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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Statement-1 : For an element order of successive ionisation energy is \(\mathrm{IE}_{1}<\mathrm{IE}_{2}<\) \(\mathrm{IE}_{3}<\mathrm{IE}_{4} \ldots \ldots .\) Statement-2 : Removal of an electron from cation is difficult than neutral atom because effective nuclear charge increases. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (3) Statement-1 is True, Statement-2 is False. (4) Statement- 1 is False, Statement-2 is True.

A circular disc of radius \(\mathrm{R}\) rolls without slipping on a rough horizontal surface. At the instant shown its linear velocity is \(\mathrm{V}\), linear acceleration a, angular velocity \(\omega\) and angular acceleration \(\alpha\). Four points \(\mathrm{A}\), B, C and D lie on its circumference such that the diameter AC is vertical \& BD horizontal then choose the correct options. (1) \(\mathrm{V}_{\mathrm{B}}=\sqrt{\mathrm{V}^{2}+(\mathrm{R} \omega)^{2}}\) (2) \(V_{C}=V+R \omega\) (3) \(\mathrm{a}_{\mathrm{A}}=\sqrt{(\mathrm{a}-\mathrm{R} \alpha)^{2}+\left(\omega^{2} \mathrm{R}\right)^{2}}\) (4) \(\mathrm{a}_{\mathrm{D}}=\sqrt{\left(\mathrm{a}+\omega^{2} \mathrm{R}\right)^{2}+(\mathrm{R} \alpha)^{2}}\)

Choose the right statement from the following: (1) The average KE of a molecule of any ideal gas is the same at the same temperature (2) The average translational KE of a molecule of any ideal gas is the same at the same temperature. (3) The average translational KE of a molecule of diatomic gas is \(1.5\) times the average rotational KE of a molecule at a given moderate temperature. (4) The average rotational KE of a molecule of a monoatomic gas increases linearly with increase in temperature.

Statement-1 : Benzene and ethene both give reactions with electrophilic reagents. Statement-2: Benzene and ethene both have loosely bound \(\pi\) electrons, which can be donated to vacant orbital of the electrophile. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement- 1 . (2) Statement-1 is True, Statement- 2 is True; Statement-2 is NOT a correct explanation for Statement-1. (3) Statement-1 is True, Statement-2 is False. (4) Statement- 1 is False, Statement-2 is True.

There are five different oranges and three different apples. Number of ways they can be divided into two groups of four fruits if each group must contain atleast one apple is (1) 95 (2) 65 (3) 60 (4) 35 (5) 30

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free