JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 10)

Let $$f, g:(0, \infty) \rightarrow \mathbb{R}$$ be two functions defined by $$f(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d t$$ and $$g(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d t$$. Then, the value of $$9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)$$ is equal to :
10
9
8
6

Explanation

$$\begin{aligned} & \mathrm{f}(\mathrm{x})=\int_\limits{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\ & \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\ & \mathrm{g}(\mathrm{x})=\int_\limits0^{\mathrm{x}^2} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-t} \mathrm{dt} \\ & \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}^2}(2 \mathrm{x})-0 \\ & \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{xe}^{-\mathrm{x}^2}-2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}+2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2} \end{aligned}$$

Integrating both sides w.r.t. $$\mathrm{x}$$

$$\begin{aligned} & \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\int_\limits0^\alpha 2 \mathrm{xe}^{-\mathrm{x}^2} \mathrm{dx} \\ & \mathrm{x}^2=\mathrm{t} \\ & \Rightarrow \int_0^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_0^{\sqrt{\alpha}} \\ & =-\mathrm{e}^{\left(\log _c(9)^{-1}\right)+1} \\ & \Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8 \end{aligned}$$

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