Problem 115
Question
If the equation \(a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots \ldots \ldots . .+a_{1} x=0\) \(a_{1} \neq 0, n \geq 2\), has a positive root \(x=\alpha\), then the equation \(n a_{n} x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots \ldots .+a_{1}=0\) has a positive root, which is (a) greater than \(\alpha\) (b) smaller than \(\alpha\) (c) greater than or equal to \(\alpha\) (d) equal to \(\alpha\)
Step-by-Step Solution
Verified Answer
(b) smaller than \(\alpha\)
1Step 1: Understand the given polynomial
The given polynomial is \(a_{n} x^{n}+a_{n-1} x^{n-1} + \ldots + a_{1} x = 0\). It is specified that \(a_{1} eq 0\) and the polynomial has a positive root \(x = \alpha\).
2Step 2: Differentiate the polynomial
To find the derivative of the given polynomial, differentiate each term with respect to \(x\):\[\frac{d}{dx}(a_{n} x^{n}) = n a_{n} x^{n-1},\frac{d}{dx}(a_{n-1} x^{n-1}) = (n-1) a_{n-1} x^{n-2},\ldots\frac{d}{dx}(a_{1} x) = a_{1}\] Therefore, the derivative is \[n a_{n} x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_{1} = 0\].
3Step 3: Apply Rolle's Theorem
Rolle's Theorem states if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), with \(f(a) = f(b)\), then there is at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\). Since the original polynomial has a positive root \(\alpha\), and the function is continuous and differentiable, it implies there must be a positive root to the derivative which lies between two positive roots of the original polynomial.
4Step 4: Determine nature of the roots
Since the derivative has a positive root different from \(\alpha\) by Rolle's Theorem, this new root will be less than \(\alpha\), as it lies in the interval where the original function crosses from positive to negative or vice versa before reaching \(\alpha\).
5Step 5: Conclusion
Based on the application of Rolle's Theorem and the properties of differentiable functions, the new polynomial formed by differentiating has a root smaller than the positive root \(\alpha\) of the original polynomial.
Key Concepts
Rolle's TheoremPositive Root AnalysisPolynomial Roots
Rolle's Theorem
Rolle's Theorem is a fundamental concept in calculus, often used to find roots of polynomial derivatives. Imagine it this way: if you have a smooth curve that starts and ends at the same point on a graph, there must be at least one spot along the curve where the slope, or tangent, is completely flat. This is the crux of Rolle's Theorem. It provides a bridge for understanding the behavior of polynomial functions.
- The theorem applies to functions that are continuous and differentiable.
- The key condition is that the start and end points of the interval must have the same function value.
Positive Root Analysis
Positive root analysis involves examining where a polynomial equation can have solutions that are greater than zero. To unwrap the mystery of positive roots, start by recognizing that if a polynomial has a positive root, the function must curve up through the x-axis somewhere in the positive part of the graph.
- Since the polynomial's behavior around roots can indicate the presence of additional roots, evaluating the derivative helps us understand the nature around these points.
- Through analysis, the presence of a positive root can suggest sign changes in the graph and potential additional roots in close proximity.
Polynomial Roots
Understanding polynomial roots is key to delving into polynomial functions. Roots are where the polynomial equals zero; visually, they're points where the graph of the polynomial touches or crosses the x-axis.
- In our case, polynomial roots are especially interesting when they are positive since they craft specific characteristics of the function curve.
- Polynomial roots are influenced by the coefficients and the degree of the equation.
Other exercises in this chapter
Problem 113
A value of \(c\) for which conclusion of Mean Value Theorem holds for the function \(f(x)=\log _{e} x\) on the interval \([1,3]\) is (a) \(\log _{3} e\) (b) \(\
View solution Problem 114
Let \(f\) be differentiable for all \(x\). If \(f(1)=-2\) and \(f^{\prime}(x) \geq 2\) for \(x \in[1,6]\), then (a) \(f(6) \geq 8(\) b \() f(6)
View solution Problem 116
If \(2 a+3 b+6 c=0\), then at least one root of the equation \(a x^{2}+b x+c=0\) lies in the interval (a) \((1,3)\) (b) \((1,2)\) (c) \((2,3)\) (d) \((0,1)\)
View solution Problem 118
Let \(f(a)=g(a)=k\) and their nth derivatives \(f^{n}(a), g^{n}(a)\) exist and are not equal for some \(n\). Further if \(\lim _{x \rightarrow a} \frac{f(a) g(x
View solution