Let f : (0,∞) → (0,∞) be a differentiable function such that f(1) = e and lim t→x (t²f²(x)-x²f²(t))/

Q: Wait, why do we treat $x$ as a constant in the limit? Isn't $x$ a variable?

That is a brilliant question! In the expression $\lim_{t \to x}$, the subscript $t \to x$ tells us exactly who the 'active' variable is. It is $t$. As $t$ moves closer and closer to $x$, $x$ itself is just a fixed point on the number line. If you treat $x$ as a variable, you will end up with a mess of partial derivatives. Always look at the limit notation first—it is your compass!

Q: Why did we get a $\frac{0}{0}$ form? How do I know for sure?

Always test the limit by direct substitution first! If you plug $t=x$ into the numerator, you get $x^2f^2(x) - x^2f^2(x)$, which is $0$. Plugging $t=x$ into the denominator gives $x-x$, which is also $0$. Since we have $0$ divided by $0$, the limit is indeterminate, and L'Hospital's Rule is our golden ticket to resolve it.

Q: How do I know $f(x)$ is not zero? Can I just divide by it?

Great catch! In mathematics, we must always justify division by zero. The problem explicitly states that the function $f : (0, \infty) \to (0, \infty)$ maps to the interval $(0, \infty)$. This means $f(x)$ is strictly positive for all $x$ in the domain. Since $f(x) > 0$, it is never zero, and we are perfectly safe to divide by it.

Question

Let $f : (0,\infty) 	o (0,\infty)$ be a differentiable function such that $f(1) = e$ and $\lim_{t 	o x} \frac{t^2f^2(x)-x^2f^2(t)}{t-x} = 0$ If $f(x) = 1$, then $x$ is equal to :

Options:

(A) $2e$

(B) $e$

(C) $\frac{1}{2e}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with a seemingly intimidating expression:

$$\lim_{t 	o x} \frac{t^2f^2(x)-x^2f^2(t)}{t-x} = 0$$

When you see a limit like this, do not panic. Your first instinct should always be to check the form. By substituting $t = x$, we see the numerator becomes $x^2f^2(x) - x^2f^2(x) = 0$, and the denominator becomes $x - x = 0$. We have arrived at the classic $\frac{0}{0}$ indeterminate form.

Here is the secret: in the expression $\lim_{t 	o x}$, $t$ is the variable, and $x$ is a constant. This distinction is vital. If you try to differentiate with respect to $x$, you will lose your way. We must differentiate with respect to $t$. This is where **L'Hospital's Rule** becomes our most powerful tool.

### The L'Hospital's Dance

Applying L'Hospital's Rule, we differentiate the numerator and the denominator with respect to $t$. Let's take it step by step. For the numerator, we have two terms. The first term, $t^2f^2(x)$, treats $f^2(x)$ as a constant. Thus, its derivative with respect to $t$ is simply $2tf^2(x)$.

The second term, $x^2f^2(t)$, treats $x^2$ as a constant. Using the chain rule, the derivative of $f^2(t)$ is $2f(t)f'(t)$. So, the derivative of the second term is $x^2 \cdot 2f(t)f'(t)$. The denominator, $t-x$, is the easiest part—its derivative is simply $1$.

Now, we substitute $t=x$ back into our differentiated expression:

$$\frac{2xf^2(x) - 2x^2f(x)f'(x)}{1} = 0$$

This simplifies to $2xf^2(x) - 2x^2f(x)f'(x) = 0$. We have successfully stripped away the limit, leaving us with a clean, algebraic relationship between the function and its derivative.

### The Differential Equation

Look at the equation $2xf^2(x) - 2x^2f(x)f'(x) = 0$. We can factor out $2xf(x)$. Since the problem defines the range of $f$ as $(0, \infty)$, we know $f(x) 
eq 0$, and since the domain is $(0, \infty)$, $x 
eq 0$. Therefore, we can safely divide by $2xf(x)$ without fear. This leaves us with:

$$f(x) - xf'(x) = 0$$

Rearranging this, we get $xf'(x) = f(x)$, or more elegantly:

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

This is a first-order separable differential equation. It is the heartbeat of the problem. To solve it, we integrate both sides with respect to $x$:

$$\int \frac{f'(x)}{f(x)} dx = \int \frac{1}{x} dx$$

This yields $\ln(f(x)) = \ln(x) + \ln(C)$, which simplifies to $f(x) = Cx$.

### Final Calculation

We are almost there. We have the general form $f(x) = Cx$. We are given the initial condition $f(1) = e$. Substituting $x=1$, we get $e = C(1)$, which means $C = e$. Our function is fully revealed: **$f(x) = ex$**.

The final question asks us to find $x$ when $f(x) = 1$. Setting $ex = 1$, we find:

$$x = \frac{1}{e}$$

Look at what we have achieved. We started with a limit, navigated through the derivative, solved a differential equation, and arrived at a precise value. This is the elegance of JEE Advanced mathematics. It is not about memorizing formulas; it is about understanding the flow of logic.

Animated Solution for Mathematics - Differential Equations: Let $f : (0, \infty) \to (0, \infty)$ be a differentiable function such that $f (1) = e$ and $lim_{t \to x} \frac{t ^{2} f ^{2} ( x ) - x ^{2} f ^{2} ( t )}{t - x} = 0$ If $f (x) = 1$ , then $x$ is equal to :

Select Answer:

Visualized Solution

Analyze the Limit Form

Apply L'Hospital's Rule

Differentiate Numerator and Denominator

Evaluate the Limit

Simplify the Equation

Form the Differential Equation

Integrate to find $f (x)$

Apply Initial Condition

Solve for $f (x) = 1$

Conclusion and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The L'Hospital's Dance

The Differential Equation

Final Calculation

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