# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Miscellaneous Exercise

1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f =
f o g = IR.

Solution:
Firstly, Find the inverse of f.
Let say, g is inverse of f and
y = f(x) = 10x + 7
y = 10x + 7
or x = (y-7)/10
or g(y) = (y-7)/10; where g : Y → N
Now, gof = g(f(x)) = g(10x + 7) 2. Let f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is
even. Show that f is invertible. Find the inverse of f. Here, W is the set of all
whole numbers.

Solution:
f : W → W be defined as f(n) = n – 1, if n is odd and f(n) = n + 1, if n is
even.
Function can be defined as: f is invertible, if f is one-one and onto.
For one-one:
There are 3 cases:
for any n and m two real numbers:
Case 1: n and m : both are odd
f(n) = n + 1
f(m) = m + 1
If f(n) = f(m)
=> n + 1 = m + 1
=> n = m
Case 2: n and m : both are even
f(n) = n – 1
f(m) = m – 1
If f(n) = f(m)
=> n – 1 = m – 1
=> n = m
Case 3: n is odd and m is even
f(n) = n + 1
f(m) = m – 1
If f(n) = f(m)
=> n + 1 = m – 1
=> m – n = 2 (not true, because Even – Odd ≠ Even )
Therefore, f is one-one

Check for onto: Say f(n) = y, and y ∈ W
Case 1: if n = odd
f(n) = n – 1
n = y + 1
Which show, if n is odd, y is even number.
Case 2: If n is even
f(n) = n + 1
y = n + 1
or n = y – 1
If n is even, then y is odd.
In any of the cases y and n are whole numbers.
This shows, f is onto.
Again, For inverse of f
f
-1
: y = n – 1
or n = y + 1 and y = n + 1
n = y – 1 Therefore, f-1
(y ) = y. This show inverse of f is f itself.     F-1 = {(3, a), (2, b), (1, c)}
(ii) F = {(a, 2), (b, 1), (c, 1)}
Since element b and c have the same image 1 i.e. (b, 1), (c, 1).
Therefore, F is not one-one function.

12. Consider the binary operations * : R × R → R and o : R × R → R defined as a * b = |a
– b| and a o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is
associative but not commutative. Further, show that ∀ a, b, c ∈ R, a * (b o c) = (a * b) o (a
* c). [If it is so, we say that the operation * distributes over the operation o]. Does o
Solution:
Step 1: Check for commutative and associative for operation *.
a * b = |a – b| and b * a = |b – a| = (a, b)
Operation * is commutative.
a*(b*c) = a*|b-c| = |a-(b-c)| = |a-b+c| and
(a*b)*c = |a-b|*c = |a-b-c|
Therefore, a*(b*c) ≠ (a*b)*c
Operation * is associative.
Step 2: Check for commutative and associative for operation o.
aob = a ∀ a, b ∈ R and boa = b
This implies aob boa
Operation o is not commutative.
Again, a o (b o c) = a o b = a and (aob)oc = aoc = a
Here ao(boc) = (aob)oc
Operation o is associative.    