JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the maximum value of \(a\), for which the function \(f_{a}(x)=\tan ^{-1} 2 x-3 a x+7\) is non-decreasing in \(\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)\), is \(\bar{a}\), then \(f_{a}\left(\frac{\pi}{8}\right)\) is equal to
- A \(8-\frac{\pi}{4}\)
- B \(8-\frac{4 \pi}{9\left(4+\pi^{2}\right)}\)
- C \(8\left(\frac{1+\pi^{2}}{9+\pi^{2}}\right)\)
- D \(8-\frac{9 \pi}{4\left(9+\pi^{2}\right)}\)
Answer & Solution
Correct Answer
(D) \(8-\frac{9 \pi}{4\left(9+\pi^{2}\right)}\)
Step-by-step Solution
Detailed explanation
\(Bonus\) \(f _{ a }( x )=\tan ^{-1} 2 x -3 ax +7\) \(f _{ a }^{\prime}( x )=\frac{2}{1+4 x ^{2}}-3 a \geq 0\) \(a \leq\left(\frac{2}{3\left(1+4 x ^{2}\right)}\right)_{\text {min. }}\) at \(x =\pm \frac{\pi}{6}\) \(a _{\max }=\overline{ a }=\frac{6}{9+\pi^{2}}\)…
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