Given f(x)=|1-x+|x|| , pick the statements that are true. (A) f(x) is…

The correct answer is Option (3) → (A) and (B) only

The form of f depends on the sign of x, through the inner modulus |x|.

For \(x \ge 0\), |x| = x:

    \(f(x)=|1-x+x|=1\)

For \(x \lt 0\), |x| = −x:

    \(f(x)=|1-2x|=1-2x\quad(\text{since } 1-2x \gt 1 \gt 0)\)

So

    \( f(x)=\begin{cases}1-2x, & x \lt 0\\ 1, & x \ge 0\end{cases} \)

Both pieces are continuous, and at x = 0 both give 1, so f is continuous everywhere.

The one-sided derivatives at x = 0:

    \(Lf'(0) = -2,\quad Rf'(0) = 0\)

They differ, so f is not differentiable at x = 0. Near \(x=\frac{1}{2}\) (a positive value), f(x) = 1 is constant, so \(f'\left(\frac{1}{2}\right)=0\).

So f is continuous everywhere, is not differentiable at x = 0 (left derivative −2, right derivative 0), and is differentiable at \(x=\frac{1}{2}\).

Checking the statements

(A) \(f(x)\) is continuous — Correct. Both pieces give 1 at x = 0.

(B) \(f(x)\) is not differentiable at \(x=0\) — Correct. The left derivative is −2, the right derivative is 0.

(C) \(f(x)\) is differentiable everywhere — Incorrect. It fails at x = 0.

(D) \(f(x)\) is not differentiable at \(x=1/2\) — Incorrect. f(x) = 1 for x ≥ 0, so \(f'\left(\frac{1}{2}\right)=0\).