JEE Main 2025 — 3 April, Evening Shift — Question 38

Simplify the Boundaries The inequality $|x-y| \leq y$ can be expanded as:

$$-y \leq x-y \leq y$$

Taking the right side:

$$x-y \leq y \implies x \leq 2y \implies y \geq \frac{x}{2}$$

The upper boundary is given as $y \leq 4\sqrt{x}$. Thus, the region is bounded by: Upper Curve: $y = 4\sqrt{x}$ Lower Line: $y = \frac{x}{2}$ Find Intersection Points Set the two equations equal to find the bounds of integration:

$$4\sqrt{x} = \frac{x}{2} \implies 8\sqrt{x} = x$$

Squaring both sides:

$$64x = x^2 \implies x^2 - 64x = 0 \implies x(x - 64) = 0$$

The intersection points are $(0, 0)$ and $(64, 32)$.

Setup the Integral The area $A$ is given by:

$$A = \int_{0}^{64} \left( 4\sqrt{x} - \frac{x}{2} \right) dx$$

Evaluate the Integral

$$A = \left[ 4 \cdot \frac{x^{3/2}}{3/2} - \frac{1}{2} \cdot \frac{x^2}{2} \right]_{0}^{64}$$ $$A = \left[ \frac{8}{3}x^{3/2} - \frac{x^2}{4} \right]_{0}^{64}$$

Substituting the upper limit $x = 64$:

$$A = \frac{8}{3}(64)^{3/2} - \frac{(64)^2}{4}$$ $$A = \frac{8}{3}(512) - \frac{4096}{4}$$ $$A = \frac{4096}{3} - 1024$$ $$A = \frac{4096 - 3072}{3} = \frac{1024}{3}$$

Final Answer: The area of the region is $\frac{1024}{3}$ square units.