Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (IF): For a first-order linear differential equation, the integrating factor is given by $IF = e^{\int P(x) dx}$. The solution is then $y \cdot IF = \int Q(x) \cdot IF \, dx + C$.
• L'Hôpital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the limit exists.

Step-by-Step Solution

Step 1: Rewrite the given differential equation in the standard form.

The given differential equation is $f'(x) = 7 - \frac{3}{4} \frac{f(x)}{x}$. We rewrite it in the standard form of a first-order linear differential equation: $f'(x) + \frac{3}{4x} f(x) = 7$ Here, $P(x) = \frac{3}{4x}$ and $Q(x) = 7$.

Step 2: Calculate the integrating factor (IF).

The integrating factor is given by: $IF = e^{\int P(x) dx} = e^{\int \frac{3}{4x} dx} = e^{\frac{3}{4} \int \frac{1}{x} dx} = e^{\frac{3}{4} \ln|x|} = e^{\ln|x|^{3/4}} = x^{3/4}$ Since $x > 0$, we can write $IF = x^{3/4}$.

Step 3: Find the general solution of the differential equation.

The general solution is given by: $f(x) \cdot IF = \int Q(x) \cdot IF \, dx + C$ $f(x) \cdot x^{3/4} = \int 7 \cdot x^{3/4} \, dx + C$ $f(x) \cdot x^{3/4} = 7 \int x^{3/4} \, dx + C$ $f(x) \cdot x^{3/4} = 7 \cdot \frac{x^{7/4}}{7/4} + C$ $f(x) \cdot x^{3/4} = 4x^{7/4} + C$ $f(x) = 4x + Cx^{-3/4}$

Step 4: Use the given condition f(1) ≠ 4 to find the constant C.

We are given that $f(1) \neq 4$. Substituting $x = 1$ in the general solution, we get: $f(1) = 4(1) + C(1)^{-3/4}$ $f(1) = 4 + C$ Since $f(1) \neq 4$, we have $4 + C \neq 4$, which implies $C \neq 0$.

Step 5: Evaluate the limit $\lim_{x \to 0^+} x f(\frac{1}{x})$.

We need to find $\lim_{x \to 0^+} x f(\frac{1}{x})$. Substituting $f(x) = 4x + Cx^{-3/4}$, we get: $f\left(\frac{1}{x}\right) = 4\left(\frac{1}{x}\right) + C\left(\frac{1}{x}\right)^{-3/4} = \frac{4}{x} + Cx^{3/4}$ Now, we find the limit: $\lim_{x \to 0^+} x f\left(\frac{1}{x}\right) = \lim_{x \to 0^+} x \left(\frac{4}{x} + Cx^{3/4}\right)$ $\lim_{x \to 0^+} x f\left(\frac{1}{x}\right) = \lim_{x \to 0^+} (4 + Cx^{7/4})$ $\lim_{x \to 0^+} x f\left(\frac{1}{x}\right) = 4 + C \lim_{x \to 0^+} x^{7/4} = 4 + C(0) = 4$

However, the correct answer is that the limit does not exist. Let's reconsider the problem statement. We have $f(x) = 4x + Cx^{-3/4}$ and $C \ne 0$.

We want to find $\lim_{x \to 0^+} x f(\frac{1}{x})$. We have $f(\frac{1}{x}) = \frac{4}{x} + Cx^{3/4}$. Then $x f(\frac{1}{x}) = x (\frac{4}{x} + Cx^{3/4}) = 4 + Cx^{7/4}$. Thus, $\lim_{x \to 0^+} x f(\frac{1}{x}) = \lim_{x \to 0^+} (4 + Cx^{7/4}) = 4$.

The given answer is that the limit does not exist. This suggests there may be an error in the problem statement or the given answer. Since we have rigorously derived that the limit is 4, let's analyze the situation where the limit doesn't exist.

If we consider the limit as $x \to 0^+$, the term $Cx^{7/4}$ will approach 0. However, if the function $f(x)$ was not differentiable, the problem would be different.

Let's reconsider the question. We have $f(x) = 4x + Cx^{-3/4}$ with $C \neq 0$. Then $f'(x) = 4 - \frac{3}{4}Cx^{-7/4}$. The given equation is $f'(x) = 7 - \frac{3}{4}\frac{f(x)}{x} = 7 - \frac{3}{4} \frac{4x + Cx^{-3/4}}{x} = 7 - \frac{3}{4} (4 + Cx^{-7/4}) = 7 - 3 - \frac{3}{4}Cx^{-7/4} = 4 - \frac{3}{4}Cx^{-7/4}$.

This confirms our solution for $f(x)$. The limit $\lim_{x \to 0^+} x f(1/x) = 4$ exists. There must be an error in the question or the correct answer provided.

Common Mistakes & Tips

• Remember to correctly calculate the integrating factor and use it to solve the differential equation.
• Pay attention to the given conditions and use them to find the constant of integration.
• Be careful when evaluating limits involving functions with fractional exponents.

Summary

We solved the first-order linear differential equation to find $f(x) = 4x + Cx^{-3/4}$, where $C \neq 0$. Then, we evaluated the limit $\lim_{x \to 0^+} x f(\frac{1}{x})$ and found it to be equal to 4. Since the given answer is that the limit does not exist, there appears to be an error in the problem statement or the provided answer.

Final Answer

The final answer is \boxed{4}, which corresponds to option (C).