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

BITSATBITSAT 2010

Solution:

$f(x)=\tan x-4 x \Rightarrow f'(x)=\sec ^{2} x-4$
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
$\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$