Question

Let $f\left(x\right)=\{\begin{array}{cc}(1+|\sin x|{)}^{\frac{a}{|\sin x|}},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\frac{\tan 2x}{\tan 3x}},& 0<x<\frac{\pi }{6}\end{array}.$
Let $a$ and $b$ be such that $f$ is continuous at $x=0.$ Then $3(a+\log b)$ equals

Answer 1

$f\left(x\right)=\{\begin{array}{cc}(1+|\sin x|{)}^{\frac{a}{|\sin x|}},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\frac{\tan 2x}{\tan 3x}},& 0<x<\frac{\pi }{6}\end{array}$
Since, $f\left(x\right)$ is continuous at $x=0$
$\Rightarrow \underset{h\to 0}{lim}f(0-h)=f\left(0\right)=\underset{h\to 0}{lim}f(0+h)$
Now,
$\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}(1+|\sin (-h)|{)}^{\frac{a}{|\sin (-h)|}}$
$=\underset{h\to 0}{lim}(1+\sin h{)}^{\frac{a}{\sin h}}$
$=e^{(1+\sin h-1)\frac{a}{\sin h}}=e^a$
So,
$e^a=b$

Also, $\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{e}^{\frac{\tan 2h}{\tan 3h}}$
$=\underset{h\to 0}{lim}{e}^{\frac{\tan 2h}{2h}\times \frac{3h}{\tan 3h}\times \frac{2}{3}}$
$=e^{2/3}$

$\Rightarrow e^a=b=e^{2/3}$
$\Rightarrow a=\frac{2}{3}$

$\therefore 3(a+\log b)=3(\frac{2}{3}+\frac{2}{3})=4$