Question

Given \(t \gt 1\).
Quantity A: \(\frac{1}{\left(1 + \frac{1}{t}\right)^2}\)
Quantity B: \(\frac{1}{\left(1 + \frac{1}{\sqrt{t}}\right)^2}\)

• A. if Quantity A is greater;
• B. if Quantity B is greater;
• C. if the two quantities are equal;
• D. if the relationship cannot be determined from the information given.
• E.

Hint:

When dealing with abstract inequalities, plugging in a simple test value is a great way to quickly determine the relationship and confirm your algebraic reasoning. For \(t=4\): Quantity A = \(\frac{1}{(1 + 1/4)^2} = \frac{1}{(5/4)^2} = \frac{1}{25/16} = \frac{16}{25} = 0.64\). Quantity B = \(\frac{1}{(1 + 1/\sqrt{4})^2} = \frac{1}{(1 + 1/2)^2} = \frac{1}{(3/2)^2} = \frac{1}{9/4} = \frac{4}{9} \approx 0.44\). Since \(0.64 \textgreater 0.44\), Quantity A is greater.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to compare two fractions. Since the numerators are both 1, the fraction with the smaller denominator will have the larger value. Therefore, the problem reduces to comparing the denominators.
Step 2: Key Formula or Approach:
The comparison will rely on the properties of inequalities. Specifically, how inequalities change when taking reciprocals or squaring positive numbers.
Step 3: Detailed Explanation:
Let's compare the terms within the denominators. We are given that \(t \gt 1\).
For any number greater than 1, its square root is smaller than the number itself.
For example, if \(t=4\), then \(\sqrt{t}=2\), and \(4 \gt 2\). So, we can state that:
\[ t \gt \sqrt{t} \]
Since \(t\) and \(\sqrt{t}\) are both positive, taking the reciprocal of both sides will reverse the inequality sign:
\[ \frac{1}{t} \lt \frac{1}{\sqrt{t}} \]
Adding 1 to both sides does not change the direction of the inequality:
\[ 1 + \frac{1}{t} \lt 1 + \frac{1}{\sqrt{t}} \]
Since \(t \gt 1\), both expressions in the inequality are positive. Squaring both sides of an inequality involving positive numbers preserves the direction of the inequality:
\[ \left(1 + \frac{1}{t}\right)^2 \lt \left(1 + \frac{1}{\sqrt{t}}\right)^2 \]
This shows that the denominator of Quantity A is smaller than the denominator of Quantity B.
Step 4: Final Answer:
We have Quantity A = \(\frac{1}{\text{Denominator A}}\) and Quantity B = \(\frac{1}{\text{Denominator B}}\), where Denominator A = \(\left(1 + \frac{1}{t}\right)^2\) and Denominator B = \(\left(1 + \frac{1}{\sqrt{t}}\right)^2\).
Since Denominator A \lt Denominator B and both are positive, taking the reciprocal reverses the inequality:
\[ \frac{1}{\text{Denominator A}} \gt \frac{1}{\text{Denominator B}} \]
Therefore, Quantity A is greater than Quantity B.