Continuity And Differentiability Question 2
Question 2 - 25 January - Shift 2
If the function
$f(x)=\begin{array}{cc}(1+|\cos x|) \frac{\lambda}{|\cos x|}, & 0<x<\frac{\pi}{2} \\ \quad \quad \mu & , x=\frac{\pi}{2} \\ e^{\frac{\cot 6 x}{\cot 4 x}} & , \frac{\pi}{2}<x<\pi\end{array}$
is continuous at $x=\frac{\pi}{2}$, then
$9 \lambda+6 \log _e \mu+\mu^{6}-e^{6 \lambda}$ is equal to
(1) 11
(2) 8
(3) $2 e^{4}+8$
(4) 10
Show Answer
Answer: (4)
Solution:
Formula: Continuity of a Function at a Point, Differentiability at a point
$ \begin{aligned} & \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|}}=e^\lambda \\ & \Rightarrow f(\pi / 2)=\mu \end{aligned} $
For continuous function $\Rightarrow e^{2 / 3}=e^\lambda=\mu$ $ \lambda=\frac{2}{3}, \mu=e^{2 / 3} $
Now, $9 \lambda+6 \log _e \mu+\mu^6-e^{6 \lambda}=10$
