Question

A function $f\left(x\right)$ is continuous in interval $[0,2]$. It is known that $f\left(0\right)=f\left(2\right)=-1$ and $f\left(1\right)=1$. Which one of the following statements must be true?

• A. There exists a value 'y' in the interval $(0,1)$ such that $f\left(y\right)=f(y+1)$
• B. For every value 'y' in the interval $(0,1),f\left(y\right)=f(z-y)$
• C. The maximum value of the function in the interval $(0,2)$ is $1$
• D. There exists a value 'y' in the interval $(0,1)$ such that $f\left(y\right)=-f(2-y)$

Answer 1

Answer: (A), (D)

There exists a value 'y' in the interval $(0,1)$ such that $f\left(y\right)=-f(2-y)$

Let $g\left(y\right)=f\left(y\right)-f(y+1)$, where $yϵ[0,1]$
Then
$g\left(0\right)=f\left(0\right)-f\left(1\right)=1-1-=-ve$
$g\left(1\right)=f\left(1\right)-f\left(2\right)=1+1=ve$
$\because g\left(0\right)$ and $g\left(1\right)$ have opposite signs so there must exist $yϵ[0,1]$ such that
$g\left(y\right)=0$
$\Rightarrow f\left(y\right)=f(y+1)$
Hence there exists a value 'y' in the interval $(0,1)$ such that $f\left(y\right)=f(y+1)$
So option (a) is true
By similar logic, option (d) is also true