IIT-JEE 1981Question 1
For all $x$ in $[0,1]$, let the second derivative $f''(x)$ of a function $f(x)$ exist and satisfy $|f''(x)|<1$. If $f(0)=f(1)$, then show that $|f'(x)|<1$ for all $x$ in $[0,1]$.
Differentiation and its Application
Answerproved
Solution
Concept. Rolle gives a point $c$ where $f'(c)=0$; the mean value theorem on $f'$ then bounds $f'(x)$ by $|x-c|$. Formulas used. (1) Rolle's theorem(2) MVT: $f'(x)-f'(c)=f''(\xi)(x-c)$ for some $\xi$ between $x$ and $c$Step 1 - A zero of $f'$. $f(0)=f(1)$, so by (1) $f'(c)=0$ for some $c\in(0,1)$. Step 2 - Bound. For $x\ne c$ in $[0,1]$, by (2): $$|f'(x)|=|f''(\xi)|\,|x-c|<1\cdot|x-c|\le1.$$ For $x=c$, $|f'(c)|=0<1$. Hence proved.