Question

Given $f\left(x\right)=\frac{1}{[1-x]}$, $g\left(x\right)=f(f(x\left)\right)$ and $h\left(x\right)=f(f(f\left(x\right)\left)\right)$. Then the value of $f\left(x\right).g\left(x\right).h\left(x\right)$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A

$-1$

Explanation for the correct option:

Find the value of $f\left(x\right).g\left(x\right).h\left(x\right)$:

$g\left(x\right)=f(f(x\left)\right)$

$g\left(x\right)=\frac{1}{1–\left(\frac{1}{(1–x)}\right)}$

$=\frac{1–x}{-x}$

$=\frac{x–1}{x}$

$h\left(x\right)=f(f(f\left(x\right)\left)\right)$

$=\frac{1}{1-\left({\displaystyle \frac{x-1}{x}}\right)}$

$=\frac{x}{x-x+1}$

$=x$

$\therefore f\left(x\right).g\left(x\right).h\left(x\right)=\left(\frac{1}{1-x}\right)\times \left(\frac{x-1}{x}\right)\times \left(x\right)$

$=\left(-\frac{1}{x-1}\right)\times \left(\frac{x-1}{x}\right)\times \left(x\right)$

$\mathbf{=}\mathbf{}\mathbf{-}\mathbf{1}$

Hence, Option ‘A’ is Correct.