Problem 43
Question
If \(f_{j}=\sum_{i=0}^{2} a_{i j} x^{i}, j=1,2,3\) and if \(f_{j}^{\prime} f_{j}^{\prime \prime}\) denote \(\frac{d f_{j}}{d x}, \frac{d^{2} f_{j}}{d x^{2}}\) respectively, then \(g(x)=\left|\begin{array}{lll}f_{1} & f_{2} & f_{3} \\\ f_{1}^{\prime} & f_{2}^{\prime} & f_{3}^{\prime} \\ f_{1}^{\prime \prime} & f_{2}^{\prime \prime} & f_{3}^{\prime \prime}\end{array}\right|\) is (A) a cubic in \(x\) (B) a quadratic in \(x\) (C) linear in \(x\) (D) a constant
Step-by-Step Solution
Verified Answer
(A) a cubic in \(x\).
1Step 1: Understand the Given Functions
The function \( f_j = \sum_{i=0}^{2} a_{ij} x^{i} \) for \( j = 1, 2, 3 \) means each \( f_j \) is a quadratic polynomial in \( x \). This implies \( f_j = a_{0j} + a_{1j}x + a_{2j}x^2 \).
2Step 2: Differentiate Each Function
Calculate the first and second derivatives of \( f_j \):- First derivative: \( f_j' = \frac{d}{dx}(a_{0j} + a_{1j}x + a_{2j}x^2) = a_{1j} + 2a_{2j} x \)- Second derivative: \( f_j'' = \frac{d^2}{dx^2}(a_{0j} + a_{1j}x + a_{2j}x^2) = 2a_{2j} \).
3Step 3: Construct the Matrix
Construct the matrix for \( g(x) \):\[\begin{bmatrix} f_1 & f_2 & f_3 \ f_1' & f_2' & f_3' \ f_1'' & f_2'' & f_3''\end{bmatrix}= \begin{bmatrix} a_{01} + a_{11} x + a_{21} x^2 & a_{02} + a_{12} x + a_{22} x^2 & a_{03} + a_{13} x + a_{23} x^2 \ a_{11} + 2a_{21} x & a_{12} + 2a_{22} x & a_{13} + 2a_{23} x \ 2a_{21} & 2a_{22} & 2a_{23}\end{bmatrix}\]
4Step 4: Evaluate the Determinant
The function \( g(x) \) is defined as the determinant of the constructed matrix. We have 3 rows and 3 columns, with each involving powers of \( x \) up to squared terms. Therefore, expanding this determinant will involve polynomial multiplication resulting in an overall degree of at most 3.
5Step 5: Conclusion on the Nature of g(x)
Given that the highest terms in the polynomial come from the products of the maximum degrees in each row, \( g(x) \) will be a cubic polynomial due to terms involving \( x^3 \) appearing in the determinant calculation.
Key Concepts
Quadratic PolynomialsDerivativesMatrix Algebra
Quadratic Polynomials
A quadratic polynomial is a mathematical expression of the form \( ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants, and \( x \) represents the variable. Quadratic polynomials have a degree of 2, which means the highest power of \( x \) is squared. These polynomials are important in various branches of mathematics, including algebra and calculus.
Quadratic polynomials can be expressed in different forms, such as:
Quadratic polynomials can be expressed in different forms, such as:
- Standard form: \( ax^2 + bx + c \)
- Factored form: \( a(x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the roots of the polynomial.
Derivatives
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. For a given function \( f(x) \), the derivative \( f'(x) \) represents the rate of change or the slope of the function at a given point. When differentiating quadratic polynomials, you'll notice that the operation reduces the degree of the polynomial by one.
In our exercise, we needed to calculate both the first and second derivatives of the function \( f_j(x) = a_{0j} + a_{1j}x + a_{2j}x^2 \):
In our exercise, we needed to calculate both the first and second derivatives of the function \( f_j(x) = a_{0j} + a_{1j}x + a_{2j}x^2 \):
- First derivative: \( f_j'(x) = a_{1j} + 2a_{2j}x \). Here, the constant term \( a_{0j} \) disappears, and \( x^2 \) becomes \( 2x \).
- Second derivative: \( f_j''(x) = 2a_{2j} \). This derivative is constant, reflecting that the original function was a simple quadratic.
Matrix Algebra
Matrix algebra involves operations with matrices, which are rectangular arrays of numbers, symbols, or expressions, arranged in rows and columns. Matrices are used in various fields of mathematics to solve systems of equations, transform geometric figures, and more.
In our exercise, we form a specific 3x3 matrix composed of the quadratic polynomials and their derivatives:
The steps involved in determinant calculation, such as pairing terms from each row according to specific rules, lead to a conclusion that \( g(x) \) is a cubic polynomial, owing to a maximum of degree 3.
In our exercise, we form a specific 3x3 matrix composed of the quadratic polynomials and their derivatives:
- The first row contains the original quadratic functions \( f_1, f_2, f_3 \).
- The second row contains their first derivatives \( f_1', f_2', f_3' \).
- The third row contains their second derivatives \( f_1'', f_2'', f_3'' \).
The steps involved in determinant calculation, such as pairing terms from each row according to specific rules, lead to a conclusion that \( g(x) \) is a cubic polynomial, owing to a maximum of degree 3.
Other exercises in this chapter
Problem 41
The value of the determinant of \(n\)th order, being given by \(\left|\begin{array}{cccc}x & 1 & 1 & \ldots \\ 1 & x & 1 & \ldots \\ 1 & 1 & x & \ldots \\ \ldot
View solution Problem 42
The value of the determinant \(\left|\begin{array}{ccc}\sqrt{x}+\sqrt{y} & 2 \sqrt{z} & \sqrt{z} \\ \sqrt{y z}+\sqrt{2 x} & z & \sqrt{2 z} \\ y+\sqrt{x z} & \sq
View solution Problem 45
\((b+c)(y+z)-a x=b-c\), \((c+a)(z+x)-b y=c-a\), \((a+b)(x+y)-c z=a-b\), where \(a+b+c \neq 0\), then \(x=\) (A) \(\frac{c-b}{a+b+c}\) (B) \(\frac{a-c}{a+b+c}\)
View solution Problem 46
If \(x \neq 0, y \neq 0, z \neq 0\) and \(\left|\begin{array}{ccc}1+x & 1 & 1 \\\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right|=0\), then \(x^{-1}+y^{-
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