Question

Let $f:(-\frac{\pi }{4},\frac{\pi }{4})\to R$ be defined as $f\left(x\right)=\{\begin{array}{cc}(1+|\sin x|{)}^{3a/|\sin x|},& -\frac{\pi }{4}<x<0\\ b,& x=0\\ e^{\cot 4x/\cot 2x},& 0<x<\frac{\pi }{4}\end{array}.$ If $f$ is continuous at $x=0,$ then the value of $6a+b^2$ is equal to

• A. $1+e$
• B. $1-e$
• C. $e$
• D. $e-1$

Answer 1

The correct option is A $1+e$

Since $f\left(x\right)$ is continuous at $x=0$
$\therefore R.H.L.=L.H.L.=f\left(0\right)=b$
Now,
$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{x\to 0^{+}}{lim}{e}^{\cot 4x/\cot 2x}$
$=\underset{h\to 0}{lim}{e}^{\cot 4h/\cot 2h}$
$=\underset{h\to 0}{lim}{e}^{\frac{\cos 4h\sin 2h}{\sin 4h\cos 2h}}$
$=\underset{h\to 0}{lim}{e}^{\frac{\cos 4h}{2{\cos }^{2}2h}}=e^{1/2}$
$\therefore b=e^{1/2}$

$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{x\to 0^{-}}{lim}(1+|\sin x|{)}^{3a/|\sin x|}$
$=\underset{h\to 0}{lim}(1+\sin h{)}^{3a/\sin h}$ $\left(1^{\infty }\text{Form}\right)$
$=e^{3a}$
$\therefore e^{3a}=b=e^{1/2}$
$\Rightarrow a=\frac{1}{6}$
Hence, $6a+b^2=1+e$