JEE PYQ: Application Of Derivatives Question 21
Question 21 - 2021 (26 Feb Shift 1)
The maximum slope of the curve $y = \frac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$ occurs at the point:
(1) $(2, 9)$
(2) $(2, 2)$
(3) $\left(3, \frac{21}{2}\right)$
(4) $(0, 0)$
Show Answer
Answer: (2)
Solution
Slope $= \frac{dy}{dx} = 2x^3 - 15x^2 + 36x - 19$. Setting $\frac{d^2y}{dx^2} = 6x^2 - 30x + 36 = 0 \Rightarrow x = 2, 3$. At $x = 2$: $\frac{d^3y}{dx^3} = 12(2) - 30 < 0$, so maximum. $y(2) = 8 - 40 + 72 - 38 = 2$. Point is $(2, 2)$.
BETA
