Question

For $xϵR-\{0,1\}$, 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 $\left(f_2^{\circ }{J}^{\circ }{f}_1\right)\left(x\right)=f_3\left(x\right)$ then $J\left(x\right)$ is equal to

• A. $f_3\left(x\right)$
• B. $f_1\left(x\right)$
• C. $f_2\left(x\right)$
• D. $\frac{1}{x}{f}_3\left(x\right)$

Answer 1

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

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

$(f_2\cdot J\cdot f_1)\left(x\right)=f_3\left(x\right)$
$f_2\cdot \left(J\right(f_1\left(x\right)\left)\right)=f_3\left(x\right)$

$f_2\cdot \left(J\left(\frac{1}{x}\right)\right)=\frac{1}{1-x}$
$1-J\left(\frac{1}{x}\right)=\frac{1}{1-x}$

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

Now $x\to \frac{1}{x}$

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