Continuity

We define a new function $g (x)$ to test the condition. Let $g (x) = f (x) - (1 - x)$ . We know continuous functions on a closed interval satisfy the Intermediate Value Theorem (IVT).

Check values at endpoints: $g (0) = f (0) - (1 - 0) = 0 - 1 = - 1$ . $g (1) = f (1) - (1 - 1) = 1 - 0 = 1$ .

Since $g (0) < 0$ and $g (1) > 0$ , and $g$ is continuous, there must exist at least one $c$ in $(0, 1)$ such that $g (c) = 0$ . $⟹ f (c) - (1 - c) = 0$ $⟹ f (c) = 1 - c$ .

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

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Continuity

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Let $f (x) = [x]^{2} + {x}$ , where $[.]$ denotes the greatest integer function and ${.}$ den...