JEE PYQ: Application Of Derivatives Question 54
Question 54 - 2020 (09 Jan Shift 2)
Let a function $f : [0, 5] \to \mathbb{R}$ be continuous, $f(1) = 3$ and $F$ be defined as:
$$F(x) = \int_1^x t^2 g(t),dt, \text{ where } g(t) = \int_1^t f(u),du$$
Then for the function $F$, the point $x = 1$ is:
(1) a point of local minima
(2) not a critical point
(3) a point of local maxima
(4) a point of inflection
Show Answer
Answer: (1)
Solution
$F’(x) = x^2 g(x)$. At $x = 1$: $g(1) = 0$, so $F’(1) = 0$. $F’’(x) = 2xg(x) + x^2 f(x)$. $F’’(1) = 0 + f(1) = 3 > 0$. So $x = 1$ is a local minimum.
BETA
