Problem 168 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The correct answer is (B) \(f(1) = -g(1)\).

Step 1: Understanding Polynomial Division

For a polynomial \(h(x)\) to be divisible by another polynomial \(x^2 + x + 1\), \(h(x)\) must yield a remainder of 0 when divided by \(x^2 + x + 1\). This means that the roots of \(x^2 + x + 1\) should satisfy \(h(x) = 0\).

Step 2: Roots of the Divisor

The polynomial \(x^2 + x + 1\) can be factored or identified to have roots \(\omega\) and \(\omega^2\), where \(\omega\) is a primitive cube root of unity. These roots satisfy \(\omega^3 = 1\) and \(\omega eq 1\).

Step 3: Apply Root Conditions

For \(h(x)\) to be divisible by \(x^2 + x + 1\), we have \(h(\omega) = 0\) and \(h(\omega^2) = 0\). Specifically, this results in: \[ \omega f(\omega^3) + \omega^2 g(\omega^6) = 0 \]\[ \omega^2 f((\omega^2)^3) + (\omega^2)^2 g((\omega^2)^6) = 0 \]

Step 4: Simplifying Conditions

Using the properties that \(\omega^3 = 1\) and \(\omega^6 = 1\), the above equations simplify to:\[ \omega f(1) + \omega^2 g(1) = 0 \]\[ \omega^2 f(1) + \omega g(1) = 0 \]

Step 5: Solving the System of Equations

From the equations:Equation 1: \( \omega f(1) + \omega^2 g(1) = 0 \)Equation 2: \( \omega^2 f(1) + \omega g(1) = 0 \)To solve this, equate the coefficients of \(\omega\) and \(\omega^2\). This results in the following system:1. \(f(1) = -g(1)\) from both equations being zero.

Key Concepts

Divisibility Conditions

Divisibility conditions in polynomials play a crucial role in determining how one polynomial can divide another. When we say that a polynomial \( h(x) \) is divisible by another polynomial like \( x^2 + x + 1 \), this implies that when you divide \( h(x) \) by this divisor, there should be no remainder. In simpler terms, if \( h(x) \) is divisible by \( x^2 + x + 1 \), the polynomial must equate to zero at all the roots of the divisor.
Understanding this concept is key in polynomial division problems since solving them often revolves around checking divisibility by verifying that the given conditions are met. These conditions help in reducing larger polynomial problems into simpler ones by focusing on specific values of \( x \) that simplify the polynomial expressions further.

Roots of Unity

Roots of unity are particularly fascinating in mathematics due to their cyclical nature. Primitive roots of unity are complex numbers that, when raised to a certain power, return to 1.
If \( \omega \) is a primitive cube root of unity, then it satisfies \( \omega^3 = 1 \) but \( \omega eq 1 \).The polynomial \( x^2 + x + 1 \) has roots which can be expressed as \( \omega \) and \( \omega^2 \) because solving \( x^2 + x + 1 = 0 \) using the quadratic formula gives us these complex roots.
Understanding these roots is instrumental in finding whether a given polynomial\( h(x) \) holds true under these conditions. When dealing with roots of unity, especially in polynomial equations, they provide useful identities that simplify equations significantly, as evidenced in dividing by \( x^2 + x + 1 \) where you explore these roots and their properties.

System of Equations

A system of equations is essentially a set of equations with multiple variables that are solved together. In our context, after addressing divisions and the roots of unity, we end up with a system of two equations in terms of \( f(x) \) and \( g(x) \).
These equations are:

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

You solve such systems by trying to find values of \( f(1) \) and \( g(1) \) that satisfy both equations simultaneously.In this case, the magic lies in recognizing the interrelationship between coefficients that eventually led to the condition that \( f(1) = -g(1) \).
Systems like these, often arising from polynomial conditions, help encapsulate complex relationships in polynomial equations by breaking them down into solvable parts.