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

Question

If the equation \(a_{n} x^{n}+a_{n-1} x^{n-1}+-+a_{1} x=0, a_{1} \neq 0\), \(n \geq 2\), has a positive root \(x=\alpha\), then the equation \(n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_{1}=0\) has a positive root, which is (A) greater than a (B) smaller than a (C) greater than or equal to a (D) equal to a

Step-by-Step Solution

Answer

The answer is (B) smaller than a.

Step 1: Understand the problem

The problem involves two polynomial equations. We start with an equation of degree \(n\) and know it has a positive root \(x = \alpha\). The task is to find out what happens with the positive roots of the derived polynomial equation of degree \(n-1\).

Step 2: Analyze the first equation

The given equation is \( a_{n}x^{n} + a_{n-1}x^{n-1} + \ldots + a_{1}x = 0 \). We know that \(x = \alpha\) is a positive root of this equation.

Step 3: Differentiate the equation

Differentiate the first polynomial equation with respect to \(x\). The derivative is \( n a_{n}x^{n-1} + (n-1) a_{n-1}x^{n-2} + \ldots + a_{1} = 0 \), which is the given second equation.

Step 4: Apply Rolle's Theorem

Since \(x = \alpha\) is a root of the original equation, according to Rolle's Theorem, there must exist a root \(x = \beta\) of the derivative equation \(n a_{n} x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_{1} = 0\) such that \(\beta\) lies in between consecutive roots of the original polynomial.

Step 5: Conclusion on root location

\( \alpha \), being a positive root in the original polynomial of degree \(n\), implies that the derived polynomial's positive root \( \beta \) must be less than \( \alpha \) because it is situated between consecutive positive roots of the original polynomial.

Key Concepts

Roots of Equations

Polynomials can take various forms, and understanding their roots is crucial in algebra. A "root" of an equation is the value of the variable that makes the entire equation equal to zero. For a polynomial equation like \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 = 0\), roots are particularly important because they reveal where the polynomial touches or crosses the x-axis on a graph.
For the given equation in the exercise, \(x = \alpha\) is already established as a positive root. This means that if you substitute \(\alpha\) for \(x\), the equation results in zero. Finding roots is often the first step in understanding more complex behaviors of polynomials, as these roots define critical points on their graphs.

• Polynomials of degree \(n\) can have up to \(n\) roots.
• Roots can be complex, but in this case, we're focusing on positive real roots.
• The properties of these roots are used to further analyze polynomial behavior.

Rolle's Theorem

Rolle's Theorem is a crucial concept in calculus, particularly helpful when dealing with differentiable functions on closed intervals. It states that if a function is continuous on \([a, b]\), differentiable on \((a, b)\), and if \(f(a)=f(b)\), then there is at least one \(c\) in \((a, b)\) where the derivative \(f'(c) = 0\).
In the context of this exercise, we utilize Rolle’s Theorem to predict the existence of another root of the derivative of a polynomial. Since \(x = \alpha\) is a root, and given that polynomials are continuous and differentiable everywhere, there exists a point \(\beta\) such that the derivative of the polynomial is zero at that point. In simpler terms, there's a smooth turning point on the curve between consecutive roots, which aligns with our task of locating the positive root for the new derived equation.

• Useful for finding stationary points (maxima, minima)
• Ensures at least one zero exists for the derivative in a given range
• Only applicable under specific continuity and differentiability conditions

Differentiation in Polynomials

Differentiation is the calculus method of finding the rate at which a function changes. When applied to polynomials, it helps us generate other polynomial equations known as derivatives. Differentiating a polynomial like \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1=0\) results in reducing its degree by one. This process is what leads to the second equation in our exercise.
The rules for differentiating polynomials are straightforward:

• For each term \(a_n x^n\), the derivative is \(n a_n x^{n-1}\).
• Repeat this until all terms are differentiated.

The newly formed polynomial by differentiation, \(n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 = 0\), provides insights into the behavior of its parent polynomial, especially near its roots. It captures the slope of the original polynomial’s curve at any point \(x\), marking critical points that delineate intervals of increase or decrease.
Understanding differentiation not only facilitates finding additional roots in polynomials but also enhances our capability of analyzing their graphical properties.

Higher Degree Polynomials

Higher degree polynomials, such as those with degrees greater than two, possess intriguing characteristics and complexities. These polynomials can exhibit multiple turns, intersect the x-axis at several points, and thus have multiple roots, both real and complex.
When dealing with higher degree polynomials in the context of roots, key features include:

• A degree \(n\) polynomial can have up to \(n\) real roots.
• The number of turns or bends in the graph is \(n-1\), which results from the nature of their derivatives.

In our exercise, the polynomial of degree \(n\) is differentiated to yield a polynomial of degree \(n-1\). This step shifts our focus from the curve's intersections with the x-axis to critical points that are pivotal between consecutive roots of the higher degree polynomial. Essentially, working with higher degree polynomials in this manner opens a window into their tangent behaviors and modifications of slope, guiding us from maximum to minimum values along their curves. These characteristics are fundamental in predicting the behavior of complicated dynamic outputs expressed through polynomial equations.