JEE Mathematics 2025 Question with Solution

Given: The curves are $y = 4 - \frac{x^2}{4}$ and $y = \frac{x - 4}{2}$.

Find: The value of $6\alpha$, where $\alpha$ is the area bounded by the two curves.

First, find the points of intersection by equating the two curves:

$$4 - \frac{x^2}{4} = \frac{x - 4}{2}$$

Multiplying by $4$:

$$16 - x^2 = 2(x - 4)$$ $$16 - x^2 = 2x - 8$$

Rearranging:

$$x^2 + 2x - 24 = 0$$

Solving the quadratic equation:

$$x = \frac{-2 \pm \sqrt{4 + 96}}{2} = \frac{-2 \pm 10}{2}$$ $$x = 4 \quad \text{or} \quad x = -6$$

Now determine the upper and lower curves. At $x = 0$:

$$\begin{aligned} \text{Parabola:} &\quad y = 4 - 0 = 4 \\ \text{Line:} &\quad y = \frac{0 - 4}{2} = -2 \end{aligned}$$

So the parabola lies above the line on $[-6, 4]$.

Hence,

$$\alpha = \int_{-6}^{4} \left[\left(4 - \frac{x^2}{4}\right) - \left(\frac{x - 4}{2}\right)\right] dx$$

Simplifying the integrand:

$$6 - \frac{x^2}{4} - \frac{x}{2}$$

Therefore,

$$\alpha = \int_{-6}^{4} \left(6 - \frac{x^2}{4} - \frac{x}{2}\right) dx$$

Break the integral into parts:

$$\int_{-6}^{4} 6 \, dx = 6(4 - (-6)) = 60$$ $$\int_{-6}^{4} \frac{x^2}{4} \, dx = \frac{1}{12}[x^3]_{-6}^{4} = \frac{1}{12}(64 - (-216)) = \frac{280}{12} = \frac{70}{3}$$ $$\int_{-6}^{4} \frac{x}{2} \, dx = \frac{1}{4}[x^2]_{-6}^{4} = \frac{1}{4}(16 - 36) = -5$$

Combining these results:

$$\alpha = 60 - \frac{70}{3} - (-5) = 65 - \frac{70}{3} = \frac{125}{3}$$

Now calculate:

$$6\alpha = 6 \times \frac{125}{3} = 250$$

Therefore, the correct option is A.

Given: The bounded area between $y = 4 - \frac{x^2}{4}$ and $y = \frac{x - 4}{2}$.

Find: The value of $6\alpha$.

Using the intersection points $x = -6$ and $x = 4$, the area is

$$\alpha = \int_{-6}^{4} \left\{\left(4 - \frac{x^2}{4}\right) - \left(\frac{x - 4}{2}\right)\right\} dx$$

Simplify:

$$\alpha = \int_{-6}^{4} \left\{-\frac{x^2}{4} - \frac{x}{2} + 6\right\} dx$$

Now integrate directly:

$$\alpha = \left[-\frac{x^3}{12} - \frac{x^2}{4} + 6x\right]_{-6}^{4}$$ $$\alpha = \frac{125}{3}$$

Therefore,

$$6\alpha = 6 \times \frac{125}{3} = 250$$

So, the correct option is A. This works because the required area is the integral of upper curve minus lower curve over the interval between the intersection points.