Answer
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Hint: A function is said to be injection if and only if every element in the domain has a unique image in its co-domain. A function is said to be surjective if each element of the co-domain is mapped by at least one element of the domain.
Complete step-by-step answer:
(1) \[f:N\to N\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
Calculate \[f\left( {{x}_{1}} \right)\]
Calculate \[f\left( {{x}_{2}} \right)\]
Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\] and \[{{x}_{2}}\] are natural numbers, they are always positive
Hence, \[{{x}_{1}}={{x}_{2}}\]
So, the given function is one-one (injective)
Checking onto function or else called as surjective
Let \[f(x)=y\], such that \[y\in N\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
\[y=2\]
\[x=\pm \sqrt{2}\]
Since, x is not a natural number the given function is not onto function.
So f is one-one function and not onto function.
(2) \[f:Z\to Z\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
(1) Calculate \[f\left( {{x}_{1}} \right)\]
(2) Calculate \[f\left( {{x}_{2}} \right)\]
(3) Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\]and \[{{x}_{2}}\] are integers numbers,
Hence, \[{{x}_{1}}\]does not have a unique image
\[f\left( -1 \right)=1\]
\[f\left( 1 \right)=1\]
So, the given function is not one-one (injective)
Checking onto function or else called surjective.
Let \[f(x)=y\], such that \[y\in Z\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
Note that y is an integer, it can be negative also
Putting y=-2
\[x=\pm \sqrt{-2}\]
Which is not possible because root of negative numbers s not defined
Hence x is not an integer
The given function is not an onto function.
So f is neither one-one function nor onto function.
(3) \[f:R\to R\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
(1) Calculate \[f\left( {{x}_{1}} \right)\]
(2)Calculate \[f\left( {{x}_{2}} \right)\]
(3) Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\] and \[{{x}_{2}}\] are real numbers,
Hence, \[{{x}_{1}}\] does not have a unique image
\[f\left( -1 \right)=1\]
\[f\left( 1 \right)=1\]
So, the given function is not one-one (injective)
Checking onto function or else called as surjective
Let \[f(x)=y\], such that \[y\in R\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
Note that y is a real number, it can be negative also
Putting y=-2
\[x=\pm \sqrt{-2}\]
Which is not possible because root of negative numbers s not defined
Hence x is not a real number
the given function is not onto function
so f is neither one-one function nor onto function
(4) \[f:N\to N\]given by \[f\left( x \right)={{x}^{3}}\]
Given \[f\left( x \right)={{x}^{3}}\]
Checking one-one or injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{3}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{3}\]
\[{{x}_{1}}={{x}_{2}}\]
Since, \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\], then \[{{x}_{1}}={{x}_{2}}\]
Hence, it is one-one or injective function
Checking onto or surjective function
\[f\left( x \right)={{x}^{3}}\]
Let \[f(x)=y,y\in N\]
\[x={{y}^{\dfrac{1}{3}}}\]
Here y is a natural number
Let y=2
\[x={{2}^{\dfrac{1}{3}}}\]
So, x is not a natural number
So, the given function is not onto
So f is one-one function and not onto function.
(5)\[f:Z\to Z\]given by \[f\left( x \right)={{x}^{3}}\]
Given \[f\left( x \right)={{x}^{3}}\]
Checking one-one or injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{3}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{3}\]
\[{{x}_{1}}={{x}_{2}}\]
Since, \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\], then \[{{x}_{1}}={{x}_{2}}\]
Hence, it is one-one or injective function
Checking onto or surjective function
\[f\left( x \right)={{x}^{3}}\]
Let \[f(x)=y,y\in Z\]
\[x={{y}^{\dfrac{1}{3}}}\]
Here y is an integer
Let y=2
\[x={{2}^{\dfrac{1}{3}}}\]
So, x is not an integer
So, the given function is not onto
So f is one-one function and not onto function.
Note: Injective function is also called as one-one function and surjective function is also called as onto function. A function is called a bijective function if it is both one-one and onto function, it is also called as bijection. Please check carefully whether the elements in domain has unique image or not and note the elements in domain and codomain to check whether it is one-one function or onto function
Complete step-by-step answer:
(1) \[f:N\to N\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
Calculate \[f\left( {{x}_{1}} \right)\]
Calculate \[f\left( {{x}_{2}} \right)\]
Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\] and \[{{x}_{2}}\] are natural numbers, they are always positive
Hence, \[{{x}_{1}}={{x}_{2}}\]
So, the given function is one-one (injective)
Checking onto function or else called as surjective
Let \[f(x)=y\], such that \[y\in N\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
\[y=2\]
\[x=\pm \sqrt{2}\]
Since, x is not a natural number the given function is not onto function.
So f is one-one function and not onto function.
(2) \[f:Z\to Z\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
(1) Calculate \[f\left( {{x}_{1}} \right)\]
(2) Calculate \[f\left( {{x}_{2}} \right)\]
(3) Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\]and \[{{x}_{2}}\] are integers numbers,
Hence, \[{{x}_{1}}\]does not have a unique image
\[f\left( -1 \right)=1\]
\[f\left( 1 \right)=1\]
So, the given function is not one-one (injective)
Checking onto function or else called surjective.
Let \[f(x)=y\], such that \[y\in Z\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
Note that y is an integer, it can be negative also
Putting y=-2
\[x=\pm \sqrt{-2}\]
Which is not possible because root of negative numbers s not defined
Hence x is not an integer
The given function is not an onto function.
So f is neither one-one function nor onto function.
(3) \[f:R\to R\]given by \[f\left( x \right)={{x}^{2}}\]
Given \[f\left( x \right)={{x}^{2}}\]
To check whether it is a injectivity function or not the steps involved are
(1) Calculate \[f\left( {{x}_{1}} \right)\]
(2)Calculate \[f\left( {{x}_{2}} \right)\]
(3) Putting \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
We have to prove that \[{{x}_{1}}={{x}_{2}}\]
Checking one- one function or else called as injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{2}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{2}\]
\[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\]
\[{{x}_{1}}^{2}\]\[={{x}_{2}}^{2}\]
\[{{x}_{1}}={{x}_{2}}\]or \[{{x}_{1}}=-{{x}_{2}}\]
Since, \[{{x}_{1}}\] and \[{{x}_{2}}\] are real numbers,
Hence, \[{{x}_{1}}\] does not have a unique image
\[f\left( -1 \right)=1\]
\[f\left( 1 \right)=1\]
So, the given function is not one-one (injective)
Checking onto function or else called as surjective
Let \[f(x)=y\], such that \[y\in R\]
\[{{x}^{2}}=y\]
\[x=\pm \sqrt{y}\]
Note that y is a real number, it can be negative also
Putting y=-2
\[x=\pm \sqrt{-2}\]
Which is not possible because root of negative numbers s not defined
Hence x is not a real number
the given function is not onto function
so f is neither one-one function nor onto function
(4) \[f:N\to N\]given by \[f\left( x \right)={{x}^{3}}\]
Given \[f\left( x \right)={{x}^{3}}\]
Checking one-one or injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{3}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{3}\]
\[{{x}_{1}}={{x}_{2}}\]
Since, \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\], then \[{{x}_{1}}={{x}_{2}}\]
Hence, it is one-one or injective function
Checking onto or surjective function
\[f\left( x \right)={{x}^{3}}\]
Let \[f(x)=y,y\in N\]
\[x={{y}^{\dfrac{1}{3}}}\]
Here y is a natural number
Let y=2
\[x={{2}^{\dfrac{1}{3}}}\]
So, x is not a natural number
So, the given function is not onto
So f is one-one function and not onto function.
(5)\[f:Z\to Z\]given by \[f\left( x \right)={{x}^{3}}\]
Given \[f\left( x \right)={{x}^{3}}\]
Checking one-one or injective function
\[f\left( {{x}_{1}} \right)={{x}_{1}}^{3}\]
\[f\left( {{x}_{2}} \right)={{x}_{2}}^{3}\]
\[{{x}_{1}}={{x}_{2}}\]
Since, \[f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\], then \[{{x}_{1}}={{x}_{2}}\]
Hence, it is one-one or injective function
Checking onto or surjective function
\[f\left( x \right)={{x}^{3}}\]
Let \[f(x)=y,y\in Z\]
\[x={{y}^{\dfrac{1}{3}}}\]
Here y is an integer
Let y=2
\[x={{2}^{\dfrac{1}{3}}}\]
So, x is not an integer
So, the given function is not onto
So f is one-one function and not onto function.
Note: Injective function is also called as one-one function and surjective function is also called as onto function. A function is called a bijective function if it is both one-one and onto function, it is also called as bijection. Please check carefully whether the elements in domain has unique image or not and note the elements in domain and codomain to check whether it is one-one function or onto function
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