JEE Main 2026 — 28 January, Evening Shift — Question 23

$f(x)=1-2 x+e^{x} \int_{0}^{x} e^{-t} f(t) d t$

$e^{-x} f(x)=(1-2 x) e^{-x}+\int_{0}^{x} e^{-t} f(t) d t$ $e^{-x} f^{\prime}(x)-e^{-x} f(x)=-2 e^{-x}+(1-2 x) e^{-x}(-1)+e^{-x} f(x)$

$\mathrm{f}^{\prime}(\mathrm{x})-2 \mathrm{f}(\mathrm{x})=2 \mathrm{x}-3$

$\frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}=2 \mathrm{x}-3$

$\Rightarrow \mathrm{y} . \mathrm{e}^{-2 \mathrm{x}}=\int \mathrm{e}^{-2 \mathrm{x}}(2 \mathrm{x}-3) \mathrm{dx}$

On solving we get $\mathrm{f}(\mathrm{x})=1-\mathrm{x}$

$g(x)=\int_{0}^{x}(3-t)^{15}(t-4)^{6}(t+12)^{17} d t$

$g^{\prime}(x)=(3-x)^{15}(x-4)^{6}(x+12)^{17}$

$=-(x-3)^{15}(x-4)^{6}(x+12)^{17}$

Local maxima $\Rightarrow \mathrm{q}=3$

Local minima $\Rightarrow p=-12=|p+q|=9$