JEE Mathematics 2023 Question with Solution

Given: The region is defined by $y \geq x^2$, $y \leq 8-x^2$, and $y \leq 7$.

Find: The area of the region.

From the given inequalities, the boundary curves are $y = x^2$, $y = 8-x^2$, and the line $y = 7$. The intersection points mentioned in the solution are $(-2,4)$ and $(2,4)$ for $y=x^2$ with $y=8-x^2$, and $(-1,7)$ and $(1,7)$ for $y=7$ with $y=8-x^2$.

The area is set up as

Now evaluate the first integral:

So,

Next evaluate the second integral:

Substitute these values into the area expression:

As simplified in the provided solution,

Therefore, the area of the region is $20$. The correct option is C.

Given: The inequalities are $y \geq x^2$, $y \leq 8-x^2$, and $y \leq 7$.

Find: The enclosed area.

The key idea is to identify where the upper boundary changes. The curves $y=x^2$ and $y=8-x^2$ intersect at $x=\pm 2$, giving points $(-2,4)$ and $(2,4)$. The line $y=7$ intersects $y=8-x^2$ at $x=\pm 1$, giving points $(-1,7)$ and $(1,7)$.

The provided solution uses symmetry and writes the area in the form

Then each integral is computed directly.

For $\int_{-1}^{1} (8-2x^2) \, dx$,

For $\int_{0}^{1} x^2 \, dx$,

Substituting back and simplifying as shown in the solution gives

Hence, the correct option is C.