Mathematics - Continuity - Differentiability Question with Solution | TestHub

If $f\left(x\right)$ is continuous at point $x=a,$ then $\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }f\left(x\right)$

From the question, we can say that $\underset{x\to 0}{\lim }f\left(x\right)=b$

And $\underset{x\to 0^{+}}{\lim }{e}^{\frac{\cot 4x}{\cot 2x}}=\underset{x\to 0^{+}}{\lim }{e}^{\frac{\tan 2x}{\tan 4x}}=\underset{x\to 0^{+}}{\lim }{e}^{\frac{{\displaystyle \frac{\tan 2x}{2\times 2x}}}{{\displaystyle \frac{\tan 4x}{4x}}}}=e^{\frac{1}{2}}=b$

And $\underset{x\to 0^{-}}{\lim }{\left(1+\left|\sin x\right|\right)}^{\frac{3a}{\left|\sin x\right|}}=e^{\underset{x\to 0^{-}}{\lim }\left[\left(\left|\sin x\right|\right)\left(\frac{3a}{\left|\sin x\right|}\right)\right]}=e^{3a}$

From the above we can say that

$e^{3a}=e^{\frac{1}{2}}$

$\Rightarrow a=\frac{1}{6}\Rightarrow 6a=1$

So, $\left(6a+b^2\right)=\left(1+e\right)$