Question

Given $f\left(x\right)=\sqrt{\frac{8}{1-x}+\frac{8}{1+x}}$ and $g\left(x\right)=\frac{4}{f\left(\sin x\right)}+\frac{4}{f\left(\cos x\right)}$ then $g\left(x\right)$ is

• A. periodic with fundamental period $\frac{\pi }{2}$
• B. periodic with fundamental period $\pi$
• C. periodic with fundamental period $2\pi$
• D. aperiodic

Answer 1

The correct option is A periodic with fundamental period $\frac{\pi }{2}$

$f\left(x\right)=\sqrt{8\left(\frac{1+x+1-x}{1-x^2}\right)}$

$=\sqrt{\frac{16}{1-x^2}}$

$f\left(x\right)=\frac{4}{\sqrt{1-x^2}}$

Hence
$f\left(sin\right(x\left)\right)=4|sec\left(x\right)|$

$f\left(cos\right(x\left)\right)=4|cosec\left(x\right)|$

Therefore $g\left(x\right)$ will be

$g\left(x\right)=|sin\left(x\right)|+|cos\left(x\right)|$

Now for the period of $g\left(x\right)$

$g(x+T)=g\left(x\right)$

Where T is the period of $g\left(x\right)$

$g(\frac{\pi }{2}+x)=|cos\left(x\right)|+|-sin\left(x\right)|$ [ $\because |-x|=|x|$ ]

$g(\frac{\pi }{2}+x)=|cos\left(x\right)|+|sin\left(x\right)|$

$g(\frac{\pi }{2}+x)=g\left(x\right)$

Hence period is $\frac{\pi }{2}$