Question

If $f\left(x\right)=\frac{1-x}{1+x},x\ne 0,-1$ and $\alpha =f\left(f\right(x\left)\right)+f\left(f\right(1/x\left)\right),$ then

• A. $\alpha >2$
• B. $\alpha <-2$
• C. $|\alpha |>2$
• D. $\alpha =2$

Answer 1

The correct option is C $|\alpha |>2$

We have, $f\left(x\right)=\frac{1-x}{1+x}$
$\Rightarrow f\left(f\right(x\left)\right)=f\left(\frac{1-x}{1+x}\right)$
$=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}=x$
Again, $f\left(x\right)=\frac{1-x}{1+x}$
$\Rightarrow f\left(1/x\right)=\frac{1-\frac{1}{x}}{1+\frac{1}{x}}$
$=\frac{x-1}{x+1}$
$\Rightarrow f\left(f\right(1/x\left)\right)=f\left(\frac{x-1}{x+1}\right)$
$=\frac{1-\frac{x-1}{x+1}}{1+\frac{x-1}{x+1}}$
$=1/x$
Therefore, $\alpha =f\left(f\right(x\left)\right)+f\left(f\right(1/x\left)\right)$
$\Rightarrow \alpha =x+\frac{1}{x}$
$\Rightarrow |\alpha |=|x+\frac{1}{x}|\ge 2$
$\Rightarrow |\alpha |\ge 2$