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IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

Let $$f:(0,1) \to R$$ be defined by $$f(x) = {{b - x} \over {1 - bx}}$$, where be is a constant such that $$0 < b < 1$$. Then

A
f is not invertible on (0, 1).
B
f $$\ne$$ f$$-$$1 on (0, 1) and $$f'(b) = {1 \over {f'(0)}}$$.
C
f = f$$-$$1 on (0, 1) and $$f'(b) = {1 \over {f'(0)}}$$.
D
f$$-$$1 is differentiable on (0, 1).

Explanation

Here, $$f(x) = {{b - x} \over {1 - bx}}$$

where, 0 < b < 1, 0 < x < 1

For function to be invertible it should be one-one onto.

$$\therefore$$ Check range :

Let $$f(x) = y \Rightarrow y = {{b - x} \over {1 - bx}}$$

$$ \Rightarrow y - bxy = b - x \Rightarrow x(1 - by) = b - y$$

$$ \Rightarrow x = {{b - y} \over {1 - by}}$$

where, 0 < x < 1

$$\therefore$$ $$0 < {{b - y} \over {1 - by}} < 1$$

$${{b - y} \over {1 - by}} > 0$$ and $${{b - y} \over {1 - by}} < 1$$

$$ \Rightarrow y < b$$ or $$y > {1 \over b}$$ .... (i)

$${{(b - 1)(y + 1)} \over {1 - by}} < - 1 < y < {1 \over b}$$ ..... (ii)

From Eqs. (i) and (ii), we get

$$y \in \left( { - 1,{1 \over b}} \right) \subset $$ Codomain

Thus, f(x) is not invertible.

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