JEE Main 2021 (Online) 24th February Morning ShiftQuestion 1
The function $f(x)=\frac{4x^{3}-3x^{2}}{6}-2\sin x+(2x-1)\cos x$ :
Differentiation and its Application
Solution
Concept. $f'$ factorises as $(2x-1)(x-\sin x)$, and $x-\sin x$ has the sign of $x$. Formulas used. (1) $\dfrac{d}{dx}\left[(2x-1)\cos x\right]=2\cos x-(2x-1)\sin x$(2) $x-\sin x>0$ for $x>0$ and $<0$ for $x<0$Step 1 - $f'$. By (1), $$f'(x)=2x^2-x-2\cos x+2\cos x-(2x-1)\sin x=(2x-1)(x-\sin x).$$ Step 2 - Signs. By (2): for $x\ge\frac12$ both factors are $\ge0$, so $f'\ge0$. For $0<x<\frac12$, $f'<0$; for $x<0$, $f'>0$. Step 3 - Conclude. $f$ increases on $\left[\frac12,\infty\right)$; it is not monotonic on $\left(-\infty,\frac12\right]$. Answer: (A)