Let a function ƒ: [0, 5] → R be continuous, ƒ(1) = 3 and F be defined as: F(x) = ∫₁ˣ t²g(t)dt, where g(t) = ∫₁^(t) f(u)du Then for the function F, the point x = 1 is:

Integral Calculus: JEE Main 2020 Mathematics question

JEE Main 2020 · 9 Jan, Evening shift · Q3307

JEE Main 20209 Jan, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

Let a function ƒ : [0, 5]

\to

R be continuous, ƒ(1) = 3 and F be defined as :

F(x) = \int\limits_1^x {{t^2}g(t)dt}

, where

g(t) = \int\limits_1^t {f(u)du}

Then for the function F, the point x = 1 is :