Let f : R rightarrow R be a twice differentiable function such that f(2)=1 . If F ( x )= x f( x ) for all x in R , int 0 2 x F prime (x) d x=6 and int 0 2 x 2 F prime prime (x) d x=40 , then F prime (2)+int 0 2 F(x) d x is equal to :

Mathematics · Definite Integration

JEE Main 2025 — 28 January, Evening Shift — Question 12

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a twice differentiable function such that $f(2)=1$ . If $\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x})$ for all $\mathrm{x} \in \mathrm{R}$ , $\int_{0}^{2} x F^{\prime}(x) d x=6$ and $\int_{0}^{2} x^{2} F^{\prime \prime}(x) d x=40$ , then $F^{\prime}(2)+\int_{0}^{2} F(x) d x$ is equal to :

Option A:
11
Correct
Option B:
15
Option C:
9
Option D:
13

Answer: A

Step-by-step solution

$\int_{0}^{2} x F^{\prime}(x) d x=6$

$=\left.x F(x)\right|_{0} ^{2}-\int_{0}^{2} F(x) d x=6$

$=2 \mathrm{~F}(2)-\int_{0}^{2} \mathrm{xf}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{F}(2)=2 \mathrm{~f}(2)=2]$

$\int_{0}^{2} x f(x) d x=-2$

$\Rightarrow \int_{0}^{2} \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2$

Also $\int_{0}^{2} x^{2} F^{\prime \prime}(x) d x=\left.x^{2} F^{\prime}(x)\right|_{0} ^{2}-2 \int_{0}^{2} x F^{\prime}(x) d x=40$

$=4 \mathrm{~F}^{\prime}(2)-2 \times 6=40$

$F^{\prime}(2)=13$

$\therefore \mathrm{F}^{\prime}(2)+\int_{0}^{2} \mathrm{~F}(\mathrm{x})=13-2=11$

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Definite Integration
Topic: Evaluation of Definite Integrals

Let f be a real valued continuous function defined on the positive real axis such that g( x)=int 0^ x t f( t) dt . If g (x^3 )=x^6+x^7 …

Answer: A

Step-by-step solution

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