Let f: (- (π)/(4), (π)/(4)) → R be defined as, f(x) = \ cases (1 +...

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

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

And $\lim_{x \to {0}^{+}}e\dfrac{\cot 4x}{\cot 2x} =\lim_{x \to {0}^{+}}e\dfrac{\tan 2x}{\tan 4x} =\lim_{x \to {0}^{+}}e\dfrac{\dfrac{\tan 2x}{2 \times 2x}}{\dfrac{\tan 4x}{4x}} = e\dfrac{1}{2} =b$

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

From the above we can say that

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

$⇒ a = \dfrac{1}{6} ⇒ 6a = 1$

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