Let f: R arrow R be a twice differentiable function such that f(2)=1 . If F(x)=x f(x) for all x ∈ R , ∫_0^2 x ~F^ (x) d x=6 and ∫_0^2 x^2 ~F^ (x) d x=40 , then F^ (2)+∫_0^2 ~F(x) d x is equal to :

Correct answer: A. 11 aligned & ∫_0^2 xF^ (x) dx=6 & =.xF(x)|_0 ^2-∫_0^2 f(x) dx=6 & =2 ~F(2)-∫_0^2 xF(x) dx=6[ f(2)=2 ~F(2)=2] & ∫_0^2 xF(x) dx=-2 ...(1) & ∫_0^2 ~F(x) dx=-2 ...(2) aligned Also ∫_0^2 x^2 F^ (x) d x=.x^2 F^ (x)|_0 ^2-2 ∫_0^2 x^ F^ (x) d x=40 aligned & =4 F^ (2)-2 × 6=40 & F^ (2)=13 & F^ (2)+∫_0^2 F(x)=13-2=11 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Definite Integration

Let f: R arrow R be a twice differentiable function such that...

Let

\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}

be a twice differentiable function such that

f(2)=1

. If

\mathrm{F}(x)=x f(x)

for all

x \in \mathbf{R}

\int_0^2 x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6

and

\int_0^2 x^2 \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40

, then

\mathrm{F}^{\prime}(2)+\int_0^2 \mathrm{~F}(x) \mathrm{d} x

is equal to :

Options

A $11$
B $13$
C $15$
D $9$

Correct answer

A. $11$

Step-by-step solution

\begin{aligned} & \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\ & =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\ & =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\ & \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 ...(1)\\ & \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2 ...(2) \end{aligned}

Also

\int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x^{\prime} F^{\prime}(x) d x=40

\begin{aligned} & =4 F^{\prime}(2)-2 \times 6=40 \\ & F^{\prime}(2)=13 \\ & \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11\end{aligned}

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