NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117067
How many functions \(f: Z \rightarrow Z\) are there such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in Z\) ?
1 1
2 2
3 3
4 Infinitely many
Explanation:
D Given condition, \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) ; \mathrm{x}, \mathrm{y} \in \mathrm{Z}\) \(\mathrm{f}(\mathrm{x})=\mathrm{k}(\mathrm{x})\) So, there are infinitely many bijections.
Shift-II
Sets, Relation and Function
117076
Let \(S, T, U\) be three non-void sets and \(f: S \rightarrow\) \(T, g: T \rightarrow U\) and composed mapping g.f : \(S \rightarrow\) \(U\) be defined. Let g.f be injective mapping. Then
1 f, g both are injective
2 neither \(f\) nor \(g\) is injective
3 f is obviously injective
4 g is obviously injective
Explanation:
D gof : \(\mathrm{S} \rightarrow \mathrm{U}\) gof is injective or (one - one) function: So, \(g\left(f_1\right)=g\left(f_2\right)\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\therefore \mathrm{g}\) is obviously injective.
WB JEE-2022
Sets, Relation and Function
117134
The function \(f: A \rightarrow B\) given by \(f(x)=x, x \in A\), is one to one but not onto. Then,
Exp: (D) : Given, \(f(x)=x, x \in A\) As \(f(x)\) is one to one but not onto. It means some element in B has no pre image in A is proper set B i.e., \(\mathrm{A} \subset \mathrm{B}\).
[Kerala CEE-2016]
Sets, Relation and Function
117052
If a set A contains 5 elements, then the total number of injective functions from \(A\) onto itself is
1 \(5^5\)
2 \(2^5\)
3 \(5^2\)
4 5 !
Explanation:
D A one to one function means every input maps to a distinct output. So, if the domain has \(\mathrm{m}\) element the range should also have \(\mathrm{m}\) elements. So, the total number of injective from A onto itself is \(5 !\).
117067
How many functions \(f: Z \rightarrow Z\) are there such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in Z\) ?
1 1
2 2
3 3
4 Infinitely many
Explanation:
D Given condition, \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) ; \mathrm{x}, \mathrm{y} \in \mathrm{Z}\) \(\mathrm{f}(\mathrm{x})=\mathrm{k}(\mathrm{x})\) So, there are infinitely many bijections.
Shift-II
Sets, Relation and Function
117076
Let \(S, T, U\) be three non-void sets and \(f: S \rightarrow\) \(T, g: T \rightarrow U\) and composed mapping g.f : \(S \rightarrow\) \(U\) be defined. Let g.f be injective mapping. Then
1 f, g both are injective
2 neither \(f\) nor \(g\) is injective
3 f is obviously injective
4 g is obviously injective
Explanation:
D gof : \(\mathrm{S} \rightarrow \mathrm{U}\) gof is injective or (one - one) function: So, \(g\left(f_1\right)=g\left(f_2\right)\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\therefore \mathrm{g}\) is obviously injective.
WB JEE-2022
Sets, Relation and Function
117134
The function \(f: A \rightarrow B\) given by \(f(x)=x, x \in A\), is one to one but not onto. Then,
Exp: (D) : Given, \(f(x)=x, x \in A\) As \(f(x)\) is one to one but not onto. It means some element in B has no pre image in A is proper set B i.e., \(\mathrm{A} \subset \mathrm{B}\).
[Kerala CEE-2016]
Sets, Relation and Function
117052
If a set A contains 5 elements, then the total number of injective functions from \(A\) onto itself is
1 \(5^5\)
2 \(2^5\)
3 \(5^2\)
4 5 !
Explanation:
D A one to one function means every input maps to a distinct output. So, if the domain has \(\mathrm{m}\) element the range should also have \(\mathrm{m}\) elements. So, the total number of injective from A onto itself is \(5 !\).
117067
How many functions \(f: Z \rightarrow Z\) are there such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in Z\) ?
1 1
2 2
3 3
4 Infinitely many
Explanation:
D Given condition, \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) ; \mathrm{x}, \mathrm{y} \in \mathrm{Z}\) \(\mathrm{f}(\mathrm{x})=\mathrm{k}(\mathrm{x})\) So, there are infinitely many bijections.
Shift-II
Sets, Relation and Function
117076
Let \(S, T, U\) be three non-void sets and \(f: S \rightarrow\) \(T, g: T \rightarrow U\) and composed mapping g.f : \(S \rightarrow\) \(U\) be defined. Let g.f be injective mapping. Then
1 f, g both are injective
2 neither \(f\) nor \(g\) is injective
3 f is obviously injective
4 g is obviously injective
Explanation:
D gof : \(\mathrm{S} \rightarrow \mathrm{U}\) gof is injective or (one - one) function: So, \(g\left(f_1\right)=g\left(f_2\right)\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\therefore \mathrm{g}\) is obviously injective.
WB JEE-2022
Sets, Relation and Function
117134
The function \(f: A \rightarrow B\) given by \(f(x)=x, x \in A\), is one to one but not onto. Then,
Exp: (D) : Given, \(f(x)=x, x \in A\) As \(f(x)\) is one to one but not onto. It means some element in B has no pre image in A is proper set B i.e., \(\mathrm{A} \subset \mathrm{B}\).
[Kerala CEE-2016]
Sets, Relation and Function
117052
If a set A contains 5 elements, then the total number of injective functions from \(A\) onto itself is
1 \(5^5\)
2 \(2^5\)
3 \(5^2\)
4 5 !
Explanation:
D A one to one function means every input maps to a distinct output. So, if the domain has \(\mathrm{m}\) element the range should also have \(\mathrm{m}\) elements. So, the total number of injective from A onto itself is \(5 !\).
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
117067
How many functions \(f: Z \rightarrow Z\) are there such that \(f(x+y)=f(x)+f(y)\) for all \(x, y \in Z\) ?
1 1
2 2
3 3
4 Infinitely many
Explanation:
D Given condition, \(\mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) ; \mathrm{x}, \mathrm{y} \in \mathrm{Z}\) \(\mathrm{f}(\mathrm{x})=\mathrm{k}(\mathrm{x})\) So, there are infinitely many bijections.
Shift-II
Sets, Relation and Function
117076
Let \(S, T, U\) be three non-void sets and \(f: S \rightarrow\) \(T, g: T \rightarrow U\) and composed mapping g.f : \(S \rightarrow\) \(U\) be defined. Let g.f be injective mapping. Then
1 f, g both are injective
2 neither \(f\) nor \(g\) is injective
3 f is obviously injective
4 g is obviously injective
Explanation:
D gof : \(\mathrm{S} \rightarrow \mathrm{U}\) gof is injective or (one - one) function: So, \(g\left(f_1\right)=g\left(f_2\right)\) \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) \(\therefore \mathrm{g}\) is obviously injective.
WB JEE-2022
Sets, Relation and Function
117134
The function \(f: A \rightarrow B\) given by \(f(x)=x, x \in A\), is one to one but not onto. Then,
Exp: (D) : Given, \(f(x)=x, x \in A\) As \(f(x)\) is one to one but not onto. It means some element in B has no pre image in A is proper set B i.e., \(\mathrm{A} \subset \mathrm{B}\).
[Kerala CEE-2016]
Sets, Relation and Function
117052
If a set A contains 5 elements, then the total number of injective functions from \(A\) onto itself is
1 \(5^5\)
2 \(2^5\)
3 \(5^2\)
4 5 !
Explanation:
D A one to one function means every input maps to a distinct output. So, if the domain has \(\mathrm{m}\) element the range should also have \(\mathrm{m}\) elements. So, the total number of injective from A onto itself is \(5 !\).