If the maximum value of a , for which the function f _ a (x) = ^ -...

Given ${f}_{a}\left(x\right) = \tan^{- 1}2x - 3ax + 7$ is non-decreasing function in $\left(- \dfrac{\pi}{6}, \dfrac{\pi}{6}\right)$, so

${{f}_{a}}^{'}\left(x\right) = \dfrac{2}{1 + 4{x}^{2}} - 3a \ge 0$

i.e. a \le {\left(\dfrac{2}{3\left(1 + 4{x}^{2}\right)}\right)}_\min

Now, maximum value of $a$ is at $x = 0$

i.e. {a}_\max = \overset{¯}{a} = \dfrac{6}{9 + {\pi}^{2}}

Hence, $f\overset{¯}{a}\left(\dfrac{\pi}{8}\right) = \tan^{- 1}\dfrac{\pi}{4} - 3\left(\dfrac{6}{9 + {\pi}^{2}}\right)\left(\dfrac{\pi}{8}\right) + 7$

$= \tan^{- 1}\dfrac{\pi}{4} - \dfrac{9\pi}{4\left({\pi}^{2} + 9\right)} + 7 = 8 - \dfrac{9\pi}{4\left({\pi}^{2} + 9\right)}$