Question

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

• A. $2 e^4+8$
• B. 11
• C. 8
• D. 10

Answer 1

The Correct Option is D

The correct answer is (D) : 10
$\Rightarrow \underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\cot 6x}{\cot 4x}}=\underset{x\to \frac{{\pi }^{+}}{2}}{\lim }{e}^{\frac{\sin 4x\times \cos 6x}{\sin 6x\cdot \cos 4x}}=e^{2/3}$
$\Rightarrow \underset{x\to \frac{{\pi }^{-}}{2}}{\lim }(1+|\cos x|{)}^{\frac{\lambda }{\cos x}\mid }=e^{\lambda }$
$\Rightarrow f(\pi /2)=\mu$
For continuous function $\Rightarrow e^{2/3}=e^{\lambda }=\mu$
$\lambda =\frac{2}{3},\mu =e^{2/3}$
Now, $9\lambda +6{\log }_e\mu +{\mu }^6-e^{6\lambda }=10$