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

Updated On: Jan 29, 2026

f (1/2)<1/2 and f ( 1/3)>1/3
f (1/2)>1/2 and f ( 1/3)<1/3
f ( 4/3)<4/3 and f ( 2/3)<2/3
f (4/3)>4/3 and f ( 2/3)>2/3

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The Correct Option is C

Solution and Explanation

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