Let f be a non-negative function defined on the interval [0, 1]. If ∫₀ˣ √(1 - (f'(t))²) dt = ∫₀ˣ f(t

Q: Wait, why can we just differentiate the integral? Is that always allowed?

Great question! The Fundamental Theorem of Calculus guarantees that if the integrand is continuous, the derivative of the integral $\int_{0}^{x} g(t) dt$ is simply $g(x)$. It is the bridge between integration and differentiation, and it is perfectly valid here.

Q: Why did we pick the positive root when we took the square root of $(f'(x))^2$?

The problem states $f$ is non-negative and $f(0)=0$. If the derivative were negative, the function would immediately become negative, violating the non-negativity constraint. Thus, the function must increase from zero, requiring the positive root.

Q: How do we know for sure that $\sin x 0$?

Think of the unit circle or the Taylor series expansion. The Taylor series for $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$. For small positive $x$, the terms subtract from $x$, making $\sin x$ strictly less than $x$.

Question

Let $f$ be a non-negative function defined on the interval [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 Options: (A) $f(1/2) < 1/2$ and $f(1/3) > 1/3$ (B) $f(1/2) > 1/2$ and $f(1/3) > 1/3$ (C) $f(1/2) < 1/2$ and $f(1/3) < 1/3$ (D) $f(1/2) > 1/2$ and $f(1/3) < 1/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a mathematical detective story. We are given an integral equation:

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

At first glance, this looks like a formidable wall of symbols. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Our first task is to strip away this mask.

### The Newton-Leibniz Magic

Whenever you see variable limits in an integral, the Newton-Leibniz rule is your best friend. It is the key that unlocks the door.

By applying $\frac{d}{dx}$ to both sides, we effectively differentiate the integral with respect to its upper limit. The integral signs vanish, leaving us with the core relationship:

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

This is the moment of clarity. We have successfully moved from the realm of integral calculus into the domain of differential equations.

### The Birth of a Differential Equation

Now, the problem transforms. We have a radical equation, which can be cumbersome. To make things easier to handle, let's get rid of that square root.

Squaring both sides gives us $1 - (f'(x))^2 = (f(x))^2$. With a little algebraic rearrangement, we isolate the derivative term:

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

This is beautiful. It is a classic separable differential equation. Since the problem states that $f$ is a non-negative function starting at $f(0) = 0$, the function must be increasing initially. Therefore, we take the positive root:

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

### The Trigonometric Revelation

We are now in the home stretch. We rearrange the equation to separate the variables:

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

Integrating both sides is a standard procedure. The integral on the left is the derivative of the inverse sine function, yielding:

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

To find the constant $C$, we use our initial condition $f(0) = 0$. Plugging this in, we get $\sin^{-1}(0) = 0 + C$, which means $C = 0$. Substituting $C$ back into our equation, we get $\sin^{-1}(f(x)) = x$, or simply:

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

### The Final Verdict

Finally, we compare $\sin x$ and $x$. For any $x > 0$, the sine curve lies strictly below the line $y = x$. This is a fundamental geometric property.

Therefore, at $x = 1/2$, we have $f(1/2) = \sin(1/2) < 1/2$. Similarly, at $x = 1/3$, we have $f(1/3) = \sin(1/3) < 1/3$.

Both conditions are satisfied, leading us directly to the correct option. You have conquered the problem, not by brute force, but by peeling back the layers of calculus to reveal the **elegant truth beneath**.

Animated Solution for Mathematics - Definite Integration: Let $f$ be a non-negative function defined on the interval [0, 1]. If $\int_{0}^{x} 1 - (f^{'} (t))^{2} d t = \int_{0}^{x} f (t) d t$ , $0 \leq x \leq 1$ , and $f (0) = 0$ , then

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Visualized Solution

The Integral Equation

Differentiating the Equation

Simplifying the Integrals

Squaring Both Sides

Rearranging Terms

Finding $f^{'} (x)$

Setting up the Differential Equation

Integrating Both Sides

Applying Initial Conditions

Revealing $f (x)$

Comparing $sin x$ and $x$

Checking the First Point

Checking the Second Point

Selecting the Correct Option

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Newton-Leibniz Magic

The Birth of a Differential Equation

The Trigonometric Revelation

The Final Verdict

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