Let a function f : 0 , 5 → R be continuous, f 1 = 3 and F be defined as: F x = ∫ 1 x t 2 g t d t…

Let a function f:0,5→R be continuous, f1=3 and F be defined as:
Fx=∫1xt2gtdt, where gt=∫1tfudu.

Then for the function Fx, 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

Solution

Given, Fx=∫1xt2gtdt

By Leibnitz rule we get, 

F'x=x2gx 

⇒F'1=1.g1=0 ∵g1=0

Now F''x=2xgx+x2g'x

⇒F''x=2xgx+x2fx ∵g'x=fx

⇒F''1=0+1×3

⇒F''1=3

Fx has a local minimum at x=1.

Asked in: JEE Main 2020 (09 Jan Shift 2)

Practice more Applications of Derivatives questions on Aicharya