JEE Mathematics 2026 Question with Solution

Given: The region is bounded by $y = 1 + x^2$ and $y = 3 - x$, and the line $x = -1$ divides it into two parts.

Find: If the ratio is $m:n$ in lowest terms, find $m+n$.

First find the intersection points of the two curves:

So the region extends from $x=-2$ to $x=1$.

On this interval, the upper curve is $y=3-x$ and the lower curve is $y=1+x^2$. Hence the vertical strip height is

The area to the left of $x=-1$ is

and the area to the right of $x=-1$ is

An antiderivative is

Now evaluate the left part:

At $x=-1$, the value is $-2 - \frac{1}{2} + \frac{1}{3} = -\frac{13}{6}$. At $x=-2$, the value is $-4 - 2 + \frac{8}{3} = -\frac{10}{3}$. Therefore,

Now evaluate the right part:

At $x=1$, the value is $2 - \frac{1}{2} - \frac{1}{3} = \frac{7}{6}$. At $x=-1$, the value is $-\frac{13}{6}$. Therefore,

Hence the ratio of left area to right area is

So the two parts are in the ratio $7:20$. Therefore, taking $m:n = 20:7$ or $7:20$ changes only the order, but in either case

The solution lists Option B but its own computation ends at $27$. The worked integrals support $m+n=27$, so the correct option is A.

The bounded region exists where the line lies above the parabola. Solve

which reduces to

with roots $-2$ and $1$. So the enclosed part is between these two $x$-values.

Because the dividing line is vertical, split the area at $x=-1$. Then use upper minus lower in each definite integral:

Thus,

Multiply both by $6$ to write the ratio without fractions:

Hence the sum of the coprime terms is