Question

If the maximum value of $a$, for which the function $f _{ a }( x )=\tan ^{-1} 2 x -3 ax +7$ is non-decreasing in $\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$, is $\overline{ a }$, then $f _{\bar{a}}\left(\frac{\pi}{8}\right)$ is equal to

• A. $8-\frac{9 \pi}{4\left(9+\pi^2\right)}$
• B. $8-\frac{4 \pi}{9\left(4+\pi^2\right)}$
• C. $8\left(\frac{1+\pi^2}{9+\pi^2}\right)$
• D. $8-\frac{\pi}{4}$

Answer 1

Answer: (A), (B), (C), (D)

$f_a(x)=\tan ^{-1} 2 x-3 a x+7$
$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. }} \text { at } x =\pm \frac{\pi}{6} $
$a _{\max }=\overline{ a }=\frac{6}{9+\pi^2} $
$f _{ a }\left(\frac{\pi}{8}\right)=\tan ^{-1} \frac{\pi}{4}-3 \frac{6}{9+\pi^2} \frac{\pi}{8}+7=\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{4\left(\pi^2+9\right)}+7$