Question:
The value of k for which the function \(f(x) = \begin{cases} \left(\frac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}, & 0 \lt x \lt \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}\) is continuous at \(x = \frac{\pi}{2}\), is :
The value of k for which the function \(f(x) = \begin{cases} \left(\frac{4}{5}\right)^{\frac{\tan 4x}{\tan 5x}}, & 0 \lt x \lt \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}\) is continuous at \(x = \frac{\pi}{2}\), is :
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Find the limit of the exponent tan4x / tan5x as x tends to pi/2.
Updated On: Nov 11, 2026
- \(\frac{17}{20}\)
- \(\frac{2}{5}\)
- \(\frac{3}{5}\)
- \(-\frac{2}{5}\)
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The Correct Option is C
Solution and Explanation
Let \(x = \frac{\pi}{2} - h\) with \(h \to 0^+\).
\(\tan 4x = \tan(2\pi - 4h) = -\tan 4h\) and \(\tan 5x = \tan\left(\frac{5\pi}{2} - 5h\right) = \cot 5h\).
\(\frac{\tan 4x}{\tan 5x} = -\tan 4h \cdot \tan 5h \to 0\).
So \(\lim_{x \to \pi/2} f(x) = \left(\frac45\right)^0 = 1\).
Continuity requires \(k + \frac25 = 1 \Rightarrow k = \frac35\).
Answer: (3) \(\frac{3}{5}\)
\(\tan 4x = \tan(2\pi - 4h) = -\tan 4h\) and \(\tan 5x = \tan\left(\frac{5\pi}{2} - 5h\right) = \cot 5h\).
\(\frac{\tan 4x}{\tan 5x} = -\tan 4h \cdot \tan 5h \to 0\).
So \(\lim_{x \to \pi/2} f(x) = \left(\frac45\right)^0 = 1\).
Continuity requires \(k + \frac25 = 1 \Rightarrow k = \frac35\).
Answer: (3) \(\frac{3}{5}\)
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