Question

f the maximum value of a, for which the function
$fa(x)=\tan^{−1}\ ⁡2x−3ax+7$
is non-decreasing in $(−\frac{π}{6},\frac{π}{6})$, is a―, then $f\overline{a}(\frac{π}{8})$
is equal to

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

Answer 1

Answer: (A)

$8-\frac{9π}{4(9+π^2)}$

Answer 2

$fa(x) = \tan^{-1}\ 2x -3ax+7$
because $fa(x)$ is non decreasing in $(- \frac{π}{6},\frac{π}{6})$
Therefore , $f'a(x) >= 0$
$⇒ \frac{2}{1+4x^2}-3a≥0$
$⇒ 3a ≤ \frac{2}{1+4x^2}$
So, $a_{max} = \frac{2}{3}(\frac{1}{1+4\times \frac{π^2}{36}})$
$= \frac{6}{9+π^2} = \overline{a}$
$\therefore fa(\frac{\pi}{8}) = \tan^{-1} \frac{\pi}{4}-3. \frac{\pi}{8} . \frac{6}{9+ \pi^2}+7$