JEE Mathematics 2024 Question with Solution

Given: $y^2 \le 4x$, $x < 4$ and the sign condition from the solution is treated as

Find: the area of the required region.

From $y^2 \le 4x$, the boundary is the parabola

so the vertical height of the enclosed strip is

Using sign analysis shown in the solution:

• For $y > 0$, valid intervals are $x \in (0,1) \cup (2,3)$.
• For $y < 0$, valid intervals are $x \in (1,2) \cup (3,4)$. Hence, over the whole region the allowed strips together cover $x \in (0,4)$ exactly once in area calculation.

Therefore, the area is

Now,

so

Therefore, the total area is $\frac{32}{3}$. The correct option is D.

Given: $y^2 \le 4x$, $x < 4$ and the sign condition used in the extracted solution. Find: the required area by splitting according to the sign of $y$.

Because the sign condition contains a factor of $y$, consider two cases.

For Case I, $y > 0$, the solution gives

which holds for

For Case II, $y < 0$, the solution gives

which holds for

These intervals partition the full range from $0$ to $4$ between the upper and lower halves of the parabola. Since the parabola is

each vertical strip contributes height $2\sqrt{x}$ on the chosen half, and combining both halves gives

Now evaluate:

Hence the correct option is D.