The value of k,for which the function f(x) = | vedclass

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(B) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $f(\frac{\pi}{2}) = \lim_{x 	o \frac{\pi}{2}} f(x)$.First,evaluate the limit: $\lim_{x

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$\frac{3}{5}$

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(B) Since $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $f(\frac{\pi}{2}) = \lim_{x 	o \frac{\pi}{2}} f(x)$.First,evaluate the limit: $\lim_{x 	o \frac{\pi}{2}} (\frac{4}{5})^{\frac{	an 4x}{	an 5x}}$.Let $x = \frac{\pi}{2} + h$,where $h 	o 0$. Then $	an 4x = 	an(4(\frac{\pi}{2} + h)) = 	an(2\pi + 4h) = 	an 4h \approx 4h$.And $	an 5x = 	an(5(\frac{\pi}{2} + h)) = 	an(\frac{5\pi}{2} + 5h) = \cot 5h \approx \frac{1}{5h}$.Thus,the exponent is $\frac{	an 4x}{	an 5x} = \frac{	an 4h}{\cot 5h} = 	an 4h \cdot 	an 5h 	o 0 \cdot \infty$ form.Actually,$\lim_{x 	o \frac{\pi}{2}} \frac{	an 4x}{	an 5x} = \lim_{h 	o 0} \frac{	an 4h}{\cot 5h} = \lim_{h 	o 0} 	an 4h 	an 5h = 0 \cdot 0 = 0$.Therefore,$\lim_{x 	o \frac{\pi}{2}} f(x) = (\frac{4}{5})^0 = 1$.Equating to $f(\frac{\pi}{2})$,we get $k + \frac{2}{5} = 1$.$k = 1 - \frac{2}{5} = \frac{3}{5}$.

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is:

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