Verified Solution • 2025–26 Session moderate

QJEEM20Maf327: JEE Main 2020 Mathematics - Class 12 JEEM20Maf327 Solution: Let a function ƒ : [0, 5] R be continuous, ƒ(1) = 3 and F be…

Let a function ƒ : [0, 5] →\to R be continuous, ƒ(1) = 3 and F be defined as :

F(x)=∫1xt2g(t)dtF(x) = \int\limits_1^x {{t^2}g(t)dt} , where g(t)=∫1tf(u)dug(t) = \int\limits_1^t {f(u)du}

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

Multiple Choice Options

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

Step-by-Step Verified Solution

F(x)=∫1xt2g(t)dtF(x) = \int\limits_1^x {{t^2}g(t)dt}

⇒\Rightarrow F'(x) = x2g(x) = x2 ∫1tf(u)du\int\limits_1^t {f(u)du}

∴\therefore 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

∴\therefore For the function f(x), x = 1 is a point of local minima.

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