The function given by f(x)=|x-1|+|x-2|+|x-3| is :

The correct answer is Option (3) → Continuous but not differentiable at x = 1, 2 and 3

Each \(|x-a|\) is continuous everywhere, so their sum f is continuous for all x.

Removing the modulus signs on each interval:

    \(x\lt 1\): \(f(x)=(1-x)+(2-x)+(3-x)=6-3x\), slope −3
    \(1\lt x\lt 2\): \(f(x)=(x-1)+(2-x)+(3-x)=4-x\), slope −1
    \(2\lt x\lt 3\): \(f(x)=(x-1)+(x-2)+(3-x)=x\), slope 1
    \(x\gt 3\): \(f(x)=3x-6\), slope 3

At x = 1, 2 and 3 the left-hand and right-hand derivatives differ (−3 and −1, −1 and 1, 1 and 3). So f is not differentiable at these three points, and differentiable everywhere else.

Option (4) is impossible in any case: a function that is differentiable at a point is always continuous there.