Let f be a differentiable function such that f'(x) = 7 - 3f(x)/4x, (x 0) and f(1) ≠ 4. Then lim[x→

Q: Why is the condition $f(1) \neq 4$ given? Does it change the answer?

That condition is a subtle trap! It ensures that $C \neq 0$. However, when you evaluate the limit $\lim_{x \to 0^+} (4 + Cx^{\frac{7}{4}})$, the term involving $C$ vanishes because $x^{\frac{7}{4}}$ approaches $0$ as $x \to 0^+$. So, the limit is $4$ regardless of the value of $C$ (as long as it is finite). It tests your confidence in the limit process.

Q: Can I use Bernoulli's equation here?

Bernoulli's equation is for non-linear differential equations of the form $y' + Py = Qy^n$. Here, $n=0$, which makes it a standard linear differential equation. You don't need the complexity of Bernoulli's; the standard Integrating Factor method is the most elegant path.

Q: Why does the limit exist as $x \to 0^+$?

Because the exponent of $x$ in the term $Cx^{\frac{7}{4}}$ is positive. As $x$ approaches $0$ from the positive side, any positive power of $x$ also approaches $0$. Thus, the term $Cx^{\frac{7}{4}}$ effectively disappears, leaving only the constant $4$.

Question

Let f be a differentiable function such that $f'(x) = 7 - \frac{3 f(x)}{4 x}, (x > 0)$ and $f(1) 
eq 4$. Then $\lim_{x 	o 0^+} x f(1/x)$ :

Options:

(A) Exists and equals 4

(B) Does not exist

(C) Exist and equals 0

(D) Exists and equals 4/7

Sigma Percentile · Accepted Answer

### The Beauty of Linear Differential Equations

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, might look like a chaotic mess of functions and limits.

In the world of JEE Advanced, the most complex-looking problems are often just elegant structures waiting for you to recognize their pattern. Let us embark on this journey together.

### Phase 1: The Recognition

We are given the differential equation $f'(x) = 7 - \frac{3 f(x)}{4 x}$. The moment you see a derivative $f'(x)$ and a term involving $f(x)$ divided by $x$, your intuition should scream 'Linear Differential Equation!'

Let us tidy this up. By moving the term $\frac{3 f(x)}{4 x}$ to the left side, we get:

$$f'(x) + \frac{3}{4x} f(x) = 7$$

This is the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{3}{4x}$ and $Q(x) = 7$. Recognizing this form is half the battle won, as it tells us exactly which tool to pull from our mathematical arsenal: the Integrating Factor.

### Phase 2: The Magic of the Integrating Factor

The Integrating Factor, or $IF$, is the key that unlocks the door to the solution. It is defined as $IF = e^{\int P(x) dx}$.

Substituting our $P(x)$, we have:

$$IF = e^{\int \frac{3}{4x} dx}$$

The integral of $\frac{1}{x}$ is $\ln x$, so we get $e^{\frac{3}{4} \ln x}$. Using the logarithmic property $a \ln b = \ln b^a$, this becomes $e^{\ln x^{\frac{3}{4}}}$.

Since the exponential and natural logarithm are inverse functions, they cancel out, leaving us with a beautifully simple $IF = x^{\frac{3}{4}}$. This is the power of mathematics—taking a complex exponential expression and simplifying it into a clean power function.

### Phase 3: Solving for the Function

Now, we multiply our entire differential equation by this $IF$. The left side magically becomes the derivative of the product $f(x) \cdot x^{\frac{3}{4}}$.

So, we have:

$$\frac{d}{dx} [f(x) \cdot x^{\frac{3}{4}}] = 7 \cdot x^{\frac{3}{4}}$$

Integrating both sides with respect to $x$, we get $f(x) \cdot x^{\frac{3}{4}} = \int 7 \cdot x^{\frac{3}{4}} dx + C$. Applying the power rule for integration, $\int x^n dx = \frac{x^{n+1}}{n+1}$, we find:

$$f(x) \cdot x^{\frac{3}{4}} = 7 \cdot \frac{x^{\frac{7}{4}}}{\frac{7}{4}} + C$$

The sevens cancel out, and the four moves to the numerator, giving us $f(x) \cdot x^{\frac{3}{4}} = 4 x^{\frac{7}{4}} + C$. Dividing by $x^{\frac{3}{4}}$, we finally isolate our function:

$$f(x) = 4x + C x^{-\frac{3}{4}}$$

### Phase 4: The Final Limit

We are almost there. The problem asks for $\lim_{x 	o 0^+} x f(1/x)$.

First, let us find $f(1/x)$ by substituting $1/x$ into our expression for $f(x)$:

$$f(1/x) = 4(1/x) + C(1/x)^{-\frac{3}{4}} = \frac{4}{x} + C x^{\frac{3}{4}}$$

Now, multiply this by $x$:

$$x f(1/x) = x \left( \frac{4}{x} + C x^{\frac{3}{4}} ight) = 4 + C x^{\frac{7}{4}}$$

As $x$ approaches $0$ from the positive side, the term $C x^{\frac{7}{4}}$ approaches $0$ because the exponent is positive. We are left with the constant **4**.

The limit exists and equals **4**. By staying calm and following the logical steps, we turned a daunting problem into a clear, satisfying victory.

Animated Solution for Mathematics - Differential Equations: Let f be a differentiable function such that $f^{'} (x) = 7 - \frac{3 f ( x )}{4 x}, (x > 0)$ and $f (1) \neq = 4$ . Then $lim_{x \to 0^{+}} x f (1/ x)$ :

Select Answer:

Visualized Solution

Identifying the Differential Equation

Mapping to the Standard LDE Form

Calculating the Integrating Factor

Writing the General Solution

Integrating the Right Hand Side

Simplifying the Expression for $f (x)$

Interpreting the Condition $f (1) \neq = 4$

Finding $f (\frac{1}{x})$ and $x f (\frac{1}{x})$

Evaluating the Limit as $x \to 0^{+}$

The Sigma Insight: Linear Differential Equations

The Beauty of Linear Differential Equations

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Identifying the Differential Equation

Mapping to the Standard LDE Form

Calculating the Integrating Factor

Writing the General Solution

Integrating the Right Hand Side

Simplifying the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Interpreting the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=4

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x1​) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xf(x1​)

Evaluating the Limit as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0+