Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be an infinitely differentiable function such that for all \(a, b \in \mathbb{R}\) with \(a<b\), \[ \frac{f(b) - f(a)}{b - a} = f''\left(\frac{a+b}{2}\right). \] Then

• A. \(f\) must be a polynomial of degree less than or equal to 2.
• B. \(f\) must be a polynomial of degree greater than 2.
• C. \(f\) is not a polynomial.
• D. \(f\) must be a linear polynomial.

Hint:

Whenever a differentiable functional equation holds for all intervals \([a,b]\), it often constrains \(f\) to a low-degree polynomial.

Answer 1

The Correct Option is A

Step 1: Interpret the given condition.
The condition relates the average rate of change \(\frac{f(b)-f(a)}{b-a}\) to the second derivative at the midpoint. Let \(a = x - h\), \(b = x + h\). Then \[ \frac{f(x+h) - f(x-h)}{2h} = f''(x). \]
Step 2: Expand using Taylor series.
Using Taylor expansion around \(x\): \[ f(x \pm h) = f(x) \pm f'(x)h + \frac{f''(x)}{2}h^2 \pm \frac{f'''(x)}{6}h^3 + \frac{f^{(4)}(x)}{24}h^4 + \cdots \] Subtracting gives \[ \frac{f(x+h) - f(x-h)}{2h} = f'(x) + \frac{f'''(x)}{6}h^2 + \cdots \] The condition states this equals \(f''(x)\) for all \(h\), implying \[ f'(x) = f''(x), \quad f'''(x)=0. \]
Step 3: Solve the differential constraints.
From \(f'''(x)=0\), we get \(f\) is a quadratic polynomial.
Step 4: Conclusion.
Hence, \(f(x)\) must be a polynomial of degree \(\le 2\).