Question

STATEMENT - 1 : If the roots of equation \(a x^{4}+b x^{3}+c x^{2}+d x+e=0\) are in H.P then the roots of the equation \(e x^{4}+d x^{3}+c x^{2}+b x+a=0\) are in A.P. STATEMENT- 2 : The equation having reciprocals of the roots of the equation \(f(x)=0\), as roots is \(f\left(\frac{1}{x}\right)=0\) (1) Statement \(-1\) is True, Statement \(-2\) is True ; Statement \(-2\) is a correct explanation for Statement \(-1\) (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement- 1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement \(-1\) is False, Statement \(-2\) is True

Short answer

Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1.

Step-by-step solution

Understand Statement 1

Statement 1 asserts that if the roots of the equation \( ax^{4} + bx^{3} + cx^{2} + dx + e = 0 \) are in Harmonic Progression (H.P.), then the roots of the equation \( ex^{4} + dx^{3} + cx^{2} + bx + a = 0 \) are in Arithmetic Progression (A.P.).

Review Harmonic and Arithmetic Progressions

Recall that in a Harmonic Progression (H.P.), the reciprocals of the roots form an Arithmetic Progression (A.P.). Conversely, if we take roots that are in A.P., their reciprocals will be in H.P.

Reciprocal roots transformation

Given that the roots of \(f(x) = ax^{4} + bx^{3} + cx^{2} + dx + e = 0 \) are in H.P., let's denote the roots by \( y_1, y_2, y_3, y_4 \) such that their reciprocals \(\frac{1}{y_1}, \frac{1}{y_2}, \frac{1}{y_3}, \frac{1}{y_4}\) are in A.P.

Form the reciprocal roots equation

The equation having roots \( \frac{1}{y_1}, \frac{1}{y_2}, \frac{1}{y_3}, \frac{1}{y_4} \) is obtained by substituting \( x \rightarrow \frac{1}{x} \) in the given equation. Therefore, \( a(\frac{1}{x})^4 + b(\frac{1}{x})^3 + c(\frac{1}{x})^2 + d(\frac{1}{x}) + e = 0 \). Simplifying this results in \( ex^4 + dx^3 + cx^2 + bx + a = 0 \).

Analyze Statement 2

Statement 2 claims that the equation having reciprocals of the roots of \( f(x) = 0 \) is \( f(\frac{1}{x}) = 0 \). This is a fundamental property of polynomial equations.

Determine the correctness of statements

Both Statement 1 and Statement 2 are correct. Additionally, since transforming the roots to their reciprocals shows that the roots of the new polynomial will be in A.P. if the originals were in H.P., Statement 2 correctly explains Statement 1.

Key concepts

Harmonic Progression

In mathematics, a Harmonic Progression (H.P.) is a sequence of numbers derived from the reciprocals of an Arithmetic Progression (A.P.). To understand this better, let's first look at what makes up an arithmetic progression.
In an Arithmetic Progression, each term after the first is the sum of the previous term plus a constant difference (d). For example, if we have an A.P.: a, a+d, a+2d, a+3d, and so on.
In a Harmonic Progression, the elements are the reciprocals of the numbers in an A.P. Hence, if we start with: a, a+d, a+2d... in A.P., it results in: 1/a, 1/(a+d), 1/(a+2d)... for H.P.
A common example could be if the A.P. is 3, 6, 9... then the H.P. would be 1/3, 1/6, 1/9... This sequence's key property is that their reciprocals form an A.P.
So, if for a polynomial equation, the roots form an H.P., their reciprocals would form an A.P.

Arithmetic Progression

An Arithmetic Progression (A.P.) is one of the most fundamental types of progressions in mathematics. You might have come across sequences like 2, 4, 6, 8... where each successive number is obtained by adding a constant difference (d). This constant can be positive or negative.
In the context of polynomial roots, if the roots of an equation are in A.P., it means they follow this structured order. For example, if the roots are y1, y2, y3, y4, in A.P., we will have y2 = y1 + d, y3 = y1 + 2d, y4 = y1 + 3d.
Understanding how A.P. works helps solve complex polynomial equations because knowing the structured nature of the roots provides deep insights.

Reciprocal Roots

Reciprocal roots play an essential role in understanding relationships between polynomial equations. Suppose you have a polynomial equation where the roots are y1, y2, y3, y4. The reciprocals of these roots are simply 1/y1, 1/y2, 1/y3, 1/y4.
A crucial property you need to remember is that if you want to find a polynomial whose roots are the reciprocals of another polynomial, you need to transform the equation appropriately. The given polynomial is f(x) = 0, then the polynomial with reciprocal roots is f(1/x) = 0.
For example, if the polynomial equation is ax^4 + bx^3 + cx^2 + dx + e = 0, and you need a polynomial with reciprocal roots. First, you substitute x with 1/x. After simplification, you end up with ex^4 + dx^3 + cx^2 + bx + a = 0.
Recognizing the relationship between reciprocals and H.P. helps understand the deeper connections within polynomial roots and their behavior.