# International Mathematics Competition for University Students 2019

Select Year:

IMC 2019
 Information Problems & Solutions Results: Individuals Teams Leaders Schedule Contact

## IMC2019: Day 2, Problem 6

Problem 6. Let $\displaystyle f,g: \RR\longrightarrow \RR$ be continuous functions such that $\displaystyle g$ is differentiable. Assume that $\displaystyle \bigl(f(0)-g^{\prime} (0)\bigr)\bigl(g^{\prime} (1)-f(1)\bigr)>0$. Show that there exists a point $\displaystyle c\in(0,1)$ such that $\displaystyle f(c)=g^\prime (c)$.

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

Solution. Define $\displaystyle F(x)=\int_{0}^{x}f(t)\mathrm{d}t$ and let $\displaystyle h(x)=F(x)-g(x)$. By the continuouity of $\displaystyle f$ we have $\displaystyle F'=f$, so $\displaystyle h'=f-g'$.

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

 IMC1994 IMC1995 IMC1996 IMC1997 IMC1998 IMC1999 IMC2000 IMC2001 IMC2002 IMC2003 IMC2004 IMC2005 IMC2006 IMC2007 IMC2008 IMC2009 IMC2010 IMC2011 IMC2012 IMC2013 IMC2014 IMC2015 IMC2016 IMC2017 IMC2018 IMC 2019