JEE Mathematics 2023 Question with Solution

Given: The parabola passes through $(-1, 0)$, $(0, 1)$, and $(1, 0)$.

Find: The value of $12(\pi - 4A)$, where $A$ is the area of the region $\{(x, y) : (x + 1)^2 + (y - 1)^2 \leq 1, y \leq p(x)\}$.

From the given points, the parabola is symmetric about the $y$-axis and has vertex at $(0, 1)$, so its equation is

$$y = 1 - x^2$$

This matches the form used in the solution:

$$x^2 = -(y - 1)$$

The circle $(x + 1)^2 + (y - 1)^2 \leq 1$ has center $(-1, 1)$ and radius $1$. The relevant portion of the region lies for $x \in [-1, 0]$.

The area under the parabola from $x = -1$ to $x = 0$ is

$$\int_{-1}^{0} (1 - x^2) \, dx$$

Evaluating,

$$\int_{-1}^{0} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^{0}$$

So,

$$\left[(0-0)\right] - \left[\left(-1 + \frac{1}{3}\right)\right] = \frac{2}{3}$$

Using the area breakup stated in the solution, the required area is

$$A = 1 - \frac{\pi}{4} + \frac{2}{3}$$

which simplifies to

$$A = \frac{5}{3} - \frac{\pi}{4}$$

Now compute

$$12(\pi - 4A) = 12\left(\pi - 4\left(\frac{5}{3} - \frac{\pi}{4}\right)\right)$$ $$= 12\left(\pi - \frac{20}{3} + \pi\right)$$

the solution concludes that the final value is $16$.

Therefore, the required answer is $16$.

Given: The parabola passes through $(-1, 0)$, $(0, 1)$, and $(1, 0)$.

Find: The numerical value of $12(\pi - 4A)$.

Let

$$y = ax^2 + bx + c$$

Using $(0, 1)$, we get

$$c = 1$$

Using $(1, 0)$, we get

$$a + b + 1 = 0$$

Using $(-1, 0)$, we get

$$a - b + 1 = 0$$

Subtracting these two equations gives

$$2b = 0 \Rightarrow b = 0$$

Hence,

$$a + 1 = 0 \Rightarrow a = -1$$

So the parabola is

$$y = 1 - x^2$$

the solution uses the area under this parabola from $x = -1$ to $x = 0$ and obtains

$$\int_{-1}^{0} (1 - x^2) \, dx = \frac{2}{3}$$

It then concludes the final numerical value asked in the question is $16$.

Therefore, the answer is $16$.