Question

Let $f\left(x\right)=\{\begin{array}{cc}{(1+\left|\sin x\right|)}^{\frac{a}{\left|\sin x\right|}},& -\frac{\pi }{6}<x<0\\ b,& x=0\\ e^{\frac{\tan 2x}{\tan 3x}},& 0<x<\frac{\pi }{6}\end{array}$
If $f$ is continuous at $x=0$, then which of the following option(s) is correct

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

Answer 1

Answer: (A), (B)

$b=e^{\frac{2}{3}}$

For $f$ to be continuous at $x=0$, $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$
Now,
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(1+\left|\sin x\right|{)}^{\frac{a}{\left|\sin x\right|}}(1^{\infty }\text{form})$
$=\underset{x\to 0^{-}}{lim}{e}^{(1-\sin x-1)\cdot \frac{a}{-\sin x}}=e^a$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}{e}^{\frac{\tan 2x}{\tan 3x}}$
$=e^{\underset{x\to 0^{+}}{lim}\frac{\tan 2x}{2x}\cdot \frac{3x}{\tan 3x}\cdot \frac{2}{3}}=e^{\frac{2}{3}}$
$\therefore e^a=b=e^{\frac{2}{3}}$
$\Rightarrow a=\frac{2}{3},b=e^{\frac{2}{3}}$