Question

If the function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{(1-\left|\tan x\right|)}^{a/\left|\tan x\right|},& -\frac{\pi }{4}<x<0\end{array}\\ \begin{array}{cc}b,& x=0\end{array}\\ \begin{array}{cc}{e}^{\frac{\sin 3x}{\sin 2x}},& 0<x<\frac{\pi }{4}\end{array}\end{array}$ is continuous at $x=0$, then

• A. $a=-\frac{3}{2},b=\frac{3}{2}$
• B. $a=\frac{3}{2},b=e^{3/2}$
• C. $a=-\frac{3}{2},b=e^{3/2}$
• D. None of these

Answer 1

Answer: (C)

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

Since the function $f\left(x\right)$ is continuous at $x=0,$ therefore,

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

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

$\Rightarrow \underset{h\to 0}{lim}{\left[{(1-\left|\tan h\right|)}^{-1/\left|\tan h\right|}\right]}^{-a}=b=\underset{h\to 0}{lim}{e}^{\frac{\sin 3h/3h}{\sin 2h/2h}.\frac{3}{2}}$

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