Question

For $x\in R-\{0,1\},$ let $f_1\left(x\right)=\frac{1}{x}$, $f_2\left(x\right)=1-x$ and $f_3\left(x\right)=\frac{1}{1-x}$ be three given functions. If a function, $J\left(x\right)$ satisfies $(f_2\circ J\circ f_1)\left(x\right)=f_3\left(x\right)$ then $J\left(x\right)$ is equal to :

• A. $f_1\left(x\right)$
• B. $f_2\left(x\right)$
• C. $f_3\left(x\right)$
• D. $\frac{1}{x}\cdot f_3\left(x\right)$

Answer 1

The correct option is C $f_3\left(x\right)$

Given :
$f_1\left(x\right)=\frac{1}{x}$, $f_2\left(x\right)=1-x$ and $f_3\left(x\right)=\frac{1}{1-x}$
$(f_2\circ J\circ f_1)\left(x\right)=f_3\left(x\right)$.........(1)

Substituting the values of $f_1\left(x\right)$ and $f_3\left(x\right)$ in equation (1), we will get
$(f_2\circ J)\left(\frac{1}{x}\right)=\frac{1}{1-x}$ $\therefore \left(f_2\right(J\left(\frac{1}{x}\right))=\frac{1}{1-x}$

Substituting $J\left(\frac{1}{x}\right)$ in the function $f_2\left(x\right)$ we will get,
$1-J\left(\frac{1}{x}\right)=\frac{1}{1-x}$
$\Rightarrow 1-\frac{1}{1-x}=J\left(\frac{1}{x}\right)$
$\Rightarrow 1-\frac{\frac{1}{x}}{\frac{1}{x}-1}=J\left(\frac{1}{x}\right)$

Putting $\frac{1}{x}=t$,
$\Rightarrow 1-\frac{t}{t-1}=J\left(t\right)$
$\Rightarrow 1-\frac{t}{t-1}=J\left(t\right)$
$\Rightarrow \frac{t-1-t}{t-1}=J\left(t\right)$
$\Rightarrow \frac{-1}{t-1}=J\left(t\right)$ $\therefore J\left(t\right)=\frac{1}{1-t}$

So , $J\left(x\right)=\frac{1}{1-x}=f_3\left(x\right)$