If the function f(x) = \ cases (1 + | x |)(λ)/(| x |), & 0

Given,

$f\left(x\right)$ is continuous at $x = \dfrac{\pi}{2}$

Now, solving $L . H . L .$ at $x = \dfrac{\pi}{2}$ we get,

$\lim_{x \to \dfrac{{\pi}^{+}}{2}}e\dfrac{\cot 6x}{\cot 4x} = \lim_{x \to \dfrac{{\pi}^{+}}{2}}e\dfrac{\sin 4x \cdot \cos 6x}{\sin 6x \cdot \cos 4x} = {e}^{2 / 3}$

Similarly, on simplification for $R . H . L .$ we get

$\lim_{x \to \dfrac{\pi}{2} -}\left(1 + | \cos x |\right)\dfrac{\lambda}{| \cos x |} = {e}^{\lim_{x \to \dfrac{\pi}{2} -}\lambda} = {e}^{\lambda}$

$∵ f\left(\dfrac{\pi}{2}\right) = \mu$

For continuous function,

$f\left(\dfrac{\pi}{2} -\right) = f\left(\dfrac{\pi}{2}\right) = f\left(\dfrac{\pi}{2} +\right)$

$⇒ e\dfrac{2}{3} = {e}^{\lambda} = \mu$

$⇒ \lambda = \dfrac{2}{3}, \mu = e\dfrac{2}{3}$

Now,

$9\lambda + 6\log_{e}\mu + {\mu}^{6} - {e}^{6\lambda}$

$= 9 ( \dfrac{2}{3} ) + 6 ( \dfrac{2}{3} ) + {\left(e\dfrac{2}{3}\right)}^{6} - {e}^{6 ( \dfrac{2}{3} )}$

$= 6 + 4 + {e}^{4} - {e}^{4}$

$= 10$