Question

The function $f(x) = \tan x - 4x$ is strictly decreasing on

• A. $\left( - \frac{\pi}{3} , \frac{\pi}{3} \right)$
• B. $\left( \frac{\pi}{3} , \frac{\pi}{2} \right)$
• C. $\left( - \frac{\pi}{3} , \frac{\pi}{2} \right)$
• D. $\left( \frac{\pi}{2} ,\pi \right)$

Answer 1

Answer: (A)

$\left( - \frac{\pi}{3} , \frac{\pi}{3} \right)$

Answer 2

The correct option is (A): $\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$.
$f(x)=\tan x-4 x \Rightarrow f'(x)=\sec ^{2} x-4$
When $\frac{-\pi}{3} < x < \frac{\pi}{3}, 1 < \sec x < 2$
Therefore, $1 < \sec ^{2} x < 4$
$\Rightarrow-3 < \left(\sec ^{2} x-4\right) < 0$
Thus, for $\frac{-\pi}{3} < x < \frac{\pi}{3}, f'(x) < 0$
Hence, $f$ is strictly decreasing on
$\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$