JEE Main 2026 — 22 January, Morning Shift — Question 3

The region is bounded by $y = 1+x^2$ (parabola) and $y = 3-x$ (line). Find intersection: $1+x^2 = 3-x \Rightarrow x^2+x-2=0 \Rightarrow x=-2,1$. So the region is for $x \in [-2,1]$. Area to the left of $x=-1$: $A_1 = \int_{-2}^{-1} [(3-x)-(1+x^2)]\,dx = \int_{-2}^{-1} (2-x-x^2)\,dx$. Area to the right of $x=-1$: $A_2 = \int_{-1}^{1} [(3-x)-(1+x^2)]\,dx = \int_{-1}^{1} (2-x-x^2)\,dx$. Compute $A_1$: $\left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{-1} = \left(-2 - \frac{1}{2} + \frac{1}{3}\right) - \left(-4 -2 + \frac{8}{3}\right) = \frac{7}{6}$. Compute $A_2$: $\left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-1}^{1} = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-2 - \frac{1}{2} + \frac{1}{3}\right) = \frac{20}{6} = \frac{10}{3}$. Ratio $A_1 : A_2 = \frac{7}{6} : \frac{10}{3} = 7:20$. Thus $m=7, n=20$ (or vice versa) and $m+n=27$.