Let x be a real number between 0 and 1. select the correct answer using the code given below:
Which of the following statements is/are correct?
I. x² > x³
II. x > √x
I only
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.
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.
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.
Based on the analysis, only Statement I is correct.
| 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 |
Understanding how numbers between 0 and 1 behave when operated upon is crucial for solving inequalities like these. Here are some key properties:
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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When a child reaches adolescence, there is apt to be a conflict between the parents and the child, since
the latter considers himself to be by now quite capable of managing his own affairs, while the former
are filled with parental solicitude, which is often a disguise for love of power. Parents consider, usually,
that the various moral problems which arise in adolescence are peculiarly their province. The options
they express, however, are so dogmatic that the young seldom confide in them, and usually go their
own way in secret.