Q47 / 75 · JEE Main 2020 (09 Jan Shift 2) · Mathematics

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:

  • Correct answer
    A
    a point of local minima
  • B
    not a critical point
  • C
    a point of local maxima
  • D
    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.

Hence, the correct answer is option: 'A'.

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