Question:
If
\[
f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right],
\]
then the value of \(f'(e)\) is equal to
If
\[ f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right], \]
then the value of \(f'(e)\) is equal to
\[ f(x) = \cos^{-1} \left[ \frac{1 - (\log x)^2}{1 + (\log x)^2} \right], \]
then the value of \(f'(e)\) is equal to
Show Hint
Convert inverse trigonometric expressions using standard identities.
Updated On: Mar 23, 2026
- 1
- \(\dfrac{1}{e}\)
- \(\dfrac{2}{e}\)
- (2)/(e²)
Show Solution
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The Correct Option is B
Solution and Explanation
Using identity:
\[ \cos^{-1} \left( \frac{1 - t^2}{1 + t^2} \right) = 2 \tan^{-1} t, \quad \text{with } t = \log x. \]
\[ f(x) = 2 \tan^{-1} (\log x) \]
\[ f'(x) = \frac{2}{1 + (\log x)^2} \cdot \frac{1}{x} \]
At \(x = e\),
\[ f'(e) = \frac{2}{1 + 1} \cdot \frac{1}{e} = \frac{1}{e}. \]
\[ \cos^{-1} \left( \frac{1 - t^2}{1 + t^2} \right) = 2 \tan^{-1} t, \quad \text{with } t = \log x. \]
\[ f(x) = 2 \tan^{-1} (\log x) \]
\[ f'(x) = \frac{2}{1 + (\log x)^2} \cdot \frac{1}{x} \]
At \(x = e\),
\[ f'(e) = \frac{2}{1 + 1} \cdot \frac{1}{e} = \frac{1}{e}. \]
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