JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Application of Derivatives

Differentiating and factoring,

On $\left[-\frac{\pi }{6},\frac{\pi }{2}\right]$ the range of $\sin x$ is $\left[-\frac{1}{2},1\right]$, so $2\sin x+1\ge 0$ and $\sin x+2>0$, while $\cos x>0$ throughout the open interval. The sign of $f^{\prime }$ is thus the sign of $\sin x$.
For $0<x<\frac{\pi }{2}$, $\sin x>0$ and $f^{\prime }(x)>0$, so $f$ rises there. For $-\frac{\pi }{6}<x<0$, $\sin x<0$ and $f^{\prime }(x)<0$: $f$ is decreasing in $\left(-\frac{\pi }{6},0\right)$.