Question:medium

The values of infimum and supremum of the set \(S=\{x\in \mathbb{R}:x^3\leq x\}\) are

Show Hint

To find infimum and supremum, first solve the inequality to determine the complete set.
  • \(0\) and \(1\)
  • \(-1\) and \(1\)
  • \(0\) and \(3\)
  • \(1\) and \(3\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
We need to find the infimum (greatest lower bound) and supremum (least upper bound) of a set S defined by an inequality. This involves solving the inequality to determine the interval(s) that define the set.

Step 2: Key Formula or Approach:

1. Solve the inequality \(3x^2 < x^3\). 2. Rewrite the inequality as \(x^3 - 3x^2 > 0\). 3. Factor the polynomial and analyze its sign to find the values of x for which the inequality holds. 4. Identify the infimum and supremum from the resulting set.

Step 3: Detailed Explanation:

We need to solve the inequality: \[ 3x^2 < x^3 \] Move all terms to one side: \[ x^3 - 3x^2 > 0 \] Factor out the common term \(x^2\): \[ x^2(x - 3) > 0 \] For this product to be positive, both factors must have the same sign. We know that \(x^2\) is always non-negative. - If \(x=0\), then \(x^2=0\), and the inequality \(0 > 0\) is false. So \(x \neq 0\). - If \(x \neq 0\), then \(x^2 > 0\). Since \(x^2\) is positive, for the product \(x^2(x-3)\) to be positive, the other factor \((x-3)\) must also be positive. \[ x - 3 > 0 \] \[ x > 3 \] So, the set S is the open interval \((3, \infty)\). \[ S = (3, \infty) = \{x \in \mathbb{R} \mid x > 3\} \] Now we find the infimum and supremum of this set. - The infimum (greatest lower bound) of the interval \((3, \infty)\) is 3. - The supremum (least upper bound) of the interval \((3, \infty)\) does not exist, as the set is unbounded above. We say the supremum is \(\infty\). Note on Discrepancy: None of the given options match the result (infimum=3, supremum=\(\infty\)). This indicates a severe error in the question. The OCR has transcribed the question as \(3x^2 < x^3\). If the question was intended to be, for example, \(3x < x^2\), then \(x^2-3x > 0 \implies x(x-3)>0\), which gives \(S = (-\infty, 0) \cup (3, \infty)\), which is also unbounded. If the question was \(x^3 < 3x^2\), then \(x^2(x-3)<0\), which gives \(x<3\) and \(x \neq 0\), so \(S = (-\infty, 0) \cup (0, 3)\), which is unbounded below. The question seems irredeemably flawed as stated and transcribed.

Step 4: Final Answer:

As per the inequality \(3x^2 < x^3\), the set is \((3, \infty)\), which has an infimum of 3 and is unbounded above (supremum is \(\infty\)). None of the options are correct.
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