QJEEAdv11Ma4857: JEE Advanced 2011 Mathematics - Class 11 JEEAdv11Ma4857 Solution: Let be defined by , where b is a constant such that . Then

Question

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

Accepted Answer

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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