Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Let f : (0, 1) → R be defined by f(x) = 
, where b is a constant such that 0 < b < 1. Then
A
f is not invertible on (0, 1)
B
f ≠ f –1 on (0, 1) and f ′ = 
= 
=
C
f = f –1 on (0, 1) and f ′
D
f –1 is differentiable on (0, 1)
Text Solution
Verified by ExpertsThe correct answer is:
A
Sol. f(x) = 
= 
+ 
f ′ (x) = 
b, f ′ (x) < 0 ∀ x (0, 1)
Range of f(x) is (–1, b)
so range ≠ co-domain
so f is not invertible
f –1 doesnot exist
No comparison with f –1
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