Question

Let \(a, b \in \mathbb{R}\) and \(0 \le a < b\), then:

• A. \(\frac{1}{a} < \frac{1}{b}\)
• B. \(ab > b^2\)
• C. \(ab < a^2\)
• D. \(a^2 < ab\)

Hint:

To quickly verify algebraic inequalities, plug in simple positive values that satisfy the condition.
Let \(a = 2\) and \(b = 3\). Then:
- \(a^2 = 4\) and \(ab = 6\).
Since \(4 < 6\) is true, \(a^2 < ab\) is the correct inequality.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves order properties and inequalities of real numbers.
When working with inequalities, multiplying both sides by a positive number preserves the direction of the inequality, whereas multiplying by a negative number reverses it.

Step 3: Detailed Explanation:
We are given the inequality:
\[ 0 \le a < b \]
This means that both \(a\) and \(b\) are non-negative, and \(b\) is strictly greater than \(a\).
Let us analyze each of the given options:
- Option (A): \(\frac{1}{a} < \frac{1}{b}\)
If \(a = 0\), the term \(\frac{1}{a}\) is undefined.
If \(a > 0\), taking the reciprocal of both sides reverses the inequality: \(\frac{1}{a} > \frac{1}{b}\).
Thus, Option (A) is incorrect.
- Option (B): \(ab > b^2\)
We start with our given inequality:
\[ a < b \]
Since \(0 \le a < b\), the number \(b\) must be strictly positive (\(b > 0\)).
Multiply both sides of the inequality by the positive number \(b\):
\[ a \cdot b < b \cdot b \implies ab < b^2 \]
Thus, Option (B) is incorrect.
- Option (C): \(ab < a^2\)
Let us test this by multiplying both sides of our inequality by \(a\).
If \(a > 0\), multiplying both sides of \(a < b\) by the positive number \(a\) preserves the inequality:
\[ a \cdot a < a \cdot b \implies a^2 < ab \]
Thus, Option (C) is incorrect.
- Option (D): \(a^2 < ab\)
As shown above, multiplying both sides of the strict inequality \(a < b\) by the positive real number \(a\) (assuming \(a > 0\)) gives:
\[ a^2 < ab \]
This inequality holds true for all positive real numbers \(a\) and \(b\) where \(a < b\).

Step 4: Final Answer:
Therefore, the correct option is (D).