Question

If $f\left(x\right)=\{\begin{array}{cc}(1+|\sin x|{)}^{a/|\sin x|},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\tan 2x/\tan 3x},& 0<x<-\frac{\pi }{6}\end{array}$, then the value of $a$ and $b$, if $f$ is continuous at $x=0$ are respectively

• A. $\frac{2}{3},\frac{3}{2}$
• B. $\frac{2}{3},e^{2/3}$
• C. $\frac{3}{2},e^{3/2}$
• D. None of these

Answer 1

The correct option is B $\frac{2}{3},e^{2/3}$

$f\left(x\right)=\{\begin{array}{cc}(1+|\sin x|{)}^{a/|\sin x|},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\tan 2x/\tan 3x},& 0<x<-\frac{\pi }{6}\end{array}$

For $f\left(x\right)$ to be continuous at $x=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$\underset{x\to 0^{-}}{lim}(1+|\sin x|{)}^{a/|\sin x|}$

$=e^{\underset{x\to 0}{lim}\left\{|\sin x|\frac{a}{|\sin x|}\right\}}=e^a$

Now, $\underset{x\to 0^{+}}{lim}{e}^{\tan 2x/\tan 3x}$

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

Since, $f\left(x\right)$ is continuous at $x=0$.

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