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## 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?
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. 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?
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 Question 4.
A graph representing the function f(x) is given in it is clear that f(9) = 2. (i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(a) f (0) = 9
(b) f (7) = 6
(c) f (2) = 6
(d) f(10) = 0 (ii) For what value of x is f(x) = 1 ?
When f(x) = 1 the value of x is 9.5

(iii) Describe the following
(i) Domain
(ii) Range.
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} (iv) What is the image of 6 under f?
The image of 6 under f is 5.

Question 5.
Let f (x) = 2x + 5. If x ≠ 0 then find
$$\frac{f(x+2)-f(2)}{x}$$
f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
= 2x + 4 + 5
= 2x + 9  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).
(i) f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1 (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 = $$\frac { 2 }{ 4 }$$ = $$\frac { 1 }{ 2 }$$ 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. Solution: 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 Question 8.
A function f is defined by f(x) = 3 – 2x. Find x such that f(x2) = (f (x))2.
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 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 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

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. Types of function

1. One – One function (injection) 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 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) 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 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. 5. Bijection A function f: A → B is both one – one and onto, then f is called a bijection from A to B. 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 A function f: A → B is called a constant function if the range of f contains only one element. 2. Identity function 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.