JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Application of Derivatives

Differentiating, $f^{\prime }(x)=\frac{12x^2-6x}{6}-2\cos x+[2\cos x+(2x-1)(-\sin x)]=2x^2-x-(2x-1)\sin x$. The first two terms factor as $x(2x-1)$, so $f^{\prime }(x)=(2x-1)(x-\sin x)$. The factor $x-\sin x$ has the same sign as $x$, positive for $x>0$ and negative for $x<0$. For $x\ge \frac{1}{2}$ both $2x-1\ge 0$ and $x-\sin x>0$, so $f^{\prime }(x)\ge 0$ and the function increases throughout $\left[\frac{1}{2},\infty \right)$. On the other side the two factors change sign at different points, so the function is not monotonic over the whole of $\left(-\infty ,\frac{1}{2}\right]$, and the valid statement is that it increases on $\left[\frac{1}{2},\infty \right)$.