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

To solve this problem, we need to determine the conditions under which the function \( fa(x) = \tan^{-1} 2x - 3ax + 7 \) is non-decreasing in the interval \((- \frac{\pi}{6}, \frac{\pi}{6})\) and then evaluate \( f\overline{a}(\frac{\pi}{8}) \) for the maximum value of \( a \). Let's go through this step-by-step:

1. First, compute the derivative of \( fa(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left( \tan^{-1} 2x - 3ax + 7 \right) \] Using the chain rule, the derivative of \( \tan^{-1}(2x) \) is: \[ \frac{d}{dx}(\tan^{-1}(2x)) = \frac{2}{1+(2x)^2} = \frac{2}{1+4x^2} \] And the derivative of \( -3ax \) is: \[ \frac{d}{dx}(-3ax) = -3a \] So, the total derivative is: \[ f'(x) = \frac{2}{1+4x^2} - 3a \]
2. For \( fa(x) \) to be non-decreasing, \( f'(x) \geq 0 \): \[ \frac{2}{1+4x^2} - 3a \geq 0 \Rightarrow \frac{2}{1+4x^2} \geq 3a \] Solving for \( a \), we have: \[ a \leq \frac{2}{3(1+4x^2)} \] The function should be non-decreasing for all \( x \) in \((- \frac{\pi}{6}, \frac{\pi}{6})\). Thus, consider the endpoint \( x = \pm \frac{\pi}{6} \): \[ a \leq \frac{2}{3\left(1 + 4 \left(\frac{\pi}{6}\right)^2\right)} \] \[ a \leq \frac{2}{3\left(1 + \frac{4\pi^2}{36}\right)} = \frac{2}{3\left(1 + \frac{\pi^2}{9}\right)} \]
3. The maximum value of \( a \) is thus: \[ \overline{a} = \frac{2}{3\left(1 + \frac{\pi^2}{9}\right)} \] We then calculate \( f(\frac{\pi}{8}) \): \[ f\overline{a}(\frac{\pi}{8}) = \tan^{-1}(2 \times \frac{\pi}{8}) - 3\overline{a}\times \frac{\pi}{8} + 7 \] Substitute \(\overline{a}\): \[ f\overline{a}(\frac{\pi}{8}) = \tan^{-1}\left(\frac{\pi}{4}\right) - \frac{\pi}{8} \times \frac{2}{3(1 + \frac{\pi^2}{9})} \times 3 + 7 \] \[ = \tan^{-1}\left(\frac{\pi}{4}\right) - \frac{\pi}{4(9+\pi^2)} + 7 \] \[ = \frac{\pi}{4} - \frac{\pi}{4(9+\pi^2)} + 7 \] \[ = 8 - \frac{9\pi}{4(9+\pi^2)} \]

Therefore, the correct answer is \( 8-\frac{9\pi}{4(9+\pi^2)} \).