The function $f(x)=|x-1|$ is strictly increasing in the interval .
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (1, ∞)
$f(x)=|x-1|=\left\{\begin{matrix}x-1,&x≥1\\1-x,&x<1\end{matrix}\right.$
$f'(x)=\left\{\begin{matrix}1,&x≥1\\-1,&x<1\end{matrix}\right.$
f(x) is strictly increasing in (1, ∞)