Question

If $f\left(x\right)=\frac{1-x}{1+x}$ then the value of $\left(fof\right)\left(x\right)$.

• A. $\frac{1-x}{1+x}$
• B. $x$
• C. $\frac{1}{x}$
• D. none of these

Answer 1

Answer: (B)

$x$

Given,

$f\left(x\right)=\frac{1-x}{1+x}$......(1)

Now,

$\left(fof\right)\left(x\right)$

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

$=f\left(\frac{1-x}{1+x}\right)$ [ Using (1)]

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

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

$=\frac{2x}{2}$

$=x$.