Question

Let the function $f : [0, 1] \rightarrow \mathbb{R}$ be defined as

$$f(x) = \max \left\{ \frac{|x - y|}{x + y + 1} : 0 \leq y \leq 1 \right\}, \text{ for } 0 \leq x \leq 1.$$

Then which of the following statements is correct?

• A. $f$ is strictly increasing on $[0, \tfrac{1}{2}]$ and strictly decreasing on $[\tfrac{1}{2}, 1]$
• B. $f$ is strictly decreasing on $[0, \tfrac{1}{2}]$ and strictly increasing on $[\tfrac{1}{2}, 1]$
• C. $f$ is strictly increasing on $\bigl[0, \tfrac{\sqrt{3}-1}{2}\bigr]$ and strictly decreasing on $\bigl[\tfrac{\sqrt{3}-1}{2}, 1\bigr]$
• D. $f$ is strictly decreasing on $\bigl[0, \tfrac{\sqrt{3}-1}{2}\bigr]$ and strictly increasing on $\bigl[\tfrac{\sqrt{3}-1}{2}, 1\bigr]$

Answer 1

Answer: (D)

$f$ is strictly decreasing on $\bigl[0, \tfrac{\sqrt{3}-1}{2}\bigr]$ and strictly increasing on $\bigl[\tfrac{\sqrt{3}-1}{2}, 1\bigr]$

Answer 2

Step 1. For fixed $x$, split the maximisation over $y\in[0,1]$ at $y=x$:

• $y\le x$: $\displaystyle \frac{|x-y|}{x+y+1}=\frac{x-y}{x+y+1}$ decreases in $y$, so its maximum is at $y=0$:
  $\;A(x)=\frac{x}{x+1}.$
• $y\ge x$: $\displaystyle \frac{|x-y|}{x+y+1}=\frac{y-x}{x+y+1}$ increases in $y$, so its maximum is at $y=1$:
  $\;B(x)=\frac{1-x}{x+2}.$

Thus

$$f(x)=\max\bigl\{A(x),\,B(x)\bigr\}.$$

Step 2. Find the switch point by solving $A(x)=B(x)$:

$$\frac{x}{x+1}=\frac{1-x}{x+2} \;\Longrightarrow\; 2x^2+2x-1=0 \;\Longrightarrow\; x=\frac{\sqrt3-1}{2}.$$

Step 3. Check monotonicity on each region:

• For $0\le x<\tfrac{\sqrt3-1}2$, $B(x)$ dominates and
  $\;B'(x)=\frac{-3}{(x+2)^2}<0$ ⇒ strictly decreasing.
• For $\tfrac{\sqrt3-1}2<x\le1$, $A(x)$ dominates and
  $\;A'(x)=\frac1{(x+1)^2}>0$ ⇒ strictly increasing.

Hence the correct behaviour is strictly decreasing on $[0,\tfrac{\sqrt3-1}{2}]$ and strictly increasing on $[\tfrac{\sqrt3-1}{2},1]$.