Question

If $f\left(x\right)=\underset{n\to \infty }{lim}{e}^{x\tan \left(1/n\right)ln\left(1/n\right)}$ and $\int \frac{f\left(x\right)}{\sqrt[3]{3\left({\sin }^{11}x\cos x\right)}}dx=g\left(x\right)+c$, then

• A. $g\left(\frac{\pi }{4}\right)=\frac{3}{2}$
• B. $g\left(x\right)$ is continous for all $x\in R$
• C. $g\left(\frac{\pi }{4}\right)=-\frac{15}{8}$
• D. $g\left(\frac{\pi }{4}\right)=\frac{1}{2}$

Answer 1

The correct option is C $g\left(\frac{\pi }{4}\right)=-\frac{15}{8}$

$f\left(x\right)=\underset{n\to \infty }{lim}{e}^{x\tan \left(\frac{1}{n}\right)\log \left(\frac{1}{n}\right)}$

But $\underset{n\to \infty }{lim}\tan \left(\frac{1}{n}\right)\log \left(\frac{1}{n}\right)=\underset{n\to \infty }{lim}\left[\frac{\log n}{n}\frac{\tan \frac{1}{n}}{\frac{1}{n}}\right]=0$

So $f\left(x\right)=e^0=1$

Hence
$g\left(x\right)=\int \frac{f\left(x\right)}{\sqrt[3]{3{\sin }^{11}x\cos x}}dx=\int \frac{1}{\sqrt[3]{3{\sin }^{11}x\cos x}}dx$

Substitute $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$g\left(x\right)=\int (t^{\frac{-11}{3}}+t^{\frac{-5}{3}})dt=\frac{-3}{8}{t}^{\frac{-8}{3}}-\frac{3}{2}{t}^{\frac{-2}{3}}+c$

$=\frac{-3}{8}\frac{(1+4{\tan }^{2}x)}{{\tan }^{2}x\sqrt[3]{3{\tan }^{2}x}}+c$

Thus $g\left(\frac{\pi }{4}\right)=-\frac{15}{8}$

Clearly $g$ is not defined at $x=0$ and odd multiple of $x=\frac{\pi }{2}$