JEE Mathematics 2025 Question with Solution

Given: The region is defined by $x^2 + 4x + 2 \leq y \leq |x| + 2$.

Find: The area enclosed between the two curves.

The solution states that the upper curve is $y = |x| + 2$ and the lower curve is $y = x^2 + 4x + 2$.

Write $|x|$ piecewise:

So the curve $y = |x| + 2$ becomes $y = -x + 2$ for $x < 0$ and $y = x + 2$ for $x \ge 0$.

Find the intersection points as shown in the extracted working.

For the left branch,

which gives

so

and hence $x=-5$ or $x=0$.

For the right-side form used in the working,

which gives

so

and hence $x=-3$ or $x=0$.

Using the interval stated in the solution, split the area as

First integral:

Second integral:

According to the solution, after evaluating and summing these contributions, the final area is

Therefore, the correct option is C.

Note: the second provided approach refers to $|x+2|$, which does not match the question text. The first approach matches the given question, so it is used to determine the answer.