Question

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

$\frac{2}{5}$

$f\left(x\right)={\left(\frac{4}{5}\right)}^{\frac{\tan 4x}{\tan 5x}}$
$\underset{x\to \frac{\pi }{2}}{lim}{\left(\frac{4}{5}\right)}^{\frac{\tan 4x}{\tan 5x}}=k+\frac{2}{5}$

$\Rightarrow \underset{x\to \frac{\pi }{2}}{lim}{\left(\frac{4}{5}\right)}^{\tan 4x\cot \left(5x\right)}=k+\frac{2}{5}$

$\Rightarrow {\left(\frac{4}{5}\right)}^{\underset{x\to \frac{\pi }{2}}{lim}(\tan 4x.\cot (5x\left)\right)}=k+\frac{2}{5}$

$\Rightarrow {\left(\frac{4}{5}\right)}^{\tan \left(2\pi \right)\cot \frac{5\pi }{2}}=k+\frac{2}{5}$

$\Rightarrow {\left(\frac{4}{5}\right)}^{0\times \cos (2\pi +\frac{\pi }{2})}=k+\frac{2}{5}$

$\Rightarrow {\left(\frac{4}{5}\right)}^0=k+\frac{2}{5}$

$\Rightarrow k+\frac{2}{5}=1$

$\Rightarrow k=1-\frac{2}{5}$

$\Rightarrow k=\frac{3}{5}$