117127
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^4\), then
1 \(f\) is one-one but not onto
2 \(f\) is neither one-one nor onto
3 \(f\) is one-one and onto
4 f may be one-one and onto
Explanation:
Exp: (B) : Given, \(f: R \rightarrow R\) be defined by \(f(x)=x^4\) Let \(a, b \in R\) such that \(f(a)=f(b)\) \(a^4=b^4\) \(a= \pm b\) Then, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) does not implies that \(\mathrm{x}_1=\mathrm{x}_2\) So, it is not one-one function. And, let an element 3 in co-domain \(\mathrm{R}\). Then, there does not contain any \(\mathrm{x}\) in domain \(\mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=3\) So, it is not onto. Hence, function \(f(x)=x^4\) is neither one-one nor onto.
[Karnataka CET 2017]
Sets, Relation and Function
117128
Let \(f: R \rightarrow R\) be defined by \(f(x)=\frac{1}{x} \forall x \in R\), then \(f\) is
1 onto
2 not defined
3 one-one
4 bijective
Explanation:
Exp: (B) : Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be defined by \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}, \forall \mathrm{x} \in \mathrm{R}\) Then, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\), is not defined for \(\mathrm{x}=0\). Since the given question, domain is given \(\mathrm{R}\) but it is not defined. So, the given function \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\) is not defined.
[Karnataka CET-2015]
Sets, Relation and Function
117129
On the set of integers \(Z\), define \(f: Z \rightarrow Z\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) then \(f\) is
1 surjective but not injective
2 bijective
3 injective but not surjective
4 neither injective nor surjective
Explanation:
Exp: (A) : Given, \(\mathrm{Z}=\text { set of integers }\) And \(\quad \mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) Then, \(\mathrm{f}(2)=\frac{2}{2}=1\) And \(f(1)=0\) So, \(\mathrm{f}(2) \neq \mathrm{f}(1)\) and \(1 \neq 2\) Then, \(f\) is not one-one function or \(f\) is not injective function. And, for every even values of \(n\), we get a set of integers \(Z=(-\infty, \infty)\). Then, \(f\) is onto or surjective. So, function \(f\) is surjective onto but not injective (oneone) .
[Karnataka CET 2009]
Sets, Relation and Function
117130
Let \(f: R \rightarrow R\) be defined as \(f(x)=2 x-1\) and \(g\) : \(R-\{1\} \rightarrow R\) be defined as \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Then, the composition function \(f(g(x))\) is
1 one-one but not onto
2 onto but not one-one
3 Neither one-one nor onto
4 Both one-one and onto
Explanation:
Exp: (A) : : We have, \(f(x)=2 x-1\) \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Now, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))=2 \mathrm{~g}(\mathrm{x})-1\) \(=2 \frac{\left(\mathrm{x}-\frac{1}{2}\right)}{(\mathrm{x}-1)}-1\) \(=\frac{2(2 \mathrm{x}-1)-2(\mathrm{x}-1)}{2(\mathrm{x}-1)}\) \(=\frac{4 \mathrm{x}-2-2 \mathrm{x}+2}{2 \mathrm{x}-2}\) \(=\frac{2 \mathrm{x}}{2 \mathrm{x}-2}=\frac{\mathrm{x}}{\mathrm{x}-1}\) So, the range of \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is \(\mathrm{R}-\{1\}\) Co domain is \(\mathrm{R}\) Hence, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is not onto as the range and co domain are not same. We know that, if the function is one-one, then the function is always increasing or decreasing in its domain. \(f(g(x))= \frac{x}{x-1}\) \(f^{\prime}(g)(x) =\frac{(x-1)-x(1)}{(x-1)^2}\) \(=\frac{-1}{(x-1)^2}\) Therefore we can conclude that \(\mathrm{f}^{\prime}(\mathrm{g})(\mathrm{x})\) is always decreasing as there is a negative sign. So, the function is one-one Hence, \(\mathrm{f}[\mathrm{g}(\mathrm{x})]\) is one-one but not onto function.
117127
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^4\), then
1 \(f\) is one-one but not onto
2 \(f\) is neither one-one nor onto
3 \(f\) is one-one and onto
4 f may be one-one and onto
Explanation:
Exp: (B) : Given, \(f: R \rightarrow R\) be defined by \(f(x)=x^4\) Let \(a, b \in R\) such that \(f(a)=f(b)\) \(a^4=b^4\) \(a= \pm b\) Then, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) does not implies that \(\mathrm{x}_1=\mathrm{x}_2\) So, it is not one-one function. And, let an element 3 in co-domain \(\mathrm{R}\). Then, there does not contain any \(\mathrm{x}\) in domain \(\mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=3\) So, it is not onto. Hence, function \(f(x)=x^4\) is neither one-one nor onto.
[Karnataka CET 2017]
Sets, Relation and Function
117128
Let \(f: R \rightarrow R\) be defined by \(f(x)=\frac{1}{x} \forall x \in R\), then \(f\) is
1 onto
2 not defined
3 one-one
4 bijective
Explanation:
Exp: (B) : Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be defined by \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}, \forall \mathrm{x} \in \mathrm{R}\) Then, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\), is not defined for \(\mathrm{x}=0\). Since the given question, domain is given \(\mathrm{R}\) but it is not defined. So, the given function \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\) is not defined.
[Karnataka CET-2015]
Sets, Relation and Function
117129
On the set of integers \(Z\), define \(f: Z \rightarrow Z\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) then \(f\) is
1 surjective but not injective
2 bijective
3 injective but not surjective
4 neither injective nor surjective
Explanation:
Exp: (A) : Given, \(\mathrm{Z}=\text { set of integers }\) And \(\quad \mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) Then, \(\mathrm{f}(2)=\frac{2}{2}=1\) And \(f(1)=0\) So, \(\mathrm{f}(2) \neq \mathrm{f}(1)\) and \(1 \neq 2\) Then, \(f\) is not one-one function or \(f\) is not injective function. And, for every even values of \(n\), we get a set of integers \(Z=(-\infty, \infty)\). Then, \(f\) is onto or surjective. So, function \(f\) is surjective onto but not injective (oneone) .
[Karnataka CET 2009]
Sets, Relation and Function
117130
Let \(f: R \rightarrow R\) be defined as \(f(x)=2 x-1\) and \(g\) : \(R-\{1\} \rightarrow R\) be defined as \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Then, the composition function \(f(g(x))\) is
1 one-one but not onto
2 onto but not one-one
3 Neither one-one nor onto
4 Both one-one and onto
Explanation:
Exp: (A) : : We have, \(f(x)=2 x-1\) \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Now, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))=2 \mathrm{~g}(\mathrm{x})-1\) \(=2 \frac{\left(\mathrm{x}-\frac{1}{2}\right)}{(\mathrm{x}-1)}-1\) \(=\frac{2(2 \mathrm{x}-1)-2(\mathrm{x}-1)}{2(\mathrm{x}-1)}\) \(=\frac{4 \mathrm{x}-2-2 \mathrm{x}+2}{2 \mathrm{x}-2}\) \(=\frac{2 \mathrm{x}}{2 \mathrm{x}-2}=\frac{\mathrm{x}}{\mathrm{x}-1}\) So, the range of \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is \(\mathrm{R}-\{1\}\) Co domain is \(\mathrm{R}\) Hence, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is not onto as the range and co domain are not same. We know that, if the function is one-one, then the function is always increasing or decreasing in its domain. \(f(g(x))= \frac{x}{x-1}\) \(f^{\prime}(g)(x) =\frac{(x-1)-x(1)}{(x-1)^2}\) \(=\frac{-1}{(x-1)^2}\) Therefore we can conclude that \(\mathrm{f}^{\prime}(\mathrm{g})(\mathrm{x})\) is always decreasing as there is a negative sign. So, the function is one-one Hence, \(\mathrm{f}[\mathrm{g}(\mathrm{x})]\) is one-one but not onto function.
117127
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^4\), then
1 \(f\) is one-one but not onto
2 \(f\) is neither one-one nor onto
3 \(f\) is one-one and onto
4 f may be one-one and onto
Explanation:
Exp: (B) : Given, \(f: R \rightarrow R\) be defined by \(f(x)=x^4\) Let \(a, b \in R\) such that \(f(a)=f(b)\) \(a^4=b^4\) \(a= \pm b\) Then, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) does not implies that \(\mathrm{x}_1=\mathrm{x}_2\) So, it is not one-one function. And, let an element 3 in co-domain \(\mathrm{R}\). Then, there does not contain any \(\mathrm{x}\) in domain \(\mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=3\) So, it is not onto. Hence, function \(f(x)=x^4\) is neither one-one nor onto.
[Karnataka CET 2017]
Sets, Relation and Function
117128
Let \(f: R \rightarrow R\) be defined by \(f(x)=\frac{1}{x} \forall x \in R\), then \(f\) is
1 onto
2 not defined
3 one-one
4 bijective
Explanation:
Exp: (B) : Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be defined by \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}, \forall \mathrm{x} \in \mathrm{R}\) Then, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\), is not defined for \(\mathrm{x}=0\). Since the given question, domain is given \(\mathrm{R}\) but it is not defined. So, the given function \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\) is not defined.
[Karnataka CET-2015]
Sets, Relation and Function
117129
On the set of integers \(Z\), define \(f: Z \rightarrow Z\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) then \(f\) is
1 surjective but not injective
2 bijective
3 injective but not surjective
4 neither injective nor surjective
Explanation:
Exp: (A) : Given, \(\mathrm{Z}=\text { set of integers }\) And \(\quad \mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) Then, \(\mathrm{f}(2)=\frac{2}{2}=1\) And \(f(1)=0\) So, \(\mathrm{f}(2) \neq \mathrm{f}(1)\) and \(1 \neq 2\) Then, \(f\) is not one-one function or \(f\) is not injective function. And, for every even values of \(n\), we get a set of integers \(Z=(-\infty, \infty)\). Then, \(f\) is onto or surjective. So, function \(f\) is surjective onto but not injective (oneone) .
[Karnataka CET 2009]
Sets, Relation and Function
117130
Let \(f: R \rightarrow R\) be defined as \(f(x)=2 x-1\) and \(g\) : \(R-\{1\} \rightarrow R\) be defined as \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Then, the composition function \(f(g(x))\) is
1 one-one but not onto
2 onto but not one-one
3 Neither one-one nor onto
4 Both one-one and onto
Explanation:
Exp: (A) : : We have, \(f(x)=2 x-1\) \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Now, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))=2 \mathrm{~g}(\mathrm{x})-1\) \(=2 \frac{\left(\mathrm{x}-\frac{1}{2}\right)}{(\mathrm{x}-1)}-1\) \(=\frac{2(2 \mathrm{x}-1)-2(\mathrm{x}-1)}{2(\mathrm{x}-1)}\) \(=\frac{4 \mathrm{x}-2-2 \mathrm{x}+2}{2 \mathrm{x}-2}\) \(=\frac{2 \mathrm{x}}{2 \mathrm{x}-2}=\frac{\mathrm{x}}{\mathrm{x}-1}\) So, the range of \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is \(\mathrm{R}-\{1\}\) Co domain is \(\mathrm{R}\) Hence, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is not onto as the range and co domain are not same. We know that, if the function is one-one, then the function is always increasing or decreasing in its domain. \(f(g(x))= \frac{x}{x-1}\) \(f^{\prime}(g)(x) =\frac{(x-1)-x(1)}{(x-1)^2}\) \(=\frac{-1}{(x-1)^2}\) Therefore we can conclude that \(\mathrm{f}^{\prime}(\mathrm{g})(\mathrm{x})\) is always decreasing as there is a negative sign. So, the function is one-one Hence, \(\mathrm{f}[\mathrm{g}(\mathrm{x})]\) is one-one but not onto function.
117127
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^4\), then
1 \(f\) is one-one but not onto
2 \(f\) is neither one-one nor onto
3 \(f\) is one-one and onto
4 f may be one-one and onto
Explanation:
Exp: (B) : Given, \(f: R \rightarrow R\) be defined by \(f(x)=x^4\) Let \(a, b \in R\) such that \(f(a)=f(b)\) \(a^4=b^4\) \(a= \pm b\) Then, \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right)\) does not implies that \(\mathrm{x}_1=\mathrm{x}_2\) So, it is not one-one function. And, let an element 3 in co-domain \(\mathrm{R}\). Then, there does not contain any \(\mathrm{x}\) in domain \(\mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=3\) So, it is not onto. Hence, function \(f(x)=x^4\) is neither one-one nor onto.
[Karnataka CET 2017]
Sets, Relation and Function
117128
Let \(f: R \rightarrow R\) be defined by \(f(x)=\frac{1}{x} \forall x \in R\), then \(f\) is
1 onto
2 not defined
3 one-one
4 bijective
Explanation:
Exp: (B) : Given, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be defined by \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}, \forall \mathrm{x} \in \mathrm{R}\) Then, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\), is not defined for \(\mathrm{x}=0\). Since the given question, domain is given \(\mathrm{R}\) but it is not defined. So, the given function \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}}\) is not defined.
[Karnataka CET-2015]
Sets, Relation and Function
117129
On the set of integers \(Z\), define \(f: Z \rightarrow Z\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) then \(f\) is
1 surjective but not injective
2 bijective
3 injective but not surjective
4 neither injective nor surjective
Explanation:
Exp: (A) : Given, \(\mathrm{Z}=\text { set of integers }\) And \(\quad \mathrm{f}: \mathrm{Z} \rightarrow \mathrm{Z}\) as \(f(n)=\left\{\begin{array}{l}\frac{n}{2}, n \text { is even } \\ 0, n \text { is odd }\end{array}\right.\) Then, \(\mathrm{f}(2)=\frac{2}{2}=1\) And \(f(1)=0\) So, \(\mathrm{f}(2) \neq \mathrm{f}(1)\) and \(1 \neq 2\) Then, \(f\) is not one-one function or \(f\) is not injective function. And, for every even values of \(n\), we get a set of integers \(Z=(-\infty, \infty)\). Then, \(f\) is onto or surjective. So, function \(f\) is surjective onto but not injective (oneone) .
[Karnataka CET 2009]
Sets, Relation and Function
117130
Let \(f: R \rightarrow R\) be defined as \(f(x)=2 x-1\) and \(g\) : \(R-\{1\} \rightarrow R\) be defined as \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Then, the composition function \(f(g(x))\) is
1 one-one but not onto
2 onto but not one-one
3 Neither one-one nor onto
4 Both one-one and onto
Explanation:
Exp: (A) : : We have, \(f(x)=2 x-1\) \(g(x)=\frac{x-\frac{1}{2}}{x-1}\) Now, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))=2 \mathrm{~g}(\mathrm{x})-1\) \(=2 \frac{\left(\mathrm{x}-\frac{1}{2}\right)}{(\mathrm{x}-1)}-1\) \(=\frac{2(2 \mathrm{x}-1)-2(\mathrm{x}-1)}{2(\mathrm{x}-1)}\) \(=\frac{4 \mathrm{x}-2-2 \mathrm{x}+2}{2 \mathrm{x}-2}\) \(=\frac{2 \mathrm{x}}{2 \mathrm{x}-2}=\frac{\mathrm{x}}{\mathrm{x}-1}\) So, the range of \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is \(\mathrm{R}-\{1\}\) Co domain is \(\mathrm{R}\) Hence, \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is not onto as the range and co domain are not same. We know that, if the function is one-one, then the function is always increasing or decreasing in its domain. \(f(g(x))= \frac{x}{x-1}\) \(f^{\prime}(g)(x) =\frac{(x-1)-x(1)}{(x-1)^2}\) \(=\frac{-1}{(x-1)^2}\) Therefore we can conclude that \(\mathrm{f}^{\prime}(\mathrm{g})(\mathrm{x})\) is always decreasing as there is a negative sign. So, the function is one-one Hence, \(\mathrm{f}[\mathrm{g}(\mathrm{x})]\) is one-one but not onto function.
