JEE Mathematics 2023 Question with Solution

Given: $y(x + 1) \, dx - x^2 \, dy = 0$ and $y(1) = e$.

Find: $\lim_{x \to 0^+} f(x)$.

Rearrange the differential equation:

$$\frac{dy}{dx} = \frac{y(x+1)}{x^2}$$

So,

$$\frac{dy}{y} = \frac{x+1}{x^2} \, dx = \left(\frac{1}{x} + \frac{1}{x^2}\right) \, dx$$

Integrating both sides,

$$\int \frac{1}{y} \, dy = \int \left(\frac{1}{x} + \frac{1}{x^2}\right) \, dx$$

Hence,

$$\ln |y| = \ln |x| - \frac{1}{x} + C$$

Exponentiating,

$$y = C_1 x e^{-\frac{1}{x}}$$

Using $y(1) = e$,

$$e = C_1 \cdot 1 \cdot e^{-1}$$

Therefore,

$$C_1 = e^2$$

So the solution is

$$y = e^2 x e^{-\frac{1}{x}}$$

Now evaluate the required limit:

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^2 x e^{-\frac{1}{x}} = 0$$

Therefore, the required limit is $0$ and the correct option is A.

The solution also contains an inconsistent statement claiming option B, but the worked solution clearly gives the value $0$, which matches option A.

From the obtained solution,

$$f(x) = e^2 x e^{-\frac{1}{x}}$$

As $x \to 0^+$, we have $\frac{1}{x} \to +\infty$, so

$$e^{-\frac{1}{x}} \to 0$$

and it approaches $0$ faster than any positive power of $x$. Hence the product

$$x e^{-\frac{1}{x}}$$

also tends to $0$. Therefore,

$$\lim_{x \to 0^+} f(x) = 0$$