Question

The function $f ( x )=\tan\, x -4 x$ 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)

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