Problem 168 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)\) and \(g(x)\) are two polynomials such that the polynomial \(h(x)=x f\left(x^{3}\right)+x^{2} g\left(x^{6}\right)\) is divisible by \(x^{2}+x+1\), then (A) \(f(1)=g(1)\) (B) \(f(1)=-g(1)\) (C) \(h(1)=0\) (D) \(h(-1)=0\)

Step-by-Step Solution

Answer

(B) \(f(1) = -g(1)\)

Step 1: Understand the Problem

We have the function \(h(x)=x f(x^3)+x^2 g(x^6)\), which needs to be divisible by \(x^2+x+1\). We are tasked with finding conditions related to the functions \(f(x)\) and \(g(x)\) based on this divisibility.

Step 2: Roots of the Divisor

Identify the roots of the polynomial \(x^2+x+1\). They are the complex numbers \(\omega\) and \(\omega^2\), where \(\omega = e^{2\pi i/3}\) and \(\omega^2 = e^{-2\pi i/3}\). These are called the cube roots of unity and satisfy the equation \(\omega^3=1\) and \(1+\omega+\omega^2=0\).

Step 3: Applying the Roots to h(x)

Since \(h(x)\) is divisible by \(x^2+x+1\), it must be zero when evaluated at \(\omega\) and \(\omega^2\). Thus, \(h(\omega) = \omega f(\omega^3) + \omega^2 g(\omega^6) = 0\) and \(h(\omega^2) = \omega^2 f((\omega^2)^3) + (\omega^2)^2 g((\omega^2)^6) = 0\).

Step 4: Simplifying with Powers of \(\omega\)

Note that \(\omega^3 = 1\) and \((\omega^2)^3 = 1\), which means \(h(\omega) = \omega f(1) + \omega^2 g(1) = 0\) and \(h(\omega^2) = \omega^2 f(1) + \omega g(1) = 0\). Solve the following system of linear equations:\[ \omega f(1) + \omega^2 g(1) = 0 \]\[ \omega^2 f(1) + \omega g(1) = 0 \]

Step 5: Solving the System of Equations

Multiply the first equation by \(\omega\) and the second by \(\omega^2\) (or vice-versa) to align coefficients:1. \( \omega^2 f(1) + \omega^3 g(1) = 0 \)2. \( \omega^4 f(1) + \omega^3 g(1) = 0 \)Since \(\omega^3 = 1\), simplify to:\[ f(1) + \omega g(1) = 0 \]\[ \omega f(1) + g(1) = 0 \]Solving gives \(f(1) = - g(1)\).

Step 6: Identify the Correct Option

From our solutions, we derive that \(f(1) = -g(1)\). This corresponds to option (B) in the given options.

Key Concepts

Roots of Unity

The roots of unity are complex numbers that satisfy the equation \(x^n = 1\). When \(n = 3\), these are called the cube roots of unity, which include the numbers \(1\), \(\omega\), and \(\omega^2\). Here, \(\omega = e^{2\pi i/3}\) and \(\omega^2 = e^{-2\pi i/3}\). These numbers have the special property that \(\omega^3 = 1\) and the sum \(1 + \omega + \omega^2 = 0\).

Understanding the roots of unity is essential in various fields of mathematics, including polynomial theory and complex analysis. They provide a foundation for finding solutions to polynomial equations, particularly those with complex coefficients or operating within circular symmetries.

• **\(\omega\)** represents a rotation by 120 degrees in the complex plane.
• The identity \(1 + \omega + \omega^2 = 0\) is crucial for simplifying expressions involving roots of unity.
• These roots rearrange polynomial expressions into simpler forms, aiding divisibility tests and factorization.

System of Equations

A system of equations consists of multiple equations that are solved together, as they share common variables. In this problem, we have a system of linear equations involving the values \(f(1)\) and \(g(1)\).

The given equations are:

• \(\omega f(1) + \omega^2 g(1) = 0\)
• \(\omega^2 f(1) + \omega g(1) = 0\)

To solve such systems, we can use substitution, elimination, or matrix methods. Here, the properties of \(\omega\) were utilized to simplify and solve the system.

By multiplying and substituting these forms, we learn:

• Simplification with known values like \(\omega^3 = 1\) is critical.
• Aligning terms or coefficients makes equations easier to manage.
• The solution found was \(f(1) = -g(1)\), which came from calculating these aligned equations.

Cube Roots of Unity

The cube roots of unity frequently appear in mathematics, especially in problems involving polynomial division, as seen in this exercise. They are earmarked by their symmetry and cyclical properties, providing powerful tools for simplification.

**Role in Division:** For a polynomial \(x^n + ax + b\) to be divisible by another, a set of roots can often help verify and simplify the division.

**In Complex Planes:** With the cube roots of unity, observe how values revert every three steps, making calculations cyclic and predictable.

**Real World Applications:** Their properties drive insights in signal processing, quantum mechanics, and solving cubic equations.

To harness the cube roots effectively:

• Recognize cyclical patterns: \(\omega^3 = 1\) is key.
• Utilize \(\omega\) relationships: \(f(\omega^3)\) and \(g(\omega^6)\) cycle back to \(f(1)\) and \(g(1)\), respectively.
• Leverage their summative property for zero: As shown, \(1 + \omega + \omega^2 = 0\) helps collapse and navigate terms efficiently.