Question

If $f\left(x\right)={lim}_{n\to \infty }{e}^{xtan\left(1/n\right)ln\left(1/n\right)}$ and $\int \frac{f\left(x\right)}{\sqrt[3]{3si{n}^{11}xcosx}}dx=g\left(x\right)+C$, then

• A. $g\left(\frac{\pi }{4}\right)=\frac{3}{2}$
• B. g(x) is continuous for all x
• C. $g\left(\frac{\pi }{4}\right)=-\frac{15}{8}$
• D. g(x) is non-differentiable at infinitely many points

Answer 1

Answer: (C), (D)

The correct options are
C

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

D

g(x) is non-differentiable at infinitely many points

${lim}_{n\to \infty }tan\left(\frac{1}{n}\right)ln\left(\frac{1}{n}\right)=-{lim}_{n\to \infty }\frac{tan\left(\frac{1}{n}\right)}{\left(\frac{1}{n}\right)}.\frac{ln\left(n\right)}{n}=-1.0=0$
Then $f\left(x\right)=e^{\circ }=1$
$\therefore \int \frac{f\left(x\right)}{\sqrt[3]{3si{n}^{11}xcosx}}dx=\int \frac{d\left(x\right)}{si{n}^{\frac{11}{3}}xco{s}^{\frac{1}{3}}x}=\int si{n}^{-\frac{11}{3}}xco{s}^{-\frac{1}{3}}Xdx=\int ta{n}^{-\frac{11}{3}}Xco{s}^{-4}xdx=\frac{(tanx{)}^{-\frac{11}{3}+1}}{(-\frac{11}{3}+1)}+\frac{(tanx{)}^{-\frac{2}{3}}}{(-\frac{2}{3})}+C=-\frac{3}{8}ta{n}^{-\frac{8}{3}}x-\frac{3}{2}ta{n}^{-\frac{2}{3}}x+C=g\left(X\right)+C$