Step 1: Solve the Differential Equation

The given differential equation is $f'(x) = 2 - \frac{f(x)}{x}$. We can rewrite this as a first-order linear differential equation: $f'(x) + \frac{1}{x}f(x) = 2$ This is in the form $y' + P(x)y = Q(x)$, with $y = f(x)$, $P(x) = \frac{1}{x}$, and $Q(x) = 2$.

The integrating factor (I.F.) is calculated as: $I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$ (since $x \in (0, \infty)$).

Multiply the differential equation by the integrating factor: $x f'(x) + x \cdot \frac{1}{x} f(x) = 2x$ $x f'(x) + f(x) = 2x$ The left side is the derivative of the product of $f(x)$ and the integrating factor $x$: $\frac{d}{dx}(x f(x)) = 2x$

Integrate both sides with respect to $x$: $\int \frac{d}{dx}(x f(x)) dx = \int 2x dx$ $x f(x) = x^2 + C$ where $C$ is the constant of integration.

The general solution for $f(x)$ is: $f(x) = x + \frac{C}{x}$

To verify, let's find the derivative of our solution: $f'(x) = 1 - \frac{C}{x^2}$ Substitute $f(x)$ and $f'(x)$ into the original equation: $1 - \frac{C}{x^2} = 2 - \frac{x + C/x}{x} = 2 - (1 + \frac{C}{x^2}) = 1 - \frac{C}{x^2}$. The solution is correct.

Step 2: Interpret the Initial Condition

The condition given is $f(1) \text{ e } 1$. This is likely a typographical error for $f(1) \neq 1$. Let's proceed with this assumption. Using our solution, we find $f(1)$: $f(1) = 1 + \frac{C}{1} = 1 + C$ The condition $f(1) \neq 1$ implies $1 + C \neq 1$, which means $C \neq 0$.

So we have the function $f(x) = x + \frac{C}{x}$ and its derivative $f'(x) = 1 - \frac{C}{x^2}$, with the constraint that $C$ is a non-zero constant.

Step 3: Evaluate Each Option

A: $\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = 1$

First, find the expression for $f'(\frac{1}{x})$: $f'\left( \frac{1}{x} \right) = 1 - \frac{C}{(1/x)^2} = 1 - Cx^2$ Now, we take the limit as $x \to 0^+$: $\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - Cx^2) = 1 - C(0) = 1$ This limit is 1, regardless of the value of $C$. So, Option A is correct.

B: $\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = 2$

First, find the expression for $xf(\frac{1}{x})$: $f\left( \frac{1}{x} \right) = \frac{1}{x} + \frac{C}{1/x} = \frac{1}{x} + Cx$ $x f\left( \frac{1}{x} \right) = x \left( \frac{1}{x} + Cx \right) = 1 + Cx^2$ Now, we take the limit as $x \to 0^+$: $\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + Cx^2) = 1 + C(0) = 1$ The option states the limit is 2. Our calculated limit is 1. So, Option B is incorrect.

C: $\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = 0$

First, find the expression for $x^2 f'(x)$: $x^2 f'(x) = x^2 \left( 1 - \frac{C}{x^2} \right) = x^2 - C$ Now, we take the limit as $x \to 0^+$: $\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} (x^2 - C) = 0 - C = -C$ From the initial condition, we know $C \neq 0$. Therefore, the limit is $-C \neq 0$. The option states the limit is 0. So, Option C is incorrect.

D: $|f(x)| \le 2$ for all $x \in (0,2)$

This statement must be true for any function $f$ that satisfies the given conditions, which means for any $C \neq 0$. Let's test this with a specific case. Let's choose a simple non-zero value for $C$, say $C=1$. This corresponds to an initial condition like $f(1)=2$, which satisfies $f(1) \neq 1$. For $C=1$, the function is $f(x) = x + \frac{1}{x}$. Let's check the value of $f(x)$ for an $x$ in the interval $(0,2)$. For instance, let $x=0.5$. $f(0.5) = 0.5 + \frac{1}{0.5} = 0.5 + 2 = 2.5$ $|f(0.5)| = 2.5$, which is not less than or equal to 2. Since we found a counterexample, the statement is not true for all possible functions $f$. So, Option D is incorrect.

Conclusion

Based on the analysis, only option A is correct. Options B, C, and D are incorrect.