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 $\mathrm{F}^{\prime}(2)+\int_{0}^{2} \mathrm{~F}(\mathrm{x}) \mathrm{dx}$ is equal to :

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Definite Integrals

Let

f : R \to 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^{'} (x) d x = 6

and

\int_{0}^{2} x^{2} F^{''} (x) d x = 40

, then

F^{'} (2) + \int_{0}^{2} F (x) dx

is equal to :

Answer: (A)

Given

F (x) = x f (x)

and

f (2) = 1

, so

F (2) = 2 f (2) = 2

.

For the first integral, using integration by parts:

\int_{0}^{2} x F^{'} (x) d x = [x F (x)]_{0}^{2} - \int_{0}^{2} F (x) d x = 2 F (2) - \int_{0}^{2} F (x) d x = 6

4 - \int_{0}^{2} F (x) d x = 6

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

For the second integral:

\int_{0}^{2} x^{2} F^{''} (x) d x = [x^{2} F^{'} (x)]_{0}^{2} - 2 \int_{0}^{2} x F^{'} (x) d x = 40

4 F^{'} (2) - 2 (6) = 40

4 F^{'} (2) = 52

F^{'} (2) = 13

Therefore:

F^{'} (2) + \int_{0}^{2} F (x) d x = 13 + (- 2) = 11

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option