Problem 109 - A Complete Resource Book in Mathematics for JEE Main

Question

If the G.M. between \(a\) and \(b\) be twice the H.M., then \(\frac{a}{b}\) is equal to (A) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) (B) \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) (C) \(\frac{4+\sqrt{3}}{4-\sqrt{3}}\) (D) \(\frac{4-\sqrt{3}}{4+\sqrt{3}}\)

Step-by-Step Solution

Answer

Option (B) \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\).

Step 1: Understanding the Problem

We are given that the geometric mean (G.M.) between numbers \(a\) and \(b\) is twice the harmonic mean (H.M.) between them. We need to find the ratio \(\frac{a}{b}\).

Step 2: Formulas for G.M. and H.M.

The geometric mean between \(a\) and \(b\) is \(\text{G.M.}=\sqrt{ab}\). The harmonic mean between \(a\) and \(b\) is \(\text{H.M.}=\frac{2ab}{a+b}\).

Step 3: Setting Up the Equation

The problem states \(\text{G.M.} = 2 \times \text{H.M.}\). Substitute the formulas: \[\sqrt{ab} = 2 \times \frac{2ab}{a+b}\].

Step 4: Squaring Both Sides

Square both sides of the equation to remove the square root: \[(ab) = 4 \times \left(\frac{2ab}{a+b}\right)^2\].

Step 5: Simplifying the Equation

First, expand the right side: \[\left(\frac{2ab}{a+b}\right)^2 = \frac{4a^2b^2}{(a+b)^2}\]. Now the equation becomes \[ab = \frac{16a^2b^2}{(a+b)^2}\].

Step 6: Cross Multiplying

Cross multiply to simplify: \[(a+b)^2 \times ab = 16a^2b^2\].

Step 7: Rearrange the Equation

Divide both sides by \(ab\) and expand \((a+b)^2\): \[a^2 + b^2 + 2ab = 16ab\]. This simplifies to \[a^2 + b^2 - 14ab = 0\].

Step 8: Using the Quadratic Formula

Recognize \(a^2 + b^2 - 14ab = 0\) as a quadratic in terms of \(a/b\). Let \(x = \frac{a}{b}\), then \[bx^2 + b - 14bx = 0\], which simplifies to \[x^2 - 14x + 1 = 0\].

Step 9: Solving the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a=1\), \(b=-14\), \(c=1\). Solving gives \[x = \frac{14 \pm \sqrt{196-4}}{2}\].

Step 10: Finding the Solutions

This simplifies to \[x = \frac{14 \pm \sqrt{192}}{2}\] which is \[x = \frac{14 \pm 8\sqrt{3}}{2}\]. Further simplifying gives \[x = 7 \pm 4\sqrt{3}\].

Step 11: Determining the Correct Option

The acceptable solution for \(\frac{a}{b}\) in the options is \(7 - 4\sqrt{3}\), which corresponds to option (B) when written as \(\frac{2-\sqrt{3}}{2+\sqrt{3}}\) after rationalizing.

Key Concepts

Geometric Mean

The geometric mean of two numbers, say \(a\) and \(b\), is a type of average that is particularly useful in various mathematical settings, especially when dealing with growth rates or proportions. It is calculated as the square root of the product of the numbers:

1. \(\text{G.M.} = \sqrt{ab}\)

The geometric mean provides a measure of central tendency by favoring smaller values and is less affected by extremely large values compared to the arithmetic mean. It is often used in contexts where the numbers have a multiplicative relationship.
In the given problem, the geometric mean between \(a\) and \(b\) is specified as twice the harmonic mean, allowing us to establish a relationship between these means in order to solve for the ratio \(\frac{a}{b}\).

Harmonic Mean

The harmonic mean is another type of average suitable for situations involving rates or ratios. It is defined for two numbers \(a\) and \(b\) as follows:

1. \(\text{H.M.} = \frac{2ab}{a+b}\)

The harmonic mean tends to emphasize smaller values and is particularly useful when the quantities being averaged are rates, such as speeds or densities. In cases where the harmonic mean is specified as half of the geometric mean, you can derive important relationships between the variables involved.
In equations or problems like the current exercise, understanding how to express one mean in terms of another allows us to derive equations that we can manipulate algebraically, which plays a critical role in solving for unknown variables or ratios.

Quadratic Equations

Quadratic equations are fundamental in algebra, involving an unknown variable raised to the power of two. The general form of a quadratic equation is:

1. \(ax^2 + bx + c = 0\)

In our problem context, we arrived at the equation \(x^2 - 14x + 1 = 0\) by manipulating the given expressions involving geometric and harmonic means. Solving quadratic equations typically involves several methods:

• Factoring
• Completing the square
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In this exercise, we used the quadratic formula to find the roots of the equation. The formula is powerful especially when equations cannot be easily factored. Here, it helped us find the solution \(x = 7 \pm 4\sqrt{3}\), which was crucial to determine the given options for \(\frac{a}{b}\) in the problem.