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,0lexle1 and f(0)=0`,then

Options

A. `f(1/2)lt1/2andf(1/3)gt1/3`
B. `f(1/2)gt1/2andf(1/3)gt1/3`
C. `f(1/2)lt1/2andf(1/3)lt1/3`
D. `f(1/2)gt1/2andf(1/3)lt1/3`

Answer

To solve the problem, we need to analyze the given equation:

\[ \int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt \]

for \( x \) in the interval \([0, 1]\) with the condition \( f(0) = 0 \).

### Step 1: Understand the Equation
The left-hand side of the equation represents the area under the curve defined by \( \sqrt{1 - (f'(t))^2} \), while the right-hand side represents the area under the curve defined by \( f(t) \).

### Step 2: Differentiate Both Sides
To analyze the equality, we can differentiate both sides with respect to \( x \):

\[ \frac{d}{dx} \left( \int_0^x \sqrt{1 - (f'(t))^2} \, dt \right) = \frac{d}{dx} \left( \int_0^x f(t) \, dt \right) \]

Using the Fundamental Theorem of Calculus, we get:

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

### Step 3: Rearranging the Equation
From the equation \( \sqrt{1 - (f'(x))^2} = f(x) \), we can square both sides to eliminate the square root:

\[ 1 - (f'(x))^2 = (f(x))^2 \]

### Step 4: Rearranging Further
Rearranging gives us:

\[ (f'(x))^2 + (f(x))^2 = 1 \]

### Step 5: Recognizing the Form
The equation \( (f'(x))^2 + (f(x))^2 = 1 \) resembles the equation of a circle with radius 1. This suggests that \( f(x) \) could be a trigonometric function.

### Step 6: Finding the Function
We can express \( f(x) \) in terms of sine or cosine. Let's assume:

\[ f(x) = \sin(x) \]

Then, differentiating gives:

\[ f'(x) = \cos(x) \]

### Step 7: Verification
Now we check if this satisfies our original equation:

1. Compute \( \sqrt{1 - (f'(x))^2} = \sqrt{1 - \cos^2(x)} = \sin(x) \).
2. Both sides of the equation become equal, confirming that \( f(x) = \sin(x) \) is indeed a solution.

### Conclusion
Thus, the function \( f(x) \) that satisfies the given conditions is:

\[ f(x) = \sin(x) \]