Question

If $f\left(x\right)=\{\begin{array}{cc}{(1+\left|sinx\right|)}^{\frac{a}{\left|sinx\right|}}& ;-\frac{\pi }{6}<x<0\\ b& ;x=0\\ e^{\left(\frac{tan2x}{tan3x}\right)}& ;0<x<\frac{\pi }{6}\end{array}$
is a continuous function on $(-\frac{\pi }{6},\frac{\pi }{6})$; then

• A. $a=\frac{2}{3},b=e^2$
• B. $a=\frac{1}{3},b=e^{1/3}$
• C. $a=\frac{2}{3},b=e^{2/3}$
• D. $a=e^{2/3},b=\frac{2}{3}$

Answer 1

The correct option is C $a=\frac{2}{3},b=e^{2/3}$

$f\left(x\right)=\{\begin{array}{c}{(1+\left|\sin x\right|)}^{\frac{a}{\left|\sin x\right|}},\frac{\pi }{6}<x<0\\ b,x=0\\ e^{\frac{\tan 2x}{\tan 3x}},0<x<\frac{\pi }{6}\end{array}$
Since, $f\left(x\right)$ is continuous at $x=0$
$\therefore RHL(atx=0)=LHL(atx=0)=f\left(0\right)$
$\Rightarrow {lim}_{h\to 0}{e}^{\frac{\tan 2h}{\tan 3h}}={lim}_{h\to 0}{\{1+\left|\sin h\right|\}}^{\frac{a}{\left|\sin h\right|}}=b$
$\Rightarrow e^{\frac{2}{3}}=e^a=b$
$\therefore a=\frac{2}{3}$ and $b=e^{\frac{2}{3}}$