Question

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

• A. $\frac{17}{20}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $ - \frac{2}{5}$

Answer 1

Answer: (C)

$k + 2 /5 = (4 /5)$
$=k+\frac{2}{4}=1$
$K=\frac{3}{5}$