MHT CET · Maths · Continuity and Differentiability
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 \(\frac{17}{20}\)
- B \(\frac{3}{5}\)
- C \(-\frac{2}{5}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{5}\)
Step-by-step Solution
Detailed explanation
Since \(\mathrm{f}(x)\) is continuous at \(x=\frac{\pi}{2}\).
\(\begin{aligned}
& \therefore \quad \mathrm{f}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}} \\
& \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\lim _{x \rightarrow \frac{\pi}{2}}(\tan 4 x \cot 5 x)} \\
& \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\left(\frac{4}{5}\right)^0=1 \\
& \quad \Rightarrow \mathrm{k}=\frac{3}{5}
\end{aligned}\)
\(\begin{aligned}
& \therefore \quad \mathrm{f}\left(\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}} \\
& \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{4}{5}\right)^{\lim _{x \rightarrow \frac{\pi}{2}}(\tan 4 x \cot 5 x)} \\
& \quad \Rightarrow \mathrm{k}+\frac{2}{5}=\left(\frac{4}{5}\right)^0=1 \\
& \quad \Rightarrow \mathrm{k}=\frac{3}{5}
\end{aligned}\)
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