Question

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

• A. $p=\frac{3}{2}$
• B. $p=\frac{2}{3}$
• C. ${\log }_{e}q=p$
• D. $q=2$

Answer 1

Answer: (A), (C)

${\log }_{e}q=p$
$p=\frac{3}{2}$

Since f(x) is continuous at $x=0$
$\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$...(1)
Now , $LHL=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}(1-tanx{)}^{-\frac{p}{tanx}}$
Since $\underset{x\to 0}{lim}(1+x{)}^{1/x}=e$
$\Rightarrow LHL=e^p$...(2)
Now, $RHL=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}{e}^{\frac{\sin 3x}{\sin 2x}}$
$=\underset{x\to 0}{lim}{e}^{\frac{3}{2}(\frac{\sin 3x}{3x}\left)\right(\frac{2x}{\sin 2x})}$
$RHL=e^{3/2}$....(3)
From (1), (2) and (3)
$e^p=e^{3/2}=q$
$\Rightarrow p=\frac{3}{2}={\log }_{e}q$