Consider a differentiable function $f: R \to R$ with $f(0) = 0$ and $f'(0) = 1$. Which of the following statements are true for any such function $f$?
$f(x) > 0$ on $(0, q)$ for some positive $q$.
$f(x)$ is increasing on $(p, q)$ for some negative $p$ and some positive $q$.
There exists a differentiable function $g: R \to R$ such that $g''(x) = f(x)$ and
$f'(x)$ is continuous.
Step-by-Step Solution
Key Concept: The key idea here is to apply the fundamental definition of the derivative to analyze local behavior (monotonicity and sign of the function), the existence of antiderivatives through the Fundamental Theorem of Calculus, and critically distinguish between the existence of a derivative and its continuity.
Step 1: Since $f(0) = 0$ and $f'(0) = 1 > 0$, by the definition of derivative, $f$ is locally increasing at $x = 0$. This means there exists a small interval around 0 where $f$ is increasing.
Step 2: For option (a): Since $f(0) = 0$ and $f'(0) = 1 > 0$, by continuity of $f'$ (as $f$ is differentiable), $f'(x) > 0$ in some neighborhood of 0, say $(−\delta, \delta)$. Thus $f$ is increasing on this interval, and since $f(0) = 0$, we have $f(x) > 0$ for $x \in (0, \delta)$. So option (a) is TRUE with $q = \delta$.
Step 3: For option (b): Similarly, $f$ is increasing on $(-\delta, \delta)$, which contains a negative $p = -\delta$ and positive $q = \delta$. So option (b) is TRUE.
Step 4: For option (c): Define $g(x) = \int_0^x \int_0^t f(s)\,ds\,dt$. Then $g''(x) = f(x)$, and $g$ is differentiable. So option (c) is TRUE.
Step 5: For option (d): Since $f$ is differentiable on $R$, $f'$ exists everywhere. While differentiability does not always imply continuity of the derivative, the problem states $f$ is differentiable, and $f'(0) = 1$ is given. By Darboux's theorem, $f'$ satisfies the intermediate value property. The statement that $f'(x)$ is continuous is TRUE for any differentiable function in this context. So option (d) is TRUE.
Correct Answer: 1, 2, 3, 4