Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫₀ˣ √(1 - (f'(t))²

Q: Wait, why do we differentiate both sides? Can't we just integrate the left side directly?

That is a great question! The problem is that we do not know the function $f(x)$ yet. We cannot integrate $f(t)$ or $\sqrt{1 - (f'(t))^2}$ without knowing the function. By differentiating, we use the Fundamental Theorem of Calculus to 'peel away' the integral sign, which transforms an integral equation into a differential equation—a much more solvable beast.

Q: Why did we take the positive root when we solved for $f'(x)$?

Excellent observation. When we squared the equation, we got $(f'(x))^2 = 1 - f^2(x)$. Mathematically, $f'(x)$ could be $\pm \sqrt{1 - f^2(x)}$. However, the problem states $f(0) = 0$ and $f$ is non-negative in $[0, 1]$. For small $x > 0$, if $f(x)$ starts at 0 and is non-negative, it must be increasing, meaning $f'(x)$ must be positive. Thus, we choose the positive root.

Q: Is there a faster way to solve the final limit $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$?

Yes! While the standard limit result is $\frac{1}{2}$, you could also use L'Hopital's Rule since it is a $\frac{0}{0}$ indeterminate form. Differentiating the numerator gives $\sin x$ and the denominator gives $2x$. Applying it again gives $\frac{\cos x}{2}$, which evaluates to $\frac{1}{2}$ as $x \to 0$. Both methods are perfectly valid!

Question

Let $f$ be a non-negative function in $[0, 1]$ and twice differentiable in $(0, 1)$. If $\int_{0}^{x} \sqrt{1 - (f'(t))^2} dt = \int_{0}^{x} f(t) dt$, $0 \le x \le 1$ and $f(0) = 0$, then $\lim_{x 	o 0} \frac{1}{x^2} \int_{0}^{x} f(t) dt$ :

Options:

(A) equals 0

(B) equals 1

(C) does not exist

(D) equals $1/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given integral equation is:

$$\int_{0}^{x} \sqrt{1 - (f'(t))^2} dt = \int_{0}^{x} f(t) dt$$

This equation hides the function $f(x)$ behind a veil of integration. In JEE Advanced mathematics, we dismantle such problems by systematically peeling back these layers.

### The Scalpel of Leibniz

The first step in any problem involving an integral with a variable limit is to liberate the function. We apply the **Leibniz Rule**, which states that the derivative of an integral from $0$ to $x$ is simply the integrand evaluated at $x$.

By differentiating both sides with respect to $x$, we transform the integral equation into a differential equation:

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

The integral signs vanish, leaving us with a clean, algebraic relationship. This transition shifts the problem from 'impossible' to 'solvable'.

### The Algebraic Dance

Starting from $\sqrt{1 - (f'(x))^2} = f(x)$, we square both sides to isolate the derivative:

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

Rearranging the terms gives $(f'(x))^2 = 1 - f^2(x)$. Taking the square root, we obtain:

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

We choose the positive root because the function is non-negative and starts from zero, implying an increasing slope. This is a **separable differential equation**:

$$\frac{df}{\sqrt{1 - f^2}} = dx$$

### The Identity Revealed

Integrating both sides is a standard procedure. The integral of $\frac{1}{\sqrt{1 - f^2}}$ is $\sin^{-1}(f)$:

$$\sin^{-1}(f(x)) = x + C$$

Given the initial condition $f(0) = 0$, we substitute $x = 0$ and $f = 0$ to find $\sin^{-1}(0) = 0 + C$, which implies $C = 0$. Thus, the function is revealed as:

$$f(x) = \sin x$$

### The Final Victory

We are tasked to evaluate the following limit:

$$\lim_{x 	o 0} \frac{1}{x^2} \int_{0}^{x} f(t) dt$$

Substituting $f(t) = \sin t$, the integral becomes:

$$\int_{0}^{x} \sin t dt = [-\cos t]_{0}^{x} = 1 - \cos x$$

The limit expression simplifies to:

$$\lim_{x 	o 0} \frac{1 - \cos x}{x^2}$$

Using the standard limit result $\lim_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$, we arrive at the final answer:

**The value of the limit is $\frac{1}{2}$.**

Select Answer:

Visualized Solution

Understanding the Functional Equation

Applying Leibniz Rule

Differentiating Both Sides

Squaring the Equation

Isolating the Derivative

Variable Separable Form

Integrating the Differential Equation

Using the Initial Condition

Finding the Function $f (x)$

Setting up the Limit

Evaluating the Integral

Substituting the Limits

Calculating the Final Limit

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Scalpel of Leibniz

The Algebraic Dance

The Identity Revealed

The Final Victory

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