Let f : (0, ∞) → ℝ be a differentiable function such that f'(x) = 2 - f(x)/x for all x ∈ (0, ∞) and

Q: Wait, why do we call this a Linear Differential Equation? It looks like a simple algebraic equation.

Great observation! It is a Linear Differential Equation because the dependent variable $f(x)$ and its derivative $f'(x)$ appear only to the first power and are not multiplied together. By rearranging it to $f'(x) + \frac{1}{x}f(x) = 2$, you can clearly see it fits the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{1}{x}$ and $Q(x) = 2$. This structure is a powerful tool in your arsenal!

Q: Why can't we just use separation of variables here?

You could try, but you would quickly find that the terms $f(x)$ and $x$ are inextricably linked by the addition in $2 - \frac{f(x)}{x}$. The LDE method is specifically designed to handle this 'sum' structure by using an Integrating Factor to turn the left side into the derivative of a product, $d(f(x) \cdot I.F.)/dx$, which makes the integration straightforward.

Q: What does the condition $f(1) \neq 1$ actually tell us?

Think of it as a boundary condition that filters out a specific solution. When we solve the differential equation, we get $f(x) = x + \frac{C}{x}$. If $C$ were $0$, then $f(x) = x$, which means $f(1) = 1$. Since the problem explicitly states $f(1) \neq 1$, it forces $C$ to be non-zero. This ensures our function is a curve that avoids the point $(1,1)$ entirely.

Question

Let $f : (0, \infty) 	o \mathbb{R}$ be a differentiable function such that $f'(x) = 2 - \frac{f(x)}{x}$ for all $x \in (0, \infty)$ and $f(1) 
eq 1$. Then

Options:

(A) $\lim_{x 	o 0^+} f'(\frac{1}{x}) = 1$

(B) $\lim_{x 	o 0^+} x f(\frac{1}{x}) = 2$

(C) $\lim_{x 	o 0^+} x^2 f''(x) = 0$

(D) |f(x)| \leq 2 	ext{ for all } x \in (0, 2)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a vast mathematical landscape. You have been given a function $f(x)$ that obeys a very specific rule:

$$f'(x) = 2 - \frac{f(x)}{x}$$

At first glance, this might look like a jumble of terms, but let's pause and look deeper. This is not just an equation; it is a story of how a function changes.

To understand it, we must first bring order to the chaos. By moving the term $\frac{f(x)}{x}$ to the left side, we reveal the hidden structure:

$$f'(x) + \frac{1}{x}f(x) = 2$$

This is the classic form of a **Linear Differential Equation (LDE)**, specifically $\frac{dy}{dx} + P(x)y = Q(x)$. Recognizing this pattern is your first step toward mastery.

### The Magic Key: The Integrating Factor

Now that we have identified our LDE, we need a way to unlock it. In the world of differential equations, the Integrating Factor (I.F.) is our magic key.

It allows us to transform the left side of our equation into a simple derivative of a product. We identify $P(x) = \frac{1}{x}$ and $Q(x) = 2$.

The formula for the I.F. is $e^{\int P(x) dx}$. Substituting our $P(x)$, we get:

$$e^{\int \frac{1}{x} dx} = e^{\ln x} = x$$

This simple $x$ is the key that will turn our complex equation into something we can easily integrate.

### Setting the Stage for Integration

With our I.F. in hand, we multiply the entire equation by $x$. This gives us:

$$x f'(x) + f(x) = 2x$$

Look closely at the left side—it is exactly the derivative of the product $x f(x)$! So, we can rewrite the equation as:

$$\frac{d}{dx}(x f(x)) = 2x$$

Now, the path is clear. We integrate both sides with respect to $x$:

$$\int d(x f(x)) = \int 2x dx$$

The integral of $2x$ is $x^2$, plus our constant of integration, $C$. Thus, we have $x f(x) = x^2 + C$. Dividing by $x$ (which is safe since $x > 0$), we arrive at our general solution:

$$f(x) = x + \frac{C}{x}$$

### The Hidden Constraint

The problem gives us a curious condition: $f(1) 
eq 1$. Why would it do that? Let's test our function at $x = 1$.

We get $f(1) = 1 + \frac{C}{1} = 1 + C$. If $f(1)$ cannot be $1$, then $1 + C 
eq 1$, which implies $C 
eq 0$.

This is a vital piece of information. It tells us that our function is not simply the line $y = x$, but a family of curves that approach this line as an asymptote without ever touching it.

### The Grand Finale: Evaluating the Limit

Finally, we are asked to evaluate a limit involving the derivative. First, let's find $f'(x)$.

Differentiating $f(x) = x + C x^{-1}$, we get:

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

The question asks for the limit of $f'(\frac{1}{x})$ as $x 	o 0^+$. Substituting $\frac{1}{x}$ into our derivative, we get:

$$f'\left(\frac{1}{x}ight) = 1 - C \left(\frac{1}{x}ight)^{-2} = 1 - C x^2$$

Now, as $x$ approaches $0$ from the positive side, the term $C x^2$ vanishes into nothingness. We are left with $1 - 0 = 1$.

The final result is **1**.

Animated Solution for Mathematics - Differential Equations: Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

Select Answer:

Visualized Solution

Identify the Differential Equation

Find the Integrating Factor

Set Up the General Solution

Integrate to Find $f (x)$

Apply the Condition $f (1) \neq = 1$

Visualize the Function Family

Differentiate to find $f^{'} (x)$

Substitute $\frac{1}{x}$ into $f^{'}$

Evaluate the Limit

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic Key

Setting the Stage for Integration

The Hidden Constraint

The Grand Finale

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