Problem 109 - A Complete Resource Book in Mathematics for JEE Main 2017

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

\( \frac{a}{b} = \frac{4-\sqrt{3}}{4+\sqrt{3}} \)

Step 1: Define Geometric Mean (G.M.)

The Geometric Mean between two numbers \(a\) and \(b\) is given by \( \sqrt{ab} \).

Step 2: Define Harmonic Mean (H.M.)

The Harmonic Mean between two numbers \(a\) and \(b\) is given by \( \frac{2ab}{a+b} \).

Step 3: Set Up the Equation from Condition

According to the problem, the G.M. is twice the H.M. This gives us the equation: \( \sqrt{ab} = 2 \times \frac{2ab}{a+b} \).

Step 4: Simplify the Equation

Simplify the equation: \[ \sqrt{ab} = \frac{4ab}{a+b} \] Multiply both sides by \( (a+b) \): \[ \sqrt{ab}(a+b) = 4ab \]

Step 5: Further Simplify the Resultant Equation

Divide both sides by \( \sqrt{ab} \): \[ a+b = 4\sqrt{ab} \] Squaring both sides to remove the square root: \[ (a+b)^2 = 16ab \]

Step 6: Substitute and Solve for \( \frac{a}{b} \)

Let \( a = xb \). Substitute into the equation: \[ ((x+1)b)^2 = 16xb^2 \] Simplifying: \[ (x+1)^2 = 16x \] Expand and rearrange: \[ x^2 + 2x + 1 = 16x \] \[ x^2 - 14x + 1 = 0 \]

Step 7: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) with \( a=1, b=-14, c=1 \): \[x = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm \sqrt{192}}{2} = \frac{14 \pm 8\sqrt{3}}{2}\]Therefore, we have two solutions: \( x = 7 + 4\sqrt{3} \) and \( x = 7 - 4\sqrt{3} \).

Step 8: Choose the Correct Option

Given \( \frac{a}{b} = x \), we have two possible values for \( x \) which match with the options: \( \frac{a}{b} = 4 + \sqrt{3} \) and \( \frac{a}{b} = 4 - \sqrt{3} \). Comparing with given options, \( x = \frac{4-\sqrt{3}}{4+\sqrt{3}} \) is Option (B), thus our solution is consistent with the given constraint.

Key Concepts

Algebraic Problem Solving

Algebraic problem solving involves using algebraic techniques to find unknown values in a given problem. At the heart of many algebraic problems is the formation of equations derived from a set of conditions or relationships. This exercise begins with the identification of the relationship between the geometric mean (G.M.) and harmonic mean (H.M.) of two numbers, which is used to set up an equation. By carefully defining and utilizing these mathematical expressions, we translate a verbal description into a solvable mathematical equation.

• The geometric mean for two numbers, given as \( \sqrt{ab} \), provides a multiplicative measure of central tendency.
• The harmonic mean provides a different perspective, being \( \frac{2ab}{a+b} \), often used in scenarios involving rates or ratios.
• Forming the equation from the given condition requires interpretation: the G.M. is twice the H.M.

Understanding the descriptions set by the problem, we write the condition as an equation \( \sqrt{ab} = 2 \times \frac{2ab}{a+b} \) and proceed to manipulate this equation. This is vital, as accurate modeling of the problem in algebraic terms is the foundation for finding the solution.

Quadratic Equations

Quadratic equations are fundamental in algebra, characterized by the form \( ax^2 + bx + c = 0 \). Solving these equations often requires methods such as factoring, completing the square, or using the quadratic formula. In our problem, simplification leads to a quadratic equation that we need to solve for the variable \( x \).

• The equation \( x^2 - 14x + 1 = 0 \) arises as a direct consequence of squaring both sides of an expression to eliminate a square root.
• To find the solutions to the quadratic equation, we utilize the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Plugging in the values from our specific equation: \( a = 1 \), \( b = -14 \), and \( c = 1 \), simplifies the process to finding: \( x = 7 \pm 4\sqrt{3} \).

Quadratic equations elegantly showcase the precision of algebraic problem-solving, transforming complex relationships into solvable mathematical statements. These solutions represent possible values for \( \frac{a}{b} \), which correlate back to the options provided in the exercise.

Simplifying Expressions

Simplifying expressions is an essential skill in algebra that involves making complex equations more manageable, often through combining like terms or reducing fractions. During this exercise, several steps involve simplification to make further calculations easier and to derive meaningful results.

• Initially, the condition \( \sqrt{ab} = \frac{4ab}{a+b} \) is transformed into a simpler form by logical operations.
• Simplification often involves multiplying both sides by a common factor, such as \( (a+b) \), helping to eliminate fractions or square roots.
• As one progresses to solve the quadratic \( (x+1)^2 = 16x \), expanding and rearranging terms like \( x^2 + 2x + 1 = 16x \) aids in isolating the unknown variable.

Effective simplification streamlines the path to a solution, removing potential complications and making operations more straightforward. It allows for the cleaner derivation of solutions, like finding \( x = 7 \pm 4\sqrt{3} \), and correlating it with the problem's multiple-choice options. Simplification not only aids in solving the equation but also in confirming the correct answer through comparison with options provided.