Question

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

• A. 1 -e
• B. e -1
• C. 1 +e
• D. e

Answer 1

Answer: (C)

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