IMC / 2019 / Problems / Day 2, P6

IMC 2019 · Day 2 · P6

medium

Let $f, g : \mathbb{R} \longrightarrow \mathbb{R}$ be continuous functions such that $g$ is differentiable. Assume that $\bigl( f(0) - g'(0) \bigr) \bigl( g'(1) - f(1) \bigr) > 0$ . Show that there exists a point $c \in (0, 1)$ such that $f(c) = g'(c)$ .

Proposed by Fereshteh Malek, K. N. Toosi University of Technology

Solution (official)

Define $F(x) = \int_0^x f(t)\,dt$ and let $h(x) = F(x) - g(x)$ . By the continuouity of $f$ we have $F' = f$ , so $h' = f - g'$ .

The assumption can be re-written as $h'(0) \bigl( -h'(1) \bigr) > 0$ , so $h'(0)$ and $h'(1)$ have opposite signs. Then, by the Mean Value Theorem For Derivatives (Darboux property of derivatives) it follows that there is a point $c$ between 0 and 1 where $h'(c) = 0$ , so $f(c) = g'(c)$ .

How the field did

contestants scored

360

average (of 10)

6.09

solved (≥ 80%)

42.5%

near-0 (≤ 10%)

1.9%

discrimination

0.46

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

IMC 2019 · Day 2 · P6

How the field did

Score distribution (field cohort)

Similar problems