Key Concepts and Formulas

• Definition of the Derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
• Squeeze Theorem: If $g(x) \le h(x) \le k(x)$ for all $x$ in an interval containing $a$ (except possibly at $a$), and $\lim_{x \to a} g(x) = \lim_{x \to a} k(x) = L$, then $\lim_{x \to a} h(x) = L$.
• Constant Function: If $f'(x) = 0$ for all $x$ in an interval, then $f(x)$ is constant on that interval.

Step-by-Step Solution

Step 1: Start with the Given Condition We are given: $|f(x) - f(y)| \le |(x - y)^2|, \quad \forall (x,y) \in \mathbb{R}$ Since $(x-y)^2$ is always non-negative, $|(x-y)^2| = (x-y)^2$. Therefore, $|f(x) - f(y)| \le (x - y)^2$ This inequality holds for all real numbers $x$ and $y$.

Step 2: Form the Difference Quotient Our goal is to relate the given inequality to the definition of the derivative. We divide both sides of the inequality by $|x - y|$ for $x \neq y$: $\frac{|f(x) - f(y)|}{|x - y|} \le \frac{(x - y)^2}{|x - y|}$ WHY: Dividing by $|x-y|$ transforms the inequality into the absolute value of a difference quotient, which is necessary to find the derivative. We must specify $x \neq y$ to avoid division by zero.

Step 3: Simplify the Inequality Using the property $\frac{|a|}{|b|} = \left|\frac{a}{b}\right|$, the left side becomes $\left|\frac{f(x) - f(y)}{x - y}\right|$. For the right side, we have $\frac{(x-y)^2}{|x-y|} = \frac{|x-y|^2}{|x-y|} = |x-y|$. Thus, the inequality simplifies to: $\left| \frac{f(x) - f(y)}{x - y} \right| \le |x - y|$

Step 4: Apply Limits and the Squeeze Theorem Now, we take the limit as $x \to y$ to find the derivative. WHY: Taking the limit as $x \to y$ directly relates the difference quotient to the definition of the derivative $f'(y)$. $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| \le \lim_{x \to y} |x - y|$ The right-hand side is: $\lim_{x \to y} |x - y| = |y - y| = 0$ Thus, $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| \le 0$ Since absolute values are non-negative, we must have $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| = 0$ This implies that $\left| \lim_{x \to y} \frac{f(x) - f(y)}{x - y} \right| = 0$ So, $|f'(y)| = 0$, which means $f'(y) = 0$. Note: This step also implicitly proves that $f'(y)$ exists for all $y \in \mathbb{R}$. Since $0 \le \left| \frac{f(x) - f(y)}{x - y} \right| \le |x - y|$, and $\lim_{x \to y} 0 = 0$ and $\lim_{x \to y} |x - y| = 0$, by the Squeeze Theorem, $\lim_{x \to y} \left| \frac{f(x) - f(y)}{x - y} \right| = 0$. This implies $\lim_{x \to y} \frac{f(x) - f(y)}{x - y} = 0$, meaning $f'(y)$ exists and is equal to $0$.

Step 5: Deduce $f'(x) = 0$ Since $|f'(y)| = 0$, we must have $f'(y) = 0$ for all $y \in \mathbb{R}$. WHY: The absolute value of any real number is always non-negative (i.e., $|A| \ge 0$). The only way for an absolute value to be zero is if the number itself is zero. Therefore, $f'(y) = 0$.

Step 6: Conclude that $f(x)$ is Constant Since $f'(x) = 0$ for all $x \in \mathbb{R}$, $f(x)$ must be a constant function. WHY: A function with a derivative of zero everywhere must be constant.

Step 7: Determine the Constant Value We are given that $f(0) = 1$. Since $f(x)$ is a constant function, $f(x) = f(0)$ for all $x \in \mathbb{R}$. Therefore, $f(x) = 1$ for all $x \in \mathbb{R}$.

Step 8: Examine the Options Now we examine the given options: (A) $f(x)$ can take any value in R (B) $f(x) < 0, \forall x \in R$ (C) $f(x) > 0, \forall x \in R$ (D) $f(x) = 0, \forall x \in R$

Since $f(x) = 1$ for all $x \in \mathbb{R}$, option (A) is the correct answer.

Common Mistakes & Tips

• Dividing by Zero: Remember to explicitly state that $x \neq y$ when dividing by $|x-y|$.
• Squeeze Theorem: Correctly applying the Squeeze Theorem is crucial for proving the existence and value of the derivative.
• Constant Function Implication: A zero derivative implies a constant function, which simplifies the problem significantly.

Summary

We started with the given inequality and used the definition of the derivative and the Squeeze Theorem to show that $f'(x) = 0$ for all $x$. This implied that $f(x)$ is a constant function. Using the given condition $f(0) = 1$, we concluded that $f(x) = 1$ for all $x$. Therefore, the correct answer is that f(x) can take any value in R.

Final Answer

The final answer is \boxed{A}.