Q77 / 120 · VITEEE 2013 · Mathematics
If $f(x)=\left\{\begin{array}{cl}(1+|\sin x|)^{a / \sin x \mid} & ,-\frac{\pi}{6} < x < 0 \\ b & , x=0 \\ e^{\tan 2 x / \tan 3 x} & , 0 < x < -\frac{\pi}{6}\end{array}\right.$, then the value of $a$ and $b$, if $\mathrm{f}$ is continuous at $x=0$, are respectively.
- A$\frac{2}{3}, \frac{3}{2}$
- Correct answerB$\frac{2}{3}, e^{2 / 3}$
- C$\frac{3}{2}, e^{3 / 2}$
- DNone of these
Solution
$f(x)=\left\{\begin{array}{cl}(1+|\sin x|)^{a /|\sin x|}, & -\frac{\pi}{6} < x < 0 \\ b & , \quad x=0 \\ e^{\tan 2 x / \tan 3 x} & , 0 < x < \frac{\pi}{6}\end{array}\right.$
For $\mathrm{f}(\mathrm{x})$ to be continuous at $\mathrm{x}=0$
$\begin{array}{l}
\lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\
\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{a /|\sin x|} \\
=\lim _{x \rightarrow 0}\left\{|\sin x| \frac{a}{|\sin x|}\right\}=e^{a}
\end{array}$
$\text { Now, } \lim _{x \rightarrow 0^{+}} e^{\tan 2 x / \tan 3 x}$
$=\lim _{x \rightarrow 0^{+}} e^{\left(\frac{\tan 2 x}{2 x} \times 2 x\right) /\left(\frac{\tan 3 x}{3 x} \times 3 x\right)}$
$=\lim _{\mathrm{x} \rightarrow 0^{+}} \mathrm{e}^{2 / 3}=\mathrm{e}^{2 / 3}$
Since, $f(x)$ is continuous at $x=0$.
$\therefore \mathrm{e}^{\mathrm{a}}=\mathrm{e}^{2 / 3} \Rightarrow \mathrm{a}=\frac{2}{3}$ and $b=e^{2 / 3}$
Hence, the correct answer is option: 'B'.