JEE Main 2021 · 24 Feb · Shift 1 · Mathematics Q67

MathematicsSingle correct+4 / -1Question 67 of 90

The function $f(x)=\dfrac{4x^3-3x^2}6-2\sin x+(2x-1)\cos x$ :

increases in

\left[\dfrac12,\infty\right)

decreases in

\left[\dfrac12,\infty\right)

increases in

\left(-\infty,\dfrac12\right]

decreases in

\left(-\infty,\dfrac12\right]

Answer and worked solution

Official answer: A

Derivative $f'(x)=2x^2-x-(2x-1)\sin x=(2x-1)(x-\sin x)$ . For $x\ge\frac12$ both factors are non-negative, so f increases there.

(A)

Topics:Monotonicity

JEE Main 2021 · 24 Feb · Shift 1, Mathematics question 67