
If \[f:R \to R\] is given by \[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\],then the function f is
A. many-one onto
B. many-one into
C. one-one into
D. one-one onto
Answer
163.5k+ views
Hint: To answer this question, keep in mind that a function f:A->B is a bijection if it is both one-one and onto. If there is only one image y ε B for each x ε A and each y ε B has a unique pre-image x ε A (i.e. no two elements of A have the same image in B), then f is a one-one function. Otherwise, f is a one-to-many function.
FORMULA USED:
One to one: \[f(x) = f(y) \Rightarrow x = y\forall x,y \in A\]
Complete step by step solution: Statements with two variables are known as linear equations, and they can be resolved through trial and error. The slope of the line (the y-intercept) can be used as the \[x\] -coordinate to solve a linear equation in one variable. Solving for \[x\] is a common term used for this purpose.
Certain polynomial functions have characteristics known as "many-one qualities," which state that every element in a set belongs to at least one specific member of the set.
The equation given is \[f:R \to R\], which is defined as
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
As, \[f(x) = \dfrac{{{{( - x)}^2} - 4}}{{{{( - x)}^2} + 2}}\]
\[\begin{array}{l}f( - x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\\ = > f( - x) = f(x)\end{array}\]
Therefore, f is many-one
Since,
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
On expansion, we get
\[f(x) = \dfrac{{({x^2} + 1) - 5}}{{{x^2} + 2}}\]
Now, we have to split the denominator for each term in the numerator, we get\[f(x) = \dfrac{{({x^2} + 1)}}{{{x^2} + 2}} - \dfrac{{ - 5}}{{{x^2} + 2}}\]
On canceling the similar terms in the above equation, we get
\[f(x) = 1 - \dfrac{5}{{{x^2} + 2}}\]
On simplifying the above we get
\[ \Rightarrow 1 - 5\]
\[ = - 4\]
Hence, f is many one and into.
So, Option A is correct.
Note: In order to define many-one onto functions, one must take the natural logarithm of both sides. You may obtain \[\left( x \right) = log2{\rm{ }}\left( {f\left( x \right)} \right) = log2{\rm{ }}\left( {4x + 1} \right)\] . This function can be determined by applying the inverse function theorem. \[f\] is said to be many-one onto from A into B if \[f:AB\] is given by \[f\left( x \right) = ax + by + c\] where \[A\] and \[B\] both contain at least one element.
You might conceive of this as saying that the domain of \[f\] is all real numbers between \[1\] and \[1\], inclusive. The function f is many-one onto if and only if there is a single real integer.
FORMULA USED:
One to one: \[f(x) = f(y) \Rightarrow x = y\forall x,y \in A\]
Complete step by step solution: Statements with two variables are known as linear equations, and they can be resolved through trial and error. The slope of the line (the y-intercept) can be used as the \[x\] -coordinate to solve a linear equation in one variable. Solving for \[x\] is a common term used for this purpose.
Certain polynomial functions have characteristics known as "many-one qualities," which state that every element in a set belongs to at least one specific member of the set.
The equation given is \[f:R \to R\], which is defined as
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
As, \[f(x) = \dfrac{{{{( - x)}^2} - 4}}{{{{( - x)}^2} + 2}}\]
\[\begin{array}{l}f( - x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\\ = > f( - x) = f(x)\end{array}\]
Therefore, f is many-one
Since,
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
On expansion, we get
\[f(x) = \dfrac{{({x^2} + 1) - 5}}{{{x^2} + 2}}\]
Now, we have to split the denominator for each term in the numerator, we get\[f(x) = \dfrac{{({x^2} + 1)}}{{{x^2} + 2}} - \dfrac{{ - 5}}{{{x^2} + 2}}\]
On canceling the similar terms in the above equation, we get
\[f(x) = 1 - \dfrac{5}{{{x^2} + 2}}\]
On simplifying the above we get
\[ \Rightarrow 1 - 5\]
\[ = - 4\]
Hence, f is many one and into.
So, Option A is correct.
Note: In order to define many-one onto functions, one must take the natural logarithm of both sides. You may obtain \[\left( x \right) = log2{\rm{ }}\left( {f\left( x \right)} \right) = log2{\rm{ }}\left( {4x + 1} \right)\] . This function can be determined by applying the inverse function theorem. \[f\] is said to be many-one onto from A into B if \[f:AB\] is given by \[f\left( x \right) = ax + by + c\] where \[A\] and \[B\] both contain at least one element.
You might conceive of this as saying that the domain of \[f\] is all real numbers between \[1\] and \[1\], inclusive. The function f is many-one onto if and only if there is a single real integer.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
