117046
The number of surjective functions from \(A\) to \(B\) where \(A=\{1,2,3,4\}\) and \(B=\{a, b\}\) is
1 14
2 12
3 2
4 15
Explanation:
:Given, \(\mathrm{A}=\{1,2,3,4\} \mathrm{B}=\{\mathrm{a}, \mathrm{b}\}\) If \(A\) and \(B\) are two sets having \(m\) and \(n\) elements such that Here, \(\mathrm{m}>\mathrm{n}\) total number of function \(\mathrm{A}\) to \(\mathrm{B}=\mathrm{m}^{\mathrm{n}}\) \(=4^2\) \(=16\) Number of surjective \(=\) Total number of function \(-n\) \(=16-2\) \(=14\)
VITEEE-2016
Sets, Relation and Function
117047
The function \(f: R \rightarrow R\) defined by \(f(x)=(x-1)(x-2)(x-3)\) is
1 one-one but not onto
2 onto but not one-one
3 both one-one and onto
4 neither one-one nor onto
Explanation:
B Given, \(f(x)=(x-1)(x-2)(x-3)\) \(f(1)=f(2)=f(3)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not one-one. For each \(y \in R\), there exists \(x \in R\) such that \(f(x)=y\) \(\therefore \mathrm{f}\) is onto. If a continuous function has more than one roots, then the function is always many-one.
Shift-II
Sets, Relation and Function
117048
The solution set of the inequality \(4^{-x+\frac{1}{2}}-7 \cdot\left(2^{-x}\right)-4\lt 0\) for \(x \in R\) is
117050
Let \(f: R-\{x\} \rightarrow R\) be a function defined by \(f(x)=\frac{x-m}{x-n}\), where \(m \neq n\). Then
1 \(f\) is one-one onto
2 \(f\) is one-one into
3 f is many one onto
4 \(f\) is many one into
Explanation:
B Given, \(\mathrm{f}: \mathrm{R}-\{\mathrm{x}\} \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}\), Where \(\mathrm{m} \neq \mathrm{n}\). Consider \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be two elements in the domain \(\mathrm{R}-\) \(\{\mathrm{x}\}\), Then - \(\frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\) \(\left(x_1-m\right)\left(x_2-n\right)=\left(x_2-m\right)\left(x_1-n\right)\) \(x_1 x_2-n x_1-m x_2+m n=x_1 x_2-n x_2-m x_1+m n\) \((m-n) x_1=(m-n) x_2\) \(x_1=x_2\) \(\therefore \mathrm{f}\) is one-one function. Again, consider y be an element in the co-domain \(\mathrm{R}\), then - \(f(x)=y\) \(\frac{x-m}{x-n}=y\) \(x-m=x y-n y\) \(x-x y=-n y+m\) \(x(1-y)=m-n y\) \(x(y-1)=n y-m\) \(x=\frac{n y-m}{y-1}\) The above result is not defined for \(\mathrm{y}=1\). So, \(1 \in \mathrm{R}\) (co-domain) has not pre-image in \(\mathrm{R}-\{\mathrm{x}\}\) Hence, \(f\) is not onto. Then, \(\mathrm{f}\) is one-one and into.
117046
The number of surjective functions from \(A\) to \(B\) where \(A=\{1,2,3,4\}\) and \(B=\{a, b\}\) is
1 14
2 12
3 2
4 15
Explanation:
:Given, \(\mathrm{A}=\{1,2,3,4\} \mathrm{B}=\{\mathrm{a}, \mathrm{b}\}\) If \(A\) and \(B\) are two sets having \(m\) and \(n\) elements such that Here, \(\mathrm{m}>\mathrm{n}\) total number of function \(\mathrm{A}\) to \(\mathrm{B}=\mathrm{m}^{\mathrm{n}}\) \(=4^2\) \(=16\) Number of surjective \(=\) Total number of function \(-n\) \(=16-2\) \(=14\)
VITEEE-2016
Sets, Relation and Function
117047
The function \(f: R \rightarrow R\) defined by \(f(x)=(x-1)(x-2)(x-3)\) is
1 one-one but not onto
2 onto but not one-one
3 both one-one and onto
4 neither one-one nor onto
Explanation:
B Given, \(f(x)=(x-1)(x-2)(x-3)\) \(f(1)=f(2)=f(3)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not one-one. For each \(y \in R\), there exists \(x \in R\) such that \(f(x)=y\) \(\therefore \mathrm{f}\) is onto. If a continuous function has more than one roots, then the function is always many-one.
Shift-II
Sets, Relation and Function
117048
The solution set of the inequality \(4^{-x+\frac{1}{2}}-7 \cdot\left(2^{-x}\right)-4\lt 0\) for \(x \in R\) is
117050
Let \(f: R-\{x\} \rightarrow R\) be a function defined by \(f(x)=\frac{x-m}{x-n}\), where \(m \neq n\). Then
1 \(f\) is one-one onto
2 \(f\) is one-one into
3 f is many one onto
4 \(f\) is many one into
Explanation:
B Given, \(\mathrm{f}: \mathrm{R}-\{\mathrm{x}\} \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}\), Where \(\mathrm{m} \neq \mathrm{n}\). Consider \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be two elements in the domain \(\mathrm{R}-\) \(\{\mathrm{x}\}\), Then - \(\frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\) \(\left(x_1-m\right)\left(x_2-n\right)=\left(x_2-m\right)\left(x_1-n\right)\) \(x_1 x_2-n x_1-m x_2+m n=x_1 x_2-n x_2-m x_1+m n\) \((m-n) x_1=(m-n) x_2\) \(x_1=x_2\) \(\therefore \mathrm{f}\) is one-one function. Again, consider y be an element in the co-domain \(\mathrm{R}\), then - \(f(x)=y\) \(\frac{x-m}{x-n}=y\) \(x-m=x y-n y\) \(x-x y=-n y+m\) \(x(1-y)=m-n y\) \(x(y-1)=n y-m\) \(x=\frac{n y-m}{y-1}\) The above result is not defined for \(\mathrm{y}=1\). So, \(1 \in \mathrm{R}\) (co-domain) has not pre-image in \(\mathrm{R}-\{\mathrm{x}\}\) Hence, \(f\) is not onto. Then, \(\mathrm{f}\) is one-one and into.
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Sets, Relation and Function
117046
The number of surjective functions from \(A\) to \(B\) where \(A=\{1,2,3,4\}\) and \(B=\{a, b\}\) is
1 14
2 12
3 2
4 15
Explanation:
:Given, \(\mathrm{A}=\{1,2,3,4\} \mathrm{B}=\{\mathrm{a}, \mathrm{b}\}\) If \(A\) and \(B\) are two sets having \(m\) and \(n\) elements such that Here, \(\mathrm{m}>\mathrm{n}\) total number of function \(\mathrm{A}\) to \(\mathrm{B}=\mathrm{m}^{\mathrm{n}}\) \(=4^2\) \(=16\) Number of surjective \(=\) Total number of function \(-n\) \(=16-2\) \(=14\)
VITEEE-2016
Sets, Relation and Function
117047
The function \(f: R \rightarrow R\) defined by \(f(x)=(x-1)(x-2)(x-3)\) is
1 one-one but not onto
2 onto but not one-one
3 both one-one and onto
4 neither one-one nor onto
Explanation:
B Given, \(f(x)=(x-1)(x-2)(x-3)\) \(f(1)=f(2)=f(3)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not one-one. For each \(y \in R\), there exists \(x \in R\) such that \(f(x)=y\) \(\therefore \mathrm{f}\) is onto. If a continuous function has more than one roots, then the function is always many-one.
Shift-II
Sets, Relation and Function
117048
The solution set of the inequality \(4^{-x+\frac{1}{2}}-7 \cdot\left(2^{-x}\right)-4\lt 0\) for \(x \in R\) is
117050
Let \(f: R-\{x\} \rightarrow R\) be a function defined by \(f(x)=\frac{x-m}{x-n}\), where \(m \neq n\). Then
1 \(f\) is one-one onto
2 \(f\) is one-one into
3 f is many one onto
4 \(f\) is many one into
Explanation:
B Given, \(\mathrm{f}: \mathrm{R}-\{\mathrm{x}\} \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}\), Where \(\mathrm{m} \neq \mathrm{n}\). Consider \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be two elements in the domain \(\mathrm{R}-\) \(\{\mathrm{x}\}\), Then - \(\frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\) \(\left(x_1-m\right)\left(x_2-n\right)=\left(x_2-m\right)\left(x_1-n\right)\) \(x_1 x_2-n x_1-m x_2+m n=x_1 x_2-n x_2-m x_1+m n\) \((m-n) x_1=(m-n) x_2\) \(x_1=x_2\) \(\therefore \mathrm{f}\) is one-one function. Again, consider y be an element in the co-domain \(\mathrm{R}\), then - \(f(x)=y\) \(\frac{x-m}{x-n}=y\) \(x-m=x y-n y\) \(x-x y=-n y+m\) \(x(1-y)=m-n y\) \(x(y-1)=n y-m\) \(x=\frac{n y-m}{y-1}\) The above result is not defined for \(\mathrm{y}=1\). So, \(1 \in \mathrm{R}\) (co-domain) has not pre-image in \(\mathrm{R}-\{\mathrm{x}\}\) Hence, \(f\) is not onto. Then, \(\mathrm{f}\) is one-one and into.
117046
The number of surjective functions from \(A\) to \(B\) where \(A=\{1,2,3,4\}\) and \(B=\{a, b\}\) is
1 14
2 12
3 2
4 15
Explanation:
:Given, \(\mathrm{A}=\{1,2,3,4\} \mathrm{B}=\{\mathrm{a}, \mathrm{b}\}\) If \(A\) and \(B\) are two sets having \(m\) and \(n\) elements such that Here, \(\mathrm{m}>\mathrm{n}\) total number of function \(\mathrm{A}\) to \(\mathrm{B}=\mathrm{m}^{\mathrm{n}}\) \(=4^2\) \(=16\) Number of surjective \(=\) Total number of function \(-n\) \(=16-2\) \(=14\)
VITEEE-2016
Sets, Relation and Function
117047
The function \(f: R \rightarrow R\) defined by \(f(x)=(x-1)(x-2)(x-3)\) is
1 one-one but not onto
2 onto but not one-one
3 both one-one and onto
4 neither one-one nor onto
Explanation:
B Given, \(f(x)=(x-1)(x-2)(x-3)\) \(f(1)=f(2)=f(3)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not one-one. For each \(y \in R\), there exists \(x \in R\) such that \(f(x)=y\) \(\therefore \mathrm{f}\) is onto. If a continuous function has more than one roots, then the function is always many-one.
Shift-II
Sets, Relation and Function
117048
The solution set of the inequality \(4^{-x+\frac{1}{2}}-7 \cdot\left(2^{-x}\right)-4\lt 0\) for \(x \in R\) is
117050
Let \(f: R-\{x\} \rightarrow R\) be a function defined by \(f(x)=\frac{x-m}{x-n}\), where \(m \neq n\). Then
1 \(f\) is one-one onto
2 \(f\) is one-one into
3 f is many one onto
4 \(f\) is many one into
Explanation:
B Given, \(\mathrm{f}: \mathrm{R}-\{\mathrm{x}\} \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}\), Where \(\mathrm{m} \neq \mathrm{n}\). Consider \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be two elements in the domain \(\mathrm{R}-\) \(\{\mathrm{x}\}\), Then - \(\frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\) \(\left(x_1-m\right)\left(x_2-n\right)=\left(x_2-m\right)\left(x_1-n\right)\) \(x_1 x_2-n x_1-m x_2+m n=x_1 x_2-n x_2-m x_1+m n\) \((m-n) x_1=(m-n) x_2\) \(x_1=x_2\) \(\therefore \mathrm{f}\) is one-one function. Again, consider y be an element in the co-domain \(\mathrm{R}\), then - \(f(x)=y\) \(\frac{x-m}{x-n}=y\) \(x-m=x y-n y\) \(x-x y=-n y+m\) \(x(1-y)=m-n y\) \(x(y-1)=n y-m\) \(x=\frac{n y-m}{y-1}\) The above result is not defined for \(\mathrm{y}=1\). So, \(1 \in \mathrm{R}\) (co-domain) has not pre-image in \(\mathrm{R}-\{\mathrm{x}\}\) Hence, \(f\) is not onto. Then, \(\mathrm{f}\) is one-one and into.
117046
The number of surjective functions from \(A\) to \(B\) where \(A=\{1,2,3,4\}\) and \(B=\{a, b\}\) is
1 14
2 12
3 2
4 15
Explanation:
:Given, \(\mathrm{A}=\{1,2,3,4\} \mathrm{B}=\{\mathrm{a}, \mathrm{b}\}\) If \(A\) and \(B\) are two sets having \(m\) and \(n\) elements such that Here, \(\mathrm{m}>\mathrm{n}\) total number of function \(\mathrm{A}\) to \(\mathrm{B}=\mathrm{m}^{\mathrm{n}}\) \(=4^2\) \(=16\) Number of surjective \(=\) Total number of function \(-n\) \(=16-2\) \(=14\)
VITEEE-2016
Sets, Relation and Function
117047
The function \(f: R \rightarrow R\) defined by \(f(x)=(x-1)(x-2)(x-3)\) is
1 one-one but not onto
2 onto but not one-one
3 both one-one and onto
4 neither one-one nor onto
Explanation:
B Given, \(f(x)=(x-1)(x-2)(x-3)\) \(f(1)=f(2)=f(3)=0\) \(\therefore \mathrm{f}(\mathrm{x})\) is not one-one. For each \(y \in R\), there exists \(x \in R\) such that \(f(x)=y\) \(\therefore \mathrm{f}\) is onto. If a continuous function has more than one roots, then the function is always many-one.
Shift-II
Sets, Relation and Function
117048
The solution set of the inequality \(4^{-x+\frac{1}{2}}-7 \cdot\left(2^{-x}\right)-4\lt 0\) for \(x \in R\) is
117050
Let \(f: R-\{x\} \rightarrow R\) be a function defined by \(f(x)=\frac{x-m}{x-n}\), where \(m \neq n\). Then
1 \(f\) is one-one onto
2 \(f\) is one-one into
3 f is many one onto
4 \(f\) is many one into
Explanation:
B Given, \(\mathrm{f}: \mathrm{R}-\{\mathrm{x}\} \rightarrow \mathrm{R}\) be a function defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}\), Where \(\mathrm{m} \neq \mathrm{n}\). Consider \(\mathrm{x}_1\) and \(\mathrm{x}_2\) be two elements in the domain \(\mathrm{R}-\) \(\{\mathrm{x}\}\), Then - \(\frac{x_1-m}{x_1-n}=\frac{x_2-m}{x_2-n}\) \(\left(x_1-m\right)\left(x_2-n\right)=\left(x_2-m\right)\left(x_1-n\right)\) \(x_1 x_2-n x_1-m x_2+m n=x_1 x_2-n x_2-m x_1+m n\) \((m-n) x_1=(m-n) x_2\) \(x_1=x_2\) \(\therefore \mathrm{f}\) is one-one function. Again, consider y be an element in the co-domain \(\mathrm{R}\), then - \(f(x)=y\) \(\frac{x-m}{x-n}=y\) \(x-m=x y-n y\) \(x-x y=-n y+m\) \(x(1-y)=m-n y\) \(x(y-1)=n y-m\) \(x=\frac{n y-m}{y-1}\) The above result is not defined for \(\mathrm{y}=1\). So, \(1 \in \mathrm{R}\) (co-domain) has not pre-image in \(\mathrm{R}-\{\mathrm{x}\}\) Hence, \(f\) is not onto. Then, \(\mathrm{f}\) is one-one and into.
