JEE Mains · Maths · STD 12 - 5. continuity and differentiation
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 \(2 e^4+8\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow \lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}=\lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\sin 4 x \cdot \cos 6 x}{\sin 6 x \cdot \cos 4 x}}=e^{2 / 3}\)…
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