JEE Mathematics 2026 Question with Solution

Given: The region is $R=\{(x,y): xy\le 8,\; 1\le y\le x^2,\; x\ge 0\}$.

Find: The area of the region.

The region is bounded by the curves $y=1$, $y=x^2$, and $y=\dfrac{8}{x}$. The upper boundary changes where $x^2=\dfrac{8}{x}$.

Find the point of intersection:

Set up the integrals:

For $1\le x\le 2$, the upper curve is $y=x^2$. For $2\le x\le 8$, the upper curve is $y=\dfrac{8}{x}$.

Evaluate the first integral:

Evaluate the second integral:

Add the results:

Therefore, the correct option is D.

Given: The same region satisfies $xy\le 8$, $1\le y\le x^2$, and $x\ge 0$.

Find: Why two separate integrals are required.

Since $xy\le 8$ and $x>0$ in the region, this becomes $y\le \dfrac{8}{x}$. Together with $1\le y\le x^2$, the actual upper boundary is the smaller of $x^2$ and $\dfrac{8}{x}$.

The switch occurs when the two upper candidates are equal:

Also, from $1\le y\le x^2$, we need $x^2\ge 1$, so $x\ge 1$. From $1\le y\le \dfrac{8}{x}$, we need $\dfrac{8}{x}\ge 1$, so $x\le 8$. Hence the interval is $1\le x\le 8$.

Thus the area is computed as

which evaluates to

So the correct option is D.