The function $2 \tan ^3 x-3 \tan ^2 x+12 \tan x+3, x \in\left(0, \frac{\pi}{2}\right)$ is
Answer & explanation
Correct answer: option 1
Let $f(x)=2 \tan ^3 x-3 \tan ^2 x+12 \tan x+3$
$f'(x)=\left(6 \tan ^2 x-6 \tan x+12\right) \sec ^2 x$
$=6 \sec ^2 x \left(\tan ^2 x-\tan x+2\right)>0$
Hence f(x) is always increasing.