MCQMediumJEE Main 2025 · 28 January, Shift 2Definite Integrals

Mathematics Question from JEE Main 2025 · 28 January, Shift 2

Let f:RRf : \mathbb{R} \to \mathbb{R} be a twice-differentiable function such that f(2)=1f(2) = 1. If F(x)=xf(x)F(x) = x f(x) for all xRx \in \mathbb{R}, and the integrals 02xF(x)dx=6\int_0^2 x F'(x) \, dx = 6 and 02x2F(x)dx=40\int_0^2 x^2 F''(x) \, dx = 40, then F(2)+02F(x)dxF'(2) + \int_0^2 F(x) \, dx is equal to:

  • A

    1111

  • B

    1515

  • C

    99

  • D

    1313

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