Problem 108 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(R=(5+2 \sqrt{6})^{\mathrm{n}}\) and \(f=\) fractional part of \(R\), then \(R(1-f)=\) (A) 1 (B) \(-1\) (C) 0 (D) none of these

Step-by-Step Solution

Answer

1

Step 1: Understand the Components of the Expression

The expression given is \( R = (5 + 2 \sqrt{6})^n \) where \( f \) is the fractional part of \( R \). Our task is to find \( R(1-f) \).

Step 2: Find the Conjugate Expression

Consider the conjugate of \( 5 + 2 \sqrt{6} \), which is \( 5 - 2 \sqrt{6} \). Note that \((5 + 2 \sqrt{6})(5 - 2 \sqrt{6}) = 1\). So, \((5 + 2 \sqrt{6})^n (5 - 2 \sqrt{6})^n = 1^n = 1\).

Step 3: Define the Integer and Fractional Parts

Let \( R = (5 + 2 \sqrt{6})^n = I + f \) where \( I \) is an integer and \( 0 \leq f < 1 \). Now consider \( (5 - 2 \sqrt{6})^n \), which is a very small number close to 0 because \( 0 < 5 - 2 \sqrt{6} < 1 \).

Step 4: Relation Between Components

Since \( R = I + f \) and the whole conjugate expression yields 1, we have:\[ I + f + (5 - 2 \sqrt{6})^n = I + (f + ((5 - 2 \sqrt{6})^n)) = 1. \] Hence, \( f + (5 - 2 \sqrt{6})^n = 0 \) because \( (5 - 2 \sqrt{6})^n \) is a tiny negative effect balancing the equation.

Step 5: Solve for \( R(1-f) \)

From Step 4, if \( f=(5-2\sqrt{6})^n \): \[ 1 - f = 1 - (5 - 2\sqrt{6})^n. \] Substituting back, we get: \[ R(1-f) = (5 + 2\sqrt{6})^n \cdot (1 - (5 - 2\sqrt{6})^n) = 1 \].

Key Concepts

Exponents and Radicals

In algebra, exponents and radicals are fundamental concepts used for expressing repeated multiplication and roots of numbers, respectively. An exponent is a small number written above and to the right of a base number, indicating how many times to multiply the base by itself. For example, in \(a^n\), \(a\) is the base, and \(n\) is the exponent. This means we multiply \(a\) by itself \(n\) times.

Radicals, on the other hand, are used to signify roots of numbers. The most common radical is the square root, expressed as \(\sqrt{x}\). When we see \(\sqrt{6}\) in the exercise, it means we are finding the number that, when squared, equals 6. Just as exponents can show growth, radicals often involve a form of reduction or division, getting the root value.

For expressions like \((5 + 2\sqrt{6})^n\), combining both exponents and radicals involves first resolving the radical component and then applying the exponentiation. This expands to mean, multiply \((5 + 2\sqrt{6})\) by itself \(n\) times, applying the rules of exponents and arithmetic operations.

Conjugates

In algebra, a conjugate refers to a binomial expression with the opposite middle sign. If you have an expression \(a + b\), its conjugate would be \(a - b\). Conjugates often simplify problems involving roots because their multiplication results in a rational number.

The expression \((5 + 2\sqrt{6})\) has a conjugate of \((5 - 2\sqrt{6})\). When these two are multiplied, the radicals disappear:

• \((5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1\)

This is known as the difference of squares and results in a neat, rational product. Conjugates are especially useful in this exercise, allowing us to express and simplify powers of numbers involving radicals effectively.

Fractional Part

The fractional part of a number, denoted as \(f\), represents the portion of the number that remains after removing the integer part. For instance, if a number \(R\) is expressed as \(R = I + f\), \(I\) would be its integer part, and \(0 \leq f < 1\) would be its fractional part.

In our exercise, we have \(R = (5 + 2\sqrt{6})^n\). The fractional part, captured by \(f\), complements the integer part such that the sum becomes a whole number in the context of conjugate expressions. We learned that \((5 - 2\sqrt{6})^n\), although a small number, behaves almost like a fractional depth balancing equation with \(f\).

Understanding the fractional part is important in distinguishing the whole number from the tiny residue in expressions, which adds depth to our solutions, assuring all components of the calculation are accounted for, like achieving the result of \(R(1-f) = 1\) in the problem.