The function f(x)=|x-1| is

The correct answer is Option (1) → Continuous at x=1 and not differentiable at x=1.

$f(x)=|x-1|=\left\{\begin{matrix}x-1&x≥1\\1-x&x<1\end{matrix}\right.$

$\underset{x→1^+}{\lim}f(x)=0=\underset{x→1^-}{\lim}f(x)=f(1)$

f(x) is continuous at x = 1

$f'(x)=\left\{\begin{matrix}1&x≥1\\-1&x<1\end{matrix}\right.$ (LHD ≠ RHD) at x = 1

⇒ Not differentiable at x = 1