JEE Mathematics 2023 Question with Solution

Given: The curve is $y^2 = 8x$ and the lines are $y = x$ and $x = 2$ in the first quadrant.

Find: The value of $3\alpha$, where $\alpha$ is the area of the larger bounded region.

First find the intersection points of $y = x$ and $y^2 = 8x$.

So, $x=0$ or $x=8$. Since $y=x$, the intersection points are $O(0,0)$ and $A(8,8)$.

The line $x=2$ meets the parabola $y^2=8x$ at

so $y=\pm 4$. In the first quadrant, the relevant point is $Q(2,4)$.

For $x$ from $2$ to $8$, the upper branch of the parabola is $y=\sqrt{8x}$ and the line is $y=x$. Hence the required area is

Now evaluate the integral:

Using $\sqrt{8x}=2\sqrt{2}\,x^{1/2}$,

Also,

Therefore,

Hence,

Therefore, the required numerical value is $22$.

The solution contains an intermediate inconsistent simplification for $\int_{2}^{8}\sqrt{8x}\,dx$, but its final conclusion $\alpha = \frac{22}{3}$ and hence $3\alpha = 22$ is correct.