# 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**

**distribute over *? Justify your answer.**

**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.