MCQ
Let $f:[0,1] \rightarrow R$ be an injective continuous function that satisifes the condition $-1 < f(0) < f(1) < 1$

Then, the number of functions $g:[-1,1] \rightarrow[0,1]$ such that $(g \circ f)(x)=x$ for all $x \in[0,1]$ is

  • A
    $0$
  • B
    $1$
  • C
    more than $1$, but finite
  • $infinite$

Answer

Correct option: D.
$infinite$
d
(d)

We have,,

$-1 < f(0) < f(1) < 1$

$g=[-1,1] \rightarrow[0,1]$

$gof(x)=x, \forall x \in[0,1]$

Only condition that $g(x)$ should satisfy for $g o f(x)=x, \forall x \in[0,1]$ is that $g(x)$ should attain all values in $[0,1]$ when range of $f(x)$ a subset of $(-1,1)$ is used as image for $g(x)$. Thus, there can be infinite such functions $g(x)$ with domain $[-1,1]$ and $\operatorname{range}[0,1]$

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