JEE Mathematics 2023 Question with Solution

Given:

$$\int_0^2 x f(x) \, dx$$

with

$$f(x) = \begin{cases} e^{\min\{x^2, x - \lfloor x \rfloor\}}, & x \in [0,1),\\ e^{x-\log_e x}, & x \in [1,2) \end{cases}$$

Find: The value of the integral and the correct option.

From the solution, the integral is split as

$$\int_0^2 xF(x)dx = \int_0^1 xe^{x^2}dx + \int_1^2 xe^x dx$$

Let

$$I_1 = \int_0^1 xe^{x^2}dx, \qquad I_2 = \int_1^2 xe^x dx$$

For $I_1$, use the substitution

$$u = x^2 \Rightarrow du = 2x \, dx$$

with limits $u=0$ at $x=0$ and $u=1$ at $x=1$. Then

$$I_1 = \frac{1}{2} \int_0^1 e^u \, du$$

so

$$I_1 = \frac{1}{2}\left[e^u\right]_0^1 = \frac{1}{2}(e-1)$$

For $I_2$, the solution states

$$I_2 = e \int_1^2 x \, dx$$

Hence

$$\int_1^2 x \, dx = \frac{x^2}{2}\bigg|_1^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$$

and therefore

$$I_2 = e \cdot \frac{3}{2} = \frac{3e}{2}$$

Now add the two parts:

$$\int_0^2 xF(x)dx = I_1 + I_2 = \frac{1}{2}(e-1) + \frac{3e}{2}$$ $$= \frac{1}{2}e - \frac{1}{2} + \frac{3}{2}e = 2e - \frac{1}{2}$$

Therefore, the integral evaluates to $2e - \frac{1}{2}$.

However, the solution explicitly says "The Correct Option is D" even though the computed value matches option C in the listed options. the answer is marked as D while noting this discrepancy.](streamdown:incomplete-link)

Given: The piecewise definition of $f(x)$ and the integral

$$\int_0^2 x f(x) \, dx$$

Find: The final value and reconcile it with the listed options.

The extracted solution works with

$$\int_0^1 xe^{x^2}dx + \int_1^2 xe^x dx$$

For the first integral,

$$u = x^2, \quad du = 2x \, dx$$

so

$$\int_0^1 xe^{x^2}dx = \frac{1}{2}\int_0^1 e^u \, du = \frac{1}{2}(e-1)$$

For the second integral, the extracted solution uses

$$\int_1^2 xe^x dx = e\int_1^2 x \, dx = \frac{3e}{2}$$

Thus the total becomes

$$\frac{1}{2}(e-1) + \frac{3e}{2} = 2e - \frac{1}{2}$$

So the numerical expression obtained in the working is $2e - \frac{1}{2}$, which matches option C exactly. But the solution labels the correct option as D. This is an internal mismatch on the solution's.

Using the instruction that the solution is the primary source, the recorded answer is D, with this discrepancy preserved in the solution.