Question

If $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$, then $fofof\left(x\right)=$

• A. $\frac{3x}{\sqrt{1+x^2}}$
• B. $\frac{x}{\sqrt{1+3x^2}}$
• C. $\frac{3x}{\sqrt{1+3x^2}}$
• D. None of these

Answer 1

The correct option is B

$\frac{x}{\sqrt{1+3x^2}}$

Explanation for the correct option:

Finding the value of $fofof\left(x\right)$:

Given that,

$f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$

Now,

$\begin{array}{rcl}fofof\left(x\right)& =& fof\left(\frac{x}{\sqrt{1+x^2}}\right)\\ & =& f\left(\frac{{\displaystyle \frac{x}{\sqrt{1+x^2}}}}{\sqrt{1+{\displaystyle \frac{x^2}{1+x^2}}}}\right)\\ & =& f\left(\frac{{\displaystyle \frac{x}{\sqrt{1+x^2}}}}{\sqrt{{\displaystyle \frac{2x^2+1}{1+x^2}}}}\right)\\ & =& f\left(\frac{{\displaystyle x}}{\sqrt{{\displaystyle 2x^2+1}}}\right)\\ & =& \frac{{\displaystyle \frac{x}{\sqrt{2x^2+1}}}}{\sqrt{1+\left({\displaystyle \frac{x^2}{2x^2+1}}\right)}}\\ & =& \frac{{\displaystyle \frac{x}{\sqrt{2x^2+1}}}}{\sqrt{{\displaystyle \frac{3x^2+1}{2x^2+1}}}}\\ & =& \frac{x}{\sqrt{3x^2+1}}\end{array}$

Hence, the correct option is B.