JEE Main 2020 — Definite Integration Question with Solution
From: JEE Main 2020 (Online) 9th January Evening Slot
Question
Let a function ƒ : [0, 5] R be continuous,
ƒ(1) = 3 and F be defined as :
, where
Then for the function F, the point x = 1 is :
, where
Then for the function F, the point x = 1 is :
Choose an option
Show full solutionCorrect option: C
Correct answer
Ca point of local minima.
Step-by-step explanation
F'(x) = x2g(x) = x2
F'(1) = (1)(0) = 0
Now, F''(x) = 2xg(x) + x2g'(x)
F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]
So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0
For the function f(x), x = 1 is a point of local minima.
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