Question

Let $f:R\to R$ be defined by $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$, then $\left(fofof\right(x\left)\right)$ is

• A. $\frac{x}{\sqrt{1+x^2}}$
• B. $\frac{x}{\sqrt{1+3x^2}}$
• C. $x$
• D. $1$

Answer 1

The correct option is B $\frac{x}{\sqrt{1+3x^2}}$

Given, $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$
$\left(fof\right)x=\left\{f\right(f\left(x\right)\left)\right\}$
$=f\left(\frac{x}{\sqrt{1+x^2}}\right)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}$
$\Rightarrow \left(fof\right)x=\frac{x}{\sqrt{1+2x^2}}$
Now, $\left(fofof\right(x\left)\right)=f\left\{\right(fof\left)x\right\}x=f\left(\frac{x}{\sqrt{1+2x^1}}\right)$
$=\frac{\frac{x}{\sqrt{1+2x^2}}}{\sqrt{1+\frac{x^2}{1+2x^2}}}=\frac{x}{\sqrt{1+3x^2}}$.