IIT-JEE 2011 · Paper 2 · Mathematics Q52

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MathematicsMultiple correct+4Question 52 of 60

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

f

is not invertible on

(0,1)

f\ne f^{-1}

(0,1)

and

f'(b)=\dfrac1{f'(0)}

f=f^{-1}

(0,1)

and

f'(b)=\dfrac1{f'(0)}

f^{-1}

is differentiable on

(0,1)

Answer and worked solution

Official answer: A

Monotonic. $f'(x)=\dfrac{b^2-1}{(1-bx)^2}<0$ , so $f$ is one-one on $(0,1)$ .

Range. $f(0^+)=b$ and $f(1^-)=-1$ , so the range is $(-1,b)$ , not $\mathbb R$ . As a map $(0,1)\to\mathbb R$ , $f$ is not onto, hence not invertible (A).

Other options.

(B) and (C) use $f'(0)$ , but 0 is not in the domain.
(D) $f^{-1}$ is defined on $(-1,b)$ , not on $(0,1)$ .

(A)

Topics:Composition and inverse functions Types of functions

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IIT-JEE 2011 · Paper 2, Mathematics question 52