JEE PYQ: Application Of Derivatives Question 10
Question 10 - 2021 (24 Feb Shift 1)
The function $f(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x$:
(1) increases in $\left[\frac{1}{2}, \infty\right)$
(2) decreases in $\left(-\infty, \frac{1}{2}\right]$
(3) increases in $\left(-\infty, \frac{1}{2}\right]$
(4) decreases in $\left[\frac{1}{2}, \infty\right)$
Show Answer
Answer: (1)
Solution
$f’(x) = (2x - 1)(x - \sin x) \ge 0$ for $x \ge \frac{1}{2}$ since $x \ge \sin x$ for $x \ge 0$.
BETA
