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

Answer: (D)

$ \Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}=\displaystyle\lim _{x \rightarrow \frac{\pi^{+}}{2}} e ^{\frac{\sin 4 x \times \cos 6 x}{\sin 6 x \cdot \cos 4 x}}=e^{2 / 3}$
$ \Rightarrow \displaystyle\lim _{x \rightarrow \frac{\pi^{-}}{2}}(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, $9 \lambda+6 \log _{ e } \mu+\mu^6- e ^{6 \lambda}=10$