Question

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

• A. $a={\log }_{e}b,a=\frac{2}{3}$
• B. $b={\log }_{e}a,a=\frac{2}{3}$
• C. $a={\log }_{e}b,b=2$
• D. none of these

Answer 1

The correct option is A $a={\log }_{e}b,a=\frac{2}{3}$

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

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

$\Rightarrow \underset{x\to 0^{-}}{lim}{(1+\left|\sin x\right|)}^{\frac{a}{\left|\sin x\right|}}=b=\underset{x\to 0^{+}}{lim}{e}^{\frac{\tan 2x}{\tan 3x}}$

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

$\therefore a=\frac{2}{3}a={\log }_{e}b$