JEE Mathematics 2024 Question with Solution

Given: The region is $\{(x,y): y^2 \le 2x,\; y \ge 4x-1\}$.

Find: The area enclosed by the parabola and the line.

From $y^2 \le 2x$, the parabola is

$$y^2 = 2x$$

which can be written as

$$x = \frac{y^2}{2}.$$

The line $y = 4x-1$ can be written as

$$x = \frac{y+1}{4}.$$

For the common region, the line lies to the left of the parabola in terms of $x$, so the area is

$$\int \left(\frac{y+1}{4} - \frac{y^2}{2}\right) dy$$

between the intersection points.

Find the intersection points using

$$y^2 = 2x, \qquad y = 4x-1.$$

Substitute $x = \frac{y+1}{4}$ into $y^2 = 2x$:

$$y^2 = 2\cdot \frac{y+1}{4} = \frac{y+1}{2}.$$

So,

$$2y^2 - y - 1 = 0.$$

Factorizing,

$$(2y+1)(y-1)=0.$$

Hence,

$$y = 1, \; -\frac{1}{2}.$$

Therefore, the required area is

$$\int_{-1/2}^{1} \left(\frac{y+1}{4} - \frac{y^2}{2}\right) dy.$$

Now integrate:

$$\int \left(\frac{y+1}{4} - \frac{y^2}{2}\right) dy = \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6}.$$

So,

$$\text{Area} = \left[\frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6}\right]_{-1/2}^{1}.$$

At $y=1$,

$$\frac{1}{8} + \frac{1}{4} - \frac{1}{6} = \frac{5}{24}.$$

At $y=-\frac{1}{2}$,

$$\frac{1}{32} - \frac{1}{8} + \frac{1}{48} = -\frac{7}{96}.$$

Thus,

$$\text{Area} = \frac{5}{24} - \left(-\frac{7}{96}\right) = \frac{27}{96} = \frac{9}{32}.$$

Therefore, the area of the region is $\frac{9}{32}$ and the correct option is D.

The solution is inconsistent with this question and appears to belong to a different problem, so the answer has been derived from the given question and options.