Let f(x) be differentiable on the interval (0, ∞) such that f(1) = 1, and lim (t→x) (t² f(x) - x² f(

Q: Wait, why do we treat $x$ as a constant when differentiating with respect to $t$?

Great question! In the limit $\lim_{t \to x}$, the variable $t$ is the one approaching $x$. Since $x$ is fixed for the duration of that specific limit evaluation, it acts as a constant parameter. Think of it like a snapshot: at any specific point $x$, the limit tells us how the function behaves locally.

Q: Why can't we just use the definition of the derivative directly?

You absolutely can! If you manipulate the expression $\frac{t^2 f(x) - x^2 f(t)}{t - x}$ by adding and subtracting $x^2 f(x)$ in the numerator, you can force the definition of $f'(x)$ to appear. L'Hopital's Rule is just a faster, more mechanical way to achieve the same result.

Q: How do I know if my Integrating Factor is correct?

A quick check: if you multiply your original differential equation by your I.F., the left-hand side must become the derivative of the product of the function and the I.F. If it doesn't simplify into $\frac{d}{dx}[f(x) \cdot I.F.]$, then your I.F. calculation needs a second look.

Q: Is it possible to get a different constant $C$ if I integrate incorrectly?

Absolutely. Integration is the heart of this problem. Always remember the constant of integration $+ C$ immediately after performing the integral. Without it, you won't be able to satisfy the initial condition $f(1) = 1$.

Q: Does the interval $(0, \infty)$ matter for the final answer?

Yes, it is crucial! It ensures that $x$ is never zero, which means we don't have to worry about division by zero when we divide by $x^2$ to put the equation into standard form. It guarantees the function is well-defined throughout the domain.

Question

Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1) = 1$, and $\lim_{t 	o x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1$ for each $x > 0$. Then $f(x)$ is

Options:

(A) $\frac{1}{3x} + \frac{2x^2}{3}$

(B) $-\frac{1}{3x} + \frac{4x^2}{3}$

(C) $-\frac{1}{x} + \frac{2}{x^2}$

(D) $\frac{1}{x}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the edge of a mathematical landscape, looking at a function $f(x)$ that seems elusive. We know it exists, we know it is differentiable, and we have one anchor point: $f(1) = 1$.

But the true mystery lies in the limit:

$$
\lim_{t 	o x} \frac{t^2 f(x) - x^2 f(t)}{t - x} = 1
$$

At first glance, this looks like a chaotic jumble of variables. In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel it back together.

### The L'Hopital Breakthrough

When you see a limit that results in a $\frac{0}{0}$ form, your intuition should immediately scream 'L'Hopital!' Let us test it. As $t 	o x$, the numerator becomes $x^2 f(x) - x^2 f(x) = 0$, and the denominator becomes $x - x = 0$.

It is a perfect candidate. We differentiate the numerator and the denominator with respect to $t$, treating $x$ as a constant.

The derivative of the numerator is $2t f(x) - x^2 f'(t)$, and the derivative of the denominator is simply $1$. Now, as we let $t$ approach $x$, the expression simplifies beautifully to:

$$
2x f(x) - x^2 f'(x) = 1
$$

We have successfully transformed a limit into a differential equation. The fog is lifting.

### The Architecture of the Differential Equation

Now, let us organize our findings. We have $x^2 f'(x) - 2x f(x) = -1$. To solve this, we need the standard form of a linear differential equation: $f'(x) + P(x)f(x) = Q(x)$.

Dividing by $x^2$, we get:

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

Here, $P(x) = -\frac{2}{x}$ and $Q(x) = -\frac{1}{x^2}$. This is the moment where the 'Integrating Factor' (I.F.) becomes our most powerful tool.

The I.F. is defined as $e^{\int P(x) dx}$. Calculating this, we find:

$$
e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = e^{\ln(x^{-2})} = \frac{1}{x^2}
$$

This factor is the key that unlocks the entire structure.

### The Final Integration

With our I.F. in hand, we multiply our differential equation by $\frac{1}{x^2}$. The left side magically collapses into the derivative of a product:

$$
\frac{d}{dx} \left[ f(x) \cdot \frac{1}{x^2} ight] = -\frac{1}{x^4}
$$

Now, we integrate both sides with respect to $x$. The integral of $-x^{-4}$ is $\frac{x^{-3}}{3}$, leading us to:

$$
\frac{f(x)}{x^2} = \frac{1}{3x^3} + C
$$

We are almost there! We use our anchor point, $f(1) = 1$, to find $C$. Substituting $x=1$, we get $1 = \frac{1}{3} + C$, which means $C = \frac{2}{3}$.

### The Victory

Finally, we substitute $C$ back into our equation:

$$
\frac{f(x)}{x^2} = \frac{1}{3x^3} + \frac{2}{3}
$$

Multiplying by $x^2$, we arrive at our destination:

$$
f(x) = \frac{1}{3x} + \frac{2x^2}{3}
$$

Look at that! What started as a terrifying limit has resolved into a clean, elegant function. You didn't just solve a problem; you navigated a logical path through the heart of calculus. Keep this confidence with you—every complex problem is just a series of simple, logical steps waiting for you to connect them.

Animated Solution for Mathematics - Differential Equations: Let $f (x)$ be differentiable on the interval $(0, \infty)$ such that $f (1) = 1$ , and $lim_{t \to x} \frac{t ^{2} f ( x ) - x ^{2} f ( t )}{t - x} = 1$ for each $x > 0$ . Then $f (x)$ is

Select Answer:

Visualized Solution

Visualizing the Problem

Identifying the $0/0$ Form

Applying L'Hopital's Rule

Evaluating the Limit

Forming the Differential Equation

Standard Linear Form

Calculating the Integrating Factor

Setting up the General Solution

Performing the Integration

Finding the Constant $C$

Final Result and Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The L'Hopital Breakthrough

The Architecture of the Differential Equation

The Final Integration

The Victory

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