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Chapter 1 Relations and Functions Ex 1.3 Solutions

Question - 11 : - Considerf: {1,2,3} –> {a, b, c} given by f (1) = a, f (2)=b and f (3)=c. Find f-1 andshow that (f-1)f-1=f.

Answer - 11 : -

f:{1,2, 3,} –> {a,b,c} so that f(1) = a, f(2) = b, f(3) = c
Now let X = {1,2,3}, Y = {a,b,c}
∴ f: X –> Y
∴ f-1: Y –> X such that f-1 (a)=1, f-1(b) = 2; f-1(c) = 3
Inverse of this function may be written as
(f-1)-1 : X –> Y such that
(f-1)-1 (1) = a, (f-1)-1 (2)= b, (f-1)-1 (3) = c
We also have f: X –> Y such that
f(1) = a,f(2) = b,f(3) = c => (f-1)-1 = f

Question - 12 : - Letf: X –> Y be an invertible function. Show that the inverse of f-1 isf, i.e., (f-1)-1 = f.

Answer - 12 : -

f:X —> Y is an invertible function
f is one-one and onto
=> g : Y –> X, where g is also one-one and onto such that
gof (x) = Ix and fog (y) = Iy => g = f-1
Now f-1 o (f-1)-1 = I
and fo[f-1o (f-1)-1] =fol
or (fof-1)-1 o (f-1)-1 =f
=> Io (f-1)-1 = f
=> (f-1)-1 = f

Question - 13 : - Iff: R –> R be given by f(x) = , then fof (x) is
(a) 
(b) x³
(c) x
(d) (3 – x³)

Answer - 13 : -

f:R-> R defined by f(x) = 
fof (x) = f[f(x)] = 
= 

= 
= x

Question - 14 : - Letf:  bea function defined as f (x) = .The inverse of f is the map g: 

Answer - 14 : - Range f–>  given by
(a) 
(b) 
(c) 
(d) 


Solution

Itis given that

Let y bean arbitrary element of Range f.

Then,there exists x ∈such that 
Let us define g:Rangeas
Now,
∴ 

Thus, g isthe inverse of f i.e., f−1 = g.

Hence,the inverse of f is the map g: Range, which is given by

The correct answer is B.

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