Question

Let $f$ be a non-negative function defined in $[0, \pi/2]$, $f'$ exists and is continuous for all $x$, and $\int_0^x \sqrt{1 - (f'(t))^2} dt = \int_0^x f(t) dt$ and $f(0) = 0$. Then

• 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 ( 4/3)<4/3 and f ( 2/3)<2/3
• D. f (4/3)>4/3 and f ( 2/3)>2/3

Answer 1

The Correct Option is C

We are given the equation: \[ \int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt \] with the condition \( f(0) = 0 \). Let's analyze this equation.

Step 1: Understanding the equation
The equation essentially relates the integral of \( \sqrt{1 - (f'(t))^2} \) to the integral of \( f(t) \) over the interval \( [0, x] \). For these two integrals to be equal, it suggests a specific relationship between \( f(t) \) and \( f'(t) \). The integrand \( \sqrt{1 - (f'(t))^2} \) implies that the function \( f'(t) \) is constrained, meaning \( f'(t) \) must be less than or equal to 1 for all \( t \), since \( \sqrt{1 - (f'(t))^2} \) must remain real and non-negative. 

Step 2: Interpreting the behavior of \( f(x) \)
The equation \( \int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt \) suggests that \( f(x) \) is a function whose growth is constrained by the geometric condition involving \( f'(x) \). This implies that \( f(x) \) grows slower than or at most as fast as \( x \), and therefore \( f(x) \leq x \) for all \( x \in [0, \pi/2] \).

 Step 3: Estimating \( f(x) \) at specific points
- At \( x = 1/2 \), since \( f(x) \) is constrained by the condition \( f(x) \leq x \), we expect \( f(1/2) \leq 1/2 \). - Similarly, at \( x = 2/3 \), \( f(2/3) \) must be less than or equal to \( 2/3 \), since \( f(x) \) cannot exceed \( x \). - Finally, for \( x = 4/3 \), \( f(4/3) \) is also constrained to be less than or equal to \( 4/3 \).

 Step 4: Conclusion
Based on the constraints of the function and the relationship between \( f'(x) \) and \( f(x) \), we conclude that: \[ f(4/3) < 4/3 \quad \text{and} \quad f(2/3) < 2/3 \]

Answer:

\[ \boxed{f(4/3) < 4/3 \text{ and } f(2/3) < 2/3} \]