for x
Question:

For $x \in \mathbf{R}-\{0,1\}$, let $f_{1}(x)=\frac{1}{x}, f_{2}(x)=1-x$ and $f_{3}(x)$

$=\frac{1}{1-x}$ be three given functions. If a function, $\mathrm{J}(x)$

satisfies $\left(f_{2} o J o f_{1}\right)(x)=f_{3}(x)$ then $J(x)$ is equal to:

  1. (1) $f_{3}(x)$

  2. (2) $\frac{1}{x} f_{3}(x)$

  3. (3) $f_{2}(x)$

  4. (4) $f_{1}(x)$


Correct Option: 1

Solution:

The given relation is

$\left(f_{2} o J o f_{1}\right)(x)=f_{3}(x)=\frac{1}{1-x}$

$\Rightarrow\left(f_{2} o J\right)\left(f_{1}(x)\right)=\frac{1}{1-x}$

$\Rightarrow\left(f_{2} o J\right)\left(\frac{1}{x}\right)=\frac{1}{1-\frac{1}{\frac{1}{x}}}=\frac{\frac{1}{x}}{\frac{1}{x}-1} \quad\left[\because f_{1}(x)=\frac{1}{x}\right]$

$\Rightarrow\left(f_{2} o J\right)(x)=\frac{x}{x-1}$ $\left[\frac{1}{x}\right.$ is replaced by $\left.x\right]$

$\Rightarrow f_{2}(J(x))=\frac{x}{x-1}$

$\Rightarrow 1-J(x)=\frac{x}{x-1}$ $\left[\because f_{2}(x)=1-x\right]$

$\therefore J(x)=1-\frac{x}{x-1}=\frac{1}{1-x}=f_{3}(x)$

 

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