The problem asks to find the maximum value $M$ of the function $f(x) = |x^2 - 2x|$ in the interval $a \le x \le a+1$, where $a \ge 0$. We need to express $M$ in terms of $a$ and then find the minimum value of $M$ when $a$ varies over the non-negative real numbers.

1. Problem Description

The problem asks to find the maximum value $M$ of the function $f(x) = |x^2 - 2x|$ in the interval $a \le x \le a+1$, where $a \ge 0$. We need to express $M$ in terms of $a$ and then find the minimum value of $M$ when $a$ varies over the non-negative real numbers.

2. Solution Steps

(1) First, let's express $f(x)$ without the absolute value sign. We have $x^2 - 2x = x(x-2)$. Thus, $x^2 - 2x \ge 0$ when $x \le 0$ or $x \ge 2$, and $x^2 - 2x < 0$ when $0 < x < 2$. Therefore,

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

The graph of $y = x^2 - 2x$ is a parabola opening upwards with vertex at $x=1$, and $x^2-2x = (x-1)^2 - 1$. The vertex is at $(1,-1)$. The graph of $y=-x^2 + 2x$ is a parabola opening downwards with vertex at $x=1$, and $-x^2+2x = -(x-1)^2 + 1$. The vertex is at $(1,1)$.

So, we have M = 0, N =

2. Thus, $f(x) = x^2 - 2x$ when $x \le 0$ or $x \ge 2$ and $f(x) = -x^2 + 2x$ when $0 < x < 2$.

Now we consider the interval $a \le x \le a+1$.

Case 1: $0 \le a \le 1$. Then the maximum value of $f(x)$ occurs at either $x=a$ or $x=a+1$ or $x=1$.

When $a=0$, $0 \le x \le 1$, $M = 1$, which occurs at $x=1$.

When $a=1$, $1 \le x \le 2$, $M=1$, which occurs at $x=1$.

When $0 \le a \le 1$, $M=1$.

O = 1, P = 1

Case 2: $1 < a \le 2$. The maximum of $f(x)$ for $x \in [a, a+1]$ is attained either at $a$ or $a+1$. In the case $a+1<2$, $M=\max(2a-a^2, 2(a+1)-(a+1)^2)= \max(2a-a^2, 1-a^2)$.

If $a+1=2$, $M=1-a^2$

$f(a) = -a^2 + 2a$, $f(a+1) = (a+1)^2-2(a+1)$. $f(a+1) = (a+1)(a+1-2) = (a+1)(a-1) = a^2 - 1$.

Since $1 < a$, consider $f(a) = -a^2+2a$. Since $a \le 2$, consider $f(a+1) = (a+1)^2 - 2(a+1) = a^2 - 1$. Thus, we want to compare $-a^2+2a$ and $a^2-1$.

If $-a^2+2a > a^2-1$, then $2a^2 - 2a - 1 < 0$. $a = \frac{2 \pm \sqrt{4+8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}$. Since $a>1$, we have $1 < a < \frac{1+\sqrt{3}}{2}$.

$-a^2 + 2a$ if $1 < a \le (1+\sqrt{3})/2$, and $a^2-1$ if $(1+\sqrt{3})/2 < a \le 2$.

If $a > 2$, then $f(x) = x^2-2x$, and $M = \max(a^2-2a, (a+1)^2-2(a+1)) = \max(a^2-2a, a^2-1)$.

$f(a) = (a-1)^2-1 = a^2 - 2a$

$f(a+1) = a^2 - 1$.

Case 3: $a \ge 2$. Then $f(x) = x^2-2x$ in $[a,a+1]$.

$f(a) = a^2-2a$, $f(a+1) = (a+1)^2-2(a+1) = (a+1)(a-1) = a^2-1$.

Then, we calculate $-a^2 + ta = a^2 - 1$ if and only if $a > \frac{1 + \sqrt{3}}{2}$.

For $x= \frac{1+\sqrt{3}}{2}$. Then, we have $a^2-2a = -1$

The maximum occurs when $x=(1+\sqrt{3})/2$, so $Q=1$, $R=3$, $S=2$.

$M = -a^2 + 2a$ when $1 < a \le (1+\sqrt{3})/2$.

$T=2$

If $a > (1+\sqrt{3})/2$, then $f(a+1) > f(a)$ and $M=a^2-1$.

U = 1

When $0\le a \le 1$, $M=1$.

When $a \ge 2$, $M = a^2 - 1$, which occurs at $a+1$

$M = a^2-1$ is minimum when $a=2$, then $M = 3$.

If $1 \le a \le 2$, $M = a^2 - 2a$.

Consider case 1, $0 \le a \le 1$. Since the function equals $1$ in this case, the minimum value of $M$ is $1$ between $0$ and $1$.

Let $a=(1+\sqrt{3})/2$. Since $1<a$, we will study two regions separately.

When $1 < a \le (1+\sqrt{3})/2$, $M = -a^2 + 2a = -(a-1)^2 + 1$. Since $a>1$, this becomes $-a^2 + 2a$.

When $a = 1$, $M = 1$.

When $a = (1+\sqrt{3})/2$, $-a^2 + 2a = -(\frac{1+\sqrt{3}}{2})^2 + 2(\frac{1+\sqrt{3}}{2}) = -\frac{1+2\sqrt{3}+3}{4} + 1+\sqrt{3} = -\frac{4+2\sqrt{3}}{4} + 1+\sqrt{3} = -1 - \frac{\sqrt{3}}{2} + 1+\sqrt{3} = \frac{\sqrt{3}}{2}$.

However, the first case is $M = 1$. When $a \rightarrow 1$, the answer is approximately $a - 1$.

Also, we consider case when $a > (1+\sqrt{3})/2$. Then $M = a^2-1$. When $a = 2$, $M = 3$.

When $1<a<(1+\sqrt{3})/2$, $M=-a^2 + 2a$ is maximum when $a=1$. At $(1+\sqrt{3})/2$, the equation is $\frac{\sqrt{3}}{2}$.

Since we know when $0 < a < 1$, $M=1$. If we consider the function $-a^2+2a$, the value is $M=-a^2+2a$ only if $M \ge 1$.

Then if $1 < a < (1+\sqrt{3})/2$, then $M$ will start at $1$ and decrease at $\frac{\sqrt{3}}{2}$.

However, for our case the smallest value of M is thus $\sqrt{3} / 2$ when $M=-1a2 + 2a$.

However, $a < 1$ is $M = 1$.

So V = $\sqrt{3}$ and W = 2 is not it.

We need to recheck:

If $a > \frac{1+\sqrt{3}}{2}$ then $a^2-1$. It decreases fast but it doesn't affect at all.

The minimum value must happen when $a = \frac{1+\sqrt{3}}{2}$. And $\frac{\sqrt{3}}{2} <1$. So the point when $\frac{\sqrt{3}}{2}=M$ does NOT exist, but the problem is to when to change, so the answer might be $V=\sqrt{3}$, and $W=2$.

3. Final Answer

M = 0, N = 2

O = 1, P = 1

Q = 1, R = 3, S = 2

T = 2

U = 1

V = sqrt(3)

W = 2