Q71 / 150 · MHT CET 2011 · Mathematics

If $f(x)=\left\{\begin{array}{cc}(1+|\sin x|)^{\frac{a}{|\sin x|}}, & -\pi / 6 < x < 0 \\ b & x=0 \\ \frac{\tan 2 x}{e^{\tan 3 x}}, & 0 < x < \pi / 6\end{array}\right.$ is continuous at $x=0$, find the values of $a$ and $b$.

  • A
    $3 / 2, e^{3 / 2}$
  • B
    $-2 / 3, e^{-3 / 2}$
  • Correct answer
    C
    $2 / 3, e^{2 / 3}$
  • D
    None of these

Solution

We have, $\lim _{x \rightarrow 0^{-}} f(x)$ $$ \begin{aligned} =& \lim _{x \rightarrow 0}\{1+|\sin x|\}^{\frac{a}{|\sin x|}} \\ =& \lim _{x \rightarrow 0}|\sin x| \cdot \frac{a}{|\sin x|}=e^{a} \\ \text { and } \quad \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} e^{\frac{\tan 2 x}{\tan 3 x}} \\ =& \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \cdot \frac{3 x}{\tan 3 x} \times \frac{2}{3} \\ =& e^{2 / 3} \end{aligned} $$ For $f(x)$ to be continuous at $x=0$, we must have $$ \begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x) &=\lim _{x \rightarrow 0^{+}} f(x) \\ & &=f(0) \\ \Rightarrow & e^{a} &=e^{2 / 3}=b \\ \Rightarrow & a &=2 / 3 \\ \text { and } & b &=e^{2 / 3} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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