Students can download Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x, y) |x,y; ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Answer:
X = {1,2,3,….}
Y = {1,2,3,….}
f = {(1,2) (2, 4) (3, 6) (4, 8) ….}
Domain = {1, 2, 3, 4 ….}
Co – Domain = {1, 2, 3, 4 ….}
Range = {2, 4, 6, 8 }
Yes this relation is a function.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation
R = {(x,f(x)) |x ∈ X, f(x) = x2 + 1}
is a function from X to N?
Answer:
f(x) = x2 + 1
f(3) = 32 + 1 = 9 + 1 = 10
f(4) = 42 + 1 = 16 + 1 = 17
f(6) = 62 + 1 = 36 + 1 = 37
f(8) = 82 + 1 = 64 + 1 = 65
yes, R is a function from X to N

Question 3.
Given the function f: x → x2 – 5x + 6, evaluate
(i) f(-1)
(ii) f(2a)
(iii) f(2)
(iv) f(x – 1)
Solution:
Give the function f: x → x2 – 5x + 6.
(i) f(-1) = (-1)2 – 5(1) + 6 = 1 + 5 + 6 = 12
(ii) f(2a) = (2a)2 – 5(2a) + 6 = 4a2 – 10a + 6
(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)2 – 5(x – 1) + 6
= x2 – 2x + 1 – 5x + 5 + 6
= x2 – 7x + 12

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 4.
A graph representing the function f(x) is given in it is clear that f(9) = 2.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 1

(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
Answer:
(a) f (0) = 9
(b) f (7) = 6
(c) f (2) = 6
(d) f(10) = 0

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

(ii) For what value of x is f(x) = 1 ?
Answer:
When f(x) = 1 the value of x is 9.5

(iii) Describe the following
(i) Domain
(ii) Range.
Answer:
Domain = {0, 1, 2, 3,… .10}
= {x / 0 < x < 10, x ∈ R}
Range = {0,1,2,3,4,5,6,7,8,9}
= {x / 0 < x < 9, x ∈ R}

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

(iv) What is the image of 6 under f?
Answer:
The image of 6 under f is 5.

Question 5.
Let f (x) = 2x + 5. If x ≠ 0 then find
\(\frac{f(x+2)-f(2)}{x}\)
Answer:
f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
= 2x + 4 + 5
= 2x + 9
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 2

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 6.
A function/is defined by f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
(ii) find x such that f(x) = 0.
(iii) find x such that/ (A:) = x.
(iv) find x such that fix) =/(l – x).
Answer:
(i) f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 3

(ii) f(x) = 0
2x – 3 = 0
2x = 3
x = \(\frac { 3 }{ 2 } \)

(iii) f(x) = x
2x – 3 = x
2x – x = 3
x = 3

(iv) f(1 – x) = 2(1 – x) – 3
= 2 – 2x – 3
= – 2x – 1
f(x) = f(1 – x)
2x – 3 = – 2x – 1
2x + 2x = 3 – 1
4x = 2
x = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 7.
square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 4
Solution:
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 44
After cutting squares we will get a cuboid,
length of the cuboid (l) = 24 – 2x
breadth of the cuboid (b) = 24 – 2x
height of the cuboid (h) = 2x
Volume of the box = Volume of the cuboid
V = (24 – 2x)(24 – 2x) (x)
= (24 – 2x)2 (x)
= (576 + 4x2 – 96x) x
= 576x + 4x3 – 96x2
V = 4x3 – 96x2 + 576x
V(x) = 4x3 – 96x2 + 576x

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 8.
A function f is defined by f(x) = 3 – 2x. Find x such that f(x2) = (f (x))2.
Answer:
f(x) = 3 – 2x
f(x2) = 3 – 2 (x2)
= 3 – 2x2
(f (x))2 = (3 – 2x)2
= 9 + 4x2 – 12x
But f(x2) = (f(x))2
3 – 2 x2 = 9 + 4x2 – 12x
-2x2 – 4x2 + 12x + 3 – 9 = 0
-6x2 + 12x – 6 = 0
(÷ by – 6) ⇒ x2 – 2x + 1 = 0
(x – 1) (x – 1) = 0
x – 1 = 0 or x – 1 = 0
x = 1
The value of x = 1

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.
Solution:
Speed = \(\frac{\text { distance covered }}{\text { time taken }}\)
⇒ distance = Speed × time
⇒ d = 500 × t [ ∵ time = t hrs]
⇒ d = 500 t

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b , where a, b are constants.
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Length ‘x’ of forehand (in cm) Height y (in inches)
35 56
45 65
50 69.5
55 74

Answer:
The relation is y = 0.9x + 24.5
(i) Yes the relation is a function.
(ii) When compare with y = ax + b
a = 0.9, b = 24.5
(iii) When the forehand length is 40 cm, then height is 60.5 inches.
Hint: y = 0.9x + 24.5
= 0.9 × 40 + 24.5
= 36 + 24.5
= 60.5 feet

(iv) When the height is 53.3 inches, her forehand length is 32 cm
Hint: y = 0.9x + 24.5
53.3 = 0.9x + 24.5
53.3 – 24.5 = 0.9 x
28.8 = 0.9 x
x = \(\frac { 28.8 }{ 0.9 } \)
x = 32 cm

Representation of functions

A function may be represented by

(a) Set of ordered pairs
(b) Table form
(c) Arrow diagram
(d) Graphical form

Vertical line test
A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Types of function

1. One – One function (injection)
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 5
A function f: A → B is called one-one function if distinct elements of A have a distinct image in B.

2. Many – One function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 6
A function f: A → B is called many-on function if two or more elements of A have same image in B.

3. Onto function (surjection)
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 7
A function f: A → B is said to be on to function if the range of f is equal to the co-domain of f.

4. Into function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 8
A function f: A → B is called an into function if there exists at least one element in B which is not the image of any element of A.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

5. Bijection
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 9
A function f: A → B is both one – one and onto, then f is called a bijection from A to B.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

Horizontal line test
A function represented in a graph is one – one, if every horizontal line intersects the curve in at most one point.

Special cases of function

1. Constant function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 10
A function f: A → B is called a constant function if the range of f contains only one element.

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3

2. Identity function
Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.3 11
A function f : A → A defined by f(x) = x for all x ∈ A is called an identity function on A and is denoted by IA.

3. Real valued function
A function f: A → B is called a real valued function if the range of f is a subset of the set of all real numbers R.