JEE Mathematics 2026 Question with Solution

Given: We need the bounded areas $A_1$ and $A_2$ in the first quadrant.

Find: The value of $A_1-A_2$.

For $A_1$, the region is bounded by $y=x^2+2$, $y=8-x$ and $x=0$. Their intersection is found from

so

which gives the first-quadrant point $x=2$.

Hence,

For $A_2$, the region is bounded by $y=x^2+2$, $x=y^2$, $x=2$ and the $y$-axis. Using the working shown, change the variable to integrate with respect to $y$:

Substituting from the shown relation gives

Now subtract:

From the provided solution, the final result is stated as

Therefore, the correct option is D.

Given: Two bounded first-quadrant regions are formed by the stated curves.

Find: The required difference of areas.

The hint says to sketch the curves first. For $A_1$, compare the line $y=8-x$ with the parabola $y=x^2+2$ between $x=0$ and their point of intersection $x=2$. In this interval, the line lies above the parabola, so area is upper curve minus lower curve.

For $A_2$, the solution uses integration with respect to $y$. The right boundary is given by the relation used in the working, and the lower-to-upper limits come from $y=0$ to $y=\sqrt{2}$ as written in the source solution. Evaluating that integral gives the stated expression for $A_2$, and subtracting from $A_1$ leads to the final option D.