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Question

Let x be a real number between 0 and 1.
Which of the following statements is/are correct?

I. x² > x³
II. x > √x

select the correct answer using the code given below:

The correct answer is

I only

Understanding Inequalities for Real Numbers Between 0 and 1

The question asks us to evaluate two mathematical statements involving a real number, 'x', that lies strictly between 0 and 1 (i.e., $0 < x < 1$). We need to determine which of these statements is correct.

Analyzing Statement I: x² > x³

Statement I is: $x^2 > x^3$.

Let's consider a real number 'x' such that $0 < x < 1$. When we multiply a positive number by a number between 0 and 1, the result is smaller than the original number.

Since $0 < x < 1$, let's multiply the inequality $x < 1$ by 'x'. Since 'x' is positive, the inequality direction remains the same:

$$x \cdot x < 1 \cdot x$$

$$x^2 < x$$

Now, we have the inequality $x^2 < x$. Let's multiply this inequality by 'x' again. Since 'x' is still positive ($0 < x < 1$), the inequality direction remains the same:

$$x \cdot x^2 < x \cdot x$$

$$x^3 < x^2$$

This inequality $x^3 < x^2$ is equivalent to $x^2 > x^3$.

Therefore, for any real number 'x' between 0 and 1, $x^2$ is indeed greater than $x^3$. Statement I is correct.

Analyzing Statement II: x > √x

Statement II is: $x > \sqrt{x}$.

Again, we are considering a real number 'x' such that $0 < x < 1$. Both 'x' and $\sqrt{x}$ will be positive in this range.

To compare two positive numbers, we can compare their squares. If $a$ and $b$ are positive, then $a > b$ if and only if $a^2 > b^2$.

So, the inequality $x > \sqrt{x}$ is equivalent to:

$$x^2 > (\sqrt{x})^2$$

$$x^2 > x$$

Now we need to determine if $x^2 > x$ is true for $0 < x < 1$.

From our analysis of Statement I, we found that for $0 < x < 1$, the inequality $x^2 < x$ holds true.

This means that $x^2 > x$ is false for $0 < x < 1$.

Therefore, the original inequality $x > \sqrt{x}$ is also false for $0 < x < 1$. Statement II is incorrect.

Summary of Findings

  • Statement I: $x^2 > x^3$ is correct for $0 < x < 1$.
  • Statement II: $x > \sqrt{x}$ is incorrect for $0 < x < 1$.

Based on the analysis, only Statement I is correct.

Revision Table: Evaluating Inequalities

Statement Inequality Analysis for $0 < x < 1$ Conclusion
I $x^2 > x^3$ Multiplying $0 < x < 1$ by $x$ gives $x^2 < x$. Multiplying $x^2 < x$ by $x$ gives $x^3 < x^2$. This is equivalent to $x^2 > x^3$. Correct
II $x > \sqrt{x}$ Equivalent to $x^2 > x$. We know $x^2 < x$ for $0 < x < 1$. So, $x^2 > x$ is false. Incorrect

Additional Information: Properties of Numbers Between 0 and 1

Understanding how numbers between 0 and 1 behave when operated upon is crucial for solving inequalities like these. Here are some key properties:

  • Multiplication: When you multiply a number between 0 and 1 by another number (positive), the result is smaller than the original number. For example, $0.5 \times 10 = 5$ (smaller than 10), $0.5 \times 0.5 = 0.25$ (smaller than 0.5).
  • Exponents: For a number 'x' between 0 and 1, as the positive exponent increases, the value of $x^{\text{exponent}}$ decreases. For example, if $x=0.5$: $0.5^1 = 0.5$, $0.5^2 = 0.25$, $0.5^3 = 0.125$. Here, $0.5 > 0.25 > 0.125$, which means $x > x^2 > x^3$ for $0 < x < 1$. This property directly supports the correctness of Statement I.
  • Square Roots: For a number 'x' between 0 and 1, its square root $\sqrt{x}$ is larger than 'x'. For example, if $x=0.25$, $\sqrt{x} = \sqrt{0.25} = 0.5$. Here, $0.25 < 0.5$, which means $x < \sqrt{x}$ for $0 < x < 1$. This property shows why Statement II ($x > \sqrt{x}$) is incorrect.

These properties are opposite to those for numbers greater than 1, where increasing positive exponents result in larger values ($2 < 2^2 < 2^3$) and the square root is smaller than the number ($\sqrt{4}=2 < 4$).

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