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MCQMediumJEE 2025Applications of Integrals (Area)

JEE Mathematics 2025 Question with Solution

The area of the region {(x,y):∣x−y∣≤y≤4x}\{(x, y): |x - y| \le y \le 4\sqrt{x}\} is

  • A

    512512

  • B

    10243\frac{1024}{3}

  • C

    5123\frac{512}{3}

  • D

    20483\frac{2048}{3}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The region is defined by ∣x−y∣≤y|x-y| \le y and y≤4xy \le 4\sqrt{x}.

Find: The area of the region.

From

−y≤x−y≤y-y \le x-y \le y

we get

−y≤x−y  ⟹  0≤x-y \le x-y \implies 0 \le x

and

x−y≤y  ⟹  x≤2y  ⟹  y≥x2x-y \le y \implies x \le 2y \implies y \ge \frac{x}{2}

So the region is bounded by y=x2y = \frac{x}{2} from below and y=4xy = 4\sqrt{x} from above, with x≥0x \ge 0.

Find the intersection points by solving

x2=4x\frac{x}{2} = 4\sqrt{x}

For x>0x>0,

(x2)2=(4x)2\left(\frac{x}{2}\right)^2 = (4\sqrt{x})^2 x24=16x\frac{x^2}{4} = 16x x2=64xx^2 = 64x x(x−64)=0x(x-64)=0

Hence, x=0x=0 or x=64x=64.

Therefore, the required area is

A=∫064(4x−x2) dxA = \int_0^{64} \left(4\sqrt{x}-\frac{x}{2}\right) \, dx

Evaluate the integral:

A=[83x3/2−x24]064A = \left[\frac{8}{3}x^{3/2} - \frac{x^2}{4}\right]_0^{64}

Substituting the limits,

A=83(64)3/2−(64)24A = \frac{8}{3}(64)^{3/2} - \frac{(64)^2}{4} A=83(83)−1024A = \frac{8}{3}(8^3) - 1024 A=40963−1024A = \frac{4096}{3} - 1024 A=4096−30723=10243A = \frac{4096-3072}{3} = \frac{1024}{3}

Therefore, the area of the region is 10243\frac{1024}{3}. The correct option is B.

Inequality to Boundary Conversion

Given: ∣x−y∣≤y≤4x|x-y| \le y \le 4\sqrt{x}

Find: The area enclosed by the corresponding curves.

Use the absolute value inequality carefully:

∣x−y∣≤y  ⟺  −y≤x−y≤y|x-y| \le y \iff -y \le x-y \le y

This gives two conditions:

−y≤x−y  ⟹  x≥0-y \le x-y \implies x \ge 0 x−y≤y  ⟹  x≤2y  ⟹  y≥x2x-y \le y \implies x \le 2y \implies y \ge \frac{x}{2}

Together with y≤4xy \le 4\sqrt{x}, the strip becomes

x2≤y≤4x,x≥0\frac{x}{2} \le y \le 4\sqrt{x}, \qquad x \ge 0

Now determine where the two boundary curves meet:

x2=4x\frac{x}{2} = 4\sqrt{x} x=8xx = 8\sqrt{x}

Squaring,

x2=64xx^2 = 64x x(x−64)=0x(x-64)=0

So the intersections occur at x=0x=0 and x=64x=64, corresponding to points (0,0)(0,0) and (64,32)(64,32).

Hence,

Area=∫064(4x1/2−12x) dx\text{Area} = \int_0^{64} \left(4x^{1/2} - \frac{1}{2}x\right) \, dx =[83x3/2−14x2]064= \left[\frac{8}{3}x^{3/2} - \frac{1}{4}x^2\right]_0^{64} =83(64)3/2−14(4096)= \frac{8}{3}(64)^{3/2} - \frac{1}{4}(4096) =83(512)−1024= \frac{8}{3}(512) - 1024 =10243= \frac{1024}{3}

Therefore, the correct option is B.

Common mistakes

  • A common mistake is to treat ∣x−y∣≤y|x-y| \le y as only x−y≤yx-y \le y and miss the condition x≥0x \ge 0 coming from −y≤x−y-y \le x-y. This gives an incomplete region. Always rewrite the absolute value inequality as a double inequality first.

  • Students often choose the wrong upper and lower curves. Here y=4xy = 4\sqrt{x} is the upper curve and y=x2y = \frac{x}{2} is the lower curve on the interval of intersection. Reversing them makes the area negative.

  • Another mistake is solving x2=4x\frac{x}{2} = 4\sqrt{x} incorrectly by squaring too early without noting x≥0x \ge 0. First note the domain, then solve carefully to get both intersections x=0x=0 and x=64x=64.

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