Q62 / 150 · MHT CET 2024 (02 May Shift 2) · Mathematics

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

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

Solution

Since $\mathrm{f}(x)$ is continuous at $x=\frac{\pi}{2}$. $\begin{aligned} & \therefore \quad \mathrm{f}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}} \\ & \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\lim _{x \rightarrow \frac{\pi}{2}}(\tan 4 x \cot 5 x)} \\ & \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\left(\frac{4}{5}\right)^0=1 \\ & \quad \Rightarrow \mathrm{k}=\frac{3}{5} \end{aligned}$

Hence, the correct answer is option: 'B'.

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