Let $\mathrm{f}:\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \rightarrow \mathbf{R}$ be defined as
$f(x)=\left\{\begin{array}{ccc}(1+|\sin x|)^{\frac{3 a}{\sin x \mid}} & , & -\frac{\pi}{4} If $f$ is continuous at $x=0$, then the value of $6 a+b^{2}$ is equal to:
Correct Option: , 3
$\lim _{x \rightarrow 0} f(x)=b$
$\lim _{x \rightarrow 0^{+}} x e^{\frac{\cot 4 x}{\cot 2 x}}=e^{\frac{1}{2}}=b$
$\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{3 a}{\sin x \mid}}=e^{3 a}=e^{\frac{1}{2}}$
$\lim _{x \rightarrow 0^{-}}(1+|\sin x|)^{\frac{3 a}{\sin x \mid}}=e^{3 a}=e^{\frac{1}{2}}$
$a=\frac{1}{6} \Rightarrow 6 a=1$
$\left(6 \mathrm{a}+\mathrm{b}^{2}\right)=(1+\mathrm{e})$
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