The function $\mathrm{f}(x)=2 x-\left|x-x^2\right|$ is

The continuity of $f(x)=2x-|x-x^2|$ depends on the absolute value term, which requires piecewise definition.

Analyzing $x - x^2 = x(1 - x)$, we observe that $|x - x^2| = x - x^2$ for $0 \le x \le 1$ where $x(1 - x) \ge 0$, and $|x - x^2| = x^2 - x$ otherwise. This gives the piecewise form:

$f(x) = \begin{cases} 3x - x^2 & \text{if } x < 0 \\ x^2 + x & \text{if } 0 \le x \le 1 \\ 3x - x^2 & \text{if } x > 1 \end{cases}$

To verify continuity at $x = 1$, compute $f(1) = 1^2 + 1 = 2$, the left-hand limit $\lim_{x \to 1^-}(x^2 + x) = 2$, and the right-hand limit $\lim_{x \to 1^+}(3x - x^2) = 2$. The equality of these values confirms continuity at $x = 1$.

At $x = 0$, the left-hand limit is $\lim_{x \to 0^-}(3x - x^2) = 0$, matching $f(0) = 0$ and the right-hand limit $\lim_{x \to 0^+}(x^2 + x) = 0$, ensuring continuity there as well.

The function is defined and continuous at $x = 1$, so the correct choice is $\boxed{\text{A}}$.