If the function f(x) = begin{cases} (1+|cos x | vedclass

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(D) For the function to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow \frac{\pi}{

Vedclass · Accepted Answer

$10$

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(D) For the function to be continuous at $x = \frac{\pi}{2}$,we must have $\lim_{x \rightarrow \frac{\pi}{2}^-} f(x) = \lim_{x \rightarrow \frac{\pi}{2}^+} f(x) = f(\frac{\pi}{2})$.First,calculate the right-hand limit:$\lim_{x \rightarrow \frac{\pi}{2}^+} e^{\frac{\cot 6x}{\cot 4x}} = \lim_{x \rightarrow \frac{\pi}{2}^+} e^{\frac{\sin 4x \cdot \cos 6x}{\sin 6x \cdot \cos 4x}}$.Using $L$'Hopital's rule or expansion,$\lim_{x \rightarrow \frac{\pi}{2}^+} \frac{\sin 4x \cdot \cos 6x}{\sin 6x \cdot \cos 4x} = \frac{4 \cos(4 \cdot \frac{\pi}{2}) \cdot \cos(6 \cdot \frac{\pi}{2}) - 6 \sin(4 \cdot \frac{\pi}{2}) \cdot \sin(6 \cdot \frac{\pi}{2})}{6 \cos(6 \cdot \frac{\pi}{2}) \cdot \cos(4 \cdot \frac{\pi}{2}) - 4 \sin(6 \cdot \frac{\pi}{2}) \cdot \sin(4 \cdot \frac{\pi}{2})} = \frac{4(1)(-1) - 0}{6(-1)(1) - 0} = \frac{-4}{-6} = \frac{2}{3}$.Thus,$\lim_{x \rightarrow \frac{\pi}{2}^+} f(x) = e^{2/3}$.Next,calculate the left-hand limit:$\lim_{x \rightarrow \frac{\pi}{2}^-} (1+|\cos x|)^{\frac{\lambda}{|\cos x|}} = e^{\lim_{x \rightarrow \frac{\pi}{2}^-} \frac{\lambda}{|\cos x|} \cdot |\cos x|} = e^\lambda$.Equating the limits to $f(\frac{\pi}{2}) = \mu$:$e^\lambda = \mu = e^{2/3}$.So,$\lambda = \frac{2}{3}$ and $\mu = e^{2/3}$.Now,substitute these values into the expression:$9\lambda + 6 \log_{e} \mu + \mu^6 - e^{6\lambda} = 9(\frac{2}{3}) + 6 \log_{e} (e^{2/3}) + (e^{2/3})^6 - e^{6(2/3)}$$= 6 + 6(\frac{2}{3}) + e^4 - e^4 = 6 + 4 + 0 = 10$.

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