## Tamilnadu State Board Class 10 Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1

Question 1.

If the ordered pairs (x^{2} – 3x,y^{2} + 4y) and (-2,5) are equal, then find x and y.

Solution:

(x^{2} – 3x, y^{2 }+ 4y) = (-2,5)

x^{2} – 3x = -2

x^{2} – 3x + 2 = 0

(x – 2)(x – 1) = 0

Question 2.

The Cartesian product A × A has 9 elements among which (-1,0) and (0,1) are found. Find the set A and the remaining elements of A×A.

Solution:

A = {-1,0, 1},B = {1,0,-1}

A × B = {(-1,1), (-1,0), (-1,-1), (0,1), (0, 0), (0,-1), (1,1), (1,0), (1,-1)}

Question 3.

Given that

(i) f(0)

(ii) f(3)

(iii) f(a + 1) in terms of a.(Given that a > 0)

Solution:

(i) f(0) = 4

(ii) f(3) = \(\sqrt { 3-1 }\) = \(\sqrt { 2 }\)

(iii) f(a+ 1) = \(\sqrt { a+1-1 }\) = \(\sqrt { a }\)

Question 4.

Let A = {9,1O,11,12,13,14,15,16,17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.

Solution:

A = {9,10,11, 12, 13, 14,15,16,17}

f: A → N

f(n) = the highest prime factor of n ∈ A

f = {(9, 3), (10, 5), (11, 11), (12, 3), (13,13), (14, 7), (15, 5), (16,2), (17, 17)}

Range = {3,5, 11, 13,7,2, 17}

= {2,3,5,7,11,13,17}

Question 5.

Find the domain of the function

Solution:

Question 6.

If f(x)= x^{2}, g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).

Solution:

f(x) = x^{2}

g(x) = 3x

h(x) = x – 2

(fog)oh = x – 2

LHS = fo(goh)

fog = f(g(x)) = f(3x) = (3x)^{2} = 9x^{2}

(fog)oh = (fog) h(x) = (fog) (x – 2)

= 9(x – 2)^{2} = 9(x^{2} – 4x + 4)

= 9x^{2} – 36x + 36 …..(1)

RHS = fo(goh)

(goh) = g(h(x)) = g(x- 2)

= 3(x – 2) = 3x – 6

fo(goh) = f(3x – 6) = (3x – 6)^{2}

= 9x^{2} – 36x + 36 ……(2)

(1) = (2)

LHS = RHS

(fog)oh = fo(goh) is proved.

Question 7.

A = {1, 2} and B = {1, 2, 3, 4} , C = {5, 6} and D = {5, 6, 7, 8} . Verify whether A × C is a subset of B × D?

Solution:

A = {1,2),B = (1,2,3,4)

C = {5,6},D = {5,6,7,8)

(A × C) ⊂ (B × D) It is proved.

Question 8.

If f(x) =\(\frac { x-1 }{ x+1 } \), x ≠ 1 show that f(f(x)) = – \(\frac { 1 }{ x } \), Provided x ≠ 0.

Solution:

Hence it is proved.

Question 9.

The function/and g are defined by f(x) = 6x + 8; g(x) = \(\frac { x-2 }{ 3 } \).

(i) Calculate the value of gg (\(\frac { 1 }{ 2 } \))

(ii) Write an expression for g f(x) in its simplest form.

Solution:

Question 10.

Write the domain of the following real functions

Solution:

(i) f(x) = \(\frac { 2x+1 }{ x-9 } \)

The denominator should not be zero as the function is a real function.

∴ The domain = R – {9}

(ii) p(x) = \(\frac{-5}{4 x^{2}+1}\)

The domain is R.

(iii) g(x) = \(\sqrt { x-2 }\)

The domain = (2, ∝)

(iv) h(x) = x + 6

The domain is R.