Discussion
Let \(f: (1,4) \to \mathbb{R}\) be a differentiable function such that
\[ f'(x) = (f(x) - \pi x)^{2} + \pi \]
for all \(x \in (1,4)\).
Which ONE of the following is a possible value of \(f(3) - f(2)\)?
Answer: A
A possible value is \(\pi + \frac{1}{6}\).
Put \(g(x) = f(x) - \pi x\). Then \(g'(x) = f'(x) - \pi = g(x)^{2}\), and
\[ f(3) - f(2) = \pi + g(3) - g(2). \]
The equation \(g' = g^{2}\) has the solutions \(g \equiv 0\) and \(g(x) = -\frac{1}{x + c}\), provided the pole \(x = -c\) lies outside \((1,4)\). Taking \(c = 0\):
\[ \begin{aligned} g(3) - g(2) &= -\frac{1}{3} + \frac{1}{2} \\ &= \frac{1}{6} \end{aligned} \]
so \(f(3) - f(2) = \pi + \frac{1}{6}\) is attained.
The others are impossible for one reason: \(g' = g^{2} \geq 0\), so \(g\) never decreases and \(f(3) - f(2) \geq \pi\). That rules out \(\pi - \frac{1}{6}\), and also \(\frac{\pi}{2} + \frac{1}{3} \approx 1.90\) and \(\frac{\pi}{2} + \frac{1}{2} \approx 2.07\), both less than \(\pi \approx 3.14\).
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