EX. NO. 1.3, RELATIONS AND FUNCTIONS, 10TH MATHS NEW SYLLABUS, TAMIL NADU, TN,
1. Let f = {(x, y)  x, y ∈ N and y = 2x} be a relation on ℕ. Find the domain, codomain and range. Is this relation a function?
Solution :
Since x and y ∈ N,
x = 1
y = 2(1)
y = 2

x = 2
y = 2(2)
y = 4

x = 3
y = 2(3)
y = 6

x = 4
y = 2(4)
y = 8

f = {(1, 2) (2, 4) (3, 6) (4, 8)................}
For each values of x, we get different values of y. So the given relation is a function.
Domain is the set of values of x
Domain = {1, 2, 3, 4, ............}
Co domain is the set of value of y. Since y ∈ N
Co domain = {1, 2, 3, 4, .............}
Range means the set of values of y, which are associated with x.
Range = {2, 4, 6, 8, .....}
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 ?
Solution :
Given that :
f (x) = x2 + 1
x ∈ X
if x = 3
f(3) = 32+1
f(3) = 10

if x = 4
f(4) = 42+1
f(4) = 17

if x = 6
f(6) = 62+1
f(6) = 37

if x = 8
f(8) = 82+1
f(8) = 65

R = { (3, 10) (4, 17) (6, 37) (8, 65) }
For each values of x, we get different values of f(x).
Hence it is a function
3. Given the function f : x > x2 −5x + 6 , evaluate
(i) f (1)
(ii) f (2a)
(iii) f (2)
(iv) f (x −1)
Solution :
Given that :
f(x) = x2 −5x + 6
(i) f (1)
here we have 1 instead of x.
f(1) = (1)2 −5(1) + 6
= 1 + 5 + 6
f(1) = 12
(ii) f (2a)
here we have 2a instead of x.
f(2a) = (2a)2 −5(2a) + 6
= 4a2  10a + 6
(iii) f (2)
here we have 2 instead of x.
f(2) = (2)2 −5(2) + 6
= 4  10 + 6
= 0
(iv) f (x −1)
here we have x  1 instead of x.
f(x1) = (x1)2 −5(x1) + 6
= x2  2x + 1  5x + 5 + 6
= x2  7x + 12
4. A graph representing the function f (x) is given figure
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)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following (i) Domain (ii) Range.
(iv) What is the image of 6 under f ?
Solution :
(i) (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 ?
For x = 9.5, we get 1.
(iii) Describe the following (i) Domain (ii) Range.
Domain = { 0 ≤ x ≤ 10 }
Range = { 0 ≤ x ≤ 9 }
(iv) What is the image of 6 under f ?
The image of 6 under f is 5.
5. Let f (x) = 2x + 5. If x ≠ 0 then find [f(x + 2)  f(2)] / x
Solution :
f (x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
f(x + 2) = 2x + 4 + 5 (1)
f(x + 2) = 2x + 9
f(2) = 2(2) + 5
= 4 + 5
f(2) = 9 (2)
(1)  (2) = 2x + 9  9
= 2x
[f(x + 2)  f(2) ]/x = 2x / x = 2
6. A function f is defined by f (x) = 2x – 3
(i) find [f(0) + f(1)]/2
(ii) find x such that f (x) = 0.
(iii) find x such that f (x) = x .
(iv) find x such that f (x) = f (1−x) .
Solution :
f(x) = 2x – 3
(i) [f(0) + f(1)]/2
f(0) = 2(0)  3 = 3
f(1) = 2(1)  3 = 1
[f(0) + f(1)]/2 = [3 + (1)] / 2
= 4/2
= 2
Hence the answer is 2.
(ii) find x such that f (x) = 0.
f(x) = 2x – 3
2x  3 = 0
2x = 3
x = 3/2
Hence the answer is 3/2.
(iii) find x such that f (x) = x .
f(x) = 2x – 3
2x  3 = x
2x  x = 3
x = 3
Hence the answer is 3
(iv) find x such that f (x) = f (1−x) .
f(1  x) = 2(1 x)  3
= 2  2x  3
= 2x  1
2x  3 = 2x  1
2x + 2x =  1 + 3
4x = 2
x = 2/4 = 1/2
Hence the answer is 1/2.
7. An open box is to be made from a 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 :
Since the original shape is square, length of all sides will be equal.
length = width = 24  2x
height = x
Volume of cuboid = length ⋅ width ⋅ height
= (24  2x) ⋅ (24  2x) ⋅ x
= x(24  2x)2
= x [242  2(24) (2x) + (2x)2]
= x [576  96x + 4x2]
V (x) = 4x3  96x2 + 576x
Hence the volume of the cuboid is 4x3  96x2 + 576x.
8. A function f is defined by f (x) = 3−2x . Find x such that f (x2) = (f (x))2 .
Solution :
Given that :
f (x) = 3 − 2x
f (x2) = 3 − 2x2 (1)
(f (x))2 = (3 − 2x)2
= 32  2(3)(2x) + (2x)2
(f (x))2 = 9  12x + 4x2 (2)
(1) = (2)
3 − 2x2 = 9  12x + 4x2
4x2 + 2x2  12x + 9  3 = 0
6x2  12x + 6 = 0
x2  2x + 1 = 0
(x  1)2 = 0
x  1 = 0 (or) x  1 = 0
x = 1 (or) x = 1
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 :
Time = Distance / speed
Speed of plane = 500 km per hour
Distance travelled = d
time taken = t in hours
t = d/500
d = 500t
Hence the required distance is 500 t.
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.
Length 'x' of forehand (in cm)

Height 'y' (in inches)

35

56

45

65

50

69.5

55

74

(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.
Solution :
y = ax + b
(i) For every values of x, we get different values of y.Hence it is a function.
(ii)
By solving these two equation, we get a and b
(1)  (2)
(45.5 a + b)  (35a + b) = 65.5  56
45.5 a  35a + b  b = 9.5
10.5a = 9.5
a = 9.5/10.5
a = 0.90
Substitute a = 0.90 in (1),
45.5(0.90) + b = 65.5
b = 65.5  40.95
b = 24.5
y = 0.9x + 24.5
(iii) Find the height of a woman whose forehand length is 40 cm.
y = ? if x = 40
y = 0.9(40) + 24.5
y = 60.5
Height of woman is 60.5 inches.
(iv) x = ? if y = 53.3
53.3 = 0.9x + 24.5
53.3  24.5 = 0.9x
28.8/0.9 = x
x = 32
Length of forehand is 32 cm.