Question

Let \( f : [0, 1] \to \mathbb{R} \) be a continuous function such that \( f\left(\dfrac{1}{2}\right) = -\dfrac{1}{2} \) and \[ |f(x) - f(y) - (x - y)| \le \sin(|x - y|^2) \] for all \( x, y \in [0, 1]. \) Then \( \int_0^1 f(x) \, dx \) is
 

• A. \(-\dfrac{1}{2}\)
• B. \(-\dfrac{1}{4}\)
• C. \(\dfrac{1}{4}\)
• D. \(\dfrac{1}{2}\)

Hint:

If \( f(x) - f(y) \approx (x - y) \), the function behaves almost linearly; check midpoint conditions to estimate constants accurately.

Answer 1

The Correct Option is A

Step 1: Analyze the constraint

From the given inequality: $$|f(x) - f(y) - (x-y)| \leq \sin(|x-y|^2)$$

For small $|x-y|$, we have $\sin(|x-y|^2) \approx |x-y|^2$ (since $\sin(t) \approx t$ for small $t$).

This means: $$f(x) - f(y) \approx x - y$$

or equivalently: $$f(x) - x \approx f(y) - y$$

Step 2: Define $g(x) = f(x) - x$

The constraint suggests that $g(x) = f(x) - x$ is approximately constant.

From the inequality: $$|g(x) - g(y)| = |f(x) - x - f(y) + y| = |f(x) - f(y) - (x-y)| \leq \sin(|x-y|^2)$$

Step 3: Show $g$ is constant

For any $x, y \in [0,1]$: $$|g(x) - g(y)| \leq \sin(|x-y|^2) \leq |x-y|^2$$

By dividing the interval $[0,1]$ into $n$ equal parts and applying the inequality repeatedly, we can show that as $n \to \infty$: $$|g(x) - g(y)| = 0$$

Therefore, $g(x) = g(y)$ for all $x, y$, meaning $g$ is constant.

Step 4: Find the constant

Since $g(x) = f(x) - x$ is constant and $f\left(\frac{1}{2}\right) = -\frac{1}{2}$:

$$g\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) - \frac{1}{2} = -\frac{1}{2} - \frac{1}{2} = -1$$

Therefore, $f(x) - x = -1$ for all $x \in [0,1]$.

This gives us: $$f(x) = x - 1$$

Step 5: Calculate the integral

$$\int_0^1 f(x),dx = \int_0^1 (x-1),dx = \left[\frac{x^2}{2} - x\right]_0^1 = \frac{1}{2} - 1 = -\frac{1}{2}$$

Answer: (A) $-\frac{1}{2}$