Question

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

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

Answer 1

Answer: (B)

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

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

${lim}_{x\to 0^{-}}f\left(x\right)=f\left(0\right)={lim}_{x\to 0^{+}}f\left(x\right)$

LHL

${lim}_{x\to 0^{-}}exp(|\sin x|\times \frac{a}{|\sin x|})=e^a$

$f\left(0\right)=b$

RHL

${lim}_{x\to 0^{+}}exp\left(\frac{\tan 2x}{\tan 3x}\right)$

$={lim}_{x\to 0^{+}}exp(\frac{\tan 2x}{2xtan3x}\times 3x\times \frac{2}{3})=e^{\frac{2}{3}}$

Thus,

$exp\left(a\right)=b=exp\left(\frac{2}{3}\right)$

Hence, $a=\frac{2}{3}$ and $b=exp\left(\frac{2}{3}\right)$