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Students can Download Chapter 1 Fundamentals of Computer Notes, Plus One Computer Application Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus One Computer Application Notes Chapter 1 Fundamentals of Computer

Data and Information
Data: It represents raw facts and figures. It may be a number, string(combination of characters), etc.
Eg: Alvis, 6, First
Information: It is meaningful and processed form of data
Eg: Alvis aged 6 years is in first standard.

Data Processing: Data processing is defined as a series of actions or operations that converts data into useful information.
Data processing phases (6)

1. Capturing data – In this step acquire or collect data from the user to input into the computer.
2. Input – It is the next step. In this step appropriate data is extracted and feed into the computer.
3. Storage – The data entered into the computer must be stored before starting the processing.
4. Processing/Manipulating data – It is a laborious work. It consists of various steps like computations, classification, comparison, summarization, etc. that converts input into output.
5. Output of information – In this stage, we will get the results as information after processing the data.
6. Distribution of information – In this phase the information(result) will be given to the concerned persons/computers.

Functional units of computer
A computer is not a single unit but it consists of many functional units(lntended to perform jobs) such as Input unit, Central Processing Unit(ALU and Control Unit), Storage (Memory) Unit and Output Unit.
1. Input Unit: Its aim is to supply data (Alphanumeric, image , audio, video, etc.) to the computer for processing. The Input devices are keyboard, mouse, scanner, mic, camera, etc

2. Central Processing Unit (CPU): It is the brain of the computer and consists of three components
Arithmetic Logic Unit(ALU) – As the name implies it performs all calculations and comparison operations.
Control Unit(CU) – It controls the overall functions of a computer.
Registers – It stores the intermediate results temporarily.

3. Storage Unit(Memory Unit): A computer has huge storage capacity. It is used to store data and instructions before starts processing. Secondly, it stores the intermediate results and thirdly it stores information(processed data), that is the final results before send to the output unit(Visual Display Unit, Printer, etc)

Two Types of the storage unit
i. Primary Storage alias Main Memory: It is further be classified into Two- Random Access Memory(RAM) and Read-Only Memory(ROM). The one and only memory that the CPU can directly access is the main memory at a very high speed. It is expensive hence storage Capacity is less. RAM is volatile (when the power is switched off the content will be erased) in nature but ROM is non-volatile (It is permanent)

ii. Secondary Storage alias Auxiliary Memory: Because of the limited storage capacity of primary memory its need arises. When a user saves a file, it will be stored in this memory hence it is permanent in nature and its, capacity is huge.
Eg: Hand Disc Drive(HDD), Compact Disc(CD), DVD, Pen Drive, Blu Ray Disc etc.

4. Output Unit: After processing the data we will get information as result, that will be given to the end user through the output unit in a human-readable form. Normally monitor and printer are used.

Computer: It is an electronic machine used to perform tasks based upon set of instructions and produce results at a high degree of accuracy and speed.
Advantages of computer:

1. Speed – It can perform operations at a high speed.
2. Accuracy – It produces result at a high degree of accuracy.
3. Diligence – Unlike human beings, a computer is free from monotony, tiredness, Jack of concentration etc. We know that it is an electronic machine. Hence it can work hours without making any errors.
4. Versatility – It is capable of performing many tasks. It is useful in many fields.
5. Power of Remembering – A computer consists of huge amount of memory. So it can store and recall any amount of information without delay. Unlike human beings it can store huge amount of data and can be retrieved whenever the need arises.

Disadvantages of computer

1. No. IQ: It has no intelligent quotient. Hence they are slaves and human beings are the masters. It can’t make its own decisions.
2. No feelings: Since they are machines they have no feelings and instincts. They can perform tasks based upon the instructions given by the humans (programmers).

Number System: It is a systematic way to represent numbers in different ways. Each number system has its own Base, that is a number and that number of symbols or digits used.
The various number systems are given below:
Most Significant Digit (MSD): The digit with the most weight is called MSD. MSD is also called Left Most Digit(LMD)

Least Significant Digit (LSD): The digit with the least weight is called LSD. LSD is also called Right Most Digit(RMD)
Eg:
1) 106 : Here MSD : 1 and LSD : 6
2) 345.78: Here MSD : 3 and LSD : 8
A Binary Digit is also called a bit.
The weight of each digit of a number can be represented by the power of its base.

Number conversions
In general, to convert a decimal number into another number system(binary or octal or hexadecimal) do the following
Divide the number successively by the base of the number system do you want to convert and write down the remainders from bottom to top.
To convert a decimal fraction into another number system .multiply the number by the base of the number system do you want to convert then integer part and fractional part are separated again multiply the fractional part by the base and do the steps repeatedly until the fractional part becomes zero. Finally, write down the integer part from top to bottom.

Decimal to Binary: Divide the number by the base 2 successively and write down the remainders from bottom to top.

Decimal fraction to binary
multiply the number by the base 2 then the integer part and fractional part are separated again multiply the fractional part by the base 2 and do the steps repeatedly until the fractional part becomes zero.
Finally, write down the integer part from top to bottom.

Decimal to Octal: Divide the number by the base 8 successively and write down the remainders from bottom to top.

Decimal fraction to octal
multiply the number by the base 8 then integer part and fractional part are separated again multiply the fractional part by the base 8 and do the steps repeatedly until the fractional part becomes zero.
Finally write down the integer part from top to bottom.

Decimal to Hexadecimal: Divide the number by the base 16 successively and write down the remainders from bottom to top.

Decimal fraction to hexadecimal
multiply the number by the base 16 then integer part and fractional part are separated again multiply- the fractional part by the base 16 and do the steps repeatedly until the fractional part becomes zero. Finally write down the integer part from top to bottom.

Converting a number from any number system into decimal: For this multiply each digit by its corresponding weight and sum it up.

Binary to decimal conversion: For this multiply each bit by its corresponding weight and sum it up. The weights are power of 2.

Octal to decimal conversion: For this multiply each bit by its corresponding weight and sum it up. The weights are the power of 8.

Hexadecimal to decimal conversion: For this multiply each bit by its corresponding weight and sum it up. The weights are the power of 16.

Octal to binary conversion: Convert each octal digit into its 3-bit binary equivalent.
Consider the following table

Hexadecimal to binary conversion: Convert each Hexadecimal digit into its 4-bit binary equivalent.
Consider the following table

Binary to octal conversion
Divide the binary number into groups of 3 bits starting from the right to left(But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part)if needed and write down the corresponding octal equivalent.
Eg: (10100110)₂= ()₈?
Insert one zero in the left side to form 3 bits group

Binary to Hexadecimal conversion
Divide the binary number into groups of 4 bits starting from the right to left (But in the fractional part start dividing from left to right). Insert necessary zeros in the left side(or right side in the case of fractional part) if needed and write down the corresponding Hexadecimal equivalent.
Eg: (100100110)₂ = ()₁₆?
Insert 3 zeros in the left side to form 4 bits group

Octal to Hexadecimal conversion
First, convert an octal number into binary, then convert this binary into hexadecimal
Eg: Convert (67)₈ = ( )₁₆
Step I. First, convert this number into the binary equivalent for this do the following:

Step II. Next, convert this number into the hexadecimal equivalent for this do the following.

So the answer is (67)₈ = (37)₁₆

Hexadecimal to octal conversion
First convert Hexadecimal to binary, then convert this binary into 0ctal
Eg: Convert (A1)₁₆ = ()₈?
Step I. First convert this number into the binary equivalent. For this do the following

Step II. Next convert this number into octal equivalent. For this do the following.
So the answer is (A1)₁₆ = (241)₈

Data representation: The data stored in the computer memory is in the form of binary.

Representation of integers
There are three ways to represent integers on computers. They are as follows:

1. Sign and Magnitude-Representation (SMR)
2. 1’s Complement Representation
3. 2’s Complement Representation

1) SMR: Normally a number has two parts sign and magnitude.
eg: Consider a number +5. Here + is the sign and 5 is the magnitude. In SMR the Most Significant Bit (MSB) js used to represent the sign. If MSB is 0 sign is +ve and MSB is 1 sign is -ve.
eg: If a computer has word size is 1 byte then +5 is represented as 0 0 0 0 0 1 0 1
-5 is represented as 1 0 0 0 0 1 0 1
Here MSB is used for a sign then the remaining 7 bits are used to represent magnitude. So we can represent 27 = 128 numbers. But there are negative and positive numbers. So 128 + 128 = 256 number. The numbers are 0 to +127 and 0 to -127. Here zero is repeated. So we can represent 256 – 1 = 255 numbers.

2) 1’s Complement Representation: To get the 1’s complement of a binary number, just replace every 0 with 1 and every 1 with 0. Negative numbers are represented using 1’s complement but +ve number has no 1’s complement.
eg: To find the 1 ’s complement of -21 + 21 = 0 0 0 1 0 1 0 1
To get the 1’s complement change all 0 to 1 and all 1 to 0.
-21 = 11101010
1’s complement of -21 is 1 1 1 0 1 0 1 0
Eg: 2) Find the 1’s complement of +21
Positive numbers are represented by using SMR.
+21 = 0 0 0 1 0 1 0 1 (No need to take the 1’s complement)

3) 2’s Complement Representation: To get the 2’s complement of a binary number, just add 1 to its 1’s complement +ve number has no 2’s complement.
eg:- To find the 2’s complement of -21
+21 = 0 0 0 1 0 1 0 1
First take the 1 ’s complement for this change all 1 to 0 and all 0 to 1
-21 = 1 1 1 0 1 0 1 0
Then add 1 = 1 1 1 0 1 0 1 1
2’s complement of -21 is 1 1 1 0 1 0 1 1

b) Representation of floating point numbers.
A real number consists of an integer part and fractional part and represent by using Exponent and Mantissa method. This method is also used to represent too big numbers as well as too small numbers.
Eg: .0000000000000000000000001 can be represented easily as 1 × 10-25. Here 1 is the Mantissa and -25 is the exponent.
A computer with 32 bit word length is used 24 bits for mantissa and the remaining 8 bits used to store exponent.
1. Representation of characters,
a) 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 the U.S. government. Each character in the keyboard has a unique number.
Eg: ASCII code of ‘a’ is 97, when you press ‘a’ on the keyboard, 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.

b) EBCDIC(Extended Binary Coded Decimal Interchange Code): It is an 8 bit code introduced by IBM(International Business Machine). 28 = 256 characters can be represented by using this.

c) ISCII(Indian Standard Code for Information Interchange): It uses 8 bits to represent data and introduced by the standardization committee and adopted by the Bureau of Indian Standards(BIS).

d) Unicode: The limitations to store more characters is solved by the introduction of Unicode. It uses 16 bits so 216 = 65536 characters (i.e, world’s all written language characters) can store by using this.

Representation of audio, image, and video: Various methods are used to represent an image, audio and video data. A file (image or audio or video) consists of two parts header information (such as file name, its size, format etc.) and image data(a compressed form of value pixels intensity). Image file formats are JPEG (Joint Picture Experts Group) Format, BMP (Bitmap), TIFF (Tagged Image File Format), GIF(Graphics Interchange Format), PNG (Portable Network Graphic).
Audio File formats are WAV,MP3,MIDI,AIFF etc.
Video File formats AVI, JPEG2, WMV etc.

Kerala State Board New Syllabus Plus One Maths Notes Chapter 10 Straight Lines.

Kerala Plus One Maths Notes Chapter 10 Straight Lines

I. Slope of Line
The slope of a line is the ‘tan’ of the angle the line makes with the positive direction of the x-axis. If θ is the angle then, slope = tan θ.

The slope of the x-axis is zero and that of the y-axis is not defined.

Parallel lines have the same slope.

The product of the slopes of perpendicular lines is -1.

The slope is positive if θ < 90°. The slope is negative if θ > 90°.

The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

If three points A, B, and C are collinear, then AB and BC have the same slope.

If m₁ and m₂ be slopes of two lines then, θ the angle between is given by tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\), 1 + m₁m₂ ≠ 0

II. Equation of a Line
Equation of x-axis is y = 0.

Equation of y-axis is x = 0.

The equation of a horizontal line is y = a. If ‘a’ is positive then the line is above the x-axis and if negative it will be below the x-axis.

The equation of a vertical line is x = a. If ‘a’ is positive then the line is to the right of the x-axis and if negative it will be to the left of the x-axis.

Point-slope form: y – y₁ = m(x – x₁), where ‘m’ is the slope and (x₁, y₁) is a point on the line.

Two-Point form:
y – y₁ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) (x – x₁) where (x₁, y₁) and(x₂, y₂) are two point on the line.

Slope intercept form:
1. y = mx + c, where ‘m’ is the slope and ‘c’ is the y-intercept.
2. y = m(x – d), where ‘m’ is the slope and ‘d’ is the x-intercept.

Intercept form: \(\frac{x}{a}+\frac{y}{b}=1\) = 1, where ‘a’ and ‘b‘ are x and y intercept respectively.

Normal form: x cos θ + y sin θ = p, where ‘p’ is the length of the normal from the origin to the line and ‘θ’ is the angle the normal makes with the positive direction of the x-axis.

General equation of a Line: ax + by + c = 0, where a, b and c are real constants.
1. Slope of the line ax + by + c = 0 is \(-\frac{a}{b}\)

2. Parallel lines differ in constant term, i.e; a line parallel to ax + by + c = 0 is ax + by + k = 0.

3. A line perpendicular to ax + by + c = 0 is bx – ay + k = 0.

4. The equation of the family of lines passing through the intersection of the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is of the form a₁x + b₁y + c₁ + k(a₂x + b₂y + c₂) = 0.

5. The perpendicular distance of a point (x1, y1) from the line ax + by + c = 0 is \(\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|\)

6. The distance between the parallel lines ax + by + c = 0 and ax + by + k = 0 is \(\left|\frac{c-k}{\sqrt{a^{2}+b^{2}}}\right|\)

7. Normal form of the equation ax + by + c = 0 is x cos θ + y sin θ = p;
Where cos θ = \(\pm \frac{a}{\sqrt{a^{2}+b^{2}}}\); sin θ = \(\pm \frac{b}{\sqrt{a^{2}+b^{2}}}\) and p = \(\pm \frac{c}{\sqrt{a^{2}+b^{2}}}\)

Proper choice of signs is made so that p should be positive.

III. Shifting of Origin
An equation corresponding to a set of points with reference to a system of coordinate axes by shifting the origin is shifted to a new point is called a translation of axes.

Let us take a point P (x, y) referred to the axes OX and OY. Let (h, k) be the coordinates of origin and P(X, Y) be the coordinate of P(x, y) with respect to the new axis. Then, the transformation relation between the old coordinates (x, y) and the new coordinates (X, Y) are given by X = x + h and Y = y + k.

Kerala State Board New Syllabus Plus One Maths Notes Chapter 9 Sequences and Series.

Kerala Plus One Maths Notes Chapter 9 Sequences and Series

I. Sequence and Series
A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3, ….., k}.
Generally denoted by a₁, a₂, …….., a_n, ………
Let a₁, a₂, …….., a_n, …….. be a sequence. Then the expression a₁ + a₂ + ……. + a_n + …….. is called the series associated with the given sequence.

II. Arithmetic Progression (AP)
A sequence a₁, a₂, ……, a_n, …….. is called an arithmetic sequence or arithmetic progression if a_n+1 = a₂ + d, n ∈ N, where a₁ is called the first term and the constant term d is called the common difference of the AP.

Standard form of an AP:
a, a + d, a + 2d, …….. where a is the first term and d is a common difference.

If a constant is added to each term of an AP, the resulting sequence is also an AP.

If a constant is subtracted to each term of an AP, the resulting sequence is also an AP.

If each term of an AP is multiplied by a constant k, the resulting sequence is also an AP. But the resulting AP will have a common difference kd.

If each term of an AP is divided by a constant k, the resulting sequence is also an AP. But the resulting AP will have a common difference \(\frac{d}{k}\).

nth term, a_n = a + (n – 1)d

Sum of n terms, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]

S_n = \(\frac{n}{2}\) [t₁ + t_n]

Arithmetic mean between a and b is \(\frac{a+b}{2}\)

III. Geometric Progression (GP):
A sequence a₁ + a₂ + ……… + a_n + …….. is called Geometric sequence or Geometric progression if \(\frac{a_{k+1}}{a_{k}}=r\), k ≥ 1, where a₁ is called the first term and the constant term r is called the common ratio of the AP.

Standard form of a GP:
a, ar, ar²,…… where a is the first term and r is a common difference.

nth term, t_n = ar^n-1

Sum of n terms,

Geometric mean between a and b is √ab

Arithmetic mean ≥ Geometric mean.

Infinite G.P, and its Sum G. P. of the form a + ar + ar² + ar³ + …… ∞ is called infinite G.P.
S_∞ = \(\frac{a}{1-r}\); |r| < 1

IV. Special Series

Kerala State Board New Syllabus Plus One Maths Notes Chapter 8 Binomial Theorem.

Kerala Plus One Maths Notes Chapter 8 Binomial Theorem

Binomial theorem gives the expansion of (a + b)n for a rational number ‘n’. In this Unit, we study the binomial theorem for positive integral indices only.

I. Binomial Theorem
The expansion of a binomial for any positive integral ‘n’ is given by the binomial theorem.

There are (n + 1) terms in the expansion of (a + b)n.

The sum of the indices of ‘a’ and ‘b’ in every term of the expansion is ‘n’.

The general term in the expansion is tr+1 = nCr an-r br

Middle term in the expansion:
1. If ‘n’ is even, \(\left(\frac{n}{2}+1\right)^{t h}\) term.

2. If ‘n’ is odd, \(\left(\frac{n+1}{2}\right)^{t h}\) and \(\left(\frac{n+1}{2}+1\right)^{b_{t}}\) term.

Kerala State Board New Syllabus Plus One Maths Notes Chapter 7 Permutation and Combinations.

Kerala Plus One Maths Notes Chapter 7 Permutation and Combinations

I. Fundamental Principle of Counting
If an event can occur in ‘m’ different ways, following which another event can occur in ‘n’ different ways, then the total number of occurrences of the events in the given order is m × n.

II. Permutation
A permutation is the arrangement of some or all of a number of different objects.

Factorial notation: The notation n! represents the product of first n natural numbers,
ie; n! = n(n – 1 )(n – 2) ….. 3.2.1
1. 1! = 1
2. 0! = 1

The number of permutation of ‘n’ different objects taken ‘r’ at a time, where the objects do not repeat is n(n – 1)(n – 2)……(n – r + 1) which is denoted by ⁿP_r.

The number of permutation of ‘n’ different objects taken ‘r’ at a time, where repetition is allowed is n^r.

Permutation when all the objects are not distinct.
1. The number of permutations of ‘n’ objects, where ‘p’ objects are of the same kind and rest all different = \(\frac{n !}{p !}\)

2. The number of permutations of ‘n’ objects, where ‘p1’ objects are of one kind, ‘p2’ objects are of the second kind, …….., ‘p_k‘ objects are of a kth kind and rest all different = \(\frac{n !}{p_{1} ! p_{2} ! \ldots p_{k} !}\)

III. Combinations
A combination is a selection of some or all of a number of different objects (the order of selection is not important). The number of selection of ‘n’ things taken ‘r’ at a time is ⁿC_r.

Kerala State Board New Syllabus Plus One Maths Notes Chapter 6 Linear Inequalities.

Kerala Plus One Maths Notes Chapter 6 Linear Inequalities

Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality. In this unit we study linear inequalities in one and two variables, their formation and solution graphically.

I. Linear Inequalities in One Variable
The solution of an inequality in one variable is a value of the variable ‘x’ which makes it a true statement.

Equal numbers can be added or subtracted from both sides of the inequation.

If we multiply or divide both sides of an inequation by a positive number, the inequality sign will not be changed.

If we multiply or divide both sides of an inequation by a negative number, the inequality sign will be reversed.

To represent x < a (or x > a) on a number line, put a circle on the number ‘a’ and a dark line to the left (or right) of the number ‘a’.

To represent x ≤ a (or x ≥ a) on a number line, put a dark circle on the number ‘a’ and a dark line to the left (or right) of the number ‘a’.

II. Linear Inequalities in two Variables
The region containing all the solutions of an inequality is called the solution region.

In order to identify the half-plane represented by inequality, it is just sufficient to take any point (a, b) [say point (0, 0)] not on the line and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half-plane and shade the region which contains the point, otherwise, the inequality represents that half-plane which does not contain the point within it.

If the inequality is of the type ax + by ≥ c or ax + by ≤ c, then the point on the line ax + by = c is also included in the solution. So draw a dark line in the solution region.

If the inequality is of the type ax + by > c or ax + by < c, then the point on the line ax + by = c are not to be included in the solution. So draw a broken or dotted line in the solution region.

Kerala State Board New Syllabus Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations.

Kerala Plus One Maths Notes Chapter 5 Complex Numbers and Quadratic Equations

we have studied linear equations in one and two variables and quadratic equations in one variable. We have seen that the equation x² + 1 = 0 has no real solution since the root of a negative number does not exist in a real number. So, we need to extend the real number system to a larger number system to accommodate such numbers.

I. Complex Numbers
A number of the form a + ib, where a and b are real numbers and i = √-1.
Usually, a complex number is denoted by z, a is the real part of z denoted by Re(z) and b is the imaginary part of z denoted by Im(z).

II. Algebra of Complex Numbers

Addition: Let z₁ = a + ib and z₂ = c + id be two complex numbers. Then the sum z₁ + z₂ is obtained by adding the real and imaginary parts.

1. z₁ + z₂ = z₂ + z₁, commutative.
2. z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃, associative.
3. 0 + i0 is the identity element.
4. -z is the inverse of z.

Multiplication: Let z₁ = a + ib and z₂ = c + id be two complex numbers.
Then the product z₁z₂ is defined as follows:
z₁z₂ = (ac – bd) + i(ad + bc).

1. z₁z₂ = z₂z₁, commutative.
2. z₁(z₂z₃) = (z₁z₂)z₃, associative.
3. 1 + i0 is the identity element.
4. \(\frac{1}{z}\) is the inverse of z.
5. z₁(z₂ + z₃) = z₁z₂ + z₁z₃, distributive law.

Power of ‘i’: i³ = -i, i⁴ = 1
In general i^4k = 1, i^4k+1 = i, i^4k+2 = -1, i^4k+3 = -i

Identities:

The Modulus and Conjugate of a complex number:
Consider a complex number z = a + ib . Then, the conjugate of z is denoted by \(\bar{z}\), defined as \(\bar{z}\) = a – ib and the modulus of z is denoted by |z|, defined as \(\sqrt{a^{2}+b^{2}}\).

Properties:

III. Representation of Complex Number

Argand Plane:

A complex number z = a + ib which corresponds to the ordered pair (a, b) can be represented geometrically as the unique point P(a, b) in the XY-plane, where the real part is taken along the x-axis and the imaginary part along the y-axis. Such a plane is called the Argand Plane or Complex plane.

Polar Form:

Let the point P represent the non-zero complex number z = x + iy. Let the directed line segment OP be of length ‘r’ and be the angle which OP makes with the positive direction of the x-axis. Then, P is determined by the unique ordered pair of a real number (r, θ) called polar coordinate of the point P, where x = r cos θ, y = r sin θ and therefore the polar form of z can be represented as z = r(cos θ + i sin θ).
The principle argument of z is value ‘θ’ such that -x ≤ θ ≤ π, denoted by arg z.
To find the principle argument, we find tan α = |\(\frac{y}{x}\)|, 0 ≤ α ≤ \(\frac{\pi}{2}\)

Kerala State Board New Syllabus Plus One Maths Notes Chapter 4 Principle of Mathematical Induction.

Kerala Plus One Maths Notes Chapter 4 Principle of Mathematical Induction

Induction means the generalization from a particular case or facts. In contrast to deductive reasoning, inductive depends on working with each case and developing a conjecture by observing incidences till we have observed each and every case. In algebra or in another discipline of mathematics, there are certain results or statements that are formulated in terms of n, where n is a positive integer. To such statements, the well-suited principle that is based on the specific technique is known as the principle of mathematical induction.

The Principle of Mathematical Induction:
Suppose there is a statement P(n) involving the natural number n such that

1. The statement is true for n = 1, i.e; P(1) is true, and

2. If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e; the truth of P(k) implies the truth of P(k+1). Then, P(n) is true for all natural numbers n.