⇒ x = ±1
The intervals are (-∞, -1), (-1, 1), (1, ∞)
f'(-2) > 0
∴ Strictly increasing in (-∞, -1)
f'(0) < 0
∴ Strictly decreasing in (-1, 1)
f'(2) > 0
∴ Strictly increasing in (1, ∞).

Question 8.
Determine whether the f(x) = x² function is strictly monotonic on the indicated interval.

1. (-1, 1)
2. (-1, 0)
3. (0, 1)

Answer:
f(x) = x²
⇒ f'(x) = 2x
⇒ f'(x) = 0 ⇒ 2x = 0 ⇒ x = 0
This turning point divides the domain into the intervals (-∞, 0); (0, ∞).

1. Interval (-1,1) f'(x) < 0 and f'(x) > 0. So f(x) is not monotonic.
2. Interval (-1,0), f'(x) < 0. ∴ f(x) is strictly monotonic.
3. Interval (0, 1) f'(x) > 0 and f(x) is strictly monotonic.

Question 9.
Determine whether the f(x) = x³ – x function is strictly monotonic on the indicated interval.

1. (-1, 0)
2. (-1, -1/2)
3. (-1, 1)

Answer:
(x) = x³ -x ⇒ f'(x) = 3x² – 1
⇒ f'(x) = 0 ⇒ 3x² – 1 = 0 ⇒ x = ±\(\frac{1}{\sqrt{3}}\)
This turning point divides the domain into the intervals (-∞, \(\frac{1}{\sqrt{3}}\)); (-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)); (\(\frac{1}{\sqrt{3}}\), ∞).

1. Interval (-1, 0), f'(x) changes sign. So not monotonic.
2. Interval (-1, -1/2), f'(x) > 0 strictly monotonic.
3. lnterval(-1, 1) not monotonic

Question 10.
Find the approximate change in the volume V of a cube of side x meters caused by increasing the side by 1%.
Answer:
We have; V = x³ and ∆x = 1% of x= 0.01x
dV = \(\frac{d V}{d x}\) ∆x = 3x²∆x
= 3x² × 0.01x = 0.03x³ = 0.03V
⇒ \(\frac{d V}{V}\) = 0.03
Therefore 3% is the approximate increase in volume.

Question 11.
If the radius of a sphere is measured as 7m with an error of 0.02m then find the approximate error in calculating its volume.
Answer:
Let r be the radius of the sphere and ∆r be the error in measuring the radius then r =7m and ∆r = 0.02 m
We have; V = \(\frac{4}{3}\) πr³
dV = \(\frac{d V}{dr}\) ∆r = \(\frac{4}{3}\) π3r² × ∆r
= 4π(7)² × 0.02 = 3.92 π m³.

Question 12.
The length of a rectangle is decreasing at the rate of 5 cm/min and the width is increasing at the rate of 4cm/min. When length is 8 cm and width is 6 cm, find the rate of change of its area.
Answer:
Let length = x and width = y

Question 13.
Find the equation of tangents and normals to the given curves y= x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
Given; y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
⇒ \(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
Slope = \(\left(\frac{d y}{d x}\right)_{x=0}\) = -10
Equation of tangent at (0, 5) is;
y – 5 =(-10)(x – 0)
⇒ y – 5 = -10x ⇒ 10x + y – 5 = 0
Equation of normal at (0, 5) is;
y – 5 = \(\frac{1}{10}\)(x – 0)
⇒ 10y – 50 = x ⇒ x – 10y + 50 = 0.

Question 14.
Find the equation of tangents and normals to the given curves y = x³ at (1, 1)
Answer:
Given; y = x³
⇒ \(\frac{d y}{d x}\) = 3x²
Slope = \(\left(\frac{d y}{d x}\right)_{x=1}\) = 3
Equation of tangent at (1, 1) is; y -1 = (3)(x – 1)
⇒ y – 1 = 3x – 3 ⇒ 3x – y – 2 = 0
Equation of normal at (1, 1) is;
y – 1 = \(-\frac{1}{3}\)(x – 1)
⇒ 3y – 3 = -x + 1 ⇒ x+ 3y – 4 = 0.

Question 15.
The volume of a cube is increasing at the rate of 8cm³/s. How fast is the surface area increasing when the length of an edge is 12cm.
Answer:
Let V be the volume of the cube of side x.
We have volume = V = x³
Rate of change of volume with respect to time ‘t’ is;
ie; differentiating w.r.t t; \(\frac{d V}{d t}\) = 3x²\(\frac{d x}{d t}\)
Given; \(\frac{d V}{d t}\) = 8 and x = 12 ⇒ 8 = 3(12)²\(\frac{d x}{d t}\)

Now let Surface area = S = 6x²
Differentiating w.r.t t;

Question 16.
Find the intervals in which the function f(x) = -2x³ – 9x² – 12x + 1 strictly increasing or decreasing.
Answer:
Given; f(x) = -2x³ – 9x² – 12x + 1
⇒ f'(x) = -6x² – 18x – 12
= – 6(x² + 3x + 2)
= – 6(x + 2)(x +1)
⇒ f'(x) = 0 ⇒ -6(x + 2)(x +1) = 0
⇒ x = -2, -1
The intervals are (-∞, -2),(- 2, -1),(-1, ∞)
f'(-3) = -(-3 + 2)(-3 + 1) < 0
∴ Strictly decreasing in (-∞, -2)
f'(-1.5) = -(-1.5 + 2)(-1.5 + 1) > 0
∴ Strictly increasing in (- 2, -1)
f'(0) = -(0 + 2)(0 + 1) < 0
Strictly decreasing 1n(-1, ∞).

Question 17.
Find the local maxima and minima of the following functions. Also find the local maximum and minimum values. (each question carry 3 score)

1. f(x) = sin x + cosx, 0 < x < \(\frac{\pi}{2}\)
2. f(x) = x³ – 3x
3. f(x) = x³ – 6x² + 9x + 15
4. g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\), x > 0
5. g(x) = \(\frac{1}{x^{2}+2}\)

Answer:
1. Given; f(x) = sinx + cosx
⇒ f'(x) = cosx – sinx
For turning point f'(x) = 0
⇒ cosx – sinx = 0
⇒ cosx = sinx
⇒ x = \(\frac{\pi}{4}\)
f”(x) = -sin x – cosx
⇒ f (\(\frac{\pi}{4}\)) = -sin\(\frac{\pi}{4}\) – cos\(\frac{\pi}{4}\) < 0
Hence f(x) has a local maximum at x = \(\frac{\pi}{4}\) and local maximum value is

2. Given; f(x) = x³ – 3x
⇒ f'(x) = 3x² – 3
For turning point f'(x) = 0
⇒ 3x² – 3 = 0
⇒ x = ±1
f”(x) = 6x
When x = -1
⇒ f”(-1) = -6 < 0
Hence f(x) has a local maximum at x = -1 and local maximum value is
f(-1) = (-1)³ – 3(-1) = -1 + 3 = 2
When x = 1
⇒ f”(1) = 6 > 0
Hence f(x) has a local minimum at x = 1 and local minimum value is
f(1) = (1)³ – 3(1) = 1 – 3 = -2.

3. Given; f(x) = x³ – 6x² + 9x + 15
⇒ f'(x) = 3x² – 12x + 9
For turning point f'(x) = 0
⇒ 3x² – 12x + 9 = 0 ⇒ 3(x² – 4x + 3) = 0
⇒ 3(x – 1)(x – 3) = 0 ⇒ x = 1, 3
f”(x) = 6x – 12
When x = 1
⇒ f”( 1) = 6 – 12 < 0
Hence f(x) has a local maximum at x = 1 and local maximum value is
f(1) = (1)³ – 6(1)² + 9(1) + 15 = 19
When x = 3
⇒ f”(3) = 6(3) – 12 > 0
Hence f(x) has a local minimum at x = 3 and local minimum value is
f(3) = (3)³ – 6(3)² + 9(3) + 15 = 15.

4. Given; g(x) = \(\frac{x}{2}\) + \(\frac{2}{x}\)
⇒ g'(x) = \(\frac{1}{2}\) – \(\frac{2}{x^{2}}\)
For turning point g'(x) = 0

Since x > 0, the acceptable value of x = 2

Hence g(x) has a local maximum at x = 2 and local maximum value is g(2) = \(\frac{2}{2}\) + \(\frac{2}{2}\) = 2

5. Given; g(x) = \(\frac{1}{x^{2}+2}\)

For turning point g'(x) = 0


Hence g(x) has a local maximum at x = 2 and maximum value is g(2) = \(\frac{1}{0+2}=\frac{1}{2}\).

Question 18.
Find the absolute maximum value and minimum value of the following functions.

1. f(x) = x³, x ∈ [-2, 2]
2. f(x) = 4x – \(\frac{x^{2}}{2}\), x ∈ \(\left[-2, \frac{9}{2}\right]\)

Answer:
1. Given; f(x) = x³ ⇒ f'(x) = 3x²
For turning point f'(x) = 0 ⇒ 3x² = 0 ⇒ x = 0
f(- 2) = (-2 )³ = -8
f( 2) = (2)³ = 8
f(0) = (0)³ = 0
Absolute maximum = max{-8, 8, 0} = 8
Absolute minimum = min {-8, 8, 0} = – 8

2. Given; f(x) = 4x – \(\frac{x^{2}}{2}\) ⇒ f'(x) = 4 – x
For turning point f'(x) = 0 ⇒ 4 – x = 0
⇒ x = 4

Absolute maximum = max{-10, 8, 7.875} = 8
Absolute minimum = min {-10, 8, 7.87} = -10.

Question 19.
A television camera at ground level is filming the lift-off of a space shuttle that is rising vertically according to position equation S = 50t².
The camera is 2000 feet from the launch pad. Find the rate of change in the angle of elevation of the camera 10 seconds after lift-off.
Answer:

Question 20.
Determine whether the f(x) = sinx function is strictly monotonic on the indicated interval.

1. (0, 2π)
2. (0, π)
3. (-π/2, π/2)

Answer:
f(x) = sinx ⇒ f'(x) = cosx changes sign.

1. Interval (0, 2π). ∴ f(x) is not monotonic.
2. f'(x) changes sign in (0, π) not monotonic.
3. f'(x) > 0 in (-π/2, π/2), ie. f(x) is strictly monotonic.

Question 21.
Find the approximate change in the Surface Area of a cube of side x meters caused by decreasing the side by 1%.
Answer:
We have;
S = 6x² and ∆x = 1% of x = -0.01x
dS = \(\frac{d S}{d x}\) ∆x = 6 × 2x × ∆x
= 6 × 2x × -0.01x = -0.02 × 6x² = -0.02S
⇒ \(\frac{d S}{S}\) = -0.02
Therefore 2% is the approximate decrease in surface area.

Plus Two Maths Application of Derivatives Four Mark Questions and Answers

Question 1.
The length ‘x’ of a rectangle is decreasing at the rate of 2 cm/s and the width ‘y’ is increasing at the rate of 2 cm/s.

1. Find the rate of change of Perimeter.
2. Find \(\frac{d A}{dt}\) when x = 12 cm and y = 5 cm.

Answer:
Since the length ‘x’ is decreasing and the width ‘y’ is increasing, we have \(\frac{d x}{dt}\) = -2 cm/s and \(\frac{d y}{dt}\)
= 2 cm/sec.
1. The Perimeter ‘P’ of the rectangle is given by
P = 2 (x + y)

2. The area ‘A’ of the rectangle ‘A’ is given by
A = x.y

= 12(2) + 5(-2)
= 24 – 10 = 14 cm2/s.

Question 2.
Find the equation of all lines having slope -1. Which are tangents to the curve?
y = \(\frac{1}{x-1}\), x ≠ 1
Answer:

⇒ x² – 2x = 0 ⇒ x(x – 2) = 0 ⇒ x = 0, x = 2
At x = 0, y = -1
Equation of tangent at (0, -1) is;
At x = 2, y = \(\frac{1}{2-1}\) = 1
Equation of tangent at (2, 1) is; y – 1 = -1(x – 2)
⇒ y – 1 = -x + 2 ⇒ x + y – 3 = 0.

Question 3.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangent are parallel to x-axis.
Answer:
Given; x² + y² – 2x – 3 = 0
Differentiating with respect to x;

Since the tangent is parallel to x-axis \(\frac{d y}{d x}\) = 0
\(\frac{1-x}{y}\) = 0 ⇒ x = 1
We have; (1)² + y² – 2(1) – 3 = 0
⇒ y² = 4 ⇒ y = ±2
Hence the points are (1, 2), (1, -2).

Question 4.
Find the equation of the tangent to the curve y = \(\sqrt{3 x-2}\) which is parallel to the line 4x – 2y + 5 = 0.
Answer:
Slope of the line 4x – 2y + 5 = 0 is 2.

acceptable since y is positive.
Hence the point is \(\left(\frac{41}{48}, \frac{3}{4}\right)\)
Equation of tangent is;


⇒ 6(4y – 3) = (48x – 41)
⇒ 24y – 18 = 48x – 41
⇒ 48x – 24y – 23 = 0.

Question 5.
Prove that the curve x = y² and xy = k cut at right angles, if 8k² = 1.
Answer:
x = y² ___(1)
⇒ 1 = 2y \(\frac{d y}{d x}\) ⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2y}\)
xy = 2k ___(2)
⇒ x \(\frac{d y}{d x}\) + y.1 = 0 ⇒ \(\frac{d y}{d x}\) = \(-\frac{y}{x}\)
The product of the slopes will be – 1.

Question 6.
The gradient at any point (x, y) of a curve is 3x² – 12 and the curve through the point (2, -7).

1. Find the equation of the tangent at the point ( 2, -7 ). (2)
2. Find the equation to the curve. (2)

Answer:
1. Given gradient as 3x² – 12 ⇒ \(\frac{d y}{d x}\) = 3x² – 12
Slope at (2, -7) is given by
\(\left(\frac{d y}{d x}\right)_{x=2}\) = 3(2)² – 12 = 0
Since slope is zero, the tangent is parallel to x – axis.
Here y = – 7 is the equation of the tangent at (2, -7).

2. Given, \(\frac{d y}{d x}\) = 3x² – 12
⇒ ∫dy = ∫(3x² – 12 )dx
y = 3\(\frac{x^{3}}{3}\) – 12x + c ⇒ y = x³ – 12x + c ____(1)
Given (2, -7) is a point on the curve.
(1) ⇒ -7 = (2)³ – 12(2) + c ⇒ -7 = 8 – 24 + c ⇒ c = 9
∴ Curve is y = x³ – 12x + 9.

Question 7.
Consider the curve x^2/3 + y^2/3 = 2

1. Find the slope of the tangent to the curve at the point (1, 1). (2)
2. Find the equation of the normal at the point (1, 1). (2)

Answer:
1. Given, x^2/3 + y^2/3 = 2,
Differentiating w.r.t. x,

⇒ slope of tangent = – 1.

2. Slope of normal = \(-\frac{1}{-1}\) = 1.
Equation of the normal is
y – 1 = 1(x – 1) ⇒ y – x = 0.

Question 8.
Find the intervals in which the function f given by f(x) = 2x³ – 3x² – 36x + 7 is

1. Strictly increasing. (2)
2. Strictly decreasing. (2)

Answer:
Given; f(x) = 2x³ – 3x² – 36x + 7
⇒ f'(x) = 6x² – 6x – 36
f'(x) = 0 ⇒ 6x² – 6x – 36 = 0
⇒ 6(x² – x – 6) = 0
⇒ 6(x + 2)(x – 3) = 0 ⇒ x = -2, 3
The intervals are (-∞, -2),(- 2, 3), (3, ∞)
f'(-3) = 6(-3 + 2)(-3 – 3) > 0.
∴ Strictly increasing in (-∞, -2).
f'(0) = 6(2)(-3) < 0.
∴ Strictly decreasing in (- 2, 3).
f'(4) = 6(4 + 2)(4 – 3) > 0
∴ Strictly increasing in (3, ∞).

Question 9.
Use differentials to find the approximate value of \(\sqrt{0.6}\) up to 3 places of decimals.
Answer:
Take y = \(\sqrt{x}\), let x = 0.64 and ∆x = -0.04
Then; f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y


(1) ⇒ \(\sqrt{0.6}\) = 0.8 – 0.025 = 0.775.

Question 10.
Use differentials to find the approximate value of (0.999)^{\(\frac{1}{10}\)} up to 3 places of decimals.
Answer:
Take = x^{\(\frac{1}{10}\)}, let x = 1 and ∆x = -0.001
Then; f(x) = y
f(x + ∆x) = y + ∆y
(0.999)^{\(\frac{1}{10}\)} = x^{\(\frac{1}{10}\)} + ∆y
(0.999)^{\(\frac{1}{10}\)} = (1)^{\(\frac{1}{10}\)} + ∆y
⇒ (0.999)^{\(\frac{1}{10}\)} = 1 + ∆y ______(1)

(1) ⇒ (0.999)^{\(\frac{1}{10}\)} = 1 – 0.0001 = 0.9999.

Question 11.
Use differentials to find the approximate value of (15)^{\(\frac{1}{4}\)} up to 3 places of decimals.
Answer:
Takey = x^{\(\frac{1}{4}\)}, let x = 16 and ∆x = -1
Then; f(x) = y
f(x + ∆x) = y + ∆y
(15)^{\(\frac{1}{4}\)} = x^{\(\frac{1}{4}\)} + ∆y
(15)^{\(\frac{1}{4}\)} = 16^{\(\frac{1}{4}\)} + ∆y
(15)^{\(\frac{1}{4}\)} = 2 + ∆y ____(1)
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x

Question 12.
Use differentials to find the approximate value of (26.57)^{\(\frac{1}{3}\)} up to 3 places of decimals.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 27 and ∆x = -0.43
Then; f(x) = y
f(x + ∆x) = y + ∆y
f(x + ∆x) = f(x) + ∆y
(26.57)^{\(\frac{1}{3}\)} = 27^{\(\frac{1}{3}\)} + ∆y .
⇒ (26.57)^{\(\frac{1}{3}\)} = 3 + ∆y ____(1)

(1) ⇒ (26.57)^{\(\frac{1}{3}\)} = 3 – 0.016 = 2.984.

Question 13.
Find the approximate value of f(5.001) where f(x) = x³ – 7x² + 15
Answer:
Let x = 5 and ∆x = 0.001
Then; f(x) = y
f(x + ∆x) = y + ∆y
f (5.001) = f(x) + ∆y
f(5.001) = f(5) + ∆y
f(5.001) = 5³ – 7(5)² + 15 + ∆y
⇒ f(5.001) = -35 + ∆y …
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (3x² – 14x) × 0.001
= (3(5)² – 14(5)) × 0.001 = (75-70)0.001 = 0.005
(1) ⇒ f(5.001) = -35 + 0.005 = -34.995.

Question 14.
Find the approximate value of f(3.02) where f(x) = 3x² + 5x + 3
Answer:
Let x = 3 and ∆x = 0.02
Then; f(x) = y
f(x + ∆x) = y + ∆y
f(3.02) = f(x) + ∆y
f(3.02) = f(3) + ∆y
f(3.02) = 3(3)² + 5(2) + 3 + ∆y
⇒ f(3.02) = 45 + ∆y ____(1)
⇒ ∆y ≈ dy = \(\frac{d y}{d x}\) ∆x ⇒ dy = (6x + 5) × 0.02
= (6(3) + 5) × 0.02 = (18 + 5)0.02 = 0.46
(1) ⇒ f(3.02) = 45 + 0.46 = 45.46.

Question 15.
Consider the function y = f\(\sqrt{x}\)

1. If x = 0.0036 and ∆x = 0.0001 find ∆y. (3)
2. Hence approximate \(\sqrt{.0037}\) using differentials. (1)

Answer:
1. Let x = .0036, ∆x = 0.0001

2. (1) ⇒ \(\sqrt{.0037}\) = .000833 + .06 = .060833.

Question 16.
Find the approximate value of \(\sqrt[3]{124}\).
Answer:
f(x) = \(\sqrt[3]{x}\) = x^1/3 ⇒ f¹(x) = 1/3x^-2/3 = \(\frac{1}{3 x^{2 / 3}}\)
Let x = 125, ∆x = -1
Then; f(x) = y

Question 17.
Find two numbers x and y such that their sum is 35 and the product is x² y⁵ a maximum.
Answer:
Given; x + y = 35 ⇒ y = 35 – x
P = x² y⁵ ⇒ P = x²(35 – x)⁵
⇒ p’ = 2x(3 5 – x)⁵ + x² 5(35 – x)⁴(-1)
⇒ P’ = x(35 – x)⁴[2(35- x) – 5x]
⇒ p’ = x(35 – x)⁴[70 – 7x]
⇒ p’ = 7x(35 – x)⁴[10 – x]
⇒ p” = 7[x(3 5 – x)⁴ [-1] + x(10 – x)4(35 – x)³ (-1) + (35 – x)⁴(10 – x)]
For turning points P’ = 0
⇒ 7x(35 – x)⁴[10 – x] = 0
⇒ x = 0, 35, 10
x = 0, 35 can be rejected since correspondingly y will be y = 35, 0
⇒ P” = 7[10(35 – 10)⁴[-1]] < 0
Therefore maximum at x = 10
Thus the numbers are 10 and 35 – 10= 10.

Question 18.
Using differentials, find the approximate value of (63)^1/3.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 64 and ∆x = 1
Then; f(x) = y
f(x + ∆x) = y + ∆y

Question 19.

1. Find the point on the curve y = x³ – 10x + 8 at which the tangent is parallel to the line y = 2x + 1. (2)
2. Is the given line tangent to the curve? Why?

Answer:
1. \(\frac{d y}{d x}\) = 3x² – 10
Slope of the line y = 2x +1 is 2
⇒ 3x² – 10 = 2 ⇒ 3x² = 12 ⇒ x = ±2
When x = 2
y = 2³ – 10 × 2 + 8 = 8 – 20 + 8 = -4
When x = – 2
y = (-2)³ -10 × (-2) + 8 = -8 + 20 + 8 = 20
Therefore the points are (2, -4); (-2, 20)

2. No. Since (2, -4); (-2, 20) does not satisfies the equation y = 2x +1.

Question 20.
Suppose that a spherical balloon is inflated and it has volume ‘v’ and radius ‘r’ at time ‘t’.

1. If the balloon is inflated by pumping 900c.c. of gas per second. Find the rate a which the radius of the balloon is increasing when the radius is 15 cm. (2)
2. Find the rate of change of its surface at the instant when it radius is 15 cm. (2)

Answer:
1. Let V be the volume of the sphere of radius r.
V = \(\frac{4}{3}\) πr³, given; \(\frac{d V}{d t}\) = 900, r = 15
Differentiating w.r.t t,

2. Let ‘s’ denote the surface area of the balloon, then
S = 4πr²
Differentiating, \(\frac{d s}{d t}\) = 4π.2r.\(\frac{d r}{d t}\) =8.π r .\(\frac{d r}{d t}\)
= 8π × 15 × \(\frac{1}{\pi}\) = 120 cm²/sec.

Question 21.
Use differentials to find the approximate value of (0.009)^{\(\frac{1}{3}\)} up to 3 places of decimals.
Answer:
Take y = x^{\(\frac{1}{3}\)}, let x = 0.008 and ∆x = 0.001
Then; F(x) = Y
f(x + ∆x) = y + ∆y

Question 22.
Find the approximate value of \(\sqrt{401}\).
Answer:
f(x) = \(\sqrt{x}\) = x^1/2
f'(x) = \(\frac{1}{2 \sqrt{x}}\)
Let x = 400 ∆x = 1
f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y

Question 23.
Consider y = \(\frac{\log x}{x}\), in (0, ∞)

1. Find the value of x at which \(\frac{d y}{d x}\) = 0 (2)
2. Find the maximum value.

Answer:
1.

2.

∴ y is maximum when x = e.
The maximum value is \(\frac{1}{e}\).

Question 24.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Answer:
Slope of the line y = x – 11 is 1.
Given; y = x³ – 11x + 5 ⇒ \(\frac{d y}{d x}\) = 3x² – 11 = 1
⇒ 3x² = 12 ⇒ x = ±2
At x = 2, ⇒ y = x – 11 = 2- 11 = -9
⇒ (2, -9)
At x = -2, ⇒ y = x – 11 = -2 – 11 = -13
⇒ (-2, -13)
But the point (-2, -13) do not lie on the curve, hence the point is (2, -9).

Question 25.
Consider the curve y = x² – 2x + 7

1. Find the slope of the tangent of the curve at x = 2. (2)
2. Write down the equation of the tangent at x =2. (2)

Answer:
1. Given, y = x² – 2x + 7 ⇒ y’ = 2x – 2
(y’)_x=2 = 2(2) – 2 = 2.

2. At x = 2 , y = 2² – 2(2) + 7 = 7.
Equation of the tangent at (2, 7) is
y – 7 = 2(x – 2) ⇒ 2x – y + 3 = 0.

Question 26.
Find the absolute maximum value and minimum value of the following functions.

1. f(x) = 2x³ – 15x² + 36x + 1, x ∈ [1, 5]
2. f(x) = 12x^{\(\frac{4}{3}\)} – 6x^{\(\frac{1}{3}\)}, x ∈ [-1, 1]

Answer:
1. Given; f(x) = 2x³ – 15x² + 36x + 1, x ∈ [1, 5]
⇒ f'(x) = 6x² – 30x + 36
For turning point f'(x) = 0 ⇒ 6x² – 30x + 36 = 0
⇒ x² – 5x + 6 = 0 ⇒ (x -3)(x – 2) = 0
⇒ x = 3, 2
f(1) = 2(1)³ – 15(1)² + 36(1) + 1 = 24
f(2) = 2(2)³ – 15(2)² + 36(2) + 1 = 29
f(3) = 2(3)³ – 15(3)² + 36(3) + 1 = 28
f(5) = 2(5)³ – 15(5)² + 36(5) + 1 = 56
Absolute maximum = max {24, 29, 28, 56} = 56
Absolute minimum = min {24, 29, 28, 56} = 24

2. Given;

f'(x) = 0 at x = \(\frac{1}{8}\) and f'(x) is not defined at x = 0. Therefore;

Absolute maximum = max {18, 0, 6, \(-\frac{9}{4}\)} = 18
Absolute minimum = min {18, 0, 6, \(-\frac{9}{4}\)} = \(-\frac{9}{4}\).

Question 27.
Consider the function y = x³ – 6x² + 3x – 1

1. Find the slope at x= -1. (1)
2. Find the minimum gradient of the above curve. (3)

Answer:
1. Given,
y = x³ – 6x² + 3x – 1 ⇒ y’ = 3x² – 12x + 3
Gradient at (x = -1) = (y’)_x=1 = 3(-1)² – 12(-1) + 3 = 18.

2. Now for minimum gradient we have to apply maxima – minima condition to the function y’ .ie, y” = 6x – 12 , for turning points of y’ is given by y” = 0.
Therefore, 6x – 12 = 0 ⇒ x = 2
Now, y”’ = 6 > 0
∴ y’ is maximum at x = 2.
Minimum gradient at (x = 2) is
= 12 – 24 + 3 = – 9.

Plus Two Maths Application of Derivatives Six Mark Questions and Answers

Question 1.
A curve passes through the origin, and its gradient function is 2x – \(\frac{x^{2}}{2}\)

1. Find its y coordinate when x= 2. (4)
2. Find the equation of the tangent at x= 2. (2)

Answer:
Given;

Integrating we have; ∫dy = ∫(2x – \(\frac{x^{2}}{2}\))dx
⇒ y = x² – \(\frac{x^{3}}{6}\) + c ___(1)
Since the curve passes through (0, 0)
(1) ⇒ 0 = 0 + c ⇒ c = 0
∴ Equation of the curve is y = x² – \(\frac{x^{3}}{6}\)
When x = 2 ⇒ y = 2² – \(\frac{2^{3}}{6}\) = \(\frac{8}{3}\)
∴ coordinate is (2, \(\frac{8}{3}\)).

2. Slope at (2, \(\frac{8}{3}\)) = 2 × 2 – \(\frac{2^{2}}{2}\) = 2
∴ Equation of the tangent at (2, \(\frac{8}{3}\)) is given by
y – \(\frac{8}{3}\) = 2(x – 2) ⇒ 3y – 8 = 6x – 12 ⇒ 3y = 6x – 4.

Question 2.
(i) Choose the correct answer from the bracket. The slope of the tangent to the curve y = x³ – 2x + 3 at x = 1 is ____(1)
(a) 0
(b) 1
(c) 2
(d) 3
(ii) Find points on the curve \(\frac{x^{2}}{25}+\frac{y^{2}}{9}\) = 1 at which the tangents are (2)
(a) Parallel to x-axis
(b) parallel to y – axis.
(iii) Use differential to approximate \(\sqrt{25.6}\). (3)
Answer:
(i) (b) 1, Since

(ii)

(a) \(\frac{d y}{d x}\) = 0, since tangents are parallel to x- axis.
\(\frac{-9x}{25}\) = 0, x = 0 ∴ y = ± 3;
The points are (0, 3) and (0, -3)

(b) \(\frac{-25 y}{9 x}\) = 0, since tangents are parallel to y-axis, slope of normal = 0; y = o
∴ x = ± 5
The points are (5, 0) and (-5, 0)

(iii) Take y = \(\sqrt{x}\) , let x = 25 and ∆x = 0.6
Then; f(x) = y = \(\sqrt{x}\)
f(x + ∆x) = y + ∆y

Question 3.
Let x and y be the length and breadth of the rectangle ABCD in a circle having radius r. Let ∠CAB = θ (Ref. figure). If ∆ represent area of the rectangle and r is a constant.

1. Write ∆ in terms of r and θ. (2)
2. Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\). (1)
3. Hence find the maximum value of ∆. (2)
4. Show that the rectangle of maximum area that can be inscribed in a circle of radius r is a square of side \(\sqrt{2} r\). (1)

Answer:
1. Area of the rectangle is ∆ = xy
From the figure y = 2r sinθ, x = 2r cosθ
∆ = xy = 4r²sinθcosθ = 2r²sin2θ

2. \(\frac{d \Delta}{d \theta}\) = 4r² cos2θ ⇒ \(\frac{d^{2} \Delta}{d \theta^{2}}\) = -8r²sin2θ

3. For turning points

Therefore local maximum at θ = \(\frac{\pi}{4}\)

4. Then;

Hence the rectangle becomes a square.

Question 4.
The second derivative of the equation of a curve is given by the equation x \(\frac{d^{2} y}{d x^{2}}\) = 1, given y = 1, \(\frac{d y}{d x}\) = 0 when x= 1.

1. Find the slope at x = e. (2)
2. Find the equation of the curve. (2)
3. Find the equation of the normal at x= e. (2)

Answer:
1. Given;

Integrating we get,

∴ Slope of the curve at x = e is given by
\(\left(\frac{d y}{d x}\right)_{x=e}\) = loge = 1.

2. We have,
\(\frac{d y}{d x}\) = logx, ⇒ dy = logxdx
Integrating we get,
∫dy = ∫logx dx ⇒ y = logx.x – ∫\(\frac{1}{x}\).x dx + c₂
⇒ y = xlogx – x + c₂ ____(2)
Given; y = 1 when x = 1
(2) ⇒ 1 = 1log1 – 1 + c₂ ⇒ 1 = 0 – 1 + c₂ ⇒ c₂ = 2
Therefore the equation of the curve is
y = xlogx – x + 2

3. We have, y = xlogx – x + 2
When x = e
⇒ y = e log e – e + 2 ⇒ y = e – e + 2 = 2
So we have to find the slope at (e, 2),
We know; \(\left(\frac{d y}{d x}\right)_{x=e}\) = log e = 1
∴ Slope of the normal at (e, 2)= -1
∴ Equation of the normal at (e, 2) is.
y – 2 = (-1) (x – e)
y – 2 = – x + e ⇒ y + x = e + 2.

Question 5.
The given figure represents a cylinder Inscribed in a sphere.

1. Find an expression for the volume V of the cylinder. (2)
2. Find the height of the cylinder when its volume V is maximum. (2)
3. Find the volume and radius of the largest cylinder. (2)

Answer:
1. From the right triangle ∆OAB,
y² = R² – x² ⇒ y = \(\sqrt{R^{2}-x^{2}}\)
Which is the radius of the cylinder
Also height = 2 x
∴ Volume = V= π y² × 2x = 2π(R² – x²)x = 2π(R²x – x³).

2. Now,

For maximum or minimum,

3. Base radius =

Question 6.
If f(x) = x³ + 3x² – 9x + 4 is a real function

1. Find the intervals in which the function is increasing or decreasing. (3)
2. Find the points of local maxima or local minima of f(x) (2)
3. Graph of a function is given in the following figure:

Which among the following represents the graph of its derivative? (1)

Answer:
1. f'(x) = 3x² + 6x – 9
For turning points f'(x) = 3x² + 6x – 9 = 0
⇒ x = 1, -3
These turning point divide the domain of f(x) in the following intervals. (-∞, -3), (-3, 1), (1, ∞) in (-∞, -3)
⇒ f'(-4) = 3(-4)² + 6(-4) – 9 > 0
Hence increasing.
In (-3, 1) ⇒ f'(0) = 3(0)² + 6(0) – 9 < 0
Hence decreasing.
In (1, ∞) ⇒ f'(2) = 3(2)² + 6(2) – 9 > 0
Hence increasing.

2. x = -3 is a local maximum point and x = 1 is a local minimum point.

3. (a)

The function passes through origin and has a local maximum at x = 2.

Question 7.
Of all the Cylinders with given surface area, show that the volume is maximum when height is equal to the diameter of the base.
Answer:
Let r be the radius, h be the height, V be the volume and S be the surface area
S = 2πr² + 2 πrh

S – 6πr² = 0
2πr² + 2πrh – 6πr² = 0
So h = 2r
So volume is maximum when h = 2r.

Question 8.
Sand is pouring from a pipe. The falling sand forms a Cone on the ground in such a way that the height of the Cone is always one-sixth of the radius of the base.

1. Establish a relation between the volume ‘v’ and height ‘h’ of the Cone using the given condition. (2)
2. lf the sand is pouring at the rate of -12 cm/sec, Find the rate of change of height of the Cone. (2)
3. Find \(\frac{\mathrm{dh}}{\mathrm{dt}}\) when h = 4cm. (2)

Answer:
1. Given that the height of the Cone is one-sixth of the radius of the base, then h = \(\frac{r}{6}\) ⇒ r = 6h
Then Volume V = \(\frac{1}{3}\) πr²h = \(\frac{1}{3}\) π(6h)2.h
V = \(\frac{1}{3}\) π 36h².h = \(\frac{1}{3}\) π 36h³
V =12 πh³ ____(1)

2. Differentiating (1) we get

3. When h = 4 cm

Question 9.
(i) Choose the correct answer from the bracket. The rate of change of the area of a circle with respect to its radius r at r = 10cm is.
(a) 10π
(b) 20π
(c) 30π
(d) 40π (1)
(ii) Find the intervals in which the function f given by f(x) = x² – 6x + 5 is (2)
(a) Strictly increasing
(b) Strictly decreasing
(iii) Find the local minimum and local maximum value, if any, of the function f(x) = x³ – 6x² + 9x + 8 (3)
Answer:

(ii) f'(x) = 2x — 6; 2x – 6 = 0; x = 3
(-∞, 3 ) is strictly decreasing
(3, ∞) is strictly increasing.

(iii) f'(x) = 3x² – 12x + 9
f¹¹ = 6x — 12
For maxima, minima
f¹ = 0 → 3x² – 12x + 9 = 0
3(x – 3)(x — 1) = 0; x = 3, x = 1
At x = 3 f¹¹(x) = 6 × 3 – 12 = 18 – 12 = 6 > 0
f is minimum, the local minimum value of f = 8
At x = 1 f¹¹(x) = 6 × 1 – 12 = -6 < 0,
f is maximum, the local maximum value of f = 12.

Question 10.
A wire of length 28m is cut into two pieces. One of the pieces is be made into a square and the other into a circle. What should be the length of the two pieces so that combined area of the square and the circle is minimum using differentiation?
Answer:
Let the length of one piece be ‘x’ and other piece be ‘28 – x’. Let from the first piece we will make a circle of radius Y and from the second piece we will make a square of side y. Then,

Let A be the combined area of the circle and square, then

Question 11.
An open box of maximum volume is to be made from a square piece of tin sheet 24cm on a side by cutting equal squares from the corners and turning of the sides.
(i) Complete the following table. (2)

(ii) Using the above table, express V as a function of x and determine its domain. (1)
(iii) Find height (x. cm) of the box when volume V is maximum by differentiation. (3)
Answer:
(i)

(ii) Generalise the above table as a function.
V = x(24-2x)², 0 < x < 12.

(iii) \(\frac{d V}{d x}\) = x.2(24 – 2x)(-2) + (24 – 2x)²
= -4x(24 – 2x) + (24 – 2x)²
= -96x + 8x² + 576 + 4x² – 96x
= 12x² – 192x + 576
For maximum or minimum,

Therefore volume is maximum when x = 4 cm.

Question 12.
A square tank of capacity 250 m³ has to be dug out. The cost of land is Rs. 50 per m². The cost of digging increases with the depth and for the whole tank is Rs. 400 × (depth)².

1. Find an expression for the cost of digging the tank. (3)
2. Find the dimension of the tank when the total cost is least. (3)

Answer:
1. Let x, x and y be the length, breadth, and depth of the tank.
Then, V = x. x. y = 250 ⇒ y = \(\frac{250}{x^{2}}\).
Area of land = x²
⇒ Cost of land = 50 x²
(∵ cost of land is Rs.50/m²)
Cost of digging = 400 × (depth)² = 400 × (y)²
∴ Total cost = C = 50 x² + 400 × (y)²

2. We have, C = 50x² + \(\frac{400 \times(250)^{2}}{x^{4}}\).
Differentiating w.r.t.x, we get,


∴ Maximum at x = 10
m ⇒ when x = 10m and
y = \(\frac{250}{10^{2}}\) = 2.5m the total cost is least.

Question 13.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) times the radius of the base.
Answer:
Volume of the cone will be, V =\(\frac{1}{3}\)πr²h
h = \(\frac{3 V}{\pi r^{2}}\) ____(1)
Curved surface area will be, S = πrl
⇒ S² = π²r²l² = P
⇒ P = π²r²(h² + r²) ⇒ P = π²r²h² + π²r⁴)

⇒ 2r² = h² ⇒ h = \(\sqrt{2} r\).

Question 14.
Let ABC be an isosceles triangle inscribed in a circle having radius r. Then by figure, area of the triangle ABC is ∆

1. Find \(\frac{d \Delta}{d \theta}\) and \(\frac{d^{2} \Delta}{d \theta^{2}}\) (2)
2. Find the maximum value of ∆. (3)
3. Show that the isosceles triangle of maximum area that can be in scribed in a given circle is an equilateral triangle. (1)

Answer:
1. Area of the isosceles triangle is ∆ = \(\frac{1}{2}\)bh
From the figure b = r sin2θ, h = r + r cos2θ

2. For turning points

Therefore local maximum a θ = \(\frac{\pi}{6}\) which means the area of the isosceles triangle is maximum When θ = \(\frac{\pi}{6}\).

3. Then; ∠OCB = 30° ⇒ ∠ACB = 2∠OCB = 60°. Therefore the isosceles triangle is an equilateral triangle.

Question 15.
(i) Using the graph of the function f (x) in the interval [ a, h ] match the following.

A – Point	B – Nature
x = a	Absolute maximum
x = b	Absolute minimum
x = e	Local maximum
x = d	Local minimum
	Point of inflexion.

(ii) Consider the function f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
(a) Find the turning points of f(x). (1)
(b) Explain the nature of the turning points (1)
(c) Find the absolute extreme values of f(x). (2)
Answer:
(i)

A – Point	B – Nature
x = a	Absolute minimum
x = b	Local maximum
x = e	Point of inflexion
x = d	Absolute maximum

(ii) (a) f(x) = 12x³ – 24x² + 24x – 48
For turning points,
f'(x) = 0 ⇒ 12x³ – 24x² + 24x – 48 = 0
⇒ x³ – 2x² + 2x – 4 = 0
⇒ (x² + 2)(x – 2) = 0 ⇒ x = ± (\(\sqrt{-2}\), 2)
We admit only x = 2 as x = \(\sqrt{-2}\) is not a real number.
Therefore at x = 2 f (x) has a turning point.

(b) f”(x) = 36x² – 48x + 24
⇒ f”(2) = 36(2)² – 48 × 2 + 24 > 0
Therefore at x = 2 f(x) has a local minimum.

(c) f(0) = 25,
f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48 × 2 + 25 = -39
f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48 × 3 + 25 = 16
Consider the set { f (0), f{2), f (3)}
⇒ {25, -39, 16}
The maximum value of the above set is the absolute maximum and it is 25 at x = 0. The minimum value of the above set is the absolute minimum and it is -39 at x = 2.

Question 16.
An open box with a square base is to be made out of a given quantity of sheet of area a².

1. If the box has side x units, then show that volume V= \(\frac{a^{2} x-x^{3}}{4}\) (2)
2. Show that the maximum volume is \(\frac{a^{3}}{6 \sqrt{3}}\) (4)

Answer:
1. Area = a² = x² + 4xh,
h = height of the box.

2. We have,

Question 17.
For the function f(x) = sin2x, 0 < x < π
(i) Find the point between 0 and π that satisfies f'(x) = 0. (2)
(ii) Find the point of local maxima and local minima. (2)
(iii) Find the local maximum and local minimum value. (2)
Answer:

Question 18.
A cylindrical can with a volume of 125m3 (about 2 litres) is to be made by cutting its top and bottom from metal squares and forming its curved side by bending a rectangular sheet of metal to match its ends. What radius ‘r’ and height ‘h’ of the can will minimize the amount of material required.
Answer:

The circular top and bottom should be cut out from a square metal sheet of side 2r. Therefore the area of squares is 8r².
Area A = 8r² + 2πrh

∴ To minimize the amount of material, r = 2.5
\(h=\frac{125}{\pi(2.5)^{2}}=6.3\).

Question 19.
A rectangle sheet of tin with adjascent sides 45cm and 24cm is to be made into a box • without top, by cutting off equal squares from the comers and folding up the flaps

1. Taking the side of the square cut off as x, express the volume of the box as the function of x. (2)
2. For what value of x, the volume of the box will be maximum. (4)

Answer:
1. Length of the box = 45 – 2x
Breadth of the box = 24 – 2x
Height of the box = x
Volume; V = (45 – 2x)(24 – 2x)x
= (1080 – 138x + 4x²)x
= 4x³ – 138x² + 1080x.

2. \(\frac{d y}{d x}\) = 12x² – 276x + 1080

12x² – 276x + 1080 = 0
x² – 23x + 90 = 0
x = 18, 5
x = 18 is impossible
∴ x = 5 when x = 5, \(\frac{d^{2} y}{d x^{2}}\) < 0

The volume of the box is maximum at x = 5.

Students can Download Chapter 4 Graphs and Charts for Business Data Questions and Answers, Plus Two Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 4 Graphs and Charts for Business Data

Plus Two Accountancy Graphs and Charts for Business Data One Mark Questions and Answers

Question 1.
_____________ are the visual representation of numerical data
Answer:
Chart/ Graph

Question 2.
Chart / Graph has at least ____________ dimenstional relationship
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(a) Two

Question 3.
____________ chart is suitable for comparing multiple.
Answer:
Bar Chart

Question 4.
In column chart, the X-axis shows
(a) Value of each category
(b) Different categories
(c) Height of the chart
(d) Depth of the value
Answer:
(b) Different categories

Question 5.
________ Chart is similar to the column chart, with the difference being that the data series are displayed horizontally
(a) Line chart
(b) Pie chart
(c) Barchart
(d) Area chart
Answer:
(c) Bar chart

Question 6.
________ chart shows data changes for a certain period of time.
Answer:
Line chart

Question 7.
_______ chart contains only one data series
Answer:
Pie chart

Question 8.
_______chart shows values as circular sectors to the total circle
Answer:
Pie chart

Question 9.
Pie chart don’t have more than __________ categories.
(a) Ten
(b) Twenty Five
(c) Seven
(d) Three
Answer:
(c) Seven

Question 10.
____________ is a pictorial representation of data, which has at least two dimensional relationships.
(a) Graph
(b) Chart
(c) Diagram
(d) All the above
Answer:
(d) All the above

Question 11.
_________ Chart is used to compare values across categories.
(a) Column chart
(b) Line chart
(c) Pie chart
(d) Barchart
Answer:
(a) Column chart

Question 12.
_________ chart is used to display trends over time.
Answer:
Line chart

Question 13.
The entire chart including all elements is termed as ________
Answer:
Chart area

Question 14.
In 3D chart, the area is bounded by three axes ie _____, _____ & _____
Answer:
X, Y, Z

Question 15.
In ______ chart, both axes display values ie they have no category axis.
Answer:
XY Chart or Scatter diagram

Question 16.
_______ specifices the colour, symbol or pattern used to mark data series.
Answer:
Legends

Question 17.
The change the location of a chart, right click the chart and select.
(a) Chart Type
(b) Source Data
(c) Move here
(d) Chart Options
Answer:
(c) Move here

Question 18.
In 3D chart X, Y & Z axes are used to show
(a) Category, Value, Total
(b) Depth, Vertical, Horizontal
(c) Length, Breadth, Depth
(d) Category, Value, Series.
Answer:
(d) Category, Value, Series

Question 19.
Legand can be repositioned on the chart
(a) Anywhere
(b) On right side only
(c) On the corner only
(d) On the left side only
Answer:
(a) Anywhere

Question 20.
Which chart element details the data values and categories below the chart?
(a) Data table
(b) Data marker
(c) Data labels
(d) Datapoint
Answer:
(c) Data labels

Question 21.

Which type of chart is this?
Answer:
Radar chart

Question 22.
The intersection of both the axis (X-axis and Y-axis) is called __________ of the graph.
Answer:
Origin (0)

Question 23.
_________ Chart display in rings, where each ring represents a data series
Answer:
Doughnut chart.

Question 24.
Name the given chart

Answer:
Bar Chart

Question 25.
In _________ chart, the area below the plotted lines is solid
Answer:
Area chart.

Question 26.
Radar chart / Net chart is also known as ______
(a) Doughnut chart
(b) Pie chart
(c) Ara chart
(d) Star chart
Answer:
(d) Star chart

Question 27.
Which among the following is the special feature of 3D chart
(a) Chart area
(b) X & Y axes
(c) Chart wall
(d) Legend
Answer:
(c) Chart wall

Question 28.
Give a suitable name to the diagram

(a) Barchart
(b) Single line graph
(c) Pie chart
(d) Area Chart
Answer:
(c) Pie Chart

Plus Two Accountancy Graphs and Charts for Business Data Two Mark Questions and Answers

Question 1.
Name the different chart formats in Libre Office Calc
Answer:
Barchart, Column Chart, Pie chart, Line chart, Area chart, Doughnut chart, etc.

Question 2.
What is the importance of charts and graphs in business?
Answer:

1. Chart and graphs covey lots of business information in a visual format
2. Different business Data variables plotted in charts and graphs show the trend of the business in an easy way

Question 3.
Identify the type of chart

Answer:
Line chart

Question 4.
Give a short note on it.

1. Barchart
2. Pie chart

Answer:
1. Bar Chart
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

2. Pie chart
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

• Normal Pie chart
• Exploded Pie chart
• Doughnut chart or Donut chart
• Exploded Doughnut chart

Question 5.
What are the special features of graphs and charts?
Answer:

1. Graphs/charts are the pictorial representation of business data
2. A chart represents tabular numeric data
3. Dimensions in the data are often displayed on axes (X, Y, & Z)

Question 6.
Differentiate between Chart area and Chart wall?
Answer:
1. Chart area:
This is the total space that is enclosed by a chart. It is the background of the chart.

2. Chart wall:
In 2D chart, the wall or area is bounded by the X and Y-axis. In the 3D chart, the wall is bounded by three axes X, Y and Z

Question 7.
Quarterly sales of a business firm is used to create a bar graph. Identify the Data variables plotted on X and Y-axis
Answer:
X-axis – I^st Quarter, II^nd Quarter, III^rd Quarter, IV^th Quarter,
Y-axis – Sales in I^st Quarter, Sales in I^nd Quarter, Sales in III^rd Quarter, Sales in IV^th Quarter

Question 8.
Identify the type of chart

Answer:
(a) 2D Chart
(b) 3D Chart

Question 9.
What is the use of Auto shapes in LibreOffice Calc?
Answer:
Auto shapes tool bar allows drawing a number of geometrical shapes, arrows; flow chart elements, etc.

Question 10.
What is PIE chart? What are the specialties of PIE chart.
Answer:
Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

1. Normal Pie chart
2. Exploded Pie chart
3. Doughnut chart or Donut chart
4. Exploded Doughnut chart

Question 11.
Choose the right statements from the following.

1. We can put on the right side of the origin positive values and on left side of the origin negative values of data on X-axis
2. The upward side of origin shows postiive values and downward side of the origin shows negative values of data on Y-axis.
3. We can put on the right side of the origin negative values and on left side of the origin positive values of data on X-axis.
4. The upward side of origin shows negative values and downward side of the origin shows positive values of data on Y-axis.

Answer:
Right statements a & b

Question 12.
Identify the type of given chart. List down its features.

Answer:

1. It is a Doughnut chart
2. Features of the doughnut chart
   • It displays data in rings
   • Each ring represents a data series
   • The first data series is displayed in the center of the chart

Question 13.
What is a 3-D chart?
Answer:
Charts can be prepared with three dimensional (3-D) effects. 3- D charts have a third axis. The third axis is called as Z-axis. So a 3-D chart has the fol¬lowing dimensions.

1. Horizontal axis – Indicate the category – known as X-axis
2. Vertical axis – Indicate the derived values – known as Y-axis
3. Depth axis – Indicate the series – known as Z-axis

Question 14
Differentiate between Data Marker and Data series.
Answer:
1. Data Marker:
Individual values plotted in a chart are called data marker or data point.

2. Data Series:
Data markers of the same colour or pattern is called data series.

Question 15.
Is there any difference between

1. A column chart and
2. A bar chart?

Substantiate your answer
Answer:
1. Column Chart:
It is the most commonly used chart type. It shows a bar chart or bar graph with vertical bars. The X-axis shows the categories and Y-axis shows the value for each category. Column chart are used to compare values across categories.

2. Bar Chart:
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

Plus Two Accountancy Graphs and Charts for Business Data Three Mark Questions and Answers

Question 1.
Match the following

A	B
(a) Area chart	(1). XY chart.
(b) Barchart	(2). Display contribution to a total.
(c) Pie chart	(3). Suited for comparing multiple values.
(d) Scatter chart	(4). Display differences between several sets of data over a period of time.

Answer:

A	B
(a) Area chart	(1). Display differences between several sets of data over a period of time.
(b) Barchart	(2). Suited for comparing multiple values.
(c) Pie chart	(3). Display contribution to a total.
(d) Scatter chart	(4). XY chart.

Question 2.
What are the advantages of using Graph/ Chart?
Answer:
Advantages in using Graph/Chart:

1. It summarises a large data set in visual form
2. Charts or graphs can clarify trends better than do tables.
3. It helps to estimate key values at a glance
4. It shows each data category in a frequency distribution.
5. It permits a visual check of the accuracy and reasonableness of calculations
6. The charts and graphs allow the investigator to draw a valid conclusion.

Question 3.
What are the elements of a Chart/ Graph
Answer:

Chart elements	Description
1. Axes Titles	Mention the names or titles for X, Y and Z axes.
2. X, Y, & Z axes	In 2D chart, the horizontal X-axis contains categories and the vertical Y-axis contains dependent values. In 3D chart, the Z-axis will also be there represents the depth which
3. Chart Area	This is the total space that is enclosed by a chart. It is the background of the chart.
4. Chart wall	In 2D chart, the wall or area is bounded by the X and Y-axis. In the 3D chart, the wall is bounded by three axes X, Y, and Z.
5. Chart floor	The chart floor is the lower area in the 3D chart.
6. Main Title/ sub Title	It is the explanatory heading of the chart. It identifies the purpose of a chart.
7. Data Marker	Individual values plotted in a chart are called data marker or data point.
8. Data Series	Data markers of the same colour or pattern is called data series.
9. Legend	It is an identifier of a piece of information shown in the chart/ graph. The legends are assigned to the data series in a chart.
10. Data Label	The value of the data series plotted in a chart is known as data label.
11. Grid Lines	These are the vertical and horizontal lines that appear in a chart. It increases the readability of a chart.

Question 4.
How to use word Art styles to format text.
Answer:

• Step 1: Click in the chart element that contains text to be changed.
• Step 2: Click on the format.
• Step 3: Click on word Art styles.
• Step 4: Choose suitable options related to text formating like text fill, text outlines, shadow, etc.

Plus Two Accountancy Graphs and Charts for Business Data Four Mark Questions and Answers

Question 1.
What are the difference between 2D charts and 3D charts
Answer:

2D Chart	3D Chart
(a) The chart represents business data with just two dimensions	(a) The chart represents business data with three dimensions
(b) The two dimensions are length and height (No width)	(b) The Three dimensions are Length and Height and width (or depth)
(c) There are X-axis and Y-axis	(c) There is X-axis, Y-axis is and Z-axis
(d) The shape of the chart may be in the form of Rectangle, Square, Triangle, Polygon, etc	(d) The shape of the chart may be Cylinder, Cube, Pyramid, etc

Question 2.
List out the steps to Rotate a chart.
Answer:

• Step 1. Select the plot area of the chart.
• Step 2. Click on the format tab.
• Step 3. Click on format selection.
• Step 4. Click on 3D Rotation and type a value of angle between 0° to 360° and then click close
• Step 5. Click on the chart area of the chart and click on format tab.
• Step 6. Click on shape effects and then click on Bevel and select a bevel option.

Question 3.
What are the different types of charts?
Answer:

1. Column chart: column chart are used to compare values across categories
2. Line chart: Line charts are used to display trends over time
3. Pie chart: Pie charts display the contribution of each value to a total
4. Bar chart: Bar charts are best suited for comparing multiple values
5. Area chart: Area chart emphasis differences between several sets of data over a period of time.
6. Scatter chart: (XY chart) This chart compares pairs of values.
7. Radar chart: Display values relative to a centre point.
8. Doughnut chart: It shows the relationship of parts to a whole. This chart display data in rings, where each ring represents a data series.

1. Column Chart:
It is the most commonly used chart type. It shows a bar chart or bar graph with vertical bars. The X-axis shows the categories and Y-axis shows the value for each category. Column chart are used to compare values across categories.

2. Line Chart:
A line chart shows values in the Y-axis and categories in X-axis. The Y values of each data series is connected by a line. Line chart shows data changes for a certain period of time.

3. Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroup of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

• Normal Pie chart
• Exploded Pie chart
• Doughnut chart or Donut chart
• Exploded Doughnut chart

4. Bar Chart:
This type of chart shows a bar graph or column chart with horizontal bars. The Y-axis shows categories and the X-axis shows the value for each category. It is suitable for comparing multiple values.

5. Area chart:
The chart shows values as points on the Y-axis. The X-axis shows categories. The Y values of each data series are connected by a line. The area between each two lines is filled with a colour.

6. Scatter chart:
Scatter chart is also known as XY chart. In this type of chart, both axes display values. This chart is used to show the relationship among two variables.

7. Radar chart:
It is also known as Net chart or Star chart. A radar chart has a separate axis for each category and the axes extend outward from the center of the chart. The value of each data point is plotted on the corresponding axis.

8. Doughnut chart:
Chart display in rings, where each ring represents a data series. The first data series is displayed in the centre of the chart.

Question 4.
Identify and explain the type of chart given below.
Answer:

1. This is scatter chart or XY chart.
2. Features:
   • Both axes display values (No category)
   • This chart is used to show the relationship among two variables
   • Generally this chart is used for scientific, statistical and engineering data

Question 5.
Match the following.

A	B
(a) Legends	(i) Background of the chart
(b) Pie chart	(ii) Specifices the colour, symbol or pattern used to mark data series
(c) Grid Lines	(iii) Displays the contribution of each value to a total
(d) Chart Area	(iv) Display lines at the major intervals on the category X-axis and/or Y-axis

Answer:

• (a) – (ii);
• (b) – (iii);
• (c) – (iv);
• (d) – (i)

Question 6.
How can we change the format of a selected chart element?
Answer:

• Step 1. Click anywhere in the chart.
• Step 2. Click format
• Step 3. Click format selection
• Step 4. Select a category (Fill border, style, etc)
• Step 5. Select formatting options

Question 7.
List down any four advantages of charts/ Graphs
Answer:
Advantages in using Graph/Chart:

1. It summarises a large data set in visual form
2. Charts or graphs can clarify trends better than do tables.
3. It helps to estimate key values at a glance
4. It shows each data category in a frequency distribution.
5. It permits a visual check of the accuracy and reasonableness of calculations
6. The charts and graphs allow the investigator to draw a valid conclusion.

Question 8.
What are the features of Charts/ Graphs in Libre Office Calc?
Answer:

1. Chart is a graphical representation of data
2. They are visual representation of numerical data
3. Charts can be read more quickly than the raw data
4. A chart has at least two axes – X and Y

Plus Two Accountancy Graphs and Charts for Business Data Five Mark Questions and Answers

Question 1.
What Pie Chart? What are the different types of Pie Chart?
Answer:
Pie chart:
A pie chart displays the contribution of each value to a total. It represents multiple subgroups of a single variable. It contains only one data series. A pie chart shows values as circular sectors of the total circle. Pie chart may be

1. Normal Pie chart
2. Exploded Pie chart
3. Doughnut chart or Donut chart
4. Exploded Doughnut chart

Question 2.
Name the different elements of given chart.

Answer:

• X-Axis Title
• Y-Axis Title
• Data label
• Main Title
• Legend
• X-Axis
• Y-Axis
• Data series

Question 3.
Write the steps of changing the chart type.
Answer:

1. First select the chart by double-clicking on it. The chart should now be surrounded by a gray bonder
2. Right-click on the chart and choose chart type.
3. Select the replacement chart type.
4. Click on [OK]

Question 4.
Write the steps for preparing a chart in Libre Office Calc.
Answer:

• Step 1: Enter the data in a worksheet with proper column and row titles
• Step 2: Select the range of data using the mouse
• Step 3: Click on Insert Tab → Object → Chart. Select a chart type from the “Choose a chart type” list in chart wizard window.
• Step 4: Naming chart, X-axis and Y-axis. Click on the chart → Right click → Insert titles (Names) → OK
• Step 5: Change the layout or styles of chart.
• Step 6: Show or hide a legend
• Step 7: Display or hide chart axes or gridlines
• Step 8: Move (resize) a chart
• Step 9: Save a chart

Plus Two Accountancy Graphs and Charts for Business Data Practical Lab Work Questions and Answers

Question 1.
Draw an Area Chart from the following

Procedure:
Step 1 – Open a new blanks worksheet in LibreOffice Calc

Step 2 – Enter the above data as follows.

Step 3 – Select the range A1: D6 which is to be shown in the chart.

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Select Area Chart → Finish
Output:

Question 2.
Quarterly sales of a product are given below. Draw a bar diagram/bar chart

Ist Quarter	25600
IInd Quarter	33400
IIIrd Quarter	28700
IVth Quarter	40400

Procedure:
Step 1 – Open a new blanks worksheet in LibreOffice Calc

Step 2 – Enter the above data as follows.

Step 3 – Select the range A1: B5 which is to be shown in the chart:

Step 4 – Click on Insert menu → Click on Chart → Chart wizard Click on Bar chart → Finish
Output:

Question 3.
Draw a 3D column chart from the following details.

Year	Result %
2010	98
2011	94
2012	100
2013	85
2014	90

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc.

Step 2 – Enter the following data in the respective cells.

Step 3-Select the range A1: B6, which is to be shown in the chart.

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Click on Column Chart → Tick 3D Look → Select Bar Charts → Finish.
Output:

Question 4.
The net profits of a firm for the last six years are given below. Draw a line chart.

Year	Net Profit
2009	125800
2010	238400
2011	186500
2012	154900
2013	251000
2014	300000

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc

Step 2 – Enter the following data in the respective cells

Step 3 – Select the range A1: B7, which is to be shown in the chart

Step4- Click on Insert menu → Click on Chart → Chart Wizard → Click on Line Chart → Finish
Output:

Question 5.
Enter the following data into a LibreOffice Calc worksheet and draw a 3D Pie chart.

Item of Expenses	Amount
Stationery	4890
Tuition Fee	850
Medical Treatment	3260
Insurance premium	1580
Petrol	3500
Vegetables	700
Bank savings	8400
Charity	1200

Procedure:
Step 1 – Open a new blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in respective cells.

Step 3 – Select the range A2: B9, which is to be shown in the chart

Step 4 – Click on Insert menu → Click on Chart → Chart wizard → Click on Pie Chart Tick on 3D Look → Finish
Output:

Question 6.
Sales for the first six months in 2 years are given below. Draw a scatter chart in a LibreOffice Calc works sheet

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in the following cells.

Step 3 – Select the range A1: C7, which is to be shown in the chart

Step 4 – Click on → Insert menu Click on → Chart → Chart Wizard → Scatter Chart→ Finish
Output:

Question 7.
The production of different items in Oct. 2015 is listed below. Draw a Radar Chart

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2 – Enter the data in the following cells.

Step 3 – Select the range A1: C6, which is to be shown in the chart

Step 4 – Click on → Insert menu → Click on Chart → Chart Wizard → Click on Radar Chart → Finish
Output:

Question 8.
The following table shows the number of students passed in the higher secondary examination. Draw a doughnut chart.

Procedure:
Step 1 – Open a blank worksheet in LibreOffice Calc.

Step 2- Enter the data in the following cells.

Step 3 – Select the range A1: D6, which is to be shown in the chart.

Step 4 – Click on → Insert menu → Click on Chart → Chart Wizard → Click on doughnut Chart → Finish.
Output:

Students can Download Chapter 6 Open Economy Macroeconomics Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 6 Open Economy Macroeconomics

Plus Two Economics Open Economy Macroeconomics One Mark Questions and Answers

Question 1.
Name the ‘balancing item’ affecting the inability to record all international transactions accurately?
Answer:
Errors and omissions

Question 2.
The amount of rupees required to buy one US$ is known as ………….
(i) Rupee dollar exchange rate
(ii) Dollar rupee exchange rate
(iii) Real exchange rate
(iv) Real effective exchange rate
Answer:
(i) Rupee dollar exchange rate

Question 3.
Which among the following is a component of BOP account?
Answer:
(i) Current account
(ii) Capital account
(iii) Official reserve
(iv) All the above
Answer:
(iv) All the above

Question 4.
WTO was formed in?
(i) 1948
(ii) 1964
(iii) 1991
(iv) 1995
Answer:
(iv) 1995

Question 5.
If export > imports, it represents
(i) Trade surplus
(ii) Trade balance
(iii) Trade deficit
(iv) None of these above
Answer:
(i) Trade surplus

Question 6.
Which of the following would be an appropriate policy to reduce a Balance of Payment / Deficit.
Answer:
(i) An increase in government spending.
(ii) A cut in the level of indirect taxes
(iii) An increase in interest rates
(iv) A decrease in interest rates
Answer:
(iii) An increase in interest rates

Plus Two Economics Open Economy Macroeconomics Two Mark Questions and Answers

Question 1.
How is the exchange rate determined under flexible exchange rate regime?
Answer:
In case of flexible exchange rates, the exchange rate changes to clear the market to equate the demand for and supply of foreign exchange.

Question 2.
List two items of the capital account of balance of payment account?
Answer:

1. Gold movement
2. Reserve, Monetary gold & SDR

Question 3.
When will balance of trade shows a deficit?
Answer:
Balance of trade shows a surplus when exports are greater than imports. That is, Surplus balance of trade = Exports > Imports

Question 4.
Name two sources of demand for foreign exchange.
Answer:
Two sources of demand for foreign exchange are:

1. To purchase goods and services from abroad.
2. To send gifts and grants to foreign countries.

Question 5.
The value of a country’s import of goods is ₹200 crore and value of export of goods is ₹250 crore. Find out its balance of trade.
Answer:
Balance of trade = Value of exports – value of imports = 250 – 200 = ₹50 crore

Question 6.
Identify the items to be included to trade balance to get current account balance.
Answer:

1. Trade services
2. Net transfers

Question 7.
Identify the situation mentioned below.

1. Rupee-dollar exchange rate change from 45 to 50.
2. Rupee-dollar exchange rate changed from 45 to 43.

Answer:

1. Depreciation
2. Appreciation

Question 8.
State the National Income identity for an open economy.
Answer:
Y = C + IG + (X – M)
Where,

• Y = National Income
• C = Comsumption
• I = Investment
• G = Government spending
• X = Export
• M = Import

Question 9.
The consumption function and import function of an economy can be given as,
C = a + b.y
M = M + m.y Identity the letters

1. b
2. a
3. m

Answer:

1. Marginal propensity to consume
2. Autonomous consumption
3. Marginal propensity to import

Question 10.
If MPC = 0.8, and increase in autonomous demand is 200, calculate multiplier and the national output.
Answer:
MPC = b – c = 0.8
Therefore multiplier is
\(\begin{array}{l}
{\frac{1}{1-C} \text { or } \frac{1}{1-b}} \\
{\text { i.e. } \frac{1}{1-0.8}=\frac{1}{0.2}} \\
{=5}
\end{array}\)
National output is 5 × 200 = 1,000

Question 11.
Calculate the open economy multiplier if c = 0.5, m = 0.3. Increase in autonomous demand = 200
Answer:

Question 12.
What happens to the aggregate demand due to

1. A leakage from the circular flow of income
2. A injection to the circular flow of income

Answer:
AD falls due to the leakage from circular flow of income and AD increases due to injection in too the circular flow of income.

Question 13.
List out the expert of services from the following……..
(a) India buys a new technology from France
(b) A Japaneese tourist visits India
(c) An Indian student registers for a UK exam
(d) an Indian doctor going to work in the US.
Answer:
b and d are examples of expert services from India.

Question 14.
Analyse the effect of the following on imports and exchange rate.

1. Appreciation of domestic currency.
2. Depreciation of domestic currency
3. Increase in foreign direct investment.
4. Increase in import duty.

Answer:

1. Increase in Imports, Fall in Exchange Rate
2. Decrease in Imports, Rise in Exchange Rate
3. Increase in Imports, Fall in Exchange Rate
4. Decrease in Imports, Fall in Exchange Rate

Plus Two Economics Open Economy Macroeconomics Three Mark Questions and Answers

Question 1.
Classify the following into visible and invisible.
Steel, computer software, shipping services, wheat, machinery, food articles, banking, IT- enabled services, crude oil, shipping, textiles, Online business.
Answer:

Visibles	Invisibles
Steel	Shipping Services
Textiles	Banking
Wheat	IT Enabled Services
Machinery	Shipping
Food Articles	Insurance
Crude Oil	Online Business

Question 2.
Classify the following into current account and capital account.
Foreign direct investment, borrowing from abroad, export earning from merchandise, export earnings from banking services, earning from tourism, foreign portfolio investment.
Answer:

Current Account	Capital Account
Export earnings from banking services	Foreign direct investment
Earning from tourism	Borrowing from abroad
Export earnings from merchandise	Foreign portfolio investment

Question 3.
The open economy multiplier is smaller than that in a closed economy. Do you agree? Give reason.
Answer:
Yes, I do agree with this statement.
The open economy multiplier is smaller than that in a closed economy because a part of domestic demand falls on foreign goods. An increase in autonomous demand thus leads to a smaller increase in output compared to a closed economy. It also results in a deterioration of the trade balance

Question 4.
Complete the following table.

• Export > Import
• Export = Import
• Export < Import

Answer:

• Export > Import – Trade surplus
• Export = Import – Trade balance
• Export < Import – Trade deficit

Question 5.
Export promotion is one of the key factors for correcting disequilibrium in BOR Is there any other measure for correcting BOP? If yes, suggest 3 measures.
Answer:
Measures to correct BOP disequilibrium,

• Increase in production
• Reduction in imports
• Encouraging foreign investment
• Promotion of exports

Question 6.
What is the MPM (Marginal propensity to import) When M = 60 + 0.67?
Answer:
Marginal propensity to import (MPM) is the fraction of an additional currency of income spent on imports. The concept of MPM is same as the marginal propensity to consume (MPC). Thus, demand for imports is to depend on income and have an autonomous component.

Question 7.
Point out the items included in current account transactions of BOP.
Answer:
The current account includes receipts and payments on account of:

1. export and import of goods and services
2. tourism services
3. foreign investment incomes and out payments
4. private transfer payments
5. inter-government transfer payments.

Question 8.
What is trade deficit? Calculate the trade deficit from the following data.

Answer:
Trade deficit is the difference between export of goods and import of goods in trade in goods in current account. It is the situation where import is greater than export.
Export of goods = 90,660
Import of goods = 1,20,364 – 90,660 = 29,704 crores

Question 9.
a. Recently the government of UK decided to relax the visa norms to Indian visitors.
b. The government of India approved a purchase of weapon for Indian defence from rest of the world for an amount of 82000 crores

1. How does these decisions affect the demand for foreign exchange?
2. Analyse the consequences in the foreign exchange market with the help of a diagram.
   (supply curve of foreign exchange remain the same)

Answer:
1. Demand for foreign exchange increases

2. diagram

Plus Two Economics Open Economy Macroeconomics Five Mark Questions and Answers

Question 1.
Find the odd one out. Justify your answer.

1. tea, coir, tourism
2. rice, banking services, insurance service, transport services
3. foreign direct investment, foreign portfolio investment, remittances, borrowings
4. foreign investment, remittances, export earning from goods, export earning from services

Answer:

1. Tourism. Others are visibles
2. Rice. Others are invisibles
3. Remittances. Others are capital receipts
4. Remittances. Others are capital receipts.

Question 2.
Match the following.

A	B
Bretton Woods system	1944
SDR	1967
Fixed Exchange Rate	Pegged Rate
Triffin Dilemma	Dollar accumulation
Flexible Exchange Rate	Floating Rate

Answer:

A	B
Bretton Woods system	Pegged Rate
SDR	Dollar accumulation
Fixed Exchange Rate	1944
Triffin Dilemma	Floating Rate
Flexible Exchange Rate	1967

Question 3.
If c (marginal propensity to consume) = 0.8 and m (marginal propensity to import) = 0.3,

1. Find the open and closed economy multiplier
2. If domestic autonomous demand increases by 100, find the output level in a closed and an open economy.

Answer:
1. Closed economy multiplier
\(=\frac{1}{1-c}=\frac{1}{1-0.8}=\frac{1}{0.2}=5\)
Open economy multiplier
\(=\frac{1}{0.5}=\frac{1}{1-0.8+0.3}=\frac{1}{1-0.5}=\frac{1}{0.5}=2\)

2. If domestic autonomous demand increases by 100, in a closed economy output increases by 500 whereas it increases by only 200 in an open economy.

Question 4.
Differentiate between fixed exchange rate and flexible exchange rate.
Answer:
In a system of flexible exchange rates (also known as floating exchange rates), the exchange rate is determined by the forces of market demand and supply. Countries have had flexible exchange rate system ever since the breakdown of the Bretton Woods system in the early 1970s. Prior to that, most countries had fixed or what is called pegged exchange rate system, in which the exchange rate is pegged at a particular level.

Sometimes, a distinction is made between fixed and pegged exchange rates. Under a fixed exchange rate system, such as the gold standard, adjustment to BoP surpluses or deficits cannot be brought about through changes in the exchange rate.

Question 5.
What do you mean by managed floating? How far it is a mixture of fixed exchange rate and flexible exchange rates?
Answer:
Without any formal international agreement, the world has moved on to what can be best described as a managed floating exchange rate system. It is a mixture of a flexible exchange rate system (the floating part) and a fixed rate system (the managed part).

Under this system, also called dirty floating, central banks intervene to buy and sell foreign currencies in an attempt to moderate exchange rate movements whenever they feel that such actions are appropriate. Official reserve transactions are, therefore, not equal to zero.

Question 6.
Distinguish between the nominal exchange rate and the real exchange rate. If you were to decide whether to buy domestic goods or foreign goods, which rate would be more relevant?
Answer:
The price of one currency in terms of the other is known as the exchange rate. Nominal exchange rates are bilateral in the sense that they are exchange rates for one currency against another and they are nominal because they quote the exchange rate in money terms, i.e. so many rupees per dollar or per pound.

However, the real exchange rate is the ratio of foreign to domestic prices, measured in the same currency. It is defined as Real exchange rate = ePf/P where P and Pf are the price levels here and abroad, respectively, and e is the rupee price of foreign exchange (the nominal exchange rate).

The real exchange rate is often taken as a measure of a country’s international competitiveness. Therefore, real exchange rate is considered to be more relevant.

Question 7.
Balance of payment is a broader concept than balance of trade. Give explanation to this view.
Answer:
Balance of trade is the record of a country’s visible export and visible imports. It includes only visible trade and excludes invisible trade of services. However, balance of payment is a more comprehensive term which denoted a country’s total monetary transactions with the rest of the world. It includes both visible and invisible trade of goods and, services.

The balance of payments (BoP) records the transactions in goods, services and assets between residents of a country with the rest of the world. There are two main accounts in the BoP the current account and the capital account.

Question 8.
The current account is differentiated from capital account. Do you agree? Give explanation.
Answer:
Yes. The current account balance is the sum of the balance of merchandise trade, services and net transfers received from the rest of the world. The capital account balance is equal to capital flows from the rest of the world, minus capital flows to the rest of the world.

Question 9.
Illustrate the method of determining equilibrium under flexible exchange rate system. Also, show the effect of increase in demand for imports in the foreign exchange markets.
Answer:
In a system of flexible exchange rates, the exchange rate is determined by the forces of market demand and supply. In this case of flexible exchange rates without central bank intervention, the exchange rate moves to clear the market, to equate the demand for and supply of foreign exchange. In the following figure equilibrium exchange rate is e* which is determined by the forces of demand and supply.

At the initial equilibrium exchange rate e*, suppose there is now an excess demand for foreign exchange. To clear the market, the exchange rate must rise to the equilibrium value e₁ as shown in the following figure.

The rise in exchange rate (depreciation) will cause the quantity of import demand to fall since the rupee price of imported goods rises with the exchange rate. Also, the quantity of exports demanded will increase since the rise in the exchange rate makes exports. less expensive to foreigners. At the new equilibrium, the supply and demand for foreign exchange is again equal.

Question 10.
Differentiate between devaluation and depreciation.
Answer:
Devaluation means increase in exchange rate. Devaluation is said to occur when the exchange rate is increased by social action under a pegged exchange rate system. Devaluation is used as a tool to bridge the gap of trade deficit.

On the other hand, change in the price of foreign exchange under flexible exchange rate, when it becomes cheaper as compared to domestic currency is known as depreciation.

Question 11.
Compare balance of trade (BOT) and balance of payments (BOP).
Answer:
Balance of trade is the difference between money value of imports and exports of material goods only whereas BOP is the difference between a country’s receipts and payments in foreign exchange. The difference between the two can be summarized as follows:

BOT	BOP
1. It records only merch­andise transactions	1. It records transactions relating to both goods and services
2. It does not record trans­actions of special nature.	2. It records transactions of capital nature.
3. It is a narrow concept because it is only one part of BOP account	3. It is wider concept because it includes balance of trade, balance of Services, balance of unrequired transfers and balance of capital transactions.
4. It may be favorable, un favorable or equilibrium	4. It always remains in balance in accounting sense because receipt side is always made to be equal to payment side

Question 12.
Complete the following flow chart.

Answer:

Question 13.
Distinguish between autonomous and accommodating transactions?
Answer:
International economic transactions are called autonomous when transactions are made independently of the state of the BOP. These items are called above the line.

On the other hand, accommodating transactions are determined by the net consequences of the autonomous items, that is whether the BOP is in surplus or deficit. These items are called ‘below the line.

Question 14.
Suppose the equilibrium exchange rate is shown in the figure. What happens to this equilibrium situation when there is increase in demand for foreign exchange?

Answer:
When the demand for foreign exchange increases, there is rise in exchange rate (depreciation). At the higher exchange rate, more quantity of foreign exchange will be transacted. This is shown below.

Question 15.
Distinguish between appreciation and depreciation. Identify what happens to the exchange rate of rupees in 2015 compared to 2014.

Year	Rupee dollar exchange rate
2014	50.
2015	60.

Answer:
Appreciation refers to the increase the exchange rate of a currency. Depreciation refers to the decrease in the rate of exchange of currency. Both appreciation depreciation of exchange rate occurs due to the changes in the supply and demand of currencies. Compared to 2014 there is a depreciation of currency exchange rate in 2015.

Question 16.
The diagram below shows how the rate of exchange is determined in a free market.

Show the effect of the following on the exchange rate.

1. The rate of interest of the country increases.
2. The rate of inflation of the nearby countries.

Answer:
1. When the rate of interest increases the rate of exchange will increase. This is because an increased rate of interest would attract more depositors into the country, the demand for the currency would increase and the rate of interest also will increase as shown in the diagram below.

2. When the inflation of the nearby countries increases the people around would prefer to buy goods from this country. So the demand for the currency would increase leading to an increase in the rate of exchange.

Question 17.
Exchange rate is determined through different methods. Diagrams related with exchange rate are given below.

1. Identify the Exchange rate system corresponding to each diagram
2. Distinguish between the two.

Answer:
1. Diagram A is Flexible Exchange Rate System and Diagram B is Fixed Exchange Rate System

2. In a system of flexible exchange rates (also known as floating exchange rates), the exchange rate is determined by the forces of market demand and supply. Countries have had flexible exchange rate system ever since the breakdown of the Bretton Woods system in the early 1970s.

Prior to that, most countries had fixed or what is called pegged exchange rate system, in which the exchange rate is pegged at a particular level. Sometimes, a distinction is made between fixed and pegged exchange rates. Under a fixed exchange rate system, such as the gold standard, adjustment to BoP surpluses or deficits cannot be brought about through changes in the exchange rate.

Plus Two Economics Open Economy Macroeconomics Eight Mark Questions and Answers

Question 1.
Determine the equilibrium level of income based on the following information.
C = 100 + 0.75 (Y – T)
I = 200 – 2000;
G = 100
T = 80 + 0.20Y
X = 50
M = 20 + 0.10Y
Answer: