JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Area Under Curves

The curves $y=1+x^2$ and $y=3-x$ meet where $x^2+x-2=0$, i.e. at $x=-2$ and $x=1$, with $3-x\ge 1+x^2$ between them. Let $g(x)=(3-x)-(1+x^2)=2-x-x^2$ with antiderivative $G(x)=2x-\frac{x^2}{2}-\frac{x^3}{3}$. The part of the area left of $x=-1$ is $G(-1)-G(-2)=\frac{7}{6}$ and the part right of it is $G(1)-G(-1)=\frac{20}{6}$. The line $x=-1$ therefore divides the area in the ratio $7:20$, in which $\gcd (7,20)=1$, so $m+n=27$.