Let $f(x)=2 \sin ^3 x-3 \sin ^2 x+12 \sin x+5$, $0 \leq x \leq \frac{\pi}{2}$. Then, f(x), is
Answer & explanation
Correct answer: option 2
We have,
$f(x)=2 \sin ^3 x-3 \sin ^2 x+12 \sin x+5$
$\Rightarrow f'(x)=6 \sin ^2 x \cos x-6 \sin x \cos x+12 \cos x$
$\Rightarrow f'(x)=6\left(\sin ^2 x-\sin x+2\right) \cos x$
Since discriminant of the quadratic $\sin ^2 x-\sin x+2$ in $\sin x$ is negative.
∴ $\sin ^2 x-\sin x+2>0$ for all $x \in R$
⇒ Sign of f'(x) is same as that of cos x
⇒ f'(x) > 0 for all $x \in(0, \pi / 2)$
⇒ f(x) is increasing on $[0, \pi / 2]$