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(D) Given the function $f_{a}(x) = 	an^{-1}(2x) - 3ax + 7$.For the function to be non-decreasing,its derivative must be non-negative: $f_{a}'(x) \geq

Vedclass · Accepted Answer

$7+	an ^{-1} \frac{\pi}{4}-\frac{9 \pi}{4\left(9+\pi^{2}ight)}$

Vedclass · Answer

(D) Given the function $f_{a}(x) = 	an^{-1}(2x) - 3ax + 7$.For the function to be non-decreasing,its derivative must be non-negative: $f_{a}'(x) \geq 0$.$f_{a}'(x) = \frac{2}{1+4x^2} - 3a \geq 0$.This implies $3a \leq \frac{2}{1+4x^2}$,or $a \leq \frac{2}{3(1+4x^2)}$.To satisfy this for all $x \in \left(-\frac{\pi}{6}, \frac{\pi}{6}ight)$,$a$ must be less than or equal to the minimum value of $\frac{2}{3(1+4x^2)}$ on the interval.The minimum value occurs at the boundaries $x = \pm \frac{\pi}{6}$.At $x^2 = \frac{\pi^2}{36}$,we have $a \leq \frac{2}{3(1 + 4(\frac{\pi^2}{36}))} = \frac{2}{3(1 + \frac{\pi^2}{9})} = \frac{2}{3(\frac{9+\pi^2}{9})} = \frac{6}{9+\pi^2}$.Thus,$\bar{a} = \frac{6}{9+\pi^2}$.Now,calculate $f_{\bar{a}}\left(\frac{\pi}{8}ight) = 	an^{-1}\left(2 \cdot \frac{\pi}{8}ight) - 3 \cdot \left(\frac{6}{9+\pi^2}ight) \cdot \frac{\pi}{8} + 7$.$f_{\bar{a}}\left(\frac{\pi}{8}ight) = 	an^{-1}\left(\frac{\pi}{4}ight) - \frac{18\pi}{8(9+\pi^2)} + 7 = 7 + 	an^{-1}\left(\frac{\pi}{4}ight) - \frac{9\pi}{4(9+\pi^2)}$.

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 $f_{\bar{a}}\left(\frac{\pi}{8}\right)$ is equal to

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