If the function $$f(x) = \left\{ {\matrix{ {(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr \mu & , & {x = {\pi \over 2}} \cr e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr } } \right.$$

is continuous at $x = {\pi \over 2}$, then $9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$ is equal to

$$ \mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^ \frac{\lambda}{|\cos x|}=e^\lambda $$ <br/><br/> And, <br/><br/>$$ \begin{aligned} &\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}\\\\ &= \mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} {e^{{{\sin 4x.\cos 6x} \over {\sin 6x.\cos 4x}}}} \\\\ & =e^{\frac{2}{3}} \end{aligned} $$ <br/><br/> Also, $f(\pi / 2)=\mu$ <br/><br/>For continuous function, <br/><br/>$ \mathrm{e}^{2 / 3}=\mathrm{e}^\lambda=\mu$ <br/><br/>$\lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3}$ <br/><br/> $\therefore$ $9 \lambda+6 \ln \mu+\mu^{6}-e^{6 \lambda}$ <br/><br/> $=6+4+e^{4}-e^{4}=10$