Question

If \(\frac{1}{\left(a^{2}-b x\right)\left(b^{1}-a x\right)}=\frac{A}{a^{2}-b x}+\frac{B}{b^{2}-a x}\), then the value of \(A\) and \(B\) respectively would be (1) \(\frac{b}{b^{3}-a^{3}}, \frac{a}{b^{3}-a^{3}}\) (2) \(\frac{b}{b^{3}-a^{3}}, \frac{a}{a^{3}-b^{3}}\) (3) \(\frac{b}{a^{3}-b^{3}}, \frac{a}{a^{3}-b^{3}}\) (4) \(\frac{-b}{b^{3}-a^{3}}, \frac{-a}{a^{3}-b^{3}}\)

Short answer

Question: Given the partial fraction decomposition $\frac{1}{(a^2-bx)(b^2-ax)} = \frac{A}{a^2-bx} + \frac{B}{b^2-ax}$, where $a$ and $b$ are constants, find the values of $A$ and $B$. Answer: The values of $A$ and $B$ are as follows: $$A= \frac{b^{3}-a^{3}}{2b^{2}}$$ $$B= \frac{b^{3}-a^{3}}{2a^{2}}$$

Step-by-step solution

Setting up the equality

First we set up the equation: $$\frac{1}{\left(a^{2}-b x\right)\left(b^{2}-a x\right)}=\frac{A}{a^{2}-b x}+\frac{B}{b^{2}-a x}$$

Clear the denominators

To get rid of the denominators, we can multiply both sides of the equation by the least common multiple of the denominators, which is \(\left(a^{2}-b x\right)\left(b^{2}-a x\right)\). This gives us: $$1 = A(b^{2}-ax) + B(a^{2}-bx)$$

Rearrange the terms to find coefficients of x

Now let's rewrite the equation in terms of powers of \(x\) by expanding the products and collecting the coefficients of \(x\): $$1=(-aB+bA)x + (a^{2}B+b^{2}A)$$

Equate the coefficients of x and the constants

Since the equation is an identity, it must be true for all values of x. Therefore, the coefficients of the powers of \(x\) must be equal on both sides of the equation. We have the following equations: 1) Coefficient of \(x\): \(-aB+bA=0\) 2) Constant term: \(a^{2}B+b^{2}A=1\)

Solve the system of equations for A and B

We'll solve the linear system of equations obtained in step 4 using the substitution method. From the first equation, we can find \(A\): $$A=\frac{aB}{b}$$ Now substitute this expression for \(A\) into the second equation: $$a^{2}B+b^{2}\left(\frac{aB}{b}\right)=1$$ Simplify: $$a^{2}B+a^{2}B=b^{3}-a^{3}$$ Combining both terms: $$2a^{2}B=b^{3}-a^{3}$$ Hence, $$B=\frac{b^{3}-a^{3}}{2a^{2}}$$ Now substitute this expression for \(B\) back into the expression for \(A\): $$A=\frac{a\left(\frac{b^{3}-a^{3}}{2a^{2}}\right)}{b}$$ Simplify: $$A= \frac{b^{3}-a^{3}}{2b^{2}}$$ Thus, the values of \(A\) and \(B\) are: $$A= \frac{b^{3}-a^{3}}{2b^{2}}$$ $$B= \frac{b^{3}-a^{3}}{2a^{2}}$$ Comparing the answers with the given options, we find that none of them match our solution. Therefore, none of the provided options are correct.

Key concepts

Algebraic Fractions

Algebraic fractions, much like their numerical counterparts, involve a numerator and a denominator containing algebraic expressions. Simplifying or solving algebraic fractions typically involves finding a common denominator in order to combine terms, or breaking down complex expressions into simpler parts using techniques such as partial fraction decomposition. This is particularly useful when integrating rational functions or solving equations involving fractions.

For instance, in the given exercise, we employed partial fraction decomposition to break down a complex fraction into simpler fractions of the form \frac{A}{a^2 - bx}\text{ and }\frac{B}{b^2 - ax}\text{.} Partial fractions make it possible to work with these simpler expressions individually, which can then be manipulated using basic algebraic principles to find the values of \(A\) and \(B\).

Linear System of Equations

In algebra, a linear system of equations is a collection of two or more linear equations involving the same variables. The goal in solving such systems is to find the values of the variables that satisfy all equations simultaneously. There are various methods to solve these systems, including graphing, substitution, elimination, and matrix operations.

In the context of the exercise, once we expressed the equations in terms of powers of \(x\), we obtained a system of linear equations which we solved to find the values for \(A\) and \(B\). This required equating coefficients for the same powers of \(x\) across the identity, which resulted in a linear system that encapsulates the mathematical relationship between \(A\) and \(B\). Solving for \(A\) and \(B\) gave us the decomposed algebraic fractions.

Mathematical Identities

Mathematical identities are equations that hold true for all values of the variables involved. They are used as tools to simplify expressions and solve equations, often serving as crucial steps when working with complex algebraic relationships. In partial fractions decomposition, the identity established by equating the original fraction to the sum of its decomposed parts must hold for all values of the variable present.

The process of clearing the denominators to find an identity, as done in our example problem, allows us to equate the coefficients of corresponding terms, which are powered by \(x\), to form a linear system of equations. This system reflects the underlying mathematical identity that we must leverage in this context to solve for \(A\) and \(B\).

IIT Foundation Mathematics

IIT Foundation Mathematics is a program designed to prepare students for the rigors of technical education, especially for entrance exams to the Indian Institutes of Technology (IITs), which are among the most competitive in the world. The foundation program covers a wide range of mathematical concepts, including algebra, geometry, calculus, and more. It focuses on developing problem-solving skills and a deep understanding of mathematical principles.

The exercise given is a type of problem that could appear in IIT Foundation Mathematics coursework or examinations, as it challenges students to apply several concepts in algebra, from understanding algebraic fractions to employing linear system solving techniques and validating mathematical identities. Mastery of such problems could be crucial for a student preparing for competitive examinations where mathematical aptitude is tested at a high level.