JEE Mathematics 2025 Question with Solution

Given: The area of the region $\{(x, y) : -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a > 0\}$ is $\frac{e^2 + 8e + 1}{e}$.

Find: The value of $a$.

The area is

$$\text{Area} = \int_{-1}^{1} \left(a + e^{|x|} - e^{-x}\right) \, dx$$

Because of $|x|$, split the integral at $x = 0$:

$$\text{Area} = \int_{-1}^{0} \left(a + e^{-x} - e^{-x}\right) \, dx + \int_{0}^{1} \left(a + e^{x} - e^{-x}\right) \, dx$$

For $x \in [-1,0]$, the integrand becomes $a$, so

$$\int_{-1}^{0} a \, dx = a$$

For $x \in [0,1]$,

$$\int_{0}^{1} \left(a + e^{x} - e^{-x}\right) \, dx = a + \left[e^{x} + e^{-x}\right]_{0}^{1}$$

Now,

$$\left[e^{x} + e^{-x}\right]_{0}^{1} = \left(e + \frac{1}{e}\right) - (1 + 1) = e + \frac{1}{e} - 2$$

Hence,

$$\text{Area} = a + a + e + \frac{1}{e} - 2 = 2a + e + \frac{1}{e} - 2$$

Given that

$$2a + e + \frac{1}{e} - 2 = \frac{e^2 + 8e + 1}{e}$$

Multiplying by $e$,

$$2ae + e^2 + 1 - 2e = e^2 + 8e + 1$$

So,

$$2ae - 2e = 8e$$ $$2a - 2 = 8$$ $$2a = 10$$ $$a = 5$$

Therefore, the value of $a$ is $5$. The correct option is D.

Use the fact that $e^{|x|} = e^{-x}$ for $x < 0$ and $e^{|x|} = e^{x}$ for $x \ge 0$.

So the area becomes

$$\int_{-1}^{0} a \, dx + \int_{0}^{1} a \, dx + \int_{0}^{1} e^{x} \, dx - \int_{0}^{1} e^{-x} \, dx$$

Now evaluate each part:

$$\int_{-1}^{0} a \, dx = a, \qquad \int_{0}^{1} a \, dx = a$$ $$\int_{0}^{1} e^{x} \, dx = e - 1$$ $$\int_{0}^{1} e^{-x} \, dx = 1 - \frac{1}{e}$$

Therefore,

$$\text{Area} = 2a + (e - 1) - \left(1 - \frac{1}{e}\right) = 2a + e - 2 + \frac{1}{e}$$

Equating with the given area,

$$2a + e - 2 + \frac{1}{e} = \frac{e^2 + 8e + 1}{e}$$

This gives

$$a = 5$$

Therefore, the correct option is D.