JEE Mathematics 2024 Question with Solution

Given: The region is defined by

Find: The value of $m+n$ when the enclosed area is $\frac{m}{n}$.

Rewrite the inequalities in boundary form:

So for a fixed $y$, the right boundary is $x = 8 - 4y^2$ and the left boundary is the larger of $x = -2y^2$ and $x = 2y - 4$.

Find where the two left boundaries intersect:

Since $y \ge 0$, the relevant value is $y=1$.

Also, the top of the feasible region comes from the intersection of

which gives

Hence the relevant upper limit is $y=\frac{3}{2}$.

Therefore, split the area into two parts:

Evaluate the first integral:

Evaluate the second integral:

Now add the two parts:

So

with $m=107$ and $n=12$.

Therefore, $m+n = 107+12 = 119$. The correct option is A.

Given: The feasible region is bounded by a line, two parabolas, and $y \ge 0$. Find: The enclosed area and then $m+n$.

For horizontal integration, compare the candidate left boundaries:

• $x = -2y^2$
• $x = 2y - 4$

For $0 \le y \le 1$, we have $-2y^2 \ge 2y-4$, so the left boundary is $x=-2y^2$. For $1 \le y \le \frac{3}{2}$, we have $2y-4 \ge -2y^2$, so the left boundary is $x=2y-4$.

The common right boundary throughout the feasible part is

Hence the strip length changes at $y=1$, which is why the integral must be split there.

Using horizontal strips,

This yields

which evaluates to

Thus $m=107$, $n=12$, and therefore $m+n=119$. The correct option is A.