Question

Let $f:[0,1]\to [0,1]$ be a continuous function. Then

• A. $f\left(x\right)=x$ for at least one $0\le x\le 1$
• B. $f\left(x\right)$ will be differentiable in $[0,1]$
• C. $f\left(x\right)+x=0$ for at least one $x$ such that $0\le x\le 1$
• D. none of these

Answer 1

Answer: (A)

$f\left(x\right)=x$ for at least one $0\le x\le 1$

Suppose $f\left(x\right)\ne x$ for any $xϵ[0,1]$. This means there is no point in this interval where the graphs of $f\left(x\right)\equiv f$ and $g\left(x\right)=x$ meet.
But since $f\left(x\right)$ is continuous, it must be always greater than $x$ or always lesser than $x$. That is, $f\left(x\right)>x$ or $f\left(x\right)<x$ for all $xϵ[0,1]$.
If $f\left(x\right)>x$ for all $xϵ[0,1]$, then the only possible value for $f\left(1\right)$ is $1$, because $f\left(x\right)ϵ[0,1]$. This shows that $f\left(x\right)=x$ for $x=1$, which contradicts our assumption that $f\left(x\right)\ne x$ for all $xϵ[0,1]$.
This can also be proved for the case $f\left(x\right)<x$ for all $xϵ[0,1]$.
So, $f\left(x\right)=x$ for at least one value of $x$ in the interval $xϵ[0,1]$.