Question

Which of the following inequalities holds true?
(A) \(\sqrt{5} + \sqrt{3}>\sqrt{6} + \sqrt{2}\)
(B) If \(a>b\) and \(c<0\), then \(\frac{a}{c}<\frac{b}{c}\)
(C) \(\frac{1}{x^2}>\frac{1}{x}>1\), if \(0<x<1\)
(D) If a and b are positive integers and \(\frac{a-b}{6.25} = \frac{4}{2.5}\) then \(b>a\)
Choose the correct answer from the options given below:

• A. (A), (B) and (D) only
• B. (A), (B) and (C) only
• C. (A) and (B) only
• D. (B) and (C) only

Hint:

When testing inequalities with variables, plugging in a simple test value that fits the condition (like \(x=0.5\) for \(0<x<1\)) is a quick way to check its validity. For inequalities with square roots, squaring both sides is often the most effective method, but remember this only works if both sides are non-negative.

Answer 1

The Correct Option is B

Step 1: Concept Review:

This task involves verifying the accuracy of four distinct mathematical propositions (inequalities and equations).

Step 2: Proposition Analysis:

Proposition (A): \( \sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2} \)

Squaring both sides for comparison:

Left-hand side squared: \( (\sqrt{5} + \sqrt{3})^2 = 5 + 3 + 2\sqrt{15} = 8 + 2\sqrt{15} \)

Right-hand side squared: \( (\sqrt{6} + \sqrt{2})^2 = 6 + 2 + 2\sqrt{12} = 8 + 2\sqrt{12} \)

Comparison: \( 8 + 2\sqrt{15} > 8 + 2\sqrt{12} \), which simplifies to \( 2\sqrt{15} > 2\sqrt{12} \). Since \( 15 > 12 \), this proposition is valid.

Proposition (B): If \( a > b \) and \( c < 0 \), then \( \frac{a}{c} < \frac{b}{c} \)

Division by a negative number in an inequality results in the reversal of the inequality sign. This is a fundamental principle, therefore this proposition is valid.

Proposition (C): \( \frac{1}{x^2} > \frac{1}{x} > 1 \), given \( 0 < x < 1 \)

Testing with \( x = 0.5 \): \( \frac{1}{(0.5)^2} = \frac{1}{0.25} = 4 \), \( \frac{1}{0.5} = 2 \). The inequality becomes \( 4 > 2 > 1 \), which is true. This proposition is valid.

Proposition (D): If \( a \) and \( b \) are positive integers and \( \frac{a-b}{6.25} = \frac{4}{2.5} \), then \( b > a \)

Solving the equation: \( \frac{a-b}{6.25} = 1.6 \). Thus, \( a-b = 1.6 \times 6.25 = 10 \). This implies \( a = b + 10 \), meaning \( a > b \). Therefore, this proposition is invalid.

Step 3: Conclusion:

Propositions (A), (B), and (C) are valid, while (D) is invalid. The correct selection includes only (A), (B), and (C).