Plus One

Reviewing Kerala Syllabus Plus One Computer Science Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus One Computer Science Previous Year Question Paper March 2023

Time: 2 Hours
Total Score: 60 Marks

Answer any 5 questions from 1 to 6. Each carries 1 score. (5 × 1 = 5)

Question 1.
The number (158)₁₀ can be represented in Hexadecimal number system as ____________ .
Answer:
(9E)₁₆

Question 2.
Some of the components in the phases of programming are given below. Write them in order of their occurrence:
(i) Translation
(ii) Documentation
(iii) Problem identificaiton
(iv) Coding of program
Answer:
Problem Identification, Coding of a program, Translation, Documentation.

Question 3.
The memory size of ‘double’ data type in C++ is ……………….. bytes.
Answer:
8 bytes

Question 4.
In C++, name the loop which can be used to ensure that the body of the loop will surely be executed at least once.
Answer:
do while loop

Question 5.
Any device which is directly connected to a network is generally known as ___________ .
Answer:
node

Question 6.
Pick the odd one from the following list:
(i) Spam
(ii) Trojan horse
(iii) Phishing
(iv) Firewall
Answer:
firewall

Answer any 9 questions from 7 to 18. Each carries 2 scores. (9 × 2 = 18)

Question 7.
If (x)₈ = (101011)₂ = (y)₁₆ find the values of x and y.
(a) Temperature
(b) Pressure
(c) Frequency
(d) Wavelength
Answer:
x = 53 and y = 2B.

Question 8.
Write the following memory devices in the order of their speed. (Fastest to slowest order):
(a) Cache
(b) RAM
(c) Hard Disk
(d) Registers
Answer:
Registers , Cache, RAM, Hard Disk.

Question 9.
Define syntax error and logical error.
Answer:
Syntax error – If the programmer makes any mistakes in syntax. Eg. 5 = x instead of x = 5;
Logical error – If the programmer makes any logical mistakes.

Question 10.
Find the invalid identifier names from the following:
Answer:
d-w: It is invalid because special character hyphen is used.
999: It is invalid because it is a number.
Z$: It is invalid because special character $ is used.
8c – It is invalid because it is started with a number.

Question 11.
Write the symbols of the following C++ operators :
(a) Conditional operator
(b) Extraction operator
(c) Modulus operator
(d) NOT operator
Answer:
a) ?:
b) >>
c) %
d) !

Question 12.
Rewrite the following C++ program code with switch statement:
if (val= =1)
cout<<“Science”;
else if (val= =5)
cout<<“Computer Science”:
else
cout<<“Not a course”;
Answer:

switch(val)
{
case 1 : cout<<“Science”;
case 5: cout<<“Computer Science”;
default: cout<<“Not a course”;
}

Question 13.
(i) An array is declared as follows:
int a[5] = {1, 2, 3, 4, 5};
What will be the value of a [2] + a [3]?
(ii) How many bytes are allocated by the above array in memory?
Answer:
(i) Here a[2] = 3 and a[3] = 4, so 3 + 4 = 7.
(ii) 4 bytes for an integer so 5*4 = 20 bytes.

Question 14.
Compare Linear search and Binary search on arrays in C++
Answer:

Linear	Binary
No need of sorted array.	Need of sorted array.
Searching is slower for larger array.	It is faster.
It starts checking from the first element and checks all elements.	Divide and conquer method used.

Question 15.
Construct the function prototypes for the following functions:
(i) The function Display () accepts one argument of type double and does not return any value. (1)
(ii) Total () accepts two arguments of type into, float respectively and return a float type value. (1)
Answer:
(i) void Display(double);
(ii) float Total(int, float);

Question 16.
Briefly explain about different types of methods used for passing arguments’to a function.
Answer:
Methods of calling functions
Two types call by value and call by reference.

1. Call by value: In call by value method the copy of the original value is passed to the function, if the function makes any change will not affect the original value.
2. Call by reference: In call by reference method the address of the original value is passed to the function, if the function makes any change will affect the original value.

Question 17.
Discuss the difference in working methods of hub and switch.
Answer:
Hub: It is a small, simple and inexpensive device used to connect computers(devices) to a network. If a computer wants to transmit data to another computer. First it sends to the hub, the hub retransmits this data to all other computers. Each and every computer gets the data and check whether it is for them or not. It increases the network traffic and hence the transmission speed is low.

Switch: It is an expensive device used to connect computers(devices) to a network. Unlike hub, switch transmit data not to all computers, it retransmits data only to the intended computer. So the traffic is less and speed is high.

Question 18.
(i) Write a note on search engines.
(ii) What is meant by Phishing?
Answer:
(i) Search engines
By using search engines we will get a variety of information. It is a-newly developed tool that helped to search the information on the internet more effectively and easily. Search engines are programs that help people to locate information from crores of website on internet using a database that consists of references. Users can interact with the search engine through the home page of the search engine. To get the information about artificial intelligence just type this in the box provided for it and click the search button. Search engines searches by using a particular search algorithm then displays the matching documents or web addresses. Search engine use soft wares called spiders or bots to search documents and their web addresses. Spiders search the internet using the directions given by the search engines and prepare an index and stores it in a database. The searching algorithm searched this database when the users submits a request and create a web page displaying the matching results as hyperlinks.
Eg: Google, Yahoo, Rediff etc.

(ii) Phishing (Fishing): It is an attempt to get others information such as usenames, passwords, bank a/c details etc by acting as the authorized website. Phishing websites have URLs and home pages similar to their original ones and mislead others, it is called spoofing.

Answer any 9 questions from 19 to 29. Each carries 3 scores. (9 × 3 = 27)

Question 19.
(i) Name the technology used in third generation of computers. (1)
(ii) Draw the John Von Neumann Architecture for functional iHnits of a computer. (2)
Answer:
(i) Integrated Circuit(IC)

Question 20.
Which gates are called universal gates? Draw their symbols.
Answer:
NAND and NOR are called universal gates.
NAND gate
The output of AND gate is inverted by NOT gate is the NAND gate

In short the NAND gate is as shown below

NOR gate
The output of OR gate is inverted by NOT gate is the NOR gate.

In short the nor gate is shown below

Question 21.
(i) How e-Waste create environmental and health problems? (1)
(ii) Name the different methods for e-Waste disposal. (2)
Answer:
(i) It contains poisonous substances such as lead, mercury, cadmium etc and may cause diseases if not properly managed.

(ii) e-Waste disposal methods
a) Reuse: Reusability has an important role of e-Waste management and can reduce the volume of e-Waste
b) Incineration: It is the process of burning e. Waste at high temperature in a chimney
c) Recycling of e-Waste: It is the process of making new products from this e-Waste.
d) Land filling: It is used to level pits and cover by thick layer of soil.

Question 22.
Draw a flowchart to find the sum of first 10 natural numbers.
Answer:

Question 23.
What is meant by Literals in C++? Explain its types.
Answer:
Literals (Constants): Its value does not change during execution
1) Integer literals :- Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal and hexadecimal.
Eg. For decimal 100, 150, etc
For octal 0100, 0240, etc
For hexadecimal 0 × 100, 0 × 1 A, etc

2) Float literals A number with fractional parts and its value does not change during execution is called floating point literals.
Eg. 3.14157, 79.78,etc

3) Character literal-: A valid C++ character enclosed in single quotes, its value does not change during execution.
Eg. ‘m’, ‘f ’ etc

4) String literal:- One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null character(‘\0’)
Eg. “Mary’s”,”India”,etc

Question 24.
(i) What is the use of keyword ‘const’ in C++?
(ii) List the different data type modifiers in C++.
Answer:
(i) It is used to declare symbolic constants.
Eg. const float pi=3.14;
(ii) short, long, signed and unsigned.

Question 25.
Briefly explain different types of type conversions in C++.
Answer:
Type conversions are of two types.
1) Implicit type conversion-: This is performed by C++ compiler internally. C++ converts all the lower sized data type to the highest sized operand. It is known as type promotion. Data types are arranged lower size to higher size is as follows.
unsigned int(2 bytes), int(4 bytes),long (4 bytes); unsigned long (4 bytes), float(4 bytes), double(8 bytes), long double(10 bytes)

2) Explicit type conversion -: It is known as type casting. This is done by the programmer. The syntax is given below.
(data type to be converted) expression
Eg. int x=10;
(float) x; This expression converts the data type of the variable from integer to float.

Question 26.
Briefly explain any three jump statements.
Answer:
1. goto statementBy using goto we can transfer the control anywhere in the program without any condition. The syntax is goto label;
Eg.

#include<iostream>
using namespace std;
int main()
{
float a, b;
cout<<"Enter 2 numbers";
cin>>a>>b;
if(b==0)
goto end;
cout<<''The quotient is "<<a/b;
return 0; ,
end:cout<<"Division by zero error";
}

2. break statement:- It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch.

Syntax :
while (expression)
{
if (condition)
break;
}
Eg.
#include<iostream>
using namespace std;
main()
{
int i=1;
while(i<10)
{
cout<<i<<endl;
if(i==5)
break;
i++;
}
} The output is
1
2
3
4
5

3. continue statement:- It bypasses one iteration of the loop.

Syntax :
while (expression)
{
if (condition)
break;
}
Eg.
#include<iostream>
using namespace std;
main()
{
int i=0;
while(i<10)
{
i++;
if(i==5) continue;
cout<<i<<endl;
}
}
The output is
1
2
3
4
6
7
8
9
10

3. exit(0) functionIt is used to terminate the program. For this the header file cstdlib must be included.

Question 27.
(i) How many bytes are required to store the string “WELCOME ALL”? (1)
(ii) What is the limitation of cin>> operator while reading a string? How can we overcome it? (2)
Answer:
(i) 12 bytes(lncluding null character).
(ii) cin reads only the characters upto the space. gets() can be used to resolve this problem

Question 28.
Explain any three network topologies with figures.
Answer:
Topology: The way in which the nodes are physically interconnected to form a network.
• Bus topology
• Star topology
• Ring topology
• Mesh topology
Bus topology- In bus topology all the nodes are connected to a main cable called bus. A small device called a terminator is attached to each end of the bus.

Star topology- In star topology each node is directly connected to a hub switch.

Ring topology -In ring topology all nodes are connected using a cable that shapes like a ring or circle Data travels only in one direction in a ring.

Mesh topology- In mesh topology every node is connected to other nodes.’There will be more than one path between two nodes.

Question 29.
What is the role of Firewall and Anti-virus scanner in computer networks?
Answer:
Firewall: It is a system that controls the incoming and out going network traffic by analyzing the data and then provides security to the computer network in an organization from other network (internet).

Antivirus scanners: It is a tool used to scan computer files for viruses, worms and Trojan horses and cure the infected system. If any fault found it stops the file from running and stores the file in a special area called Qua

Answer any 2 questions from 30 to 32. Each carries 5 scores. (2 × 5 = 10)

Question 30.
(i) Write about any two methods of representing characters in memory. (2)
(ii) Draw a logic circuit for Bollean expression (\(\overline{\mathrm{A}}\) + BC) + AB (3)
Answer:
(i) Representation of characters.
1) ASCII(American Standard Code for Information Interchange) : It is 7 bits code used to represent alphanumeric and some special characters in computer memory. It is introduced by U.S. government. Each character in the key board has a unique number. Eg: ASCII code of ‘a’ is 97, when you press ‘a’ in the key board, a signal equivalent to 1100001 (Binary equivalent of 97 is 1100001) is passed to the computer memory. 2⁷ = 128, hence we can represent only 128 characters by using ASCII. It is not enough to represent all the characters of a standard keyboard.

2) EBCDIC(Extended Binary Coded Decimal Interchange Code): It is an 8 bit code introduced by IBM(lnternational Business Machine). 2⁸=26 characters can be represented by using this.

Question 31.
(i) Define Operating System. Write any two functions of operating system. (2)
(ii) Write notes on different types of language processors. (3)
Answer:
(i) Operating System: It is collection of programs which acts as an interface between user and computer. Without an operating system computer cannot do anything. Its main function is make the computer usable and use hardware in an efficient manner, eg:- Windows XP, Windows Vista, Linux, Windows 7, etc.

Major functions of an operating System

1. Process management: It includes allocation and de allocation of processes(program in execution) as well as scheduling system resources in efficient manner
2. Memory management: It takes care of allocation and de allocation of memory in efficient manner

(ii) The different language processors are given below

1. Assembler: This converts programs written in assembly language into machine language.
2. Interpreter: This converts a HLL program into machine language by.converting and executing it line by line. The first line is converted if there is no error it will be executed otherwise you have to correct it and the second line and so on.
3. Compiler: It is same as interpreter but there is a difference it translate HLL program into machine language by converting all the lines at a time. If there is no error then only it will executed.

Question 32.
(i) Explain the difference, between entry controlled loop and exit controlled loop. (3)
Entry controlled loop, exit controlled loop (3)
(ii) Classify the following loops into entry controlled loop and exit controlled loop for loop, while loop, do while loop. (2)
Answer:
i)

Entry controlled	Exit controlled
Eg. For an while	Do while
The body will be executed only if the test expression becomes true.	But here the body executes at least once even if the conditions is false
Minimum number of execution may be zero	Here it is one

ii)

Entry Controlled	Exit Controlled
For loop	Do while loop
While loop

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf Board Model Paper 2023 helps in understanding answer patterns.

Kerala Plus One Maths Board Model Paper 2023 with Answers

Time: 2 Hours
Total Scores: 60

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
(i) The interval representing {x : x ∈ R, -2 < x ≤ 3} is (1)
(A) [-2, 3]
(B) (-2, 3]
(C) (-2, 3)
(D) [-2, 3]
(ii) Write all the subsets of {a, b, c}. (2)
Answer:
(i) B (-2, 3]
(ii) {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}, φ

Question 2.
Consider A = {1, 2, 3, 4, 6}. Let R be a relation defined by R = {(a, b): a, b ∈ A, b is exactly divisible by a}
(i) Write R in roster form. (2)
(ii) Is R a function? Justify. (1)
Answer:
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) R is not a function. Since images of different elements A are not unique.

Question 3.
(i) Solve the inequality 2(x – 1) < 3(x – 2). (2)
(ii) Show the solution of the inequality on a number line. (1)
Answer:
(i) 2x – 2 < 3x – 6
2x – 3x < -6 + 2
-x < -4 x > 4
(ii)

Question 4.
(i) If ⁿC₃ = ⁿC₇, find ⁿC₂. (2)
(ii) If there are 10 persons in a meeting and each of them shakes hands with all others, what is the total number of handshakes? (1)
Answer:
(i) ⁿC₃ = ⁿC₇
⇒ n = 3 + 7 = 10
∴ ⁿC₂ = ¹⁰C₂
= \(\frac{10 \times 9}{1 \times 2}\)
= 45
(ii) Number of shake hands = ¹⁰C₂ = 45

Question 5.
(i) The length of latus rectum of the parabola y² = 10x is _______________ (1)
(ii) Find the equation of the parabola with focus (0, -3) and directrix y = 3. (2)
Answer:
(i) 10
(ii)

Equation is x² = -4ay
Put a = 3
∴ Equation is x² = -12y

Question 6.
(i) Which of the following is a point on XZ-plane?
(A) (1, -2, 0)
(B) (1, 0, -2)
(C) (0, 1, 0)
(D) (0, 1, -2)
(ii) Verify (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle. (2)
Answer:
(i) B(1, 0, -2)
(ii) Let A(0, 7, -10), B(1, 6, -6), C(4, 9, -6)

AB = BC
∴ ABC is an isosceles triangle.

Question 7.
(i) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) = _______________ (1)
(ii) Evaluate \(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{\frac{\pi}{2}-x}\right)\) (2)
Answer:
(i) 1
(ii) \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\frac{\pi}{2}-x}=\lim _{\frac{\pi}{2}-x \rightarrow 0} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\frac{\pi}{2}-x}\) = 1

Question 8.
If P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\), find
(i) P(not A) (1)
(ii) P(A or B) (1)
(iii) P(not A and not B) (1)
Answer:
(i) P(not A) = 1 – P(A)
= 1 – \(\frac{1}{4}\)
= \(\frac{3}{4}\)

(ii) P(A or B) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\)
= \(\frac{5}{8}\)

(iii) P(not A and not B) = P(A’ ∩ B’)
= P[(A ∪ B)’]
= 1 – P(A ∪ B)
= 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

Answer any 6 questions from 9 to 16. Each carries 4 scores. (6 × 4 = 24)

Question 9.
(i) If A ⊂ B, then A ∩ B = _______________ (1)
(ii) Consider the sets U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8, 10}. Verify that (A ∪ B)’ = A’ ∩ B’. (3)
Answer:
(i) A
(ii) A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}
(A ∪ B)’ = {7, 9}
A’ = {6, 7, 8, 9, 10}
B’ = {1, 3, 5, 7, 9}
A’ ∩ B’ = {7, 9}
∴ (A ∪ B)’ = A’ ∩ B’

Question 10.
(i) Draw the graph of the function f(x) =|x| – 2 (2)
(ii) Write the range of f. (1)
(iii) Write the domain of the function g(x) = \(\sqrt{9-x}\) (1)
Answer:
(i)

(ii) Range of f = [-2, ∞)
(iii) 9 – x ≥ 0
-x ≥ -9
x ≤ 9
Domain of g(x) = (-∞, 9]

Question 11.
(i) Express z = i⁹ + i¹⁸ into a + ib form. (1)
(ii) Write the conjugate of z = i⁹ + i¹⁸. (1)
(iii) Find the multiplicative inverse of 2 – 3i. (2)
Answer:
(i) z = i⁹ + i¹⁸
= i – 1
= -1 + i
a = -1, b = 1
(ii) Conjugate of z = \(\bar{z}\) = -1 – i
(iii) z = 2 – 3i
Multiplicative inverse of z

Question 12.
Find the number of different 8 letter arrangements that can be made from the letters of the work ‘DAUGHTER’ so that
(i) all vowels occur together. (2)
(ii) all vowels do not occur together. (2)
Answer:
Total number of 8 letter words = 8! = 40,320
(i) Take the three vowels (A, U, E) as a single object. Then there are 6 objects which can be arranged in 6! ways and the vowels can be arranged in 3! ways.
∴ Required no. of ways = 6! × 3! = 4,320
(ii) Number of words in which all vowels do not occur together = 40,320 – 4,320 = 36,000

Question 13.
Find (a + b)⁴ – (a – b)⁴. Hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴. (3)
Answer:
(i) (a + b)⁴ – (a – b)⁴ = (a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴) – (a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴)
= 8a³b + 8ab³
= 8ab (a² + b²)
∴ (√3 + √2)⁴ – (√3 – √2)⁴ = 8 × √3 × √2(3 + 2) = 40√6

Question 14.
Find the sum to n terms of the sequence 3, 33, 333,….. (4)
Answer:
3 + 33 + 333 + ……….n terms
= 3(1 + 11 + 111 + ………n terms)

Question 15.
Find focii, vertices, eccentricity and length of latus rectum of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). (4)
Answer:
a = 5, b = 3, c = \(\sqrt{25-9}\) = 4
Foci = (±c, 0) = (±4, 0)
Vertices = (±a, 0) = (±5, 0)
Eccentricity = \(\frac{c}{a}=\frac{4}{5}\)
Length of Latus rectum = \(\frac{2 b^2}{a}=\frac{18}{5}\)

Question 16.
A fair coin is tossed 3 times. Write the sample space. Find the probability of getting (1)
(i) 3 heads (1)
(ii) exactly two heads (1)
(iii) one head and two tails. (1)
Answer:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(i) P(3 heads) = \(\frac{1}{8}\)
(ii) P(exactly 2 heads) = \(\frac{3}{8}\)
(iii) P(1 head and 2 tails) = \(\frac{3}{8}\)

Answer any 3 questions from 17 to 20. Each carries 6 scores. (3 × 6 = 18)

Question 17.
(i) Find the radian measure of 240°. (1)
(ii) If tan x = \(\frac{-5}{12}\) and x lies in second quadrant, find the values of sin x and cos x. (2)
(iii) Prove that \(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}\) = cot x. (3)
Answer:

Question 18.
(i) Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6). (4)
(ii) Find the distance of the point (-3, 5) from the line 3x – 4y + 26 = 0. (2)
Answer:
(i) Let P(-3, 5), A(2, 5), B(-3, 6) be the given points
Slope of AB = \(\frac{6-5}{-3-2}=\frac{1}{-5}\)
∴ Slope of required line = 5
Equation is y – 5 = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0

Question 19.
(i) Find the derivative of cos x using first principle. (3)
(ii) Find the derivative of f(x) = \(\frac{1+\sin x}{\cos x}\) (3)
Answer:

Question 20.
Calculate Arithmetic Mean, Variance and Standard Deviation of the following data. (6)

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency	3	7	12	15	8	3	2

Answer:

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf March 2019 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper March 2019

Time: 2 Hours
Total Scores: 60

Answer any 6 from questions 1 to 7. Each question carries 3 points.

Question 1.
Let A = {x: x is a prime number less than 11} and B = {x: x is an integer such that 2 ≤ x ≤ 8 }.
(i) Write C = A ∩ B.
(ii) Find the number of subsets of C which have 3 elements.
(iiii) What is the probability of getting a subset of 3 elements from the power set of C.
Answer:
(i) A = {2, 3, 5, 7}; B = {2, 3, 4, 5, 7, 8}
C = {2, 3, 5, 7} ∩ {2, 3, 4, 5, 7, 8} = {2, 3, 5, 7}
(ii) Number of subsets of C which have 3 elements = ⁴C₃ = 4
(iii) Probability of getting a subset of 3 elements from the power set of C = \(\frac{\text { favourable cases }}{\text { total number of cases }}=\frac{{ }^4 C_3}{2^4}=\frac{4}{16}=\frac{1}{4}\)

Question 2.
(i) Find (a + b)⁴ – (a – b)⁴.
(ii) Hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴.
Answer:
(i) (a + b)⁴ – (a – b)⁴ = 2 × 4a³b + 2 × 4ab³ = 8ab(a² + b²)
(ii) (√3 + √2)⁴ – (√3 – √2)⁴ = 8√3 √2 ((√3)² + (√2)²)
= 8√3 √2(3 + 2)
= 40√6

Question 3.
Find the square root of the complex number 3 + 4i.
Answer:
Let x + iy = \(\sqrt{3+4 i}\)
Then (x + iy)² = 3 + 4i
⇒ x² – y² + 2xyi = 3 + 4i
Equating real and imaginary parts, we have
x² – y² = 3 ……….(1)
2xy = 4
We know the identity
(x² + y²)² = (x² – y²)² + (2xy)²
= 9 + 16
= 25
Thus, x² + y² = 5 ………….(2)
From (1) and (2), x² = 4 and y² = 1
or x = ±2 and y = ±1
Since the product xy is positive, we have
x = 2, y = 1 or, x = -2, y = -1
Thus, the square roots of 3 + 4i are 2 + i and -2 – i.

Question 4.
The sum of the first three terms of a Geometric Progression is \(\frac{13}{12}\) and their product is -1. Find the common ratio and the terms.
Answer:

Question 5.
Find the solution of the equation sin x + sin 3x + sin 5x = 0
Answer:
sin x + sin 3x + sin 5x = 0
⇒ 2 sin 3x cos 2x + sin 3x = 0
⇒ sin 3x (2 cos 2x + 1) = 0
⇒ sin 3x = 0; 2 cos 2x + 1 = 0

Question 6.
Consider the graph of the function f(x)

(i) Identify the function f(x)
(a) f(x) = sin x
(b) f(x) = cos x
(c) f(x) = tan x
(d) f(x) = cosec x
(ii) Using the function f(x) find \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{f(x)-f\left(\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}\)
Answer:

Question 7.
(i) Find the general term in the expansion of \(\left(x^2+\frac{1}{x}\right)^5\)
(ii) If the expansion of \(\left(x^2+\frac{1}{x}\right)^n\) has a term independent of x, then which of the following can be the value of n?
(a) 18
(b) 16
(c) 22
(d) 13
Answer:

Which means n should be a multiple of 3 and divisible by 2.
Hence (a) 18.

Answer any 8 from questions 8 to 17. Each question carries 4 scores.

Question 8.
In a school, a survey among 400 students, 100 were listed as taking apple juice, 150 as taking orange juice, and 75 were listed as taking both apple juice as well as orange juice.
(i) How many students take apple juice or orange juice?
(ii) How many take apple juice alone but not orange juice?
(iii) How many students were taking neither apple juice nor orange juice?
Answer:
Let A: Apple juice, O: Orange juice.
(i) n(A ∪ O) = n(A) + n(O) – n(A ∩ O)
= 100 + 150 – 75
= 175
(ii) n(A ∩ O’) = n(A) – n(A ∩ O)
= 100 – 75
= 25
(iii) n(A’ ∩ O’) = 400 – n(A ∪ O)
= 400 – 175
= 225

Question 9.
The figure shows the graph of the function f(x).

(i) Write the domain and range of f(x).
(ii) Find f(0) and f(-0.01).
(iii) Check the existence of \(\lim _{x \rightarrow 0} f(x)\).
Answer:
(i) Domain = R
Range = (-∞, 0) ∪ {1}
(ii) f(0) = 1, f(-0.01) = -0.01
(iii) \(\lim _{x \rightarrow 0^{-}} f(x)=0, \lim _{x \rightarrow 0^{+}} f(x)=1\)
\(\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)\)
Therefore, the limit does not exist.
OR
There is a break in the graph of f(x) at x = 0.
So the limit does not exist at x = 0.

Question 10.
Consider the set A = {-1, 1}
(i) Write all elements in A × A.
(ii) How many relations are there from A to A?
(iii) Write all functions from A to A which have Range = {-1, 1}.
Answer:
(i) A × A = {(-1, -1),(-1, 1),(1, -1),(1, 1)}
(ii) Number of relations = \(2^{n(A \times A)}\) = 24 = 16
(iii) There are two functions
f1 = {(-1, -1), (1, 1)}
f2 = {(-1, 1), (1, -1)}

Question 11.
Using the principle of mathematical induction, prove that n(n + 1)(n + 5) is a multiple of 3 for all n ∈ N.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12 divisible by 3, hence true.
Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k² + 8k + 12)
= (k + 1){k² + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore, by using the principle of mathematical induction, true for all n ∈ N.

Question 12.
If z is a complex number with |z| = 2 and arg(z) = \(\frac{4 \pi}{3}\), then
(i) express z in a + ib form.
(ii) Find \(\bar{z}\)
(iii) Verify that (\(\bar{x}\))² = 2z
Answer:

Question 13.
Seven cards are drawn from a well-shuffled pack of 52 playing cards.
(i) How many ways can this be done?
(ii) What is the probability that the selection contains all kings?
(iii) What is the probability that selection does not contain a king card?
Answer:

Question 14.
(i) Write the contrapositive of the given statement.
“If a number is divisible by 9, then it is divisible by 3.”
(ii) Verify by the method of contradiction:
“p: √7 is irrational.”
Answer:
(i) If a number is not divisible by 3, it is not divisible by 9.
(ii) Assume that √7 is rational. Then √7 can be written in the form √7 = \(\frac{p}{q}\), where p and q are integers without common factors.
Squaring; 7 = \(\frac{p^2}{q^2}\)
⇒ 7q² = p²
⇒ 7 divides p²
⇒ 7 divides p
Therefore, p = 7k for some integer k.
⇒ p² = 49k²
⇒ 7q² = 49k²
⇒ q² = 7k²
⇒ 7 divides q²
⇒ 7 divides q
Hence p and q have a common factor 7, which contradicts our assumption.
Therefore, √7 is irrational.

Question 15.
Calculate the mean deviation about median for the following data:

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequencies	6	7	15	16	4	2

Answer:

Question 16.
Consider the word ASSASSINATION
(i) How many different ways can the letters of the word be arranged?
(ii) How many of these words have all vowels together?
Answer:
(i) In the word ASSASSINATION there are 13 letters, of which A appears 3 times, S appears 4 times, N appears 2 times, I appears 2 times and the rest all are different.
Therefore the total number of ways is \(\frac{13!}{3!\times 4!\times 2 \times 2!}\) = 10810800.
(ii) Vowels are A, A, A, I, I, O
Vowels can be arranged = \(\frac{6!}{3!\times 2!}\)
Vowels are taken as one unit, it can be arranged = \(\frac{8!}{4!\times 2!}\)
Total arrangement in which vowels are together = \(\frac{8!}{4 \times 2!} \times \frac{6!}{3!\times 2!}\)

Question 17.
Let A(0, 7, 10), B(-1, 6, 6,) and C(-4, 9, 6) are the vertices of a triangle.
(i) Show that is a right triangle.
(ii) Find the coordinate of the centre of the circle passing through the points A, B, C.
Answer:
(i) AB² = (-1 – 0)² + (6 – 7)² + (6 – 10)²
= 1 + 1 + 16
= 18
BC² = (-4 – (-1))² + (9 – 6)² + (6 – 6)²
= 9 + 9 + 0
= 18
AC² = (-4 – 0)² + (9 – 7)² + (6 – 10)²
= 16 + 4 + 16
= 36
∴ AB² + BC² = AC²
Therefore triangle ABC is a right triangle.
Since the triangle is a right triangle, the circumcentre will lie on the midpoint of the side AC.
Centre = \(\left(\frac{0-4}{2}, \frac{7+9}{2}, \frac{10+6}{2}\right)\) = (-2, 8, 8)

Answer any 5 from questions 18 to 24. Each question carries 6 scores.

Question 18.
The figure shows a unit circle and a line L which makes 30° with the positive direction of x-axis.

(i) Write the equation of the line L.
(ii) Write the coordinates of the points A and B.
(iii) Find the equation of the tangent line to the circle at A.
Answer:
(i) The slope of the line L is
m = tan 30° = \(\frac{1}{\sqrt{3}}\)
Equation of the line is y = mx
⇒ y = \(\frac{1}{\sqrt{3}}\)x
(ii) Point A = (r cos θ, r sin θ)
= (cos 30°, sin 30°)
= \(\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)
Point B lies in the third quadrant = \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)
(iii) Equation of tangent is y – y₁ = m(x – x₁)
⇒ \(y-\frac{1}{2}=-\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)\)
⇒ 2y – 1 = -2√3x + 3
⇒ √3x + y = 2

Question 19.
Consider two lines L₁: 2x + y = 4 and L₂: (2x – y = 2)
(i) Find the angle between L₁ and L₂.
(ii) Find the equation of the line passing through the intersection of L₁ and L₂ which makes an angle 45° with the positive direction of x-axis.
(iii) Find the x and y intercepts of the third line obtained in part (ii).
Answer:
(i) The slope of the line L₁ is -2.
The slope of the line L₂ is 2.
tan θ = \(\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{2-(-2)}{1+2 \times-2}\right|=\frac{4}{3}\)
(ii) The family of lines is of the form L₁ + λL₂ = 0
⇒ (2x + y – 4) + λ(2x – y – 2) = 0
⇒ (2 + 2λ)x + (1 – λ)y – 4 – 2λ = 0
⇒ Slope = \(-\frac{2+2 \lambda}{1-\lambda}\) = tan 45 = 1
⇒ -2 – 2λ = 1 – λ
⇒ λ = -3
Equation of the line is 2x – 2y = 1
(iii) 2x – 2y = 1
⇒ \(\frac{x}{1 / 2}+\frac{y}{-1 / 2}\) = 1
∴ x intercept is \(\frac{1}{2}\)
∴ y intercept is \(-\frac{1}{2}\)

Question 20.
If an ellipse passes through (3, 1) having foci (±4, 0), then
(i) Find the length of the major axis.
(ii) Find the standard equation of the ellipse.
(iii) Find the eccentricity and length of the latus rectum.
Answer:
(i) Length of the major axis = Sum of the distances of the point (3, 1) from the foci (±4, 0)

Question 21.
(i) Find sin 75°.
(ii) The figure shows ∆ABC with side AC = 4√2 units inscribed in a circle of radius 4 units. The length of the arc BDC is \(\frac{10 \pi}{3}\) units.

(a) Write ∠A in degree measure. (2)
(b) Find the length of the sides AB and BC. (2)
Answer:

Question 22.
(i) Solve \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\).
(ii) Solve thq inequalities 2x + 3y ≤ 12; x ≥ 1; y ≥ 2 graphically.
Answer:
(i) \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
⇒ 9(x – 2) ≤ 25(2 – x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 34x ≤ 68
⇒ x ≤ 2

Question 23.
(i) Find the derivative of y = x² using the first principle. (3)
(ii) Find \(\frac{d y}{d x}\) if y = \(\frac{x}{1+\tan x}\). (3)
Answer:

Question 24.
Consider the sequence 3, 6, 9, 12,………, 99
(i) How many terms are there in the given sequence? (1)
(ii) Find the mean of the sequence. (2)
(iii) Find the sum of squares of each term of the given sequence. (2)
(iv) Find the variance of the sequence. (1)
Answer:
(i) 3 × 1, 3 × 2, 3 × 3,………, 3 × 33
Hence there are 33 terms in the given sequence.
(ii) S₃₃ = 3 + 6 + 9 +…… 99 = 3(1 + 2 + 3 +……+ 33)

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf March 2020 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper March 2020

Time: 2 Hours
Total Scores: 60

Questions 1 to 7 carry 3 scores each. Answer any 7 questions.

Question 1.
(i) If A = {x < x is a natural number, x < 5 and x > 7}, then n(A) is
(a) 1
(b) 0
(c) 2
(d) 3
Answer:
(b) 0

(ii) The set builder of (6, 12) is
(a) {x : x ∈ R, 6 < x ≤ 12}
(b) {x : x ∈ R, 6 < x < 12}
(c) {x : x ∈ R, 6 ≤ x ≤ 12}
(d) {x : x ∈ R, 6 ≤ x < 12}
Answer:
(b) {x : x ∈ R, 6 < x < 12}

(iii) If A and B are two sets such that A ⊂ B, then A ∪ B is
(a) A
(b) Null Set
(c) B
(d) {Φ}
Answer:
(c) B

Question 2.
In a survey of 600 students in a school, 150 students were found to be taking tea, 225 students were taking coffee, and 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee.
Answer:
Let the sets be defined as follows:
Tea = A, Coffee = B
n(U) = 600, n(A) = 150, n(B) = 225
n(A ∩ B) = 100
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 150 + 225 – 100
= 275
n(neither tea nor coffee) = n(A’ ∩ B’)
= n(U) – n(A ∩ B)
= 600 – 275
= 325

Question 3.
Find the principal and general solutions of cosec x = -2.
Answer:
cosec x = -2
⇒ x = \(\frac{-1}{2}\)
x lies in the third or fourth quadrant
x = \(\pi+\frac{\pi}{6}=\frac{7 \pi}{6}\)
x = \(2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\)
Principal solution are \(\frac{7 \pi}{6}\) and \(\frac{11 \pi}{6}\)
General solution
x = nπ + (-1)ⁿ y, n ∈ Z
x = nπ + (-1)ⁿ \(\frac{7 \pi}{6}\), n ∈ Z

Question 4.
(i) If the sum of first terms of an AP is equal to the sum of first 30 terms, then the sum of first 50 terms is
(a) 50
(b) 20
(c) 0
(d) 80
Answer:
(c) 0

(ii) Find the sum to infinity terms of the G.P \(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}, \ldots\)
Answer:

Question 5.
Find the sum of n terms of the series 7 + 77 + 777 +…..
Answer:

Question 6.
Consider the following figure:

(i) Find the distance of PQ.
(ii) Find the coordinates of the point that divides the line segment by joining the points P and Q internally in the ratio 2 : 3.
Answer:

Question 7.
Find the derivative of cos x from first principles.
Answer:

Question 8.
(i) Derivative of
f(x) = 1 + x + x² + x³ +…+ x⁵⁰ at x = 1 is
(a) 50
(b) 1250
(c) 1275
(d) \(\frac{101}{2}\)
(ii) Find \(\lim _{x \rightarrow 0} f(x)\) if it exists, where
\(f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{cases}\)
Answer:
(i) (c) 1275
f'(x) = 1 + 2x + 3x² + … + 50x⁴⁹
f'(1) = 1 + 2 + 3 + ….. + 50 = \(\frac{50(50+1)}{2}\) = 1275

Questions 9 to 16 carry 4 scores each. Answer any 6.

Question 9.
Match the following:


Answer:
(a) f : R → R given by f(x) = x³
(b) f : R → R given by \(f(x)=\left\{\begin{array}{rc}
1, & x>0 \\
0, & x=0 \\
-1, & x<0
\end{array}\right.\)
(c) f : R → R given by f(x) = x
(d) f : R → R given by f(x) = \(\frac{1}{x}\), x ≠ 0

Question 10.
For every positive integer n, prove that 7ⁿ – 3ⁿ is divisible by 4 using the principle of mathematical induction.
Answer:
P(1) = 7¹ – 3¹ = 4
True for n = 1
Assume that P(k) is divisible by 4
P(k) = 7^k – 3^k = 4m, where m is a natural number
Prove for n = k + 1
P(k + 1) = 7^k+1 – 3^k+1
= 7 . 7^k – 3 . 3^k
= 7 . 7^k – (7 – 4) . 3^k
= 7 . 7^k – 7 . 3^k – 4 . 3^k
= 7(7^k – 3^k) – 4.3^k
= 7 . 4m – 4 . 3^k
= 4(m7^k – 3^k)
Hence P(k + 1) is true. Therefore P(n) is true for all natural numbers.

Question 11.
(i) Modulus of a complex number is 2 and arg(z) = \(\frac{\pi}{3}\), write the complex number in the form a + ib.
(ii) Find the square root of the above complex number.
Answer:

Question 12.
Solve graphically:
2x + y ≥ 4; x + y ≤ 3; 2x – 3y ≤ 6; x ≥ 0; y ≥ 0
Answer:

Question 13.
(i) Expand \(\left(x+\frac{1}{x}\right)^6\).
(ii) Find the middle term in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\).
Answer:

Question 14.
(i) Let A(1, 2) be a fixed point and, ‘P’ be a variable point in the same plane. P moves in the plane in such a way that its distance from A is always a constant. Suppose ‘P’ is at the point (3, 3), find the equation of the path traced by ‘P’.
(ii) Consider the following ellipse:

(a) Find the equation of the ellipse.
(b) Find the coordinates of the foci.
Answer:

Question 15.
(i) Write the contra positive of the statement:
“If a number is divisible by 9, then it is divisible by 3.”
(ii) By method of contradiction, prove that √5 is irrational.
Answer:
(i) Contrapositive statement is “If a number is not divisible by 3, then it is not divisible by 9.”
(ii) Let √5 be rational.
Then √5 = \(\frac{a}{b}\), where a and b have no common factors.
√5b = a
⇒ 5b² = a² ……….(1)
⇒ a² is divisible by 5
⇒ a is divisible by 5
Then a = 5c
⇒ a² = 25c²
(1) ⇒ 5b² = 25c²
⇒ b² = 5c²
⇒ b² is divisible by 5
⇒ b is divisible by 5.
Hence a and b have a common factor √5. Which contradicts our assumption that √5 is rational. Therefore √5 is irrational.

Question 16.
(i) If E and F are two events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\), P(E and F) = \(\frac{1}{8}\), find
(a) P(E or F)
(b) P(not E and not F)
(ii) A committee of two persons is selected from two men and two women. What is the probability that the committee will have
(a) one man?
(b) two men?
Answer:

Questions from 17 to 20 carry 3 scores each. Answer any 5.

Question 17.
(i) If tan x = \(-\frac{5}{12}\), x lies in second quadrant. Find all trigonometric functions.
(ii) Without using a triangle, find the value of \(\frac{\sin x+\cos x}{\sin x-\cos x}\) if tan x = \(\frac{3}{4}\).
(iii) Prove that \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x
Answer:

Question 18.
(i) Find the number of different 8-letter arrangements that can be made from the letters of the word ‘DAUGHTER’ so that all vowels occur together.
(ii) Find the number of ways of choosing 4 cards from a pack of 52 playing cards. How many of these
(a) Are four cards of the same suit?
(b) Four cards belong to different suits?
(c) Two are red cards and two are black cards?
Answer:
(i) The vowels are A, E, U. These vowels can be arranged in 3! ways.
Count these 3 vowels as one unit.
Then the number of letters with vowels together = 6! × 3! = 4320
(ii) Number of ways of selecting 4 cards = ⁵²C₄ = 270725
(a) 4 cards from same suite = ¹³C₄ + ¹³C₄ + ¹³C₄ + ¹³C₄ = 4 × ¹³C₄ = 2860
(b) 4 cards belong to different suits = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ = 134
(c) 2 are red cards and 2 are black cards = ²⁶C₂ × ²⁶C₂ = 105625

Question 19.
Consider the following diagram:

(i) Find equation of a line passing through the midpoint of AB and perpendicular to AB.
(ii) Find a point ‘C’ on the x-axis which is equidistant from A and B.
(iii) Find area of ΔABC.
Answer:
(i) Mid point of segment AB is (3, 2)
Slope of AB = -1
Slope of perpendicular line to AB = 1
Equation of perpendicular line is y – 2 = 1(x – 3)
⇒ x – y – 1 = 0
(ii) Let C be of the form (x, 0)
Then \(\sqrt{(x-2)^2+(0-3)^2}=\sqrt{(x-4)^2+(0-1)^2}\)
⇒ x² – 4x + 4 + 9 = x² – 8x + 16 + 1
⇒ 4x = 4
⇒ x = 1
Hence point C is (1, 0)
(iii) Area = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac{1}{2}\) [2(1 – 0) + 4(0 – 3) + 1(3 – 1)]
= \(\frac{1}{2}\) [2 – 12 + 2]
= \(\frac{1}{2}\) [-8]
= -4
∴ Area = 4 sq. units.

Question 20.
From the following table:

Classes	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	3	7	12	15	8	3	2

Find
(i) Mean
(ii) Variance
(iii) Coefficient of Variation.
Answer:

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf September 2021 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper Sep 2021

Time: 2 Hours
Total Scores: 60

Answer any 6 questions from 1 to 12. Each carries 3 scores. (6 × 3 = 18)

Question 1.
Let A = {x : x is a natural less than 6} and B = {1, 2}
(i) Write A in roster form. (1)
(ii) Find A ∩ B. (1)
(iii) Find A – B. (1)
Answer:
(i) A = {1, 2, 3, 4, 5}
(ii) A ∩ B = {1, 2}
(iii) A – B = {3, 4, 5}

Question 2.
Find the sum of all natural numbers between 100 and 1000 multiples of 5. (3)
Answer:

Question 3.
Using Binomial Theorem, expand the expression (2x + 3)⁵. (3)
Answer:

Question 4.
Find \(\lim _{x \rightarrow 0} f(x)\) where \(f(x)=\left\{\begin{array}{l}
2 x+3, x \leq 0 \\
3(x+1), x>0
\end{array}\right.\) (3)
Answer:

Question 5.
In a group of 400 students, 250 can speak Hindi and 200 can speak English. Also each can speak atleast one of these two languages. How many students can speak both Hindi and English? (3)
Answer:
H → Set of students who can speak Hindi
E → Set of students who can speak English
n(H) = 250, n(E) = 200, n(E ∪ H) = 400
n(E ∩ H) = n(E) + n(H) – n(E ∪ H)
= 250 + 200 – 400
= 50

Question 6.
Consider the line 2x + 3y – 6 = 0. Find its
(i) Slope (2)
(ii) y-intercept (1)
Answer:
(i) Slope = \(\frac{-2}{3}\)
(ii) y-intercept = \(\frac{-C}{B}=\frac{6}{3}\) = 2

Question 7.
Consider the equation y² = 12x. Find
(i) The coordinates of the focus. (1)
(ii) Equation of the directrix. (1)
(iii) Length of latus rectum. (1)
Answer:
(i) Focus = (a, 0) = (3, 0)
(ii) Directrix, x = -a
⇒ x = -3
⇒ x + 3 = 0
(iii) Length of Latus Rectum = 4a = 12

Question 8.
(i) The point (0, 2, 3) lies in
(a) XY-plane
(b) YZ-plane
(c) XZ-plane
(d) None of these (1)
(ii) Find the distance between the points P(-3, 7, 2) and Q(2, 4, -1). (2)
Answer:
(i) (b)

Question 9.
Using the principle of mathematical induction, prove that 7ⁿ – 3ⁿ is divisible by 4 for all n ∈ N. (3)
Answer:
p(n): 7ⁿ – 3ⁿ is divisible by 4
p(1): 7¹ – 3¹ = 7 – 3 = 4, divisible by 4
∴ p(1) is true.
Let p(k) be true
p(k) = 7^k – 3^k is divisible by 4
ie. 7^k – 3^k = 4m ……..(1)
p(k+1): 7^k+1 – 3^k+1
= 7^k . 7 – 3^k. 3
= 7^k . 7 – 7^k. 3 + 7^k . 3 – 3^k. 3
= 7^k (7 – 3) + 3 (7^k – 3^k)
= 7^k . 4 + 3 × 4m
= 4(7^k + 3m), divisible by 4
∴ p(k + 1) is true whenever p(k) is true.

Question 10.
Consider the expansion of (x – 2y)¹². Find its
(i) general term (2)
(ii) 4th term (1)
Answer:

Question 11.
Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the XY-plane. (3)
Answer:

Let the required ratio be k : 1
In the XY plane, Z = 0
(i.e) \(\frac{-8+10}{k+1}\) = 0
⇒ -8k + 10 = 0
⇒ -8k = -10
⇒ k = \(\frac{-10}{-8}=\frac{5}{4}\)
∴ XY plane divides in the ratio 5 : 4 internally.

Question 12.
(i) Write the negation of statement “Every natural number is greater than zero”. (1)
(ii) Write the converse and contrapositive of the statement “If a number n2 is even then n is even”. (2)
Answer:
(i) “It is false that every natural number is greater than zero”
(ii) Converse: “If n is even then n2 is even”
Contra positive: “If n is not even then n2 is not even”

Answer any 6 questions from 13 to 24. each carries 4 scores. (6 × 4 = 24)

Question 13.
Let A = {1, 2, 3}
(i) No. of subsets of A is ___________ (1)
(a) 3
(b) 6
(c) 8
(d) 9
(ii) Write all subsets of A having 2 elements. (2)
(iii) If the given set A is a subset of the universal set U = {1, 2, 3, 4, 5, 6}, then write A’. (1)
Answer:
(i) (c)
(ii) {1, 2}, {2, 3}, {1, 3}
(iii) A’ = {4, 5, 6}

Question 14.
(i) If (x + 1, y – 2) = (3, 1), find the values of x and y. (2)
(ii) Let A = {1, 2, 3} and B = {3, 4}. Find A × B. (2)
Answer:
(i) x + 1 = 3, y – 2 = 1
⇒ x = 2, y = 3
(ii) A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

Question 15.
(i) If cos x = \(\frac{-1}{2}\), x lies in 3rd quadrant, find the values of sin x and tan x. (2)
(ii) Prove that \(\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}=\frac{1}{2}\) (2)
Answer:

Question 16.
For all n ∈ N, Let P(n): 1 + 3 + 3² +……+ 3^n-1 = \(\frac{\left(3^n-1\right)}{2}\)
(i) Prove that P(1) is true. (1)
(ii) Prove that the statement P(n) is true for all natural numbers using the principle of mathematical induction. (3)
Answer:
(i) p(1): 1 = \(\frac{3^{\prime}-1}{2}=\frac{2-1}{2}\) = 1
∴ p(1) is true.
(ii) Let p(k) is true
(ie) p(k): 1 + 3 + 3²+… + 3^k-1 = \(\frac{3^k-1}{2}\)
We have to prove that p(k + 1) is true
p(k+1): 1 + 3 + 3²+…+ 3^k-1 + 3^k
= \(\frac{3^k-1}{2}+3^k=\frac{3^k-1+2 \times 3^k}{2}\)
= \(\frac{3 \cdot 3^k-1}{2}=\frac{3^{k+1}-1}{2}\)
∴ p(k + 1) is true.
Hence by P.M.I, p(n) is true for all n ∈ N

Question 17.
(i) Which of the following is the value of i⁹? (1)
(a) -i
(b) i
(c) -1
(d) 1
(ii) Express the complex number 3(7 + i7) + i(7 + i7) in a + ib form. (3)
Answer:
(i) (b)
(ii) 3(7 + i7) + i(7 + i7)
= 21 + 21i + 7i + 7i²
= 21 + 28i – 7
= 14 + 28i
∴ a = 14, b = 28

Question 18.
Represent the complex number z = 1 + i√3 in the polar form. (4)
Answer:

Question 19.
(i) If ⁿC₈ = ⁿC₂ then n is ______________ (1)
(a) 6
(b) 16
(c) 1
(d) 10
(ii) How many chords can be drawn through 21 points on a circle? (3)
Answer:
(i) n = 8 + 2 = 10
(ii) No. of chords = ²¹C₂
= \(\frac{21 \times 20}{1 \times 2}\)
= 210

Question 20.
(i) How many 3 digit numbers can be formed using the digits 1, 2, 3, 4 and 5 assuming that the repetition of the digits is not allowed? (2)
(ii) Find the number of permutations using all the letters of the word ALLAHABAD. (2)
Answer:
(i) Since repetition is not allowed,
No.of 3 digit numbers = 5 × 4 × 3 = 60
(ii) No. of permutations = \(\frac{9!}{4!2!}\) = 7560

Question 21.
Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and passing through (2, -3). (4)
Answer:
Slope of given line = \(\frac{-1}{-7}=\frac{1}{7}\)
Slope of required line = -7
(x₁, y₁) = (2, -3)
Equation is y – y₁ = m(x – x₁)
⇒ y + 3 = -7(x – 2)
⇒ y + 3 = -7x + 14
⇒ 7x + y – 11 = 0

Question 22.
Consider an ellipse whose vertices are (±5, 0) and foci (±4, 0).
(i) Write the equation of the ellipse. (3)
(ii) Find the eccentricity of the ellipse. (1)
Answer:
(i) a = 5, c = 4
b² = a² – c²
= 25 – 16
= 9
Equation is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
\(\frac{x^2}{25}+\frac{y^2}{9}\) = 1
(ii) e = \(\frac{c}{a}=\frac{4}{5}\)

Question 23.
Find the derivatives of
(i) x(x² + 2x + 1) (2)
(ii) \(\frac{x+1}{x}\) (2)
Answer:

Question 24.
Prove by the method of contradiction √5 is irrational. (4)
Answer:
Assume √5 is rational
∴ √5 = \(\frac{a}{b}\), where a and b are coprime
⇒ a = √5b
⇒ a² = 5b² ………(1)
∴ 5 divides a²
Implies 5 divides a
∴ a = 5c
Substituting in (1)
25c² = 5b²
5c² = b²
(i.e) 5 divides b²
hence 5 divides b
Which is a contradiction to the fact that a and b are coprime.
Hence our assumption is wrong.
∴ √5 is irrational.

Answer any 3 questions from 25 to 30. Each carries 6 scores. (3 × 6 = 18)

Question 25.
(i) Draw the graph of the function f: R → R defined by f(x) = |x|. (3)
(ii) Let A = {1, 2, 3, 4, 5, 6} and R is a relation defined from A to A by R = {(x, y): y = x + 1}
(a) Depict this relation using an arrow diagram. (2)
(b) Write the domain of R. (1)
Answer:
(i) f(x) = |x|

(b) Domain = {1, 2, 3, 4, 5}

Question 26.
(i) Evaluate sin 75°. (3)
(ii) Prove that \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x (3)
Answer:

Question 27.
Solve the following system of inequalities graphically: (6)
2x + y ≥ 6
3x + 4y ≤ 12
x ≥ 0, y ≥ 0
Answer:

Question 28.
(i) Find the 12th term of the geometric progression 5, 25, 125,…. (2)
(ii) Find the sum to n terms of the sequence 8, 88, 888,…… (4)
Answer:

Question 29.
Consider the following table:

Class	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	3	7	12	15	8	3	2

(1) Find the mean. (2)
(2) Find the variance. (3)
(3) Find the standard deviation. (1)
Answer:

Question 30.
(i) A coin is tossed twice. What is the probability that at least one tail occurs? (2)
(ii) If E and F are two events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\) and P(E ∩ F) = \(\frac{1}{8}\). Find
(a) P(E or F) (2)
(b) P(not E and not F) (2)
Answer:
(i) S = {HH, HT, TH, TT}
E = atleast one tail = {HT, TH, TT}
P(E) = \(\frac{n(E)}{n(S)}=\frac{3}{4}\)

(ii) P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\), P(E ∩ F) = \(\frac{1}{8}\)
(a) P(E or F) = P(E) + P(F) – P(E ∩ F)
= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}\)
(b) P(not E and not F) = 1 – P(E or F)
= 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf June 2022 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper June 2022

Time: 2 Hours
Total Scores: 60

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
(i) If A is any set, then A ∩ A’ = _______________________ (1)
(a) A
(b) φ
(c) A’
(d) U
(ii) A = {x: x is a natural number less than 3}
(a) Write A in roster form. (1)
(b) Write all the subsets of A. (1)
Answer:
(i) (b) φ
(ii) (a) A = {1, 2}
(b) {1}, {2}, {1, 2}, φ

Question 2.
(i) 25° = ______________ radian. (1)
(ii) If tan x = \(\frac{5}{12}\), x lies in 3rd quadrant, then find the value of sin x and cos x. (2)
Answer:
(i) 25° = 25 × \(\frac{\pi}{180}\) = \(\frac{5 \pi}{36}\) radian

Question 3.
(i) For what values of x, the numbers \(\frac{4}{3}\), x, \(\frac{3}{4}\) are in Geometric progression? (1)
(ii) Find the nth term of the Geometric Progression: (2)
√3, 3, 3√3,….
Answer:

Question 4.
Find the angle between the lines y – √3x – 5 = 0 and √3y – x + 6 = 0. (3)
Answer:

Question 5.
(i) Focus of the parabola y² = 8x is ______________ (1)
(a) (4, 0)
(b) (0, 2)
(c) (0, -4)
(d) (2, 0)
(ii) Find the centre and radius of the circle x² + y² + 6x – 4y – 3 = 0. (2)
Answer:
(i) (d) (2, 0)
(ii) Given circle is of the form x² + y² + 2gx + 2fy + c = 0
2g = 6, 2f = -4
⇒ g = 3, f = 2, c = -3
Centre = (-g, -f) = (-3, 2)
Radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{9+4+3}\)
= 4

Question 6.
Find the ratio in which the yz-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8). (3)
Answer:
Let the yz plane divides the line segment joining A(-2, 4, 7) and B(3, -5, 8) in the ratio k : 1

∴ yz plane divides AB internally in the ratio 2 : 3

Question 7.
Evaluate the following limits:
(i) \(\lim _{x \rightarrow 2} x^2-4\) (1)
(ii) \(\lim _{x \rightarrow 2} \frac{x^2-4}{x-2}\) (1)
(iii) \(\lim _{x \rightarrow 0} \frac{\sin 4 x}{x}\) (1)
Answer:

Question 8.
Prove by the method of contradiction that √3 is irrational. (3)
Answer:
Assume that √3 is rational
∴ √3 = \(\frac{a}{b}\) where a and b are coprime.
∴ a = √3b
a² = 3b² ________(1)
i.e 3 divides a²
∴ 3 divides a
∴ There exists an integer c such that a = 3c
∴ a² = 9c²
Substituting this in equation (1),
9c² = 3b²
b² = 3c²
∴ 3 divides b²
hence 3 divides b.
∴ 3 is a common factor of a and b.
This is a contradiction to the fact that a and b are coprime.
∴ Our assumption is wrong.
Hence √3 is irrational.

Answer any 6 questions from 9 to 17. Each carries 4 scores. (6 × 4 = 24)

Question 9.
(i) Which one of the following is equal to {x: x ∈ R, -4 < x ≤ 5}? (1)
(a) (-4, 5]
(b) (-4, 5)
(c) [-4, -5]
(d) [-4, 5)
(ii) If U = {1, 2, 3, 4, 5, 6, 7}, A = {2, 3, 4, 6}, B = {3, 4, 5}, then verify that (A ∪ B)’ = A’ ∩ B’. (3)
Answer:
(i) (a) (-4, 5]
(ii) A ∪ B = {2, 3, 4, 5, 6}
(A ∪ B)’ = {1, 7}
A’ = {1, 5, 7}
B’ = {1, 2, 6, 7}
A’ ∩ B’ = {1, 7}
∴ (A ∪ B)’ = A’ ∩ B’

Question 10.
(i) Let A = {1, 2, 3, 4, 5, 6, 7, 8}. A relation R from A to A is defined by R = {(x, y): 2x – y = 0 where x, y ∈ A}. Write down its domain and range. (2)
(ii) Draw the graph of the function f: R → R defined by f(x) = |x| + 1. (2)
Answer:
(i) R = {(1, 2), (2, 4), (3, 6), (4, 8)}
Domain = {2, 3, 4}
Range = {2, 4, 6, 8}
(ii) f(x) = |x| + 1

Question 11.
Consider the statement:
P(n): \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^n}=1-\frac{1}{2^n}\)
(i) Show that P(1) is true. (1)
(ii) Prove by principle of Mathematical Induction that P(n) is true for all n ∈ N. (3)
Answer:


∴ P(k + 1) is true.
Hence by Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Question 12.
(i) If ⁿC₉ = ⁿC₈, then n = ______________ (1)
(ii) ⁿP_r = ______________ (1)
(iii) Find the number of permutations using all the letters of the word “MATHEMATICS”. (2)
Answer:
(i) n = 9 + 8 = 17
(ii) ⁿP_r = \(\frac{n!}{(n-r)!}\)
(iii) In the given word MATHEMATICS
Total letters = 11
M → 2
A → 2
T → 2
H → 1
E → 1
I → 1
C → 1
S → 1
∴ Required number of words = \(\frac{11!}{2!2!2!}\)
= \(\frac{39916800}{8}\)
= 4989600

Question 13.
Consider the expansion of (x + 9y)¹⁰. Find its
(i) number of terms (1)
(ii) general term (2)
(iii) 5th term (1)
Answer:
(i) 11

Question 14.
Find the sum of the sequence 8, 88, 888,……… to n terms. (4)
Answer:

Question 15.
(i) Find the slope of the line x – 7y + 5 = 0. (1)
(ii) Find the equation of the line perpendicular to the above line having x-intercept 3. (3)
Answer:
(i) Slope = \(\frac{-A}{B}=\frac{-1}{-7}=\frac{1}{7}\)
(ii) Slope of required line = -7
(x₁, y₁) = (3, 0)
Equation is y – y₁ = m(x – x₁)
⇒ y – 0 = -7(x – 3)
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0

Question 16.
Find the coordinates of the foci, vertices, the length of the major axis and the length of the latus rectum of the ellipse \(\frac{x^2}{36}+\frac{y^2}{16}\) = 1 (4)
Answer:
Given \(\frac{x^2}{36}+\frac{y^2}{16}\) = 1
a = 6, b = 4
c = \(\sqrt{a^2-b^2}=\sqrt{20}\)
Foci = (±c, 0) = (±√20, 0)
Vertices = (±a, 0) = (±6, 0)
Length of major axis = 2a = 12
Length of Latus Rectum = \(\frac{2 b^2}{a}=\frac{2 \times 16}{6}=\frac{16}{3}\)

Question 17.
If A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35, then find
(i) P(A or B) (2)
(ii) P(not A and not B) (2)
Answer:
(i) P(A or B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.54 + 0.69 – 0.35
= 0.88
(ii) P(not A and not B) = P(A’ ∩ B’)
= P(A ∪ B)’
= 1 – P(A ∪ B)
= 1 – 0.88
= 0.12

Answer any 3 questions from 18 to 22. Each carries 6 scores. (3 × 6 = 18)

Question 18.
(i) Prove that \(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=\frac{-\sin 2 x}{\cos 10 x}\) (3)
(ii) Find the principal and general solution of the equation sin x = \(\frac{-\sqrt{3}}{2}\). (3)
Answer:

Question 19.
(i) Represent the complex number Z = -1 + i√3 in the polar form. (3)
(ii) Solve the equation √5x² + x + √5 = 0 (3)
Answer:

Question 20.
(i) Solve the inequality \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\). (2)
(ii) Solve the following inequalities graphically: (4)
x + 3y ≤ 9
2x + y ≤ 12
x ≥ 0; y ≥ 0
Answer:
(i) \(\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}\)
9(x – 2) ≤ 25(2 – x)
9x – 18 ≤ 50 – 25x
9x + 25x ≤ 50 + 18
34x ≤ 68
x ≤ 2
∴ Solution is (-∞, 2]
(ii) x + 3y = 9, 2x + y = 12

Question 21.
(i) Find the derivative of cos x using first principle. (3)
(ii) Find the derivative of \(\frac{x^2}{3 x-1}\) (3)
Answer:
(i) f(x) = cos x
f(x + h) = cos(x + h)

Question 22.
Consider the following table:

Classes	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	6	8	14	16	4	2

(i) Find the mean. (2)
(ii) Find the variance. (3)
(iii) Find the standard deviation. (1)
Answer:


Variance = (Standard Deviation)² = 160

Reviewing Kerala Syllabus Plus One Maths Previous Year Question Papers and Answers Pdf March 2023 helps in understanding answer patterns.

Kerala Plus One Maths Previous Year Question Paper March 2023

Time: 2 Hours
Total Scores: 60

Answer any 6 questions from 1 to 8. Each carries 3 scores. (6 × 3 = 18)

Question 1.
(i) If A and B are two sets such that A ⊂ B, then A ∪ B = _________________ (1)
(ii) Write the set {x: x is a positive integer and x2 < 40} in the Roster form. (1)
(iii) Write all the subsets of {2}. (1)
Answer:
(i) B
(ii) {1, 2, 3, 4, 5, 6}
(iii) {2}, φ

Question 2.
Solve: 3{1 – x} < 2(x + 4), Also represent the solutions on a number line. (3)
Answer:
3(1 – x) < 2(x + 4)
3 – 3x < 2x + 8
-3x – 2x < 8 – 3
-5x < 5x > -1

Question 3.
(i) If (x + 1, y – 4} = (3, 7), then find the values of x and y. (1)
(ii) The Cartesian product A × A has 9 elements among which 2 elements are (-a, 0) and (0, a). Write A. Also find A × A. (2)
Answer:
(i) x + 1 = 3, y – 4 = 7
x = 2, y = 11
(ii) A = {-a, 0, a}
A × A = {(-a, – a), (-a, 0), (-a, a) (0, -a), (0, 0), (0, a), (a, -a), (a, 0), (a, a)}

Question 4.
Find the number of arrangements of the letters of the word ‘INSTITUTE ’. How many of them begin with N? (3)
Answer:
Total no.of arrangements = \(\frac{9!}{2!3!}\) = 30,240
No. of arrangements begins with N = \(\frac{8!}{2!3!}\) = 3360

Question 5.
If f: R → R defined by \(f(x)=\left\{\begin{array}{l}
2 x+3 \text { if } x \leq 0 \\
3(x+1) \text { if } x>0
\end{array}\right.\). Evaluate \(\lim _{x \rightarrow 0} f(x)\). (3)
Answer:
LHL = \(\lim _{x \rightarrow 0^{-}} f(x)\) = 2 × 0 + 3 = 3
RHL = \(\lim _{x \rightarrow 0^{+}} f(x)\) = 3(0 + 1) = 3
LHL = RHL
∴ \(\lim _{x \rightarrow 0} f(x)\) = 3

Question 6.
(i) The point (0, 5, 7) lies in (1)
(a) XY-Plane
(b) YZ-Plane
(c) XY-Plane
(d) X-axis
(ii) Find the distance between (2, -3, -1) and (-2, 4, 3). (2)
Answer:
(i) (b) YZ plane
(ii) \(\sqrt{(-4)^2+(7)^2+(4)^2}=\sqrt{16+49+16}\)
= \(\sqrt{81}\)
= 9 units

Question 7.
If P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6, then find P(B) and P(not-B). (2)
Answer:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(B) = P(A ∪ B) + P(A ∩ B) – P(A)
= 0.6 + 0.25 – 0.35
= 05
P(not B) = P(B’)
= 1 – P(B)
= 1 – 0.5
= 0.5

Question 8.
Find the centre and radius of the circle x² + y² + 8x + 10y – 8 = 0. (3)
Answer:
2g = 8, 2f = 10, c = -8
⇒ g = 4, f = 5
Centre = (-g, -f) = (-4, -5)
r = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{16+25+8}\)
= \(\sqrt{49}\)
= 7

Answer any 6 questions from 9 to 16. Each carries 4 scores. (6 × 4 = 24)

Question 9.
Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3}, B = {3, 4, 5}
(i) Find A ∪ B. (1)
(ii) Find A’ and B’. (1)
(iii) Verify (A ∪ B)’ = A’ ∩ B’. (2)
Answer:
(i) A ∪ B = {2, 3, 4, 5}
(ii) A’ = {1, 4, 5, 6}, B’ = {1, 2, 6}
(iii) (A ∪ B)’ = {1, 6}
A’ ∩ B’ = {1, 6}
∴ (A ∪ B)’ = A’ ∩ B’

Question 10.
(i) Let f: R → R, g: R → R defined by f(x) = x + 1, g(x) = 2x – 3. Find (f+g) (x) and (fg)(x). (1)
(ii) The function h: R → R defined by h(x) = |x|. Draw the graph of h(x). Also write its domain and range. (3)
Answer:
(i) (f + g)(x) = f(x) + g(x)
= x + 1 + 2x – 3
= 3x – 2
(f.g)(x) = (x + 1)(2x – 3)
= 2x² – 3x + 2x – 3
= 2x² – x – 3
(ii)

Domain = R
Range = [0, ∞)

Question 11.
(i) i^-35 = _________________ (1)
(ii) Find the multiplicative inverse and conjugate of \(\frac{1+i}{1-i}\). (3)
Answer:

Question 12.
4 cards are drawn from a pack of 52 playing cards.
(i) In how many ways can it be done? (1)
(ii) In how many ways can these 4 cards contain 2 red and 2 black? (3)
Answer:
(i) ⁵²C₄
(ii) ²⁶C₂ × ²⁶C₂

Question 13.
(i) Number of terms in the expansion of \(\left(x-\frac{1}{x}\right)^4\). (1)
(ii) Write the expansion of \(\left(x-\frac{1}{x}\right)^4\). (3)
Answer:
(i) 5

Question 14.
Insert 3 numbers between 1 and 256 so that the resulting sequence is a G.P. (4)
Answer:
Let x1, x2, x3 be the required numbers, Such that 1, x1, x2, x3, 256 is in a G.P.
a = 1, a5 = 256
ar⁴ = 256
⇒ r4 = 256
⇒ r = 4
∴ x₁ = 4, x₂ = 16, x₃ = 64
∴ Required numbers are 4, 16, 64

Question 15.
Find the co-ordinates of foci, vertices, eccentricity and length of latus rectum of the hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\). (4)
Answer:
a = 3, b = 4
c = \(\sqrt{a^2+b^2}\)
= \(\sqrt{9+16}\)
= 5
Foci = (±c, 0) = (±5, 0)
Vertices = (±a, 0) = (±3, 0)
Eccentricity = \(\frac{c}{a}=\frac{5}{3}\)
Length of Latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 16}{3}=\frac{32}{3}\)

Question 16.
A bag contains 9 discs of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be
(i) red
(ii) yellow
(iii) blue
(iv) not blue
Answer:
(i) P(red) = \(\frac{4}{9}\)
(ii) P(yellow) = \(\frac{2}{9}\)
(iii) P(blue) = \(\frac{3}{9}\) = \(\frac{1}{3}\)
(iv) P(not blue) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Answer any 3 questions from 17 to 20. Each carries 6 scores. (3 × 6 = 18)

Question 17.
(i) 25° = _________________ radian. (1)
(ii) Find the value of sin 15°. (2)
(iii) Prove that \(\frac{\sin 3 x+\sin x}{\cos 3 x+\cos x}\) = tan 2x. (3)
Answer:
(i) 25° = 25 × \(\frac{\pi}{180}\) = \(\frac{5 \pi}{36}\) radian
(ii) sin(15°) = sin(45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
(iii) LHS = \(\frac{\sin 3 x+\sin x}{\cos 3 x+\cos x}\)
= \(\frac{2 \sin 2 x \cos x}{2 \cos 2 x \cdot \cos x}\)
= \(\frac{\sin 2 x}{\cos 2 x}\)
= tan 2x
= RHS

Question 18.
(i) Find the equation of a line passing through the point (-4, 3) with slope \(\frac{1}{2}\). (2)
(ii) Write the equation of the line passing through the points (1, -1) and (3, 5). (2)
(iii) Find the angle between the lines obtained in (i) and (ii). (2)
Answer:
(i) (x₁, y₁) = (-4, 3), m = \(\frac{1}{2}\)
Equation is y – y₁ = m(x – x₂₎
⇒ y – 3 = \(\frac{1}{2}\)(x + 4)
⇒ 2y – 6 = x + 4
⇒ x – 2y + 10 = 0
(ii) (x₁, y₁) = (1, -1), (x₂, y2) = (3, 5)
Equation is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
⇒ y + 1 = \(\frac{5+1}{3-1}\) (x – 1)
⇒ y + 1 = 3(x – 1)
⇒ y + 1 = 3x – 3
⇒ 3x – y – 4 = 0
(iii) m1 = \(\frac{1}{2}\), m2 = 3
tan θ = \(\left|\frac{m_1-m_2}{1+m_1, m_2}\right|=\left|\frac{\frac{1}{2}-3}{1+\frac{1}{2} \times 3}\right|\)
⇒ θ = 45°

Question 19.
(i) Find the derivative of tan x using 1st principles. (4)
(ii) If y = x’ sin x, find \(\frac{d y}{d x}\). (2)
Answer:
(i) f(x) = tan x
f(x + h) = tan(x + h)

(ii) y = x.sin x
\(\frac{d y}{d x}=x \cdot \frac{d}{d x}(\sin x)+\sin x \cdot \frac{d}{d x}(x)\)
= x cos x + sin x × 1
= x cos x + sin x

Question 20.
Consider the following table:

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency	5	8	15	16	6

(i) Find mean. (2)
(ii) Find variance. (3)
(iii) Find standard deviation. (1)
Answer:

= 861 – 729
= 133
(iii) Standard deviation = σ = \(\sqrt{133}\) = 11.53

Kerala Plus One Maths Improvement Question Paper 2018

Time Allowed: 2½ hours
Cool off time: 15 Minutes
Maximum Marks: 80

General Instructions to Candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to familiarize yourself with the questions and plan your answers.
• Read instructions carefully.
• Read questions carefully before you answer.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices, except non-programmable calculators, are not allowed in the Examination Hall.

Answer any 6 questions from question numbers 1 to 7. Each carries 3 scores

Question 1.
a. If A= {2, 3, 4, 5} and B = {4, 5, 6, 7}, then write :
i. A ∪ B
ii. A ∩ B
b. Which one of the following is equal to {x: x∈ R, ∂< x ≤ 4 }?
i. {2,3,4}
ii. {3,4}
iii. [2, 4]
iv.(2, 4]

Question 2.
Consider the set
A = {x : x is an integer, 0 ≤ x < 4}
a. Write A in Roser form
b. If B = {5, 6}, then write A x B
c. Write the number of possible relations from A to B

Question 3.
(a) Prove that \(\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=\frac{-\sin 2 x}{\cos 10 x}\).
(b) Evaluate \(\lim _{x \rightarrow 0} \frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}\)

Question 4.
Solve the inequality \(\frac{x}{2} \geq \frac{5 x-2}{3}-\frac{7 x-3}{5}\)

Question 5.
Find the polar form of the complex number \(\frac{1+i}{1-i}\).

Question 6.
How many terms of the GP, 3, 3/2, 3/4,…………………..are needed to give the sum \(\frac { 3069 }{ 512 } \)?

Question 7.
Consider the real-valued function f(x) = \(\frac{x-3}{x^2-x-6}\)
(a) Find the domain of f(x).
(b) Evaluate \(\lim _{x \rightarrow 3} f(x)\)

Answer any 8 from questions 8 to 17. Each question carries 4 scores.

Question 8.
a. If U = {1, 2, 3,4, 5, 6, 7, 8, 9}
A = {2, 4, 6, 8}
B = {2, 3, 5, 7}
Verify (A ∪B) = A’ ∩ B’
b. If A and B are two disjoint sets with n(A) = 4 and n(B) = 2, then n (A-B) = …………

Question 9.
Consider the statement P(n) : 1-3 + 32 +……….. \({ 3 }^{ n-1 }=\frac { { 3 }^{ n-1 } }{ 2 } \)
a. Show that P(1) is TRUE
b. Prove by principle of Mathematical induction, that P(n) is TRUE for all n ∈ N

Question 10.
Solve the following inequalities graphically
2x + y ≥ 4
x + y ≤ 3 and
2x – 3y ≤ 6

Question 11.
Find the square roots of the complex number 3 + 4i.

Question 12.
a. Insert five numbers between 8 and 26 such that the resulting sequence is an AP.
b. Find the sum to n terms of the series 1 x 2 + 2 x 3 + 3 x 4+…………..

Question 13.
a. Find the equation of the perpendicular bisector of the line joining the points (0, 0) and (-3, 4).
b. Find the coordinates of the point on the line y = 3x-2 that is equidistant from (0,0) and (-3, -4)

Question 14.
a. Reduce the equation x – y = 4 into normal form.
b. Write the distance of this line from the origin

Question 15.
a. Find the derivative of f(x) = x Sin x concerning x.
b. Find the derivative of the function y = √x concerning x by using first principles.

Question 16.
Consider the points A (3, 8, 10) and B (6, 10, -8).
a. Find the ratio in which the line segment joining A and B is divided by the YZ coordinate plane.
b. Find the coordinates of the point of division.
c. Which coordinate plane divides the line segment AB internally? Justify your answer.

Question 17.
a. Write the contrapositive of the statement: “If the integer n is odd, then n2 is odd”,
b. Prove by the method of contradiction ‘ √7 is irrational’

Answer any 5 questions from question numbers 18 to 24. Each carries 6 scores

Question 18.
(a) If \(\left(\frac{x+3}{2}, \frac{y-1}{3}\right)\) = (4, 2), find the value of x and y.
b. Consider the function f (x) = |x| = 3, draw the graph of f(x)
c. Write the domain and range of f(x)

Question 19.
a. Find the value of Sin (75°)

b. In the given figure. ∠AOB = 30° and the radius of the circle is d units. Find the length of area APB.
c. Find the length of chord AB

Question 20.
a. Find the number of words with or without meaning, which can be made by using all the letters of the word GANGA.
b. If these words are written as in a dictionary, what will be the 26th word?
c. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team should have at least 3 girls?

Question 21.
a. Write the expansion of (a + b)ⁿ.
b. Find the coefficient of x⁵ y⁷in the expansion of (x-2y)¹²
c. Show that 9^n-1 – 8n – 9 is divisible by 64.

Question 22.
Focii of the ellipse in the given figure are (± √12,0) and vertices are (±4, 0).

a. Find the equation of the ellipse.
b. Write the equation of a circle with centre (0, k) and radius r.
c. The circle in the figure passes through the points A, B, and C on the ellipse. Find the equation of a circle.

Question 23.
Consider the following table

a. Find the arithmetic mean of marks given in the above data
b. Find the standard deviation of marks in the above data.
c. Find the coefficient of variation.

Question 24.
a. Consider the experiment in which a coin is tossed repeatedly until a head comes up. Write the sample space.
b. If A and B are two events of a sample space with P (A) = 0.54, P(B) = 0.69 and P (A ∩ B) = 0.35. Find P (A’ ∩ B’).
c. 3 cards are drawn from a well shuffled pack of 52 cards. Find the prohibility that
i. all the 3 cards are diamond.
ii. at least one of the cards is non diamond
iii. one card is king and two are jacks.

Answers

Answer 1.
a. A ∪ B = {2, 3, 4, 5, 6, 7}
A∩B = {4, 5}
b. iv) (2, 4]

Answer 2.
a. A = {0, 1,2,3}
b. A x B = {(0, 5),(0, 6),(1, 5),(1, 6),(2, 5), (2, 6),(3, 5),(3, 6)}
c. No. of possible relations from A to B
= 2^mn = 2^{4 x 2} = 256

Answer 3.

Answer 4.

Answer 5.

Answer 6.

Answer 7.

Answer 8.
a. A ∪ B = {2, 3, 4, 5, 6, 7, 8}
(A ∩ B)’ = {1, 9}………….(1)
A’= {1, 3, 5, 7, 9}, B’={1, 4, 6, 8, 9} A’∩B’= {1, 9}…………….(2)
From (1) and (2), we have (A + B)’ = A’ ∩ B’
b. n (A – B) = n (A) = 4

Answer 9.

Answer 10.

Answer 11.

Answer 12.

Answer 13.

Answer 14.

Answer 15.

Answer 16.
(a) Let the ratio be k : 1
Point of division is \(\left(\frac{6 k+3}{k+1}, \frac{10 k+8}{k+1}, \frac{-8 k+10}{k+1}\right)\)
Since the line joining A and B divides the YZ plane, the point of intersection has x coordinate equal to zero.
\(\frac{6 k+3}{k+1}\) = 0
⇒ 6k + 3 = 0
⇒ k = \(-\frac{1}{2}\)
Hence YZ plane divides externally in the ratio 1 : 2.
(b) Substituting the value of k, we get the point.


(c) The point A(3, 8, 10) and B(6, 10, 4) have z coordinates opposite in sign. So the point A will lie above and point B will lie below the XY plane. Hence the XY plane divides the line segment joining A and B internally.

Answer 17.
a. If n2 is not even, then the integer n is not even.
b. Let us assume that √7 is a rational number.
√7 =a/b, where a and b are co-prime, i.e., a and b have no common factors. Squaring we have,
7b² = a² ⇒ 7 divides a.
∴ Let a = lk
∴a²= 49k²  ⇒ 7b²= 49k² ⇒ b² = 7k² ⇒7  divides b.
i.e., 7 divides both a and b, which is a contradiction to our assumption that a and b have no common factor. our supposition is wrong.
∴ √7 is an irrational number.

Answer 18.
(a) \(\frac{x+3}{2}\) = 4, \(\frac{y-1}{3}\) = 2
⇒ x + 3 = 8; y – 1 = 6
⇒ x = 5; y = 7

(c) From the graph of f(x) is the domain and range is clear.
Domain = R
Range = [-3, ∞)

Answer 19.

Answer 20.
a. Number of words = \(\frac { 5! }{ 2!2! } \) = 30
b. If A is fixed, the remaining 4 letters can 4!
be permuted in \(\frac { 4! }{ 2! } \) = 12 ways.
If G is fixed, the ramaining 4 letters
can be permuted in \(\frac { 4! }{ 2! } \) = 12 ways.
∴ the 25^th word is NAAGG.
∴ the 26^th word is NAGAG
c. No. of selections = ⁴C₃ x ⁷C₂ + ⁴C₄ + ⁷C₄ = 4 x 21 + 1 x 7 = 84 + 7 = 91

Answer 21.

Answer 22.
a. (± c ,0) = (±√2 , 0) ; (± a, 0) = (± 4, 0)
c² = a² – b² ⇒ b² = a²– c² = 16 – 12 = 4
Equation of the ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 4 } \) = 1
b.. (x – 0)²+(y – k)²= r² ⇒ x² + y² – 2ky + k² – r² = 0
c. The co-ordinates of the points are :
A(-4,0), B(4,0) and C(0, -2).
Equation of the circle is
x² + y² + 2gx + 2fy + c = 0 ………….. (1)
(1) passes through A(-4, 0)
: 16 + 0 – 8g + c = 0 …………… (2)
(1) passes through B (4, 0)
: 16 + 0 + 8g + c = 0…………. (3)
passes through C (0, -2)
: 4 + 0-4f + c = 0 ……………… (4)
(2) + (3) ⇒ 32+2c=0
⇒ 2c = -32
⇒ c = -16 when c= -16
⇒ 16 + 0-8g – 16 = 0
⇒- 8g=0 ⇒ g=0
when c=-16 ⇒ 4 + 0-4f – 16 = 0
⇒- 4f = 12 ⇒ f = -3
Equation of the circle is
x² + y² + 2 (0) x + 2(-3) y+ – 16 = 0
⇒ x² + y²-6y – 16 = 0

Answer 23.

Answer 24.
a. Sample space = {H, TH, TTH, TTTH, …………………… }
b. P(A’∩B’)= 1-P(A ∪ B)=1 -[P(A) + P(B)-P(A ∩ B)]
=1 – [0.54 + 0.69-0.35] =1-0.88=0.12
c. i. P(all cards are diamond)

Plus One Maths Previous Year Question Papers and Answers

Kerala Plus One Maths Previous Year Question Paper 2018

Time Allowed: 2½ hours
Cool off time: 15 Minutes
Maximum Marks: 80

General Instructions to Candidates:

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to familiarize yourself with the questions and plan your answers.
• Read instructions carefully.
• Read questions carefully before you answer.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices, except nonprogrammable calculators, are not allowed in the Examination Hall.

Answer any six from the question numbers 1 to 7. Each carries three answers.

Question 1.
Find the sum to n terms of the sequence
4 + 44 + 444 +…………

Question 2.
Solve : Sin 2x – Sin 4x + Sin 6x = 0

Question 3.
If A and B are events such that P(A)= 1/4;
P(B) =1/2; P(A ∩ B) = 1\6 then find: 2 6
(a) P(A or B)
(b) P(not A and not B)

Question 4.
In a ΔABC, prove that \(\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2}\)

Question 5.
(a) The maximum value of the function f(x) = Sin x is ____________________
(i) 1
(ii) √3/2
(iii) 1/2
(iv) 2
(b) Prove that, (Sin x + Cos x)2 = 1 + Sin 2x.
(c) Find the maximum value of Sin x + Cos x.

Question 6.
\(\underset { x\rightarrow 2 }{ Lim } \) [x] = __________________
(i) 2
(ii) 3
(iii) 0
(iv) does not exist
(b) Evaluate: \(\underset { x\rightarrow 2 }{ Lim } \) \(\frac { { x }^{ 3 }-\quad { 4x }^{ 2 }+\quad 4x }{ { x }^{ 2 }-4 } \)

Question 7.
Once the card is drawn randomly from a pack of 52 playing cards. Find the probability that,
(a) the card drawn is black.
(b) The card drawn is a face card.
(c) The card drawn is a black face card

Answer any eight from question numbers 8 to 17. Each carries four scores.

Question 8.
(a) If A = {a, b, c}, then write Power Set P(A).
(b) If the number of subsets with two elements of a set P is 10, then find the total number of elements in set P.
(c) Find the number of elements in the power set of P.

Question 9.
Consider a Venn diagram of the Universal Set U = {1,2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}

(a) Write sets A, B in Roster form.
(b) Verify (A ∪ B)’ = A ∩ B
(c) Find n(A ∩ B)’.

Question 10.
Consider the following graphs :


(a) Which graph does not represent a function?
(b) Identify the function f(x) = 1/x. from the above graphs.
(c) Draw the graph of the function f(x) = (x-1)².

Question 11.
The figure shows the graph of a function f(x) which is a semi-circle centred at the origin.

(a) Write the domain and range of f(x).
(b) Define the function f(x).

Question 12.
(a) If 3²ⁿ⁺² – 8n – 9 is divisible by ‘k’ for all n ∈ N is true, then which one of the following is a value of ‘k’ ?
(i) 8
(ii) 6
(iii) 3
(iv) 12
(b) Prove by using the principle of Math­ematical Induction
P(n) = 1+3+3²+…………….. +3^n-1 = \(\frac { { 3 }^{ n }-1 }{ 2 } \) is true for all n ∈ N.

Question 13.
(a) Solve the inequality \(\frac{2 x-1}{3} \geq \frac{3 x-2}{4}-\frac{2-x}{5}\).
(b) Represent the solution on a number line.

Question 14.
(a) Find the n^th term of the sequence 3, 5, 7,……………..
(b) Find the sum to n terms of the series. 3 × 1² + 5 × 2² + 7 × 3²+ ……..

Question 15.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Question 16.
Consider a point A(4, 8, 10) in space.
(a) Find the distance of the point A from the XY-plane.
(b) Find the distance of the point A from the X-axis.
(c) Find the ratio in which the line segm­entjoining the point A and B (6, 10, -8) is divided by the YZ-plane.

Question 17.
(a) Which one of the following sentences is a STATEMENT?
(i) 275 is a perfect square.
(ii) Mathematics is a difficult subject.
(iii) Answer this question.
(iv) Today is a rainy day.
(b) Verify by method of contradiction: ‘√2 is irrational’.

Answer any five from question numbers 18 to 24. Each carries six scores.

Question 18.
Consider the quadratic equation x² + x + 1 = 0
(a) Solve the quadratic equation.
(b) Write the polar form of one of the roots.
(c) If the two roots of the given quadratic are α and β. Show that α² = β.

Question 19.
The graphical solution of a system of linear inequalities is shown in the figure.

(a) Find the equation of the lines L₁, L₂, L₃
(b) Find the inequalities representing the solution region.

Question 20.
(a) Which one of the following has its middle term independent
(i) \({ \left( x+\frac { 1 }{ x } \right) }^{ 10 }\)
(ii) \({ \left( x+\frac { 1 }{ x } \right) }^{ 9 }\)
(iii) \({ \left( { x }^{ 2 }+\frac { 1 }{ x } \right) }^{ 9 }\)
(iv) \({ \left( { x }^{ 2 }+\frac { 1 }{ x } \right) }^{ 10 }\)
(b) Write the expansion of \({ \left( { x }^{ 2 }+\frac { 3 }{ x } \right) }^{ 4 }\)
(c) Determine whether the expansion of  \({ \left( { x }^{ 2 }+\frac { 2 }{ x } \right) }^{ 18 }\)  will contain a term containg x¹⁰.

Question 21.
The figure shows an ellipse \(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } \)= 1 and a line L.

(a) Find the eccentricity and focus of the ellipse.
(b) Find the equation of the line
(c) Find the equation of the line parallel to line L and passing through any one of the foci.

Question 22.
(a) Find the derivative of y = Sin x from the first principle.
(b) Find \(\frac{d y}{d x}\) if y = \(\frac{x^5-\cos x}{\sin x}\)

Question 23.
Find n, if
(a) 12 × (n-1) p₃ =5 × (n+1) p₃
(b) If ⁿp_r = 840; ⁿC_r = 35 find r.
(c) The English alphabet has 5 vowels and 21 consonants. How many 4 letter words with two different vowels and two different consonants can be formed without repetition of letters?

Question 24.
Consider the following data:

(a) Find the standard deviation of the distribution.
(b) Find the coefficient of variation of the distribution.

Answers

Answer 1.

Answer 2.

Answer 3.

Answer 4.

Answer 5.
(a) 1
(b) (sin x + cosx)² = sin²x + cos²x + 2 sinx cosx = 1 + sin2x
(c) Maximum value of sin2x = 1
Maximum value of sinx + cosx = \(\sqrt { 1+1 } =\sqrt { 2 } \)

Answer 6.
(a) (iv) Does not exist.

Answer 7.

Answer 8.
(a) P(A) = {φ, {a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}
(b) nC2 = 10
⇒ \(\frac{n(n-1)}{2}\) = 10
⇒ n2 – n – 20 = 0
⇒ (n – 5)(n + 4) = 0
⇒ n = 5, -4
Total number of elements in set P is 5
(c) Number of elements in power set of P is 25 = 32

Answer 9.
(a) A = {3,4,6,10} B = {2,3,4,5,11}
(b) (A∪B)’ = { 1, 7, 8, 9, 12, 13}
A’ = { 1, 2, 5, 7, 8, 9, 11, 12, 13}
B’ = {1, 6, 7, 8, 9, 10, 12, 13 }
A’∩B’ = {1,7, 8, 9, 12, 13}
∴ (A∪B)’ =A’ ∩B’
(c) n(A∩B)’ = n(U)-n(A∩B)=13-2=11

Answer 10.
(a) There are two answers (ii) and (iii).
(vertical line interest at more than two points)
(b) (i)
(c) The graph of f(x) = (x – 1)2 is obtained by shifting the graph of f(x) = x2 to right 1 units.

Answer 11.
(a) Domain = [-4,4]
Range = [0,4]
(b) x² + y²=16
y²= 16-x²
y= \(\sqrt { 16-{ x }^{ 2 } } \)
i.e.,f (x)= \(\sqrt { 16-{ x }^{ 2 } } \)

Answer 12.
(a) 8
(b) P(n):1 + 3 + 3² + ……..3^n-1 =\(\frac { { 3 }^{ n }-1 }{ 2 } \)
Let P(1): 1 =\(\frac { { 3 }^{ 1 }-1 }{ 2 } \)=2/2=1
Hence, P(1) is true.
P(k): 1 + 3 + 3² +…… + 3^k-1=\(\frac { { 3 }^{ k }-1 }{ 2 } \)
To prove that P(k+1) is true. P(k+1):
1 + 3 +3² + + 3^i_1 +3^k+1-1 =\(\frac { { 3 }^{ k+1 }-1 }{ 2 } \)
⇒ P(k)+3^k = \(\frac { { 3 }^{ k+1 }-1 }{ 2 } \)
⇒ \(\frac { { 3 }^{ k }-1 }{ 2 } \) + 3^k= \(\frac { { 3 }^{ k+1 }-1 }{ 2 } \)
= \(\frac { { 3 }^{ k }-1 }{ 2 } \) + 3^k= \(\frac { { { 3 }^{ k } }-1+2\times { 3 }^{ k } }{ 2 } \)
Now, LHS = \(\frac { { { 3\times 3 }^{ k } }-1 }{ 2 } \) =  \(\frac { { 3 }^{ k+1 }-1 }{ 2 } \) = RHS
Hence, P(k+1) is true.
Hence, P(n) is true for all n∈N

Answer 13.
(a) \(\frac { 2x-1 }{ 3 } \ge \frac { 5(3x-3)-4(2-x) }{ 20 } \)
40x-20 ≥ (15x-10-8+4x)³
40x-20 ≥ -54
40x-57x ≥ -54 + 20
-17x ≥-34
x ≤ 2 ⇒ x∈(-∞, 2)

Answer 14.
(a) t_n= 3+(n-1)² = 2n+1
(b) Let T_n denotes the nth term of the given series.
T_n = (n^lh term of 3, 5, 7……….. )(n^th term of 1², 2², 3²…….. )
= [3+(n-1)²].n² = (2n+1)n² = 2n³ + n²

Answer 15.
Let the equation of the circle be
(x – h)² + (y – k)² = r²
Since the circle passes through (4, 1) and (6,5), we have
(4-h)² + (1 -k)² = r²……….. (1)
and (6 -h)² + (5 – k)² = r²………… (2)
Also since the centre lies on the line 4x + y = 16, we have
4h + k = 16…………. (3)
Solving the equations (1), (2) and (3), we get
h = 3 and k = 4
(4 – 3)² + (1 – 4)² = r²
r²=10
Hence, the equation of the required circle is
(x-3)²+1 (y-4)²= 10
x²-6x + 9 + y²-8y + 16-10 = 0
x² + y² – 6x –  8y + 15 = 0

Answer 16.
(a) 10 [Z coordinate of the point]
(b) Let P(4,0,0) be a line on the X-axis.
distance = \(\sqrt { { (4-4) }^{ 2 }+{ (8-0) }^{ 2 }+{ (10-0) }^{ 2 } } \)
= \(\sqrt { 64+100 } =\sqrt { 164 } =2\sqrt { 41 } \)
(c) Since the line segment divides the YZ plane, its x coordinate is zero.
So =\(\frac { { mx }_{ 1 }+{ nx }_{ 1 } }{ m+n } \)
⇒ mx₂ + nx₁=0
\(\frac { m }{ n } =\frac { { -x }_{ 1 } }{ { x }_{ 2 } } =\frac { -2 }{ 3 } \)
⇒m : n=-2 : 3

Answer 17.
(a) 275 is a perfect square,
(b) Let -√2 is rational.
\(\sqrt { 2 } =\frac { p }{ q } \) , p and q have no common factor.
p² = 2q²
i.e., 2 divides p² is 2 divides p
∴ p = 2k, p² = 4k²
2q² = 4k², q² = 2k²
2 divides q² is 2 divides q.
p and q have a common factor 2.

Answer 18.

Answer 19.
(a) L1 line passes through (2, 2) and (4, 0)
Slope = \(\frac{2-4}{2-0}\) = -1
Equation is (y – 0) = -1(x – 4) ⇒ x + y = 4
L2 is parallel to the x-axis and passes through (0, 1).
Hence the equation is y = 1
L3 passes through (0, 0) and (2, 2).
Hence the equation is y = x
(b) The inequalities that form the shaded region are x + y ≤ 4; y ≥ 1; y ≤ x

Answer 20.

Answer 21.

Answer 22.

Answer 23.
(a) 12 x (n-1)P₃ = 5 x (n+1)P₃
12 x (n-1 )(n – 2)(n – 3) = 5 x (n+1 )n(n-1)
12n² – 6Qn + 72 – 5n² – 5n = 0
7n²-65n+72 = 0
\(\frac { 65+\sqrt { 4225-2016 } }{ 14 } =\frac { 67647 }{ 14 } \)
\(\frac { 112 }{ 14 } ,\frac { 18 }{ 14 } =8,\frac { 9 }{ 7 } \)
But n cannot be a fraction. n = 8
(b) ⁿP = r! x ⁿC_r   840=r! x 35
r! = 24 ⇒ r = 4
Two different vowels can be selected in ⁵C₂ ways. Two different consonants can be selected in ^2IC₂ Ways.
Total selection 4 letters = ⁵C₂ x ²¹C₂
Total words = 4! x ⁵C₂ x²¹C₂
= 10 x 210 x 24 = 50400

Answer 24.

Plus One Maths Previous Year Question Papers and Answers

Teachers recommend solving Kerala Syllabus Plus Two Political Science Previous Year Question Papers and Answers Pdf Board Model Paper 2022 to improve time management during exams.

Kerala Plus Two Political Science Board Model Paper 2022 with Answers.

A. Answer any questions from 1 to 6. Each carries 1 Score. (4 × 1 = 4)

Question 1.
Name the Chief Election Commissioner of free India
a) Sukumar Sen
b) T.N. Seshan
c) Sunil Arora
Answer:
a) SukumarSen

Question 2.
Identify the leader of Mizo National Front:
a) Angami Zaphu Phizo
b) Laldenga
c) Longowal
Answer:
b) Laldenga

Question 3.
The Right To Information Act was the result of a popular movement started at Rajasthan in 1990. Identifying the movement.
a) Shetkari Sangathan
b) Bhartiya Kisan Union
c) Mazdoor Kisan Sakhti Sangathan
Answer:
c) Mazdoor Kisan Sakhti Sangathan

Question 4.
Identify the first summit of NAM.
a) Belgrade
b) Yalta
c) Bandung
Answer:
a) Belgrade

Question 5.
From which country India adopted the concept of five year plan
a) USA
b) USSR
c) Japan
Answer:
b) USSR

Question 6.
Find out the leader of Soviet Union during the years of distingration
a) V.l. Lenin
b) Jospeh Stalin
c) Mikhail Gorbachev
Answer:
c) Mikhail Gorbachev

B. Answer all questions from 7 to 10. Each carries 1 score. (4 x 1 = 4)

Question 7.
Identify the leader of Narmada BachaoAndolan
a) Sunderlal Bahuguna
b) MedhaPatkar
c) Chandi Prasad Bhat
Answer:
b) MedhaPatkar

Question 8.
Identify the military alliance formed under the leadership of USSR.
a) WARSAW PACT
b) NATO
c) CENTO
Answer:
a) WARSAW PACT

Question 9.
The ‘Earth Summit’ of 1992 was held in
a) Paris
b) New York
c) Rio De Janeiro
Answer:
c) Rio De Janeiro

Question 10.
Identify the first Non-Congress Prime Minister of India.
a) Morarji Desai
b) Jawaharlal Nehru
c) Lai Bahadur Shastri
Answer:
a) Morarji Desai

Part – II

A. Answer any 3 questions from 11 to 15. Each carries 2 scores.

Question 11.
Briefly explain political consequences of globalisation.
Answer:
Politically speaking, the authority of the government gets weaker. It will have to reduce its welfare schemes with the coming of MNC’S it becomes difficult for the governments to take independent decisions. According to the advocates of globalization, no political power of the government is lost. Technology helps governments to know the details of their citizens and make the administration efficient.

Question 12.
Identify any two objectives of Nehru’s foreign policy.
Answer:
Nehru was the architect of our foreign policy. He followed a policy which was conducive to the maintenance of our sovereignty, protection of our boundaries and rapid economic growth of the country.

Question 13.
List out any two challenges immediately faced by India after Independence.
Answer:
The two major challenges faced by India after independence was:
a) Nation building
b) Establishing democracy in the country

Question 14.
Find out two NAM countries from the list given be-low: (India, China, Egypt, USA)
Answer:
India, Egypt
Hegemony as Structural Power

Question 15.
Briefly explain the context of fourth general election 1967.
Answer:

1. Inflation and price rise
2. Internal disturbances
3. Hindu-Muslim riots
4. Unemployment

B. Answer any 2 questions from 16 to 18. Each caries 2 scores. (2 × 2 = 4)

Question 16.
List out any two component of Indians security strat-egy.
Answer:

1. Increase our military strength so that we can increase our safety and solve our disputes with neighbours.
2. Protect and strengthen international laws and organisations.

Question 17.
Briefly explain‘Cuban Missile Crisis’.
Answer:
Most North-Western countries became capitalist, anti-communist nations. A big power like USA could not easily tolerate a neighbouring country becoming a close ally of Russia. In 1962 the Russian President Khrushchev wanted to deploy missiles and other armaments in Cuba. Most American cities then would come under threat from Russia. Later this was known as Cuban Missile Crisis.

Question 18.
Expand the following:
a) NATO
b) ASEAN
Answer:
NATO-North Atlantic Treaty Organisation ASEAN – Association of South East Asian Nations

Part – III

A. Answer any 3 questions from 19 to 23. Each caries 4 scores. (3 × 4 = 12)

Question 19.
Explain how Indo-china war of 1962 affected India domestically.
Answer:

1. Many top ranking officers of the military resigned.
2. Nehru’s close friend and the then defense minister V.K. Krishna Menon had to resign ’
3. A no-confidence motion was brought against Nehru government at first time.
4. In 1964, Communist party split into two, pro- Chinese and pro-Russia. One was CPI (M) and the other CPI.
5. War strengthened national feeling and prompted the nation to keep its unity and embark upon development projects.

Question 20.
Discuss the major issues which led to the split of Congress party in 1969.
Answer:
Jawaharlal Nehru had great influence in the party because of his dynamic personality. His death created a vacuum in the party. Gradually the central group known as the Syndicate established itself in the Congress. In 1969 there arose a difference between Indira Gandhi and the Syndicate. When Dr. Zakir Hussain died, the Presidential post was vacant.

The Syndicate nominated Indira Gandhi’s long time rival and the then speaker Sanjiva Reddy to the post. But Indira Gandhi preferred the then Vice-President V.V. Giri to be the next President. V.V. Giri was elected. The failure of the official candidate of the Syndicate caused the formal division of the Congress. The Congress President suspended Indira Gandhi from the Party. Indira claimed that her group was the real Congress. In 1969; the Congress led by the Syndicate was known as Congress (O) and the one led by Indira Gandhi was called Congress (R).

Question 21.
The government of India and Bangladesh have differ-ences over several issues. Write any four of such issues.
Answer:

1. The disputes over the sharing of the Ganga Brahmaputra river waters has not yet been solved.
2. Issue of illegal migration .
3. Refusal to allow Indian troops to move through its territory

Question 22.
As a citizen of India how do you support India’s candidature for permanent membership in the Security Councilof UNQ. Justify your argument.
Answer:
The demand of lndia,to be a Permanent Member of the Security Council is quite a logical and legitimate one. India fulfils all the criteria to be a permanent member. It demands permanent membership on the following grounds:

1. It has the world’s second largest population.
2. It is the largest democracy in the world.
3. India has participated in the UN activities since its inception.
4. It has long relations with the UN Peacekeeping force.
5. India’s economic situation is improving.
6. India gives regularly to the UN budget. It has never defaulted on any payment.

The above reasons are good enough for India to get a permanent membership in the UN Security Council. Permanent membership has its own significance. India’s importance will increase in world matters. Our foreign policy will influence others.

Question 23.
Match the following:

A	B
Tryst with Destiny’ Speech	Sardar Patel
Formation of Andhra State	Khan Abdul Gaffar Khan
Iron Man of India	Jawaharlal Nehru
Frontier Gandhi	Potti Sri Ramulu

Answer:

A	B
Tryst with Destiny’ Speech	Sardar Patel
Formation of Andhra State	Khan Abdul Gaffar Khan
Iron Man of India	Jawaharlal Nehru
Frontier Gandhi	Potti Sri Ramulu

B. Answer any 1 questions from 24 to 25. Each caries 4 scores. (1 × 4 = 4)

Question 24.
Write any four provisions of Rajiv Gandhi – Longowal Accord.
Answer:
Punjab Accord was an agreement signed between Rajiv Gandhi and Harchand Longowal in 1985. The main provisions are
a) Handover Chandigarh to Punjab
b) Appoint a new commission to solve border disputes
c) Establish a tribunal to solve waters diputes.
d) Compensation and better treatment
e) Withdrawal of AFSPA.

Question 25.
Compare First and Second Five year plans.
Answer:

• First five year plan
  1. A renowned economist Dr. K.N. Raj prepared the draft plan of first five year plan.
  2. Priority to agriculture
  3. Poverty eradication
  4. Slow progress of the country Second five year plan
  5. PC. Mahalanobis prepared the draft proposal of the
• Second five year plan
  1. Rapid industrialisation
  2. Priority to heavy industries
  3. Tariff on imported goods

Part – IV

A. Answer any 3 questions from 26 to 29. Each caries 6 scores. (3 × 6 = 18)

Question 26.
Explain the difficulties in relation to the process of partition of India and its consequences.
Answer:
There were three reasons for that. First of all in British India, there was not a single area which had only Muslims. There were two areas in which the majority was Muslims, one in the West and the other in the East. Therefore Pakistan was formed consisting of two areas – West Pakistan and East Pakistan. Between them there were large areas of Indian Territory.

Secondly, not all areas with majority Muslim population wanted to become part of Pakistan. Khan Abdul Gaffer Khan, who was the leader of the North-Western Province had objected to the Two-Nation Theory. But ignoring his objection, the North West- Province was included in Pakistan.

The third problem was that British India’s Punjab and Bengal were areas with Muslim majority. But in these huge provinces there were very many non- Muslims. Therefore the Provinces were divided into Districts and Panchayats depending on the religious majority of the population there. The result was that on the day of Independence many people did not know to which country they belonged India or Pakistan. It deeply wounded the people.

The problem of the minorities in each country was the worst of all. The. Hindus and Sikhs in the Pakistani areas and the Muslims in Punjab and Bengal were unfortunate preys to this division of the country. When the division was decided upon, there was large scale violence against the minorities in both the countries.

The division of British India in 1947 into India and Pakistan was a very tragic incident in history. In the border areas many people on both sides were kille because of their religion and caste. Huge cities like Lahore, Amritsar and Calcutta became religious areas. Muslims avoided going to areas of Hindus and Sikhs. Similarly Hindus and Sikhs did not want to go near the Muslim areas. People were forced to flee their homes, suffering a lot of difficulties on their way. Many of the people in the minorities in the border areas had to live in refugee camps.

The governments and the police were not there to help them. People had to walk or ride in some vehicles from their homes to their new places. During the journey, many were attacked and killed; women were raped. Many were forced to accept the majority religion and marry people against their will. In many homes women were killed by their own relatives in the name of honour. Children were separated from their parents and guardians. People who came to the new land had no houses and they had to live in refugee camps.

Not only the land, but even moveable properties like tables and chairs were divided. The government and railway workers were divided. People who were living like brethren were divided. It is believed that between 5 to 10 lakh people lost their lives in this tragic division of the country.

Question 27.
What is ‘Shock Therapy1? Briefly explain its conse-quences.
Answer:
The failure of the Soviet Union made many of its constituent countries abandon Communism and follow democratic ways. Russia, Asian countries, Easer European countries etc were attracted towards organizations like the World Bank and IMF.

1. The main purpose yvas to make communist countries follow the capitalist model.
2. Private capital would be very important.
3. Cooperative farming will give way to private farming.
4. Foreign investment, open market system and currency exchange would be possible.
5. The countries of the erstwhile Soviet Union will have the facility to have contact with Western countries and trade with them.
6. Using all these, the Western countries made the member countries of the erstwhile Soviet Union come closer to them.

RESULTS:

1. Many big industries under government control collapsed.
2. 90% industries were sold to individuals or private companies.
3. The Russian Currency Rouble was devalued.
4. Because of inflation people lost the value of their savings.
5. In Cooperative farming, people had food security.
6. But now it was not there. Russia had to import food grains.
7. In Russia, the GDP in 1999 was less than that of 1989.
8. The social welfare schemes were abandoned.
9. As subsidies were withdrawn, many people experienced poverty.

Question 28.
Three new sources of threats faced by the world are given below. Explain them.
Answer:
Security threats are of a different kind now. They can be categorized as follows:

1. Terrorism:
Political attacks make the life of ordinary citizens difficult. The terrorists want the political circumstances to change. They try to bring about changes by threats or armed attacks. By unleashing violence, they make the people restless. They try to make the dislike and discontentment of the people their weapon against governments.

Their activities include hijacking planes and bombing trains and markets and other places where people assemble. They attacked and destroyed the World Trade Centre in New York on 11 September 2001. The government and the people are more cautious now against terrorists.

2. Global Poverty:
This is another factor of security threat. It is believed that the population in the under-developed countries will triple in 50 years. In countries where the population is low, the per capita income will be high. Therefore the economically advanced countries will prosper further whereas the poor countries will grow poorer. The gap between countries of the North and South will increase. People from the South countries immigrate to the North countries for better life and earning. This also creates a threat for the security of mankind.

3. Contagious Diseases: Contagious diseases are another threat to mankind. AIDS, bird flu, SARS (Severe Acute Respiratory Syndrome) etc. are dangerous contagious diseases. These spread quickly from one country to another. A country’s success or failure in controlling these diseases affects other countries also. Recently swine-flu spread all over the world. It is when a problem becomes a threat to a nation and its people that it becomes an international security threat.

Question 29.
What you mean by ‘Common but Differentiated Re-sponsibilities’ ? List out the steps taken by India for the protection of environment.
Answer:
Common but Differentiated Responsibilities is a principle within the United Nations framework convention on climate change that acknowledges different capabilities and differeing responsibilities of individual countries in climate change. Each country is responsible for climate change and related issues. But the developed countries are more responsible for that. This principle was enshrined in the Earth Summit of 1992 in Rio-de-Janeiro.

For the protection of environment India has taken the following steps:
a. According to the auto fuel policy of India, only purified fuel is used for vehicles.
b. The Energy Act passed in 2001, aims at the appropriate use of energy.
c. The 2003 Electricity Act requires the use of renewable energy sources.
d. In 2011-12, the Bio-diesel Mission acquired 11 million hectors of land for the production of biodiesel.

Recently schemes have been worked out for the import of natural gas. It has been decided to use the latest technology for purifying coal. Thus India is doing a lot for environmental protection.

B. Answer any 2 questions from 30 to 32. Each carries 6 scores. (2 x 6 = 12)

Question 30.
Explain the origin and struggle of Narmada Bachao Andolan.
Answer:
Narmada Bachao Andolan:
This Movement came as a result of the Sardar Sarovar Project. This questioned even the economic policy of the government. It was also a movement against large scale displacement of people for huge developmental projects. People of Kerala had made similar protests against the Vallarpadam Project. Sardar Sarovar Project is a large scale project. The project was to build a number of big and small dams. The project would cover Madhya Pradesh, Gujarat and Maharashtra. This project would help Gujarat to have good irrigation and electricity.

Narmada Bachao Andolan was to protect Narmada. The Movement demanded the stoppage of constructions in this river. If this project is completed, 245 villages would be submerged underwater. Some 2.5 lakhs people will have to find new homes. Starting from these villages, the Andolan began to spread. The beginning of the Movement was by demanding to rehabilitate those who would lose their homes. The argument also came up that the local communities should have power over the land, water and forests. They ask why in a democracy some people should become victims for the benefit of some others. It also brings harm to Nature. All these were the causes for the Narmada Bachao Andolan.

Question 31.
Explain any two developments that influenced indian politics since 1989.
Answer:
a) In 1989, Congress lost the election. With that election the ‘Congress System’ ended.

b) Mandal issue in national politics: In the 1.990s,the government decided to implement the job reservations recommended by the Mandal Commission. Reservations were also made for Other Backward Classes. This resulted in a move against the Mandal Commission throughout the country.

c) The economic policy changed according to the change in the government. A new economic policy was implemented in the name of Structural Adjustment.

d) In 1992, the Babri Masjid was demolished as a result of many things. It was an incident that
threatened the unity and integration of the country. It happened because of the coming of BJP with its Hindutva agenda.

e) The Murder of Rajiv Gandhi (1991): This brought changes in the leadership of Congress. In the next election, Congress won the maximum seats.

Question 32.
Briefly explain the ideologies of Bharatiya Jana Sagh and Swathanthra Party.
Answer:
Bhartiva Jan Sanah:
It was founded in 1951. The important leaders were Shyama Prasad Mukherjee, DeenDayal Upadhyaya and Balraj Madhok. BJS is different from other political . parties because of their ideas and schemes.

1. They stress the ideology of one country, one culture and one nation.
2. For India’s progress and growth, they want to fall back upon its culture and traditions.
3. They want the reunification of India and Pakistan. Thus they have the concept of a greater, undivided India.
4. They organized many protests insisting that Hindi should be the only official language of India.
5. They do not want to give any kind of concessions to religious and cultural minorities.
6. They want India to have atomic weapons.

In the first two elections, the BJS could get only very few votes. In 1952 and 1957 they gained 3 arid 4 seats respectively. In the beginning their supporters were mainly from the Hindi-speaking States – Rajasthan, Madhya Pradesh, Delhi, and UP. The BJP originated from the BJS.

Swatantra Party
It was established in August 1959. Prominent leaders wereC. Rajagopalachari, K.M. Munshi, M.G Ranga and Minoo Masani. The Swatantra Party was formed in the light of the economic policy declared by the Nagpur Conference of the Congress. This Party was formed mainly to solve the economic problems of the country. The following were its policies:

1. The Government should not make economic controls.
2. Progress will come only through individual freedom.
3. Governmental interference in economic matters, Central Planning, Nationalization, Public Sector Undertakings etc. will be against national progress.
4. Encourage Private Sector.
5. In the agricultural sector, the Party is against land ceilings, and trading with government interference.
6. It is against increasing income tax and licensing of industries.
7. It is against Russia and the Non-Aligned Movement.
8. It wanted to establish more cordial relations with America.

As the interests of many groups were voiced by this Party, it received some strength. Many rulers of various Principalities, landowners, industrialists and businessmen supported this Party. They were not in favour of nationalization and land reforms. But because of its narrow social base and lack of sincere efforts from its members, it never grew into a popular party.

Part – V

Answer any 2 questions from 33 to 35. Each carries 8 scores. (2 × 8 = 16)

Question 33.
Explain the causes for the declaration of National Emergency in 1975.
Hints:
• Economic Context
• Gujarat and Bihar Movements
• Conflict with Judiciary
• Declaration of Emergency
Answer:
Afterthe 1971 election, Indira became a popular leader with a lot of support from the people. This time there were serious problems in the Party. There were three main reasons:
a) Economic Reasons
b) Gujarat & Bihar Movement
c) Dispute with the Judiciary

a) Economic Reasons:
The main slogan in the 1971 election was ‘garibi hatao’. But when the government came to power it could not improve the economic condition of the country. There were a number of reasons for that. First of all there was the refuge problem. Then there was the Bangladesh Crisis, followed by the Indo-Pakjstan War. All these things created financial problems. Secondly, afterthe War, America stopped its aids io India. Thirdly, there was a sharp increase in oil prices. The 4th reason was inflation which made the life of ordinary people very difficult.

The 5th problem was negative growth in industrial output. Unemployment increased, especially in the rural sector. Sixthly, the salaries of government employees had to be reduced and even stopped. Seventhly, lack of rain caused serious shortfall in foodstuff. All the above things created an economic crisis in the country. There was general discontentment in the country. This gave the Opposition Parties an opportunity to organise protests.

b) Gujarat and Bihar Movement:
The second biggest problem was the Students’ protest in Gujarat and Bihar which were Congress-ruled States. The main reason was the increase in the prices of essential commodities. Shortage of food, unemployment and corruption made the students angry. In both these States the Opposition Parties supported the students. In Gujarat, Presidential Rule was imposed. At this time the main opponent of Indira Gandhi and the leader of Congress (O), Morarji Desai, decided to go on an indefinite hunger strike. He did that for demanding elections in Gujarat.

In June 1975, because of heavy pressure from various sources, election was conducted. Congress lost the election. In Bihar the students invited Jay Prakash Narayan to lead their protest. He accepted the invitation insisting that the protest must be non-violent. He asked for the dismissal of the Bihar government. He argued that there was a need for a revolution in social, economic and political spheres. But the Bihar government refused to resign. The entire country discussed the issue. Jay Prakash Narayan wanted to spread the protest to all parts of the country.

In the meantime the railway workers went on a strike. It would make the entire country come to a stand-still. In 1975, Jay Prakash Narayan organized a march to the Parliament. It was the biggest rally tRe capital had ever seen. The Opposition saw in him an alternative to Indira Gandhi. Both the protests were anti-Congress. Voices also rose against the leadership of Mrs. Gandhi. She believed that all this was done to take revenge on her.

c) Dispute with the Judiciary : Another reason for the declaration of Emergency was Indira Gandhi’s dispute with the Judiciary. The Supreme Court said that some of the things the government did were against the Constitution. Congress argued that the Supreme Court judgement was against democracy and the authority of the Parliament. The Party said that the Court was standing against some welfare measures taken to help the poor people.

The dispute was mainly in three things. Firstly, Can the Parliament change the Fundamental Rights? The Court said no. Secondly, Can the Parliament change ownership of land? Again the Court said no. Thirdly the Parliament said that it had the right to reduce fundamental rights. It amended the Constitution. But the Supreme Court objected. All these were the reasons for the dispute between the Government and the Supreme Court.

There were two more reasons. In the Kesavananda Bharati case, the Supreme Court judged that the Parliament can’t change the basic structure of the Constitution. Soon the post of the Chief Justice became vacant. Normally the senior most judge is appointed as the Chief Justice. But keeping aside 3 eligible Judges, the government appointed A.N. Roy as the Chief Justice. This appointment became controversial. Besides, the Uttar Pradesh High Court declared the election of Indira Gandhi as ni^ll and void. All these were the reasons for Mrs Gandhi to declare Emergency in June 1975.

Question 34.
‘Afterthe end of Cold War, USA exercises hegemony on the world in many ways”. Explain the three dimensions of US hegemony’ You can use the hints’
• Hegemony as Hard Power
Answer:
World nations try to gain and maintain dominance over others by using military, economic and cultural power. During the Cold War the fight was between the Soviet Union and America. With the disintegration of the Soviet, Union America remains the only Superpower. Dominance or hegemony is attained through three things:
a) Hard Power
b) Structural Power
c) Soft Power

Hard Power: This includes military power and the relations between nations. Today America is in the forefront of military power. There is nobody to Challenge its military might. It has the capacity to reach any corner of the world any moment. They spend a major part of their budget to maintain this position. They spend huge sums of money for research and technological developments. It is technology that keeps America in the forefront With their military might they are even ready to police the world, and punish the culprits.

Structural power: This dominance is based on the economic structure. The global economic system relies on America. If America helps the global economic system, it is mainly for their benefits and profits. But America does a lot of good things for the world. For example, communication channels through the oceans. Merchant ships travel through sea routes and America has much authority on the water transport system. It is the American navy that keeps the sea-routes safe for ships.

The next is the Internet. In fact it was an American military project. It was started in 1950. Today the global network functions using satellites. Most of them belong to America. 28% of the world economy is controlled by America. 15% of the international trade is also done by them. In any economic sector, at least one of out of three biggest companies will be American. The world economic structure follows the Breton Woods style of America. The World Bank, I.M.F. and World Trade Organization etc. are examples of American supremacy in world business and finance.

Now comes another example – the MBA degree. It was America that made this course and the degree so popular. It was Americans who discovered that business is a profession that could be taught. The first Business School was established in Pennsylvania in 1881. Its name was Wharton School. Today in all Countries MBA has become a prestigious degree.

Soft Power: This is the ideological and cultural dominance. America has become the model for all other nations and they try to copy America. In weak countries, America is able to make the people like its culture. We all speak highly of the American life style and personal success. America is number one in the world. By using ‘soft power, and not force, America is able to achieve this dominance over the world.

Question 35.
China’s economic success since 1978 has made it an alternative centre of power’. Explain the growth of China as a centre of power.
Answer:
The new economic policy has helped China a lot to bring it out of the moribund economic state. Privatization in the agricultural sector made crop ‘ production increase considerably and the rural economic sector improved. Thus there was considerable increase in the rural industrial sector. Chinese economy became better as a result of the increase in industry and rural growth. In 2001, China entered the WTO and thus it was able to trade with outside countries. Its decision to work jointly with the world economic system will definitely affect its economic model. Although some of these things are against Communist ideologies, the Chinese development model has become the cornerstone of its development.

China became the fastest growing economy in the world . It is the development model in China’s planning that helped it to succeed. Heavy investment in the area of production made China grow. The biggest consumer market in the world today is China. In 2001, China became a member of the WTO. China has independent trade treaties with Australia, South Korea, ASEAN, Switzerland and Pakistan.