Verified Solution • 2025–26 Session moderate

QJEEM25Maac62: JEE Main 2025 Mathematics - Class 12 JEEM25Maac62 Solution: Let be a twice differentiable function such that . If for all , …

Let f:R→R\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R} be a twice differentiable function such that f(2)=1f(2)=1 . If F(x)=xf(x)\mathrm{F}(\mathrm{x})=\mathrm{x} f(\mathrm{x}) for all x∈R\mathrm{x} \in \mathrm{R} , ∫02xF′(x)dx=6\int\limits_0^2 x F^{\prime}(x) d x=6 and ∫02x2F′′(x)dx=40\int\limits_0^2 x^2 F^{\prime \prime}(x) d x=40 , then F′(2)+∫02F(x)dxF^{\prime}(2)+\int\limits_0^2 F(x) d x is equal to :

Multiple Choice Options

  • A 13
  • B 11
  • C 9
  • D 15

Step-by-Step Verified Solution

amp;amp;∫02xF′(x)dx=6amp;=xF(x)∣02−∫02f(x)dx=6amp;=2 F(2)−∫02xF(x)dx=6[∴f(2)=2 F(2)=2]amp;∫02xF(x)dx=−2…(1)amp;⇒∫02 F(x)dx=−2…(2)amp; Also amp;amp;∫02x2F′′(x)dx=x2F′(x)∣02−2∫02xF′(x)dx=40amp;=4F′(2)−2×6=40amp;F′(2)=13amp;∴F′(2)+∫02F(x)=13−2=11\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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