For the function f(x)=|x+1|+|x-1|+|x-2| which of the following are TR…

The correct answer is Option (1) → (A), (B) and (D) only

Removing the modulus signs interval by interval (the critical points are x = −1, 1, 2):

    \( \displaystyle f(x)=\begin{cases}-3x+2, & x\lt -1\\-x+4, & -1\le x\lt 1\\x+2, & 1\le x\lt 2\\3x-2, & x\ge 2\end{cases} \)

The slope is −3, −1, 1, 3 on the four pieces. Each |x − a| is continuous, so their sum is continuous, but the slope jumps at x = −1, 1 and 2.

So f is continuous on R, is not differentiable at x = −1, 1 and 2, decreases up to x = 1 and increases after it (a minimum at x = 1, no maximum).

Checking the statements

(A) \( f(x) \) is continuous on R — Correct. It is a sum of continuous functions |x − a|.

(B) \( f(x) \) is not differentiable at \( x=1 \) — Correct. The left-hand derivative at x = 1 is −1 and the right-hand derivative is 1.

(C) \( f(x) \) is differentiable on R — Incorrect. f is not differentiable at x = −1, 1 and 2.

(D) \( f(x) \) has no local maxima — Correct. The slope only increases (−3, −1, 1, 3), so f has only a minimum, at x = 1.