JEE MAIN - Mathematics (2025 - 28th January Evening Shift - No. 3)
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 $\mathrm{x} \in \mathrm{R}$, $\int\limits_0^2 x F^{\prime}(x) d x=6$ and $\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40$, then $F^{\prime}(2)+\int\limits_0^2 F(x) d x$ is equal to :
13
11
9
15
Explanation
$$\begin{aligned}
&\begin{aligned}
& \int_0^2 \mathrm{xF}^{\prime}(\mathrm{x}) \mathrm{dx}=6 \\
& =\left.\mathrm{xF}(\mathrm{x})\right|_0 ^2-\int_0^2 \mathrm{f}(\mathrm{x}) \mathrm{dx}=6 \\
& =2 \mathrm{~F}(2)-\int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=6[\therefore \mathrm{f}(2)=2 \mathrm{~F}(2)=2] \\
& \int_0^2 \mathrm{xF}(\mathrm{x}) \mathrm{dx}=-2 \quad \ldots(1) \\
& \Rightarrow \int_0^2 \mathrm{~F}(\mathrm{x}) \mathrm{dx}=-2\quad \ldots(2)
\end{aligned}\\
&\text { Also }\\
&\begin{aligned}
& \int_0^2 x^2 F^{\prime \prime}(x) d x=\left.x^2 F^{\prime}(x)\right|_0 ^2-2 \int_0^2 x F^{\prime}(x) d x=40 \\
& =4 F^{\prime}(2)-2 \times 6=40 \\
& F^{\prime}(2)=13 \\
& \therefore F^{\prime}(2)+\int_0^2 F(x)=13-2=11
\end{aligned}
\end{aligned}$$
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