Problem 110 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(n \in N\) such that \((7+4 \sqrt{3})^{\mathrm{n}}=I+f\), where \(I \in N\) and \(0

Step-by-Step Solution

Answer

The value of \((I+f)(I-f)\) is 1.

Step 1: Express the given term in a different form

We are given that \((7+4\sqrt{3})^n = I + f\), where \(I\) is an integer and \(0 < f < 1\). Let's express \(I\) and \(f\) using the binomial expansion or another form that makes it easier to understand in the context of this problem.

Step 2: Consider the behavior of conjugate

Recognize that the conjugate of \((7 + 4\sqrt{3})\) is \( (7 - 4\sqrt{3}) \). Thus, the product \((7+4\sqrt{3})^n (7-4\sqrt{3})^n = ((7+4\sqrt{3})(7-4\sqrt{3}))^n\).

Step 3: Simplify the conjugate product

Calculate the product \((7+4\sqrt{3})(7-4\sqrt{3}) = 49 - (4\sqrt{3})^2 = 49 - 48 = 1\). So, \(((7+4\sqrt{3})(7-4\sqrt{3}))^n = 1^n = 1\).

Step 4: Express I and f in terms of conjuguates

From the problem, we have \((7+4\sqrt{3})^n = I + f\). Therefore, \((7-4\sqrt{3})^n = I - f\).

Step 5: Determine the expression for (I+f)(I-f)

Now, consider \((I+f)(I-f) = I^2 - f^2\). Substitute from Step 2: \((I+f)(I-f) = ((7+4\sqrt{3})^n)((7-4\sqrt{3})^n) = 1\) based on Step 3.

Key Concepts

Binomial Expansion

The binomial expansion is a powerful mathematical technique used to express powers of binomials in expanded form. This is especially useful when working with expressions of the form \[(a + b)^n\]where \(n\) is a natural number. The expansion involves terms formed by combinations of the components of the binomial using binomial coefficients, which are derived from Pascal's triangle.

In our given exercise, while we don't directly expand using strict coefficients, we consider the structure of \[(7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n\]Understanding how the binomial expansion applies helps when forming products and calculating powers as it can indicate that certain terms will aid in simplification. This allows us to check conjugates intelligently to deduce a simpler form of complex expressions, providing clarity when solving intricate problems.

Conjugate Multiplication

Conjugate multiplication is a handy algebraic technique for simplifying expressions involving radicals. When dealing with numbers like \[(7 + 4\sqrt{3})\]multiplying it by its conjugate \[(7 - 4\sqrt{3})\]yields a rational number, thus removing the radical component.

In practice, by multiplying a binomial with its conjugate, we often get an integer result due to the identity \[(a+b)(a-b) = a^2 - b^2\]So, combining \[(7+4\sqrt{3}) \, \text{and} \, (7-4\sqrt{3})\]gives us 49 - (4\sqrt{3})^2 = 1. This operation simplifies our original equation drastically, especially when raised to any power \(n\). The conjugate multiplication results help in confirming simplifications that arise from complex-looking expressions, ensuring the expression is easily manageable while retaining integer form.

Integer Properties

The theory of integers revolves around numbers without fractions or decimals, just solid, whole numbers. This is fundamental in algebra, especially when verifying that solutions to problems reside within the set of integers.

In this exercise, the relationship \(I + f = (7+4\sqrt{3})^n\) captures this essential principle because \(I\) is stipulated as an integer while \(f\) is constrained to be a fraction \(0 < f < 1\). The condition that \((I+f)(I-f) = 1\) ultimately rests on these integer properties. If both expressions are perfect conjugates, then according to integer and rational number rules, their operation indeed confirms the relationship, verifying the integer integrity of \(I\).

Using integer properties makes it easier to recognize other potential natural-number solutions, guiding one through formal proofs and reasoning that go beyond numeric calculations to stepwise logical conclusions.