Mean Value Theorem

Hardmathematics

If $f (x)$ is twice differentiable in $[0, 2], f (0) = 0, f (1) > 0$ and $f (2) = 0$ , then which of the following is necessarily true?

Graph with roots at 0 and 2

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$f^{'} (c) = 0$ for some $c$ in $(0, 2)$

$f^{''} (c) < 0$ for some $c$ in $(0, 2)$

$f^{''} (c) = 0$ for some $c$ in $(0, 2)$

$f (c) = 0$ for some $c$ in $(0, 1)$

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Subject: mathematics
Chapter: limit, continuity and differentiability
Topic: mean value theorem
Difficulty: Hard
Year: 2025

Solution

Correct Answer:

$f^{'} (c) = 0$ for some $c$ in $(0, 2)$

Since $f (x)$ is twice differentiable on $[0, 2]$ , it is continuous on $[0, 2]$ and differentiable on $(0, 2)$ . Given $f (0) = 0$ and $f (2) = 0$ : By Rolle's Theorem, there exists at least one $c$ in $(0, 2)$ such that $f^{'} (c) = 0$ . (While other properties like concavity might be derived from additional data, $f^{'} (c) = 0$ is the direct and necessary consequence of $f (0) = f (2) = 0$ ).

This hard difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of mean value theorem. It appeared in the 2025 exam.

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Mean Value Theorem

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