Let's analyze each statement:

A. ∃x such that x² = x + 1

This asks if there exists a real number $x$ such that $x^2 = x + 1$. Rearrange the equation: $x^2 - x - 1 = 0$ The discriminant ($\Delta$) is: $\Delta = (-1)^2 - 4(1)(-1) = 1 + 4 = 5$ Since $\Delta > 0$, this quadratic equation has two real solutions. Therefore, this statement is true.

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B. ∀x, x² ≤ x

This asks if for all real numbers $x$, $x^2 \leq x$. Rearrange: $x^2 - x \leq 0 \quad \text{or} \quad x(x - 1) \leq 0$ The critical points are $x = 0$ and $x = 1$. Test intervals:

1. For $x \in (-\infty, 0)$: $x(x - 1) > 0$ (false).
2. For $x \in [0, 1]$: $x(x - 1) \leq 0$ (true).
3. For $x \in (1, \infty)$: $x(x - 1) > 0$ (false).

Thus, this statement is false because it is not true for all $x$.

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C. ∀x, x² = 0

This asks if for all real numbers $x$, $x^2 = 0$. Clearly, this is false because $x^2 = 0$ is only true when $x = 0$. For other values of $x$, $x^2 \neq 0$.

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D. ∃x such that x² ≤ x

This asks if there exists a real number $x$ such that $x^2 \leq x$. From the analysis in $B$:

• $x^2 \leq x$ is true for $x \in [0, 1]$.
• For example, $x = 0.5$ satisfies $x^2 = 0.25 \leq 0.5$.

Thus, this statement is true.

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Final Answer:

• True: A, D
• False: B, C

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Would you like further details or a graph to visualize these solutions?
Here are some related questions you might find interesting:

1. How do we solve quadratic inequalities like $x^2 - x \leq 0$?
2. What is the significance of the discriminant ($\Delta$) in quadratic equations?
3. Can $x^n \leq x$ hold for all $n$ and $x$?
4. What are the differences between "for all" ($\forall$) and "there exists" ($\exists$)?
5. How can graphing help verify solutions to these types of problems?

Tip: When dealing with inequalities, always analyze the critical points and test intervals.