117063
If \(A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}\) are two sets, and function \(f: A \rightarrow B\) is defined by \(f(x)=\) \(\mathbf{x}+\mathbf{2} \forall \mathbf{x} \in \mathbf{A}\), then the function \(\mathbf{f}\) is
1 bijective
2 Onto
3 One-one
4 Many-one
Explanation:
C Given, \(A=\{1,2,3,4,\}\) \(B=\{1,2,3,4,5,6\}\) \(f(x)=x+2 \text { and } f: A \rightarrow B\) \(\therefore f(1)=1+2=3, f(2)=2+2=4\) \(f(3)=3+2=5, f(4)=4+2=6\) Thus each element in \(A\) has a unique image in \(B\) and no two elements in B have the same Pre-image in A. So, \(f\) is one-one.
WB JEE-2010
Sets, Relation and Function
117064
If \(f(x)\) satisfies the relation \(2 f(x)+f(1-x)=x^2\) for all real \(x\) then, \(f(x)\) is
1 \(\frac{x^2+2 x-1}{6}\)
2 \(\frac{x^2+2 x-1}{3}\)
3 \(\frac{x^2+4 x-1}{3}\)
4 \(\frac{x^2-3 x+1}{6}\)
5 \(\frac{x^2+3 x-1}{3}\)
Explanation:
B Given, \(2 f(x)+f(1-x)=x^2\) Replacing \(x\) by \((1-x)\) we get \(2 f(1-x)+f(x)=(1-x)^2\) \(2 f(1-x)+f(x)=1+x^2-2 x\) On multiplying equation (i) by 2 and subtracting from (ii) we get, \(3 f(x)=x^2+2 x-1\) \(f(x)=\frac{x^2+2 x-1}{3}\)
Kerala CEE-2009
Sets, Relation and Function
117065
A Mapping From \(\mathrm{N}\) to \(\mathrm{N}\) is defined as follows: \(\mathbf{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+\mathbf{5})^{\mathbf{2}}, \mathrm{n} \in \mathrm{N}\) ( \(\mathrm{N}\) is the set of natural numbers). Then
1 \(f\) is not one-to-one
2 \(f\) is onto
3 \(f\) is both one-to-one and onto
4 f is one - to - one but not onto
Explanation:
D One to one function basically denotes the mapping of two sets. Given, \(\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(f: N \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2\) As we know that for, one to one condition, \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2\) \(\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1-\mathrm{n}_2+10\right)=0\) \(\mathrm{n}_1=\mathrm{n}_2\) \(\therefore \mathrm{f}\) is one to one When we put \(\mathrm{n}=1,2,3,4 \ldots \ldots \infty\), we get\(f(1)=36, f(2)=49, f(3)=64, f(4)=81\). Here we see that, we do not get any pre-image of 1,2, 3 , etc. Hence, \(\mathrm{f}\) is not onto.
WB JEE-2009
Sets, Relation and Function
117066
The function \(f(x)=x^2+b x+c\), where \(b\) and \(c\) real constants, describes
1 one-to-one mapping
2 onto mapping
3 not one- to- one but onto mapping
4 neither one-to-one nor onto mapping
Explanation:
D One to one function means every domain has distinct range i.e. mapping of elements of range and domain are unique. An onto function is a function \(f\) from set \(A\) to set \(B\) that exist for each \(B\) and has at least one an A such that \(f(a)=b\). Given, \(f(x)=x^2+b x+c\) It is a quadratic equation in \(x\). So, we will get a parabola either downward or upward. Hence, it is many one mapping and not onto mapping. Hence, it is neither one to one nor onto mapping.
117063
If \(A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}\) are two sets, and function \(f: A \rightarrow B\) is defined by \(f(x)=\) \(\mathbf{x}+\mathbf{2} \forall \mathbf{x} \in \mathbf{A}\), then the function \(\mathbf{f}\) is
1 bijective
2 Onto
3 One-one
4 Many-one
Explanation:
C Given, \(A=\{1,2,3,4,\}\) \(B=\{1,2,3,4,5,6\}\) \(f(x)=x+2 \text { and } f: A \rightarrow B\) \(\therefore f(1)=1+2=3, f(2)=2+2=4\) \(f(3)=3+2=5, f(4)=4+2=6\) Thus each element in \(A\) has a unique image in \(B\) and no two elements in B have the same Pre-image in A. So, \(f\) is one-one.
WB JEE-2010
Sets, Relation and Function
117064
If \(f(x)\) satisfies the relation \(2 f(x)+f(1-x)=x^2\) for all real \(x\) then, \(f(x)\) is
1 \(\frac{x^2+2 x-1}{6}\)
2 \(\frac{x^2+2 x-1}{3}\)
3 \(\frac{x^2+4 x-1}{3}\)
4 \(\frac{x^2-3 x+1}{6}\)
5 \(\frac{x^2+3 x-1}{3}\)
Explanation:
B Given, \(2 f(x)+f(1-x)=x^2\) Replacing \(x\) by \((1-x)\) we get \(2 f(1-x)+f(x)=(1-x)^2\) \(2 f(1-x)+f(x)=1+x^2-2 x\) On multiplying equation (i) by 2 and subtracting from (ii) we get, \(3 f(x)=x^2+2 x-1\) \(f(x)=\frac{x^2+2 x-1}{3}\)
Kerala CEE-2009
Sets, Relation and Function
117065
A Mapping From \(\mathrm{N}\) to \(\mathrm{N}\) is defined as follows: \(\mathbf{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+\mathbf{5})^{\mathbf{2}}, \mathrm{n} \in \mathrm{N}\) ( \(\mathrm{N}\) is the set of natural numbers). Then
1 \(f\) is not one-to-one
2 \(f\) is onto
3 \(f\) is both one-to-one and onto
4 f is one - to - one but not onto
Explanation:
D One to one function basically denotes the mapping of two sets. Given, \(\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(f: N \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2\) As we know that for, one to one condition, \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2\) \(\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1-\mathrm{n}_2+10\right)=0\) \(\mathrm{n}_1=\mathrm{n}_2\) \(\therefore \mathrm{f}\) is one to one When we put \(\mathrm{n}=1,2,3,4 \ldots \ldots \infty\), we get\(f(1)=36, f(2)=49, f(3)=64, f(4)=81\). Here we see that, we do not get any pre-image of 1,2, 3 , etc. Hence, \(\mathrm{f}\) is not onto.
WB JEE-2009
Sets, Relation and Function
117066
The function \(f(x)=x^2+b x+c\), where \(b\) and \(c\) real constants, describes
1 one-to-one mapping
2 onto mapping
3 not one- to- one but onto mapping
4 neither one-to-one nor onto mapping
Explanation:
D One to one function means every domain has distinct range i.e. mapping of elements of range and domain are unique. An onto function is a function \(f\) from set \(A\) to set \(B\) that exist for each \(B\) and has at least one an A such that \(f(a)=b\). Given, \(f(x)=x^2+b x+c\) It is a quadratic equation in \(x\). So, we will get a parabola either downward or upward. Hence, it is many one mapping and not onto mapping. Hence, it is neither one to one nor onto mapping.
117063
If \(A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}\) are two sets, and function \(f: A \rightarrow B\) is defined by \(f(x)=\) \(\mathbf{x}+\mathbf{2} \forall \mathbf{x} \in \mathbf{A}\), then the function \(\mathbf{f}\) is
1 bijective
2 Onto
3 One-one
4 Many-one
Explanation:
C Given, \(A=\{1,2,3,4,\}\) \(B=\{1,2,3,4,5,6\}\) \(f(x)=x+2 \text { and } f: A \rightarrow B\) \(\therefore f(1)=1+2=3, f(2)=2+2=4\) \(f(3)=3+2=5, f(4)=4+2=6\) Thus each element in \(A\) has a unique image in \(B\) and no two elements in B have the same Pre-image in A. So, \(f\) is one-one.
WB JEE-2010
Sets, Relation and Function
117064
If \(f(x)\) satisfies the relation \(2 f(x)+f(1-x)=x^2\) for all real \(x\) then, \(f(x)\) is
1 \(\frac{x^2+2 x-1}{6}\)
2 \(\frac{x^2+2 x-1}{3}\)
3 \(\frac{x^2+4 x-1}{3}\)
4 \(\frac{x^2-3 x+1}{6}\)
5 \(\frac{x^2+3 x-1}{3}\)
Explanation:
B Given, \(2 f(x)+f(1-x)=x^2\) Replacing \(x\) by \((1-x)\) we get \(2 f(1-x)+f(x)=(1-x)^2\) \(2 f(1-x)+f(x)=1+x^2-2 x\) On multiplying equation (i) by 2 and subtracting from (ii) we get, \(3 f(x)=x^2+2 x-1\) \(f(x)=\frac{x^2+2 x-1}{3}\)
Kerala CEE-2009
Sets, Relation and Function
117065
A Mapping From \(\mathrm{N}\) to \(\mathrm{N}\) is defined as follows: \(\mathbf{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+\mathbf{5})^{\mathbf{2}}, \mathrm{n} \in \mathrm{N}\) ( \(\mathrm{N}\) is the set of natural numbers). Then
1 \(f\) is not one-to-one
2 \(f\) is onto
3 \(f\) is both one-to-one and onto
4 f is one - to - one but not onto
Explanation:
D One to one function basically denotes the mapping of two sets. Given, \(\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(f: N \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2\) As we know that for, one to one condition, \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2\) \(\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1-\mathrm{n}_2+10\right)=0\) \(\mathrm{n}_1=\mathrm{n}_2\) \(\therefore \mathrm{f}\) is one to one When we put \(\mathrm{n}=1,2,3,4 \ldots \ldots \infty\), we get\(f(1)=36, f(2)=49, f(3)=64, f(4)=81\). Here we see that, we do not get any pre-image of 1,2, 3 , etc. Hence, \(\mathrm{f}\) is not onto.
WB JEE-2009
Sets, Relation and Function
117066
The function \(f(x)=x^2+b x+c\), where \(b\) and \(c\) real constants, describes
1 one-to-one mapping
2 onto mapping
3 not one- to- one but onto mapping
4 neither one-to-one nor onto mapping
Explanation:
D One to one function means every domain has distinct range i.e. mapping of elements of range and domain are unique. An onto function is a function \(f\) from set \(A\) to set \(B\) that exist for each \(B\) and has at least one an A such that \(f(a)=b\). Given, \(f(x)=x^2+b x+c\) It is a quadratic equation in \(x\). So, we will get a parabola either downward or upward. Hence, it is many one mapping and not onto mapping. Hence, it is neither one to one nor onto mapping.
117063
If \(A=\{1,2,3,4\}, B=\{1,2,3,4,5,6\}\) are two sets, and function \(f: A \rightarrow B\) is defined by \(f(x)=\) \(\mathbf{x}+\mathbf{2} \forall \mathbf{x} \in \mathbf{A}\), then the function \(\mathbf{f}\) is
1 bijective
2 Onto
3 One-one
4 Many-one
Explanation:
C Given, \(A=\{1,2,3,4,\}\) \(B=\{1,2,3,4,5,6\}\) \(f(x)=x+2 \text { and } f: A \rightarrow B\) \(\therefore f(1)=1+2=3, f(2)=2+2=4\) \(f(3)=3+2=5, f(4)=4+2=6\) Thus each element in \(A\) has a unique image in \(B\) and no two elements in B have the same Pre-image in A. So, \(f\) is one-one.
WB JEE-2010
Sets, Relation and Function
117064
If \(f(x)\) satisfies the relation \(2 f(x)+f(1-x)=x^2\) for all real \(x\) then, \(f(x)\) is
1 \(\frac{x^2+2 x-1}{6}\)
2 \(\frac{x^2+2 x-1}{3}\)
3 \(\frac{x^2+4 x-1}{3}\)
4 \(\frac{x^2-3 x+1}{6}\)
5 \(\frac{x^2+3 x-1}{3}\)
Explanation:
B Given, \(2 f(x)+f(1-x)=x^2\) Replacing \(x\) by \((1-x)\) we get \(2 f(1-x)+f(x)=(1-x)^2\) \(2 f(1-x)+f(x)=1+x^2-2 x\) On multiplying equation (i) by 2 and subtracting from (ii) we get, \(3 f(x)=x^2+2 x-1\) \(f(x)=\frac{x^2+2 x-1}{3}\)
Kerala CEE-2009
Sets, Relation and Function
117065
A Mapping From \(\mathrm{N}\) to \(\mathrm{N}\) is defined as follows: \(\mathbf{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+\mathbf{5})^{\mathbf{2}}, \mathrm{n} \in \mathrm{N}\) ( \(\mathrm{N}\) is the set of natural numbers). Then
1 \(f\) is not one-to-one
2 \(f\) is onto
3 \(f\) is both one-to-one and onto
4 f is one - to - one but not onto
Explanation:
D One to one function basically denotes the mapping of two sets. Given, \(\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}\) \(f: N \rightarrow \mathrm{N}\) \(\mathrm{f}(\mathrm{n})=(\mathrm{n}+5)^2\) As we know that for, one to one condition, \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\mathrm{f}\left(\mathrm{n}_1\right)=\mathrm{f}\left(\mathrm{n}_2\right)\) \(\left(\mathrm{n}_1+5\right)^2=\left(\mathrm{n}_2+5\right)^2\) \(\left(\mathrm{n}_1-\mathrm{n}_2\right)\left(\mathrm{n}_1-\mathrm{n}_2+10\right)=0\) \(\mathrm{n}_1=\mathrm{n}_2\) \(\therefore \mathrm{f}\) is one to one When we put \(\mathrm{n}=1,2,3,4 \ldots \ldots \infty\), we get\(f(1)=36, f(2)=49, f(3)=64, f(4)=81\). Here we see that, we do not get any pre-image of 1,2, 3 , etc. Hence, \(\mathrm{f}\) is not onto.
WB JEE-2009
Sets, Relation and Function
117066
The function \(f(x)=x^2+b x+c\), where \(b\) and \(c\) real constants, describes
1 one-to-one mapping
2 onto mapping
3 not one- to- one but onto mapping
4 neither one-to-one nor onto mapping
Explanation:
D One to one function means every domain has distinct range i.e. mapping of elements of range and domain are unique. An onto function is a function \(f\) from set \(A\) to set \(B\) that exist for each \(B\) and has at least one an A such that \(f(a)=b\). Given, \(f(x)=x^2+b x+c\) It is a quadratic equation in \(x\). So, we will get a parabola either downward or upward. Hence, it is many one mapping and not onto mapping. Hence, it is neither one to one nor onto mapping.
