JEE PYQ: Application Of Derivatives Question 13
Question 13 - 2021 (24 Feb Shift 2)
Let $f : \mathbb{R} \to \mathbb{R}$ be defined as
$$f(x) = \begin{cases} -55x, & \text{if } x < -5 \ 2x^3 - 3x^2 - 120x, & \text{if } -5 \le x \le 4 \ 2x^3 - 3x^2 - 36x - 336, & \text{if } x > 4 \end{cases}$$
Let $A = {x \in \mathbb{R} : f \text{ is increasing}}$. Then $A$ is equal to:
(1) $(-5, -4) \cup (4, \infty)$
(2) $(-5, \infty)$
(3) $(-\infty, -5) \cup (4, \infty)$
(4) $(-\infty, -5) \cup (-4, \infty)$
Show Answer
Answer: (1)
Solution
For $-5 \le x \le 4$: $f’(x) = 6(x^2 - x - 20) = 6(x-5)(x+4)$. For $x > 4$: $f’(x) = 6(x-3)(x+2) > 0$. $f$ is increasing on $(-5,-4) \cup (4, \infty)$.
BETA
