Let $f:R\to R$ be a twice differentiable function such that $f(2)=1$. If $F(x)=xf(x)$ for all $x\in R,\text{ }{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=6$ and ${\int }_{0}^{2}{x}^{2}{\mathrm{F}}^{"}\left(x\right)\mathrm{d}x=40$, then ${F}^{'}\left(2\right)+{\int }_{0}^{2}F\left(x\right)dx$ is equal to : [JEE Main 2025, 28 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: c) 11. $F\left(x\right)=xf\left(x\right)\ \Rightarrow F\left(x\right)=2f\left(2\right)=2$ $\text{Given that}{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=6\ ={\left.xF\left(x\right)\right|}_{0}^{2}-{\int }_{0}^{2}F\left(x\right)dx=6\left[\text{using integration by parts}\right]\ =2\mathrm{F}(2)-{\int }_{0}^{2}xf\left(x\right)\mathrm{dx}=6\ \Rightarrow {\int }_{0}^{2}\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=-2....\left(1\right)\ \mathrm{Also}\ {\int }_{0}^{2}{x}^{2}{F}^{"}\left(x\right)dx={\left.{x}^{2}{F}^{'}\left(x\right)\right|}_{0}^{2}-2{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=40\ =4{\mathrm{F}}^{'}\left(2\right)-2\times 6=40\ {F}^{'}\left(2\right)=13\ ∴{\mathrm{F}}^{'}\left(2\right)+{\int }_{0}^{2}\mathrm{F}\left(\mathrm{x}\right)=13-2\ =11$

Let \(f:R\to R\) be a twice differentiable function such that \(f(2)=1\). If \(F(x)=xf(x)\) for all \(x\in R,\text{ }{\i…

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