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 $f_{\overline{a}}\left(\frac{\pi}{8}\right)$ 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) $8\left(\frac{1+\pi^{2}}{9+\pi^{2}}\right)$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · 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

(- \frac{π}{6}, \frac{π}{6})

, is

\overline{a}

, then

f_{\overline{a}} (\frac{π}{8})

is equal to

Answer: (A)

For

f_{a} (x)

to be non-decreasing, we need

f_{a}^{'} (x) \geq 0

f_{a}^{'} (x) = \frac{2}{1 + 4 x ^{2}} - 3 a \geq 0

This gives

a \leq \frac{2}{3 ( 1 + 4 x ^{2} )}

.

The minimum of

\frac{2}{3 ( 1 + 4 x ^{2} )}

(- \frac{π}{6}, \frac{π}{6})

occurs at the endpoints

x = \pm \frac{π}{6}

.

Therefore,

\overline{a} = \frac{2}{3 ( 1 + 4 π ^{2} /36 )} = \frac{6}{9 + π ^{2}}

.

Now,

f_{\overline{a}} (\frac{π}{8}) = tan^{- 1} \frac{π}{4} - 3 \cdot \frac{6}{9 + π ^{2}} \cdot \frac{π}{8} + 7 = tan^{- 1} \frac{π}{4} - \frac{9 π}{4 ( π ^{2} + 9 )} + 7

.

Since

tan^{- 1} \frac{π}{4} + 7 \approx 8

, the answer is

8 - \frac{9 π}{4 ( 9 + π ^{2} )}

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option