Relationship with Option B

Oct 13, 2025 by pinkandtrash.com 65 views

Introduction

In mathematics, we often encounter various types of problems that require us to simplify complex expressions and find their values. One such problem is the given quotient, which involves square roots and logarithms. In this article, we will explore the given options and determine the correct value of the quotient.

Option A: $\frac{1}{1+\sqrt{3}}$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is $1-\sqrt{3}$ . This will help us eliminate the square root in the denominator.

$\frac{1}{1+\sqrt{3}} \cdot \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{1-\sqrt{3}}{1^2 - (\sqrt{3})^2}$

Using the difference of squares formula, we can simplify the denominator:

$1^2 - (\sqrt{3})^2 = 1 - 3 = -2$

So, the expression becomes:

$\frac{1-\sqrt{3}}{-2} = \frac{\sqrt{3}-1}{2}$

This is not one of the given options, so we can move on to the next one.

Option B: $\frac{\sqrt{3}}{4}$

This option is already in its simplest form, so we can compare it with the given expression. However, we need to find a way to relate it to the original expression.

Option C: $\frac{1+\sqrt{3}}{4}$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is $1-\sqrt{3}$ . This will help us eliminate the square root in the denominator.

$\frac{1+\sqrt{3}}{4} \cdot \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{1^2 - (\sqrt{3})^2}{4(1-\sqrt{3})}$

Using the difference of squares formula, we can simplify the numerator:

$1^2 - (\sqrt{3})^2 = 1 - 3 = -2$

So, the expression becomes:

$\frac{-2}{4(1-\sqrt{3})} = \frac{-1}{2(1-\sqrt{3})}$

This is not one of the given options, so we can move on to the next one.

Option D: $\frac{1-\sqrt{3}}{4}$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is $1+\sqrt{3}$ . This will help us eliminate the square root in the denominator.

$\frac{1-\sqrt{3}}{4} \cdot \frac{1+\sqrt{3}}{1+\sqrt{3}} = \frac{1^2 - (\sqrt{3})^2}{4(1+\sqrt{3})}$

Using the difference of squares formula, we can simplify the numerator:

$1^2 - (\sqrt{3})^2 = 1 - 3 = -2$

So, the expression becomes:

$\frac{-2}{4(1+\sqrt{3})} = \frac{-1}{2(1+\sqrt{3})}$

This is not one of the given options, so we can move on to the next one.

Option E: $-\frac{\ln \sqrt{3}}{2}$

To simplify this expression, we can use the property of logarithms that states $\ln a^b = b\ln a$ . We can rewrite the expression as:

$-\frac{\ln \sqrt{3}}{2} = -\frac{\ln 3^{1/2}}{2}$

Using the property of logarithms, we can simplify the expression:

$-\frac{\ln 3^{1/2}}{2} = -\frac{\frac{1}{2}\ln 3}{2}$

Simplifying further, we get:

$-\frac{\frac{1}{2}\ln 3}{2} = -\frac{\ln 3}{4}$

This is not one of the given options, so we can move on to the next one.

Conclusion

After analyzing all the options, we can see that none of them match the given expression. However, we can try to find a relationship between the given expression and one of the options.