Question

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

• A. $f\left(x\right)=\frac{x}{\sqrt{1-x^2}}$
• B. $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$
• C. $x^2$
• D. x

Answer 1

Answer: (B)

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

Given:-

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

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

To find:-

$fog\left(x\right)=?$

$fog\left(x\right)$

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

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

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

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

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

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

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

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

Hence the correct answer is $\frac{x}{\sqrt{1+x^2}}$.