Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minPYQ_2020
MathematicsMediumsingle choice

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:

Options:

Answer:
A
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.

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima
⏱ 2mℹ️ Source: PYQ_2020

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