Let f : (0, 1) \rightarrow R be defined by f \left(\right. x \left.\right) = \frac{b - x}{1 - bx} where b is a constant such that 0 < b

Engineering

Mathematics

Inverse of a Function

Question

Let f : (0, 1) $\to$ R be defined by $f (x) = \frac{b - x}{1 - bx}$

where b is a constant such that 0 < b < 1. Then

f = f ^–1 on (0, 1) and $f' (b) = \frac{1}{f' (0)}$

f is not invertible on (0, 1)

f $\neq$ f ^–1 on (0, 1) and $f'(b)= \frac{1}{f' (0)}$

f ^–1 is differentiable on (0, 1)

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f : (0, 1) $\to$ R

$f (x) = \frac{b - x}{1 - bx} \forall b \in (0, 1)$

$f' (x) = \frac{b^{2} - 1}{{(1 - bx)}^{2}} (-) ve$

So f(x) is monotonically decreasing for x $\in$ (0, 1)

So for x $\in$ (0, 1)

f(x) $\in$ (–1, b)

so f(x) is not onto.

So f(x) is not invertible function.

The most appropriate answer to this question is (A, B), but because of ambiguity in language, IIt has declared (A) as correct answer.