Question

If $f\left(x\right)=p\left(x\right)q\left(x\right)$ where $p\left(x\right)=\sqrt{cosx},g\left(x\right)=log\left(\frac{(1-x)}{1+x}\right)$ the ${\int }_a^{b}f\left(x\right)dx$ equals where $a=-1/2,b=1/2$

• A. $2{\int }_0^{1/2}p\left(x\right)q\left(x\right)dx$
• B. $2{\int }_0^{b}p\left(x\right)q\left(x\right)$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

$p\left(x\right)=\sqrt{cosx}$
$p(-x)=\sqrt{cosx}=p\left(x\right)\therefore p\left(x\right)$ is an even function
$q\left(x\right)=log\left(\frac{1+x}{1-x}\right)$
$q(-x)=log\left(\frac{1-x}{1+x}\right)=-q\left(x\right)\therefore q\left(x\right)$ is an odd function
Now $f\left(x\right)=p\left(x\right)q\left(x\right)=$ Product of even and odd function
=an odd function
$\therefore {\int }_a^{b}f\left(x\right)f\left(x\right)dx$ is equal 0, as a,b are opposite in sign equal in magnitude