Question

If the maximum value of $a$, for which the function $f_a(x)={\tan }^{-1}2x-3ax+7$ 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)

$f_a(x)={\tan }^{-1}2x-3ax+7$
$f_a^{\prime }(x)=\frac{2}{1+4x^2}-3a\ge 0$
$a\le {\left(\frac{2}{3\left(1+4x^2\right)}\right)}_{\text{min. }}\text{ at }x=\pm \frac{\pi }{6}$
$a_{\max }=\overline{a}=\frac{6}{9+{\pi }^2}$
$f_a\left(\frac{\pi }{8}\right)={\tan }^{-1}\frac{\pi }{4}-3\frac{6}{9+{\pi }^2}\frac{\pi }{8}+7={\tan }^{-1}\frac{\pi }{4}-\frac{9\pi }{4\left({\pi }^2+9\right)}+7$