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Maths Question 19 – JEE-MAIN 2025

The area of the region {(x,y):|x−y|≤4x} is

The absolute value inequality |A|≤B can be rewritten as −B≤A≤B.

Step 1: Rewrite the inequality to define the region✦ Active

The given inequality is |x−y|≤4x. For x to be defined, we must have x≥0. The absolute value inequality can be written as:

−4x≤x−y≤4x

This gives two inequalities for y:

x−y≤4x⟹y≥x−4x
x−y≥−4x⟹y≤x+4x

So, the region is bounded by the curves y1=x−4x (lower bound) and y2=x+4x (upper bound).

Step 2: Determine the limits of integration○ Expand

To find the limits of integration, we need to find where the lower curve y1=x−4x intersects the x-axis (i.e., y1=0):

x−4x=0
x(x−4)=0

This yields x=0⟹x=0 or x=4⟹x=16. These are the limits of integration.

💡 Teacher's Secret Hint

The region is enclosed between the two curves from their intersection points. In this case, the lower curve intersects the x-axis at the points that define the extent of the region.

Step 3: Calculate the area by integration○ Expand

The area A of the region is given by the integral of the difference between the upper and lower curves from x=0 to x=16:

A=∫016(y2−y1)dx
A=∫016((x+4x)−(x−4x))dx
A=∫016(8x)dx
A=8∫016x1/2dx
A=8[x3/23/2]016
A=8[23x3/2]016
A=163[(16)3/2−(0)3/2]
A=163[(16)3−0]
A=163[43]
A=163[64]
A=10243
💡 Teacher's Secret Hint

Remember to correctly evaluate the power x3/2 as (x)3 or (x1/2)3.

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