If the function f(x)={(1+∣cosx∣)λ∣cosx∣,0<x<π2μ,x=π2cot6xecot4xπ2<x<πf(x)=\begin{cases}(1+|\cos x|) \frac{\lambda}{|\cos x|} & , 0 < x < \frac{\pi}{2} \\\mu & , \quad x=\frac{\pi}{2} \\\frac{\cot 6 x}{e^{\cot 4 x}} & \frac{\pi}{2}< x< \pi\end{cases}f(x)=⎩⎨⎧(1+∣cosx∣)∣cosx∣λμecot4xcot6x,0<x<2π,x=2π2π<x<πis continuous at x=π2,then9λ+6logeμ+μ6−e6λx=\frac{\pi}{2}, then 9 \lambda+6 \log _{ e } \mu+\mu^6- e ^{6 \lambda}x=2π,then9λ+6logeμ+μ6−e6λ is equal to
If for p ≠ q ≠ 0, the functionf(x)=p(729+x)−37729+qx3−9f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}f(x)=3729+qx−97p(729+x)−3is continuous at x = 0, then