Question

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

• A. 1 -e
• B. e -1
• C. 1 +e
• D. e

Answer 1

Answer: (C)

$\underset{x\to 0}{\lim }f(x)=b$
$\underset{x\to 0^{-}}{\lim }x{e}^{\frac{\cot 4x}{\cot 2x}}=e^{\frac{1}{2}}=b$
$\underset{x\to 0^{-}}{\lim }(1+|\sin x|{)}^{\frac{3a}{|\sin x\mid }}=e^{3a}=e^{\frac{1}{2}}$
$\underset{x\to 0^{-}}{\lim }(1+|\sin x|{)}^{\frac{3a}{|\sin x\mid }}=e^{3a}=e^{\frac{1}{2}}$
$a=\frac{1}{6}\Rightarrow 6a=1$
$\left(6a+b^2\right)=(1+e)$