/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Let a function \(f(x), x \neq 0\... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 20

Let a function \(f(x), x \neq 0\) be such that \(\mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)\) then \(\mathrm{f}(\mathrm{x})\) can be (1) \(1-\mathrm{x}^{2013}\) (2) \(\sqrt{|x|}+1\) (3) \(\frac{\pi}{2 \tan ^{-1}|x|}\) (4) \(\frac{2}{1+\mathrm{k} \ell \mathrm{n}|\mathrm{x}|}, \mathrm{k}\) being arbitrary constant

Short Answer

Expert verified
(2) is the solution

Step by step solution

01

Understand the given function relation

Given the functional equation is: i) \( \text{f}(x) + \text{f}\bigg(\frac{1}{x}\bigg) = \text{f}(x) \times \text{f}\bigg(\frac{1}{x}\bigg) \)
02

Form hypothesis for possible solutions

We need to test which of the given options satisfies the functional equation. Let's evaluate each one by substituting into the equation.
03

Test Option 1

Option (1) is \( f(x) = 1-x^{2013} \).Substitute into the equation: Substitute \( x \) and \( \frac{1}{x} \):\( (1-x^{2013}) + (1-(\frac{1}{x})^{2013}) = (1-x^{2013})(1-(\frac{1}{x})^{2013}) \)Simplify and check if both sides are equal.
04

Test Option 2

Option (2) is \( f(x) = \sqrt{|x|} + 1 \).Substitute into the equation: \( (\sqrt{|x|} + 1) + (\sqrt{|1/x|} + 1) = (\sqrt{|x|} + 1)(\sqrt{|1/x|} + 1) \)Let's check if both sides match.
05

Test Option 3

Option (3) is \( f(x) = \frac{\pi}{2 \tan^{-1}|x|} \).Substitute into the equation: Substitute \( x \) and \( \frac{1}{x} \):\( \frac{\pi}{2 \tan^{-1}|x|} + \frac{\pi}{2 \tan^{-1}|1/x|} = \frac{\pi}{2 \tan^{-1}|x|} \cdot \frac{\pi}{2 \tan^{-1}|1/x|} \)Simplify and check if both sides are equal.
06

Test Option 4

Option (4) is \( f(x) = \frac{2}{1 + k \ln|x|} \).Substitute into the equation: Substitute \( x \) and \( \frac{1}{x} \):\( \frac{2}{1 + k \ln|x|} + \frac{2}{1 + k \ln|1/x|} = \frac{2}{1 + k \ln|x|} \cdot \frac{2}{1 + k \ln|1/x|} \)Simplify and check if both sides are equal.
07

Conclusion

Only one option will satisfy the equation after thorough verification. We need to solve step-by-step for each option until we find the matching function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mathematical functions
Mathematical functions are expressions that define the relationship between inputs and outputs. They follow specific rules and can be represented in various forms such as equations, graphs, and tables. When working with functions, understanding their properties is vital. For instance, you need to know about domain (allowed input values) and range (possible output values).
In our exercise, the function is represented symbolically as \( f(x) \). Understanding this notation is crucial before delving into the problem. Simply put, \( f(x) \) denotes a function where \( x \) is the input, and \( f(x) \) is the output.
variable substitution
Variable substitution is a technique used to replace a variable with another expression. This helps simplify complex equations and solve problems more efficiently.
In our exercise, a key step is to test given function options to see if they fit the given functional equation:
\( \text{f}(x) + \text{f}\bigg(\frac{1}{x}\bigg) = \text{f}(x) \times \text{f}\bigg(\frac{1}{x}\bigg) \)
For each option, we substitute both \( x \) and \( \frac{1}{x} \). Let's illustrate this with an example:
Option 2: \( f(x) = \text{f}(\text{x}) = \text{\text{ft}}(|x|) + 1 \)
Substituting \( x \) and \( \frac{1}{x} \) into the functional equation respectively:
\( (\text{ft}(|x|) + 1) + (\text{\text{ft}}(\text{\frac{1}{x}})|/1) + 1) \)
This calculation helps determine if the given options satisfy the functional equation.
simplification process
The simplification process involves reducing complex expressions into simpler forms that are easier to work with. This is achieved through algebraic manipulation and applying mathematical rules such as the distributive property or factoring.
In our exercise, simplification allows us to compare both sides of the functional equation to determine which function option is correct.
For example, when testing Option 1:
\( (1-x^{2013}) + \text{(\text{\frac{1}{x}}|^{2013}}) = (1-x^{2013})(1-\text{\text{\frac{1}{x}}|^{2013}}) \)
Each term is simplified by expanding and factoring until we can check if both sides are equal. This rigorous process ensures that we accurately solve the given functional equation.

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Most popular questions from this chapter

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