Question

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

• A. $2e^4+8$
• B. 11
• C. 8
• D. 10

Answer 1

Answer: (D)

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