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

The Correct Option is A

\(fa(x) = \tan^{-1}\ 2x -3ax+7\)
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, \(a_{max} = \frac{2}{3}(\frac{1}{1+4\times \frac{π^2}{36}})\)
\(= \frac{6}{9+π^2} = \overline{a}\)
\(\therefore fa(\frac{\pi}{8}) = \tan^{-1} \frac{\pi}{4}-3. \frac{\pi}{8} . \frac{6}{9+ \pi^2}+7\)