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

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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?
Solution:
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1, 2, 3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1, 2, 3…..}
Range of R = {2, 4, 6, 8, 10,…}
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 ?
Solution:
x = {3,4, 6, 8}
R = ((x, f(x))|x ∈ X, f(x) = X² + 1}
f(x) = x² + 1
f(3) = 3² + 1 = 10
f(4) = 4² + 1 = 17
f(6) = 6² + 1 = 37
f(8) = 8² + 1 = 65

R = {(3, 10), (4, 17), (6, 37), (8, 65)}
R = {(3, 10), (4, 17), (6, 37), (8, 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(2 a)
(iii) f(2)
(iv) f(x – 1)
Answer:
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 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:
From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) At x = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) The image of 6 under f is 5.

Question 5.
Let f(x) = 2x + 5. If x ≠ 0 then find \(\frac{f(x+2)-f(2)}{x}\)
Solution:
Given f(x) = 2x + 5, x ≠ 0.

Question 6.
A function fis 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 f(x) = x.
(iv) find x such that f(x) = f(1 – x).
Solution:
Given f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = -1
∴ \(\frac{f(0)+f(1)}{2}=\frac{-3-1}{2}=\frac{-4}{2}\) = -2

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

Question 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 in figure. Express the volume V of the box as a function of x.

Solution:
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4x²)x
V = 4x³ – 96x² + 576x

Question 8.
A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.
Solution:
f(x) = 3 – 2x
f(x²) = 3 – 2x²

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 r in hours.
Answer:
Speed of the plane = 500 km/hr
Distance travelled in “t” hours
= 500 × t (distance = speed × time)
= 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.
Solution:
(i) Given y = ax + b …………. (1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function.

Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches
(iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
⇒ x = \(\frac{28.8}{0.9}\) = 32 cm
∴ When y = 53.3 inches, x = 32 cm