JEE Mathematics 2025 Question with Solution

Given: The region is $R = \{(x,y): |x-5| \le y \le 4\sqrt{x}\}$.

Find: The value of $3A$, where $A$ is the area of the region.

The upper curve is $y = 4\sqrt{x}$ and the lower curve is $y = |x-5|$. First find their points of intersection by solving

Let $x = t^2$ with $t \ge 0$. Then the two cases are:

for $x < 5$, and

for $x \ge 5$.

From

we get

so $t = 1$, hence $x = 1$.

From

we get

so $t = 5$, hence $x = 25$.

Therefore, the region exists for $x \in [1,25]$. Since $|x-5|$ changes form at $x = 5$, split the area into two parts:

Now evaluate:

So,

This gives

Hence,

Therefore, the correct option is A.

Given: The same region $|x-5| \le y \le 4\sqrt{x}$.

Find: A reliable value of $3A$.

A displayed solution on the page computes

which subtracts only the left branch of $|x-5|$ and ignores the contribution of the lower curve $x-5$ on $[5,25]$.

Since

the area must be split at $x=5$. Without this split, the computed area is incorrect.

Using the correct split integral,

so the final result is

This matches option A and also the listed correct answer on the page.