If the function f(x)= cc (1+| x|) | x| & , 0 < x < 2 & , x= 2 e^ … - Vidyadip

MCQ

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

A
$11$
B
$8$
C
$2 e^4+8$
✓
$10$

✓

Answer

Correct option: D.

$10$

d
$\Rightarrow \lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}=\lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\sin 4 x \cdot \cos 6 x}{\sin 6 x \cdot \cos 4 x}}=e^{2 / 3}$

$\Rightarrow \lim _{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^{\frac{\lambda}{\cos x} \mid}=e^\lambda$

$\Rightarrow f(\pi / 2)=\mu$

For continuous function $\Rightarrow e ^{2 / 3}= e ^\lambda=\mu$

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

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

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