JEE Mathematics 2024 Question with Solution

Given: The region is $0 \le x \le 3$ and $0 \le y \le \min(x^2 + 2, 2x + 2)$.

Find: The value of $12A$, where $A$ is the area of the region.

The upper boundary is the smaller of the two curves $y = x^2 + 2$ and $y = 2x + 2$. Their intersection point is found from

$$x^2 + 2 = 2x + 2$$

which gives

$$x^2 - 2x = 0$$ $$x(x-2)=0$$

So the relevant changeover point in the interval $[0,3]$ is $x = 2$.

Hence,

$$A = \int_{0}^{2} (x^2 + 2) \, dx + \int_{2}^{3} (2x + 2) \, dx$$

Now evaluate each integral:

$$\int_{0}^{2} (x^2 + 2) \, dx = \left[ \frac{x^3}{3} + 2x \right]_0^2 = \frac{8}{3} + 4 = \frac{20}{3}$$ $$\int_{2}^{3} (2x + 2) \, dx = \left[ x^2 + 2x \right]_2^3 = (9+6) - (4+4) = 7$$

Therefore,

$$A = \frac{20}{3} + 7 = \frac{41}{3}$$

So,

$$12A = 12 \times \frac{41}{3} = 164$$

Therefore, the required value is $164$.

The solution explicitly concludes that

$$A = \frac{41}{3}$$

and therefore

$$12A = 12 \times \frac{41}{3} = 164$$

So the numerical value answer is $164$.

There is a discrepancy in the second approach shown in the working: it writes an incorrect intermediate integral limit, but its final conclusion still states $164$. The first approach is consistent and gives the correct result.