# Relations and Functions Miscellaneous Exercise Solutions – NCERT Class 11 Mathematics Chapter 2

Question 1

Question 2

Question 3

Question 4

5. Find the domain and the range of the real function f defined by f (x) = |x –1|
We know that absolute or modulus of a function is defined for all of Real values of x
Hence Domain of f(x) = |x – 1| is Real Numbers R

As f(x) = |x – 1| can only exist for non-negative values.
Hence
Range of f(x) = |x – 1| = R+ ∪ {0}

Question 6

Question 7

8. Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b.
Determine a, b

9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Are the following true?
(i) (a,a) ∈ R, for all a ∈ N

We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}

If a = 1 then a2 = 1
If a = 2 then a2 = 4
As R = {(a, b) : a, b ∈ N and a = b2} is not True for all possibles of a, b ∈ N.
Thus R is not a relation.

(ii) (a,b) ∈ R, implies (b,a) ∈ R
We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}
(a,b) ∈ R, implies (b,a) ∈ R
It is alos false as if a = b2
but b = a2 is not True
For example: 4 = 22 i which means (2, 4) ∈ R
But 2 ≠ 42 which means (4, 2) R
Hence R is not a relation.

(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R
We are given that
R : N → N defined as R = {(a, b) : a, b ∈ N and a = b2}
It is false as a = b2 and b = c2
Which means a = (c2)2 = c4
a = c4 which means a ≠ c2
Hence R is not a relation.

10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?
Let’s first find out what’s A ✕ B = ?
A ✕ B = {1, 2, 3, 4} ✕ {1, 5, 9, 11, 15, 16}
A ✕ B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

(i) f is a relation from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
As f contains corresponding ordered pair where values from set A are mapped to values from set B
Hence it’s a Relation from A to B.

(ii) f is a function from A to B
f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
As order pairs (2, 9) and (2, 11) have same value 2 from set A
Hence f is not a Function

11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

12. Let A = {9,10,11,12,13} and let f : A→N be defined by f (n) = the highest prime factor of n.
Find the range of f.