JEE Main 2024 — 31 January, Shift 2 — Question 5

$f(x)=\int_{-x}^{x}\left(|t|-t^{2}\right) e^{-t^{2}} d t$

$\Rightarrow f^{\prime}(x)=2 \cdot\left(|x|-x^{2}\right) e^{-x^{2}}$

$g(\mathrm{x})=\int_{0}^{\mathrm{x}^{2}} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}$

$g^{\prime}(x)=x e^{-x^{2}}(2 x)-0$

$f^{\prime}(x)+g^{\prime}(x)=2 x e^{-x^{2}}-2 x^{2} e^{-x^{2}}+2 x^{2} e^{-x^{2}}$

Integrating both sides w.r.t.x

$f(x)+g(x)=\int_{0}^{\alpha} 2 x e^{-x^{2}} d x$

$\mathrm{x}^{2}=\mathrm{t}$

$\Rightarrow \int_{0}^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_{0}^{\sqrt{\alpha}}$ $=-\mathrm{e}^{\left(\log _{e}(9)^{-1}\right)+1}$

$\Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8$