Problem 36
Question
If \(f\left(2 x+\frac{y}{8}, 2 x-\frac{y}{8}\right)=x y\), then \(f(m, n)+f(n, m)=0\) (A) only when \(m=n\) (B) only when \(m \neq n\) (C) only when \(m=-n\) (D) for all \(m\) and \(n\).
Step-by-Step Solution
Verified Answer
(D) for all \(m\) and \(n\).
1Step 1: Understand the Problem
We are given a function \( f \) defined as \( f(2x + \frac{y}{8}, 2x - \frac{y}{8}) = xy \). We need to find when \( f(m, n) + f(n, m) = 0 \).
2Step 2: Substitute Values into the Function
Substitute \( m = 2x + \frac{y}{8} \) and \( n = 2x - \frac{y}{8} \) to express \( x \) and \( y \) in terms of \( m \) and \( n \).
3Step 3: Solve for x and y in Terms of m and n
Add the equations \( m = 2x + \frac{y}{8} \) and \( n = 2x - \frac{y}{8} \) to find \( x \). Subtract them to find \( y \).
4Step 4: Validate Function Operation
From the equations in Step 3, use \( x = \frac{m+n}{4} \) and \( y = 2(m-n) \) to verify \( f(m, n) = \left(\frac{m+n}{4}\right) (2(m-n)) \).
5Step 5: Simplifying f(m, n)
Simplify \( f(m, n) \) to observe \( f(m,n) = \frac{(m+n)(m-n)}{2} = \frac{m^2 - n^2}{2} \).
6Step 6: Evaluate f(m, n) + f(n, m)
Using the expression from Step 5, calculate \( f(n, m) = \frac{n^2 - m^2}{2} \). Then, find \( f(m, n) + f(n, m) = \frac{m^2 - n^2}{2} + \frac{n^2 - m^2}{2} = 0 \).
7Step 7: Analyze Conditions for Zero Sum
Since \( f(m, n) + f(n, m) = 0 \) independently of \( m \) and \( n \), verify if it holds for all cases without any additional conditions.
Key Concepts
Function OperationsAlgebraic ManipulationEquationsProblem-Solving Steps
Function Operations
Function operations involve combining or manipulating different functions to achieve a particular outcome. In this exercise, the function is expressed as \( f(2x + \frac{y}{8}, 2x - \frac{y}{8}) = xy \). The operations performed on this function include substitution, simplification, and transformation.
- Substitution: We substitute into the function by expressing specific variables, like \( m \) and \( n \), in terms of other variables, \( x \) and \( y \).
- Simplification: Simplification is applied to break the function into a recognizable and more approachable form, making it easier to collaborate with these expressions.
Algebraic Manipulation
Algebraic manipulation consists of rearranging and simplifying equations to achieve a clearer form. In this problem, we are addressing expressions like \( m = 2x + \frac{y}{8} \) and \( n = 2x - \frac{y}{8} \).
- By adding the equations, we can express \( x = \frac{m+n}{4} \). This action helps to consolidate separate variables into one simplified form.
- Similarly, by subtracting the equations, we can find \( y = 2(m-n) \), giving a clear expression for \( y \) in terms of \( m \) and \( n \).
Equations
Equations are mathematical statements that assert the equivalence of two expressions, using an equal sign. They form the core of this exercise as they are used to calculate and verify variables, thereby solving the problem.In the exercise, after substituting for \( m \) and \( n \), equations are restructured and manipulated to derive new expressions and simplify to find meaningful results:
- The expression for \( f(m, n) \) is simplified to \( \frac{m^2 - n^2}{2} \), and vice versa for \( f(n, m) \).
- These equations enable the calculation and verification that \( f(m, n) + f(n, m) = 0 \) holds true.
Problem-Solving Steps
Problem-solving steps are systematic methods used to solve mathematical problems systematically. The original solution outlines these steps sequentially. Here's a recap with some insights:- **Understand the Problem**: Identify what is being asked. In this case, find when \( f(m, n) + f(n, m) = 0 \) holds true.- **Substitute and Rearrange**: Replace and manipulate expressions to isolate and solve for unknowns \( x \) and \( y \).- **Validate your Steps**: By inserting newly derived expressions for \( x \) and \( y \), ensure that the function \( f(m, n) \) matches expectations.- **Conclusion and Analysis**: After confirming calculations, analyze the results to understand under what conditions they hold true, recognizing that for this problem, the function operation's sum is zero for all values of \( m \) and \( n \). These steps help structure an approach to tackle similar challenging problems reliably and effectively, creating a blueprint for finding solutions.
Other exercises in this chapter
Problem 34
Let \(f\) be a real valued function with domain \(R\) satisfying \(0 \leq f(x) \leq \frac{1}{2}\) and for some fixed \(a>0\) \(f(x+a)=\frac{1}{2}-\sqrt{f(x)-(f(
View solution Problem 35
Let \(f(x)=\sin x+\cos x, g(x)=x^{2}-1\). Then \(g(f(x))\) is invertible for \(x \in\) (A) \(\left[-\frac{\pi}{2}, 0\right]\) (B) \(\left[-\frac{\pi}{2}, \pi\ri
View solution Problem 37
If \(f(x)\) is defined on \((0,1)\), then the domain of definition of \(f\left(e^{x}\right)+f(\ln |x|)\) is (A) \((-e,-1)\) (B) \((-e,-1) \cup(1, e)\) (C) \((-\
View solution Problem 38
The value of \(\left[\frac{1}{2}\right]+\left[\frac{1}{2}+\frac{1}{100}\right]+\left[\frac{1}{2}+\frac{2}{100}\right]+\cdots+\) \(\left[\frac{1}{2}+\frac{99}{10
View solution