If the maximum value of a, for which the function [f_a(x)= ^-1 2 x-3 a x+7] is non-decreas…

Question

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 :

Accepted Answer

$\text {Given, }$

$$
 \begin{aligned}
f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\
f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a
\end{aligned}
$$

As the function $f_a^{\prime}(x)$ is non-decreasing

$$
\begin{aligned}
& \text { in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\
& f_a^{\prime}(x) \geq 0
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \frac{2}{1+4 x^2}-3 a \geq 0 \Rightarrow \frac{2}{1+4 x^2} \geq 3 a \\
& \Rightarrow a \leq \frac{2}{3\left(1+4 x^2\right)}, \text { when } x \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \\
& \because a \text { is maximum when } x^2=\frac{\pi^2}{36},
\end{aligned}
$$

$$
\begin{aligned}
a_{\max } & =\frac{2}{3\left(1+\frac{4 \pi^2}{36}\right)}=\frac{2 \times 12}{36+4 \pi^2} \\
& =\frac{6}{9+\pi^2} \\
\therefore \bar{a} & =\frac{6}{9+\pi^2}
\end{aligned}
$$

$$
\begin{aligned}
& \therefore f_a(x)=\tan ^{-1} 2 x-\frac{18}{9+\pi^2} x+7 \\
& f_\pi\left(\frac{\pi}{8}\right)=\tan ^{-1} 2\left(\frac{\pi}{8}\right)-\frac{18}{9+\pi^2} \times \frac{\pi}{8}+7 \\
& =\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{36+4 \pi^2}+7 \\
& =8-\frac{9 \pi}{36+4 \pi^2}=8-\frac{9 \pi}{4\left(9+\pi^2\right)}
\end{aligned}
$$

Solutions · 1