If the function  $f\left(x\right)=$$\left\{\begin{array}{ll}{\left(1+\left|\cos x\right|\right)}^{\frac{\lambda }{\left|\cos x\right|}},& 0<x<\frac{\mathrm{\pi }}{2}\\ \mu ,& x=\frac{\mathrm{\pi }}{2}\\ \frac{\cot 6x}{\cot 4x},& \frac{\mathrm{\pi }}{2}<x<\mathrm{\pi }\end{array}\right.$     is continuous at
$\mathrm{x}=\frac{\pi }{2}\text{, then }9\lambda +6{\log }_{\mathrm{e}}\mu +{\mu }^6-{\mathrm{e}}^{6\lambda }\text{ is equal to }$

$\left\{\begin{array}{ll}{\left(1+\left|\cos x\right|\right)}^{\frac{\lambda }{\left|\cos x\right|}},& 0<x<\frac{\mathrm{\pi }}{2}\\ \mu ,& x=\frac{\mathrm{\pi }}{2}\\ \frac{cot6x}{cot4x},& \frac{\mathrm{\pi }}{2}<x<\mathrm{\pi }\end{array}\right.$is continuous

$\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }(1+\cos \mathrm{x}{)}^{\lambda /\cos x}=1^{\alpha }$ form

$\therefore \underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=e^{\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\left(\cos x\right)\lambda }{\cos x}}=e^{\lambda }$
${\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }}^{ }f\left(x\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }{e}^{\frac{\cot 6x}{\cot 4x}}=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\underset{ }{e^{\frac{\tan 4x}{\tan 6x}}}$
$={\mathrm{e}}^{2/3}\text{ (by using L.Hospital) }$

f (x) is continuous a$x=\frac{\mathrm{\pi }}{2}$

$\therefore \lambda =\frac{2}{3},\mu =e^{\lambda }=e^{\frac{2}{3}}$

$9\lambda +6{\log }_{\mathrm{e}}\mu +{\mu }^6-{\mathrm{e}}^{6\lambda } =6+4+e^4-e^4=10$