Q88 / 200 · Manipal (MET) 2014 · Mathematics
Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$, then $(f \circ f \circ f)(x)$ is
- A$\frac{x}{\sqrt{1+x^2}}$
- Correct answerB$\frac{x}{\sqrt{1+3 x^2}}$
- C$x$
- D1
Solution
Given, $f(x)=\frac{x}{\sqrt{1+x^2}}$
$\begin{aligned}
(f \circ f) x & =\{f(f(x))\} \\
& =f\left(\frac{x}{\sqrt{1+x^2}}\right)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}
\end{aligned}$
$\Rightarrow \quad(\text { fof }) x=\frac{x}{\sqrt{1+2 x^2}}$
Now, $(fofo) (x)=f\{(f \circ f) x\} x=f\left(\frac{x}{\sqrt{1+2 x^2}}\right)$ $=\frac{\frac{x}{\sqrt{1+2 x^2}}}{\sqrt{1+\frac{x^2}{1+2 x^2}}}=\frac{x}{\sqrt{1+3 x^2}}$
Now, $(fofo) (x)=f\{(f \circ f) x\} x=f\left(\frac{x}{\sqrt{1+2 x^2}}\right)$ $=\frac{\frac{x}{\sqrt{1+2 x^2}}}{\sqrt{1+\frac{x^2}{1+2 x^2}}}=\frac{x}{\sqrt{1+3 x^2}}$
Hence, the correct answer is option: 'B'.