117114
The number of elements in the set \(S=\left\{x \in R: 2 \cos \left(\frac{x^2+6}{6}\right)=4^x+4^{-x}\right\}\) is :
1 1
2 3
3 0
4 Infinite
Explanation:
Exp: (A) : \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=4^{\mathrm{x}}+4^{-\mathrm{x}}\) L.H.S \(\leq 2 . \&\) R.H.S. \(=2\) \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=2 \quad 4^{\mathrm{x}}+4^{-\mathrm{x}}=2\) It is possible only \(x=0\) Hence, it gives only one solution.
[JEE Main-29.07.2022
Sets, Relation and Function
117115
Let \(f(x)\) be a function such that \(f(x+y)=f(x)\). \(f(y)\) for all \(x, y \in N\). If \(f(1)=3\) and \(\sum_{k=1}^n f(k)=\) 3279 , then the value of \(n\) is
1 8
2 6
3 7
4 9
Explanation:
Exp: (C) : Given \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{n}\) Given, \(f(1)=3\) \(f(2)=f(1) . f(1)=3 \times 3=3^2\) \(f(3)=f(2) . f(1)=3^2 \times 3=3^3\) \(\therefore \quad \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots . .+\mathrm{f}(\mathrm{n})=3279\) L. H. S. is in G.P. which is \(3+3^2+3^3+\ldots \ldots f(n)=3279\) \(\Rightarrow\) Here, \(\mathrm{a}=3\) \(\mathrm{r}=3\) \(\frac{\left(\mathrm{r}^{\mathrm{n}}-1\right)}{(\mathrm{r}-1)}=3 \cdot \frac{\left(3^{\mathrm{n}}-1\right)}{(3-1)}=3279\) \(\therefore \quad \frac{3^{\mathrm{n}}-1}{2}=1093\) \(\therefore \quad 3^{\mathrm{n}}-1=2186\) \(3^{\mathrm{n}}=2187\) \(\therefore \quad 3^{\mathrm{n}}=3^7\) Hence, \(\mathrm{n}=7\)
[JEE Main-24.01.2023
Sets, Relation and Function
117116
If \(f: R \rightarrow A\) defined by \(f(x)=\frac{1}{x^2+2 x+2} \forall x \in R\) is surjective, then \(\mathbf{A}=\)
117117
Let \(\mathbf{f}: \mathbf{R}-\{\mathbf{n}\} \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\frac{x-m}{x-n}\) such that \(m \neq n\), then
1 \(f\) is one one into function
2 \(f\) is one one onto function
3 \(f\) is many one into function
4 \(f\) is many one onto function
Explanation:
Exp: (A) : Given \(f(x)=\frac{x-m}{x-n}\) where \(m \neq n\), \(\forall \mathrm{x} \in \mathrm{R}-\{\mathrm{n}\}\). Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) \(\therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1-\mathrm{m}}{\mathrm{x}_1-\mathrm{n}}=\frac{\mathrm{x}_2-\mathrm{m}}{\mathrm{x}_2-\mathrm{n}} \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is one-one. Let, \(\lambda \in \mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=\lambda\) \(\therefore \frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}=\lambda \therefore \mathrm{x}=\frac{\mathrm{m}-\mathrm{n} \lambda}{1-\lambda}\) \(\mathrm{X}\) is not defined for \(\lambda=1\), also \(\mathrm{x}\) is not real. \(\therefore \mathrm{f}(\mathrm{x})\) is not an onto function. If a function is not onto, then it is an into function. Hence, \(f\) is one-one into function.
COMEDK 2020
Sets, Relation and Function
117118
The function \(f:[0, \infty) \rightarrow \mathbf{R}\) given by \(f(x)=\frac{\mathbf{x}}{\mathbf{x}+\mathbf{1}}\) is
1 one-one and onto
2 one-one but not onto
3 onto but not one-one
4 Neither one-one nor onto
Explanation:
Exp: (B) : Given, The function \(\mathrm{f}:[0, \infty) \rightarrow R\) given by \(f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) So, \(\mathrm{f}(\mathrm{x})\) is not contains the value 1 . Now, then, \(x \in[0, \infty)\), it is a one-one function but not onto function. Because every point in the co-domain is not a value of \(\mathrm{f}(\mathrm{x})\). Hence, function \(\mathrm{f}\) is one-one but not onto
117114
The number of elements in the set \(S=\left\{x \in R: 2 \cos \left(\frac{x^2+6}{6}\right)=4^x+4^{-x}\right\}\) is :
1 1
2 3
3 0
4 Infinite
Explanation:
Exp: (A) : \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=4^{\mathrm{x}}+4^{-\mathrm{x}}\) L.H.S \(\leq 2 . \&\) R.H.S. \(=2\) \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=2 \quad 4^{\mathrm{x}}+4^{-\mathrm{x}}=2\) It is possible only \(x=0\) Hence, it gives only one solution.
[JEE Main-29.07.2022
Sets, Relation and Function
117115
Let \(f(x)\) be a function such that \(f(x+y)=f(x)\). \(f(y)\) for all \(x, y \in N\). If \(f(1)=3\) and \(\sum_{k=1}^n f(k)=\) 3279 , then the value of \(n\) is
1 8
2 6
3 7
4 9
Explanation:
Exp: (C) : Given \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{n}\) Given, \(f(1)=3\) \(f(2)=f(1) . f(1)=3 \times 3=3^2\) \(f(3)=f(2) . f(1)=3^2 \times 3=3^3\) \(\therefore \quad \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots . .+\mathrm{f}(\mathrm{n})=3279\) L. H. S. is in G.P. which is \(3+3^2+3^3+\ldots \ldots f(n)=3279\) \(\Rightarrow\) Here, \(\mathrm{a}=3\) \(\mathrm{r}=3\) \(\frac{\left(\mathrm{r}^{\mathrm{n}}-1\right)}{(\mathrm{r}-1)}=3 \cdot \frac{\left(3^{\mathrm{n}}-1\right)}{(3-1)}=3279\) \(\therefore \quad \frac{3^{\mathrm{n}}-1}{2}=1093\) \(\therefore \quad 3^{\mathrm{n}}-1=2186\) \(3^{\mathrm{n}}=2187\) \(\therefore \quad 3^{\mathrm{n}}=3^7\) Hence, \(\mathrm{n}=7\)
[JEE Main-24.01.2023
Sets, Relation and Function
117116
If \(f: R \rightarrow A\) defined by \(f(x)=\frac{1}{x^2+2 x+2} \forall x \in R\) is surjective, then \(\mathbf{A}=\)
117117
Let \(\mathbf{f}: \mathbf{R}-\{\mathbf{n}\} \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\frac{x-m}{x-n}\) such that \(m \neq n\), then
1 \(f\) is one one into function
2 \(f\) is one one onto function
3 \(f\) is many one into function
4 \(f\) is many one onto function
Explanation:
Exp: (A) : Given \(f(x)=\frac{x-m}{x-n}\) where \(m \neq n\), \(\forall \mathrm{x} \in \mathrm{R}-\{\mathrm{n}\}\). Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) \(\therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1-\mathrm{m}}{\mathrm{x}_1-\mathrm{n}}=\frac{\mathrm{x}_2-\mathrm{m}}{\mathrm{x}_2-\mathrm{n}} \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is one-one. Let, \(\lambda \in \mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=\lambda\) \(\therefore \frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}=\lambda \therefore \mathrm{x}=\frac{\mathrm{m}-\mathrm{n} \lambda}{1-\lambda}\) \(\mathrm{X}\) is not defined for \(\lambda=1\), also \(\mathrm{x}\) is not real. \(\therefore \mathrm{f}(\mathrm{x})\) is not an onto function. If a function is not onto, then it is an into function. Hence, \(f\) is one-one into function.
COMEDK 2020
Sets, Relation and Function
117118
The function \(f:[0, \infty) \rightarrow \mathbf{R}\) given by \(f(x)=\frac{\mathbf{x}}{\mathbf{x}+\mathbf{1}}\) is
1 one-one and onto
2 one-one but not onto
3 onto but not one-one
4 Neither one-one nor onto
Explanation:
Exp: (B) : Given, The function \(\mathrm{f}:[0, \infty) \rightarrow R\) given by \(f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) So, \(\mathrm{f}(\mathrm{x})\) is not contains the value 1 . Now, then, \(x \in[0, \infty)\), it is a one-one function but not onto function. Because every point in the co-domain is not a value of \(\mathrm{f}(\mathrm{x})\). Hence, function \(\mathrm{f}\) is one-one but not onto
117114
The number of elements in the set \(S=\left\{x \in R: 2 \cos \left(\frac{x^2+6}{6}\right)=4^x+4^{-x}\right\}\) is :
1 1
2 3
3 0
4 Infinite
Explanation:
Exp: (A) : \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=4^{\mathrm{x}}+4^{-\mathrm{x}}\) L.H.S \(\leq 2 . \&\) R.H.S. \(=2\) \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=2 \quad 4^{\mathrm{x}}+4^{-\mathrm{x}}=2\) It is possible only \(x=0\) Hence, it gives only one solution.
[JEE Main-29.07.2022
Sets, Relation and Function
117115
Let \(f(x)\) be a function such that \(f(x+y)=f(x)\). \(f(y)\) for all \(x, y \in N\). If \(f(1)=3\) and \(\sum_{k=1}^n f(k)=\) 3279 , then the value of \(n\) is
1 8
2 6
3 7
4 9
Explanation:
Exp: (C) : Given \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{n}\) Given, \(f(1)=3\) \(f(2)=f(1) . f(1)=3 \times 3=3^2\) \(f(3)=f(2) . f(1)=3^2 \times 3=3^3\) \(\therefore \quad \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots . .+\mathrm{f}(\mathrm{n})=3279\) L. H. S. is in G.P. which is \(3+3^2+3^3+\ldots \ldots f(n)=3279\) \(\Rightarrow\) Here, \(\mathrm{a}=3\) \(\mathrm{r}=3\) \(\frac{\left(\mathrm{r}^{\mathrm{n}}-1\right)}{(\mathrm{r}-1)}=3 \cdot \frac{\left(3^{\mathrm{n}}-1\right)}{(3-1)}=3279\) \(\therefore \quad \frac{3^{\mathrm{n}}-1}{2}=1093\) \(\therefore \quad 3^{\mathrm{n}}-1=2186\) \(3^{\mathrm{n}}=2187\) \(\therefore \quad 3^{\mathrm{n}}=3^7\) Hence, \(\mathrm{n}=7\)
[JEE Main-24.01.2023
Sets, Relation and Function
117116
If \(f: R \rightarrow A\) defined by \(f(x)=\frac{1}{x^2+2 x+2} \forall x \in R\) is surjective, then \(\mathbf{A}=\)
117117
Let \(\mathbf{f}: \mathbf{R}-\{\mathbf{n}\} \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\frac{x-m}{x-n}\) such that \(m \neq n\), then
1 \(f\) is one one into function
2 \(f\) is one one onto function
3 \(f\) is many one into function
4 \(f\) is many one onto function
Explanation:
Exp: (A) : Given \(f(x)=\frac{x-m}{x-n}\) where \(m \neq n\), \(\forall \mathrm{x} \in \mathrm{R}-\{\mathrm{n}\}\). Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) \(\therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1-\mathrm{m}}{\mathrm{x}_1-\mathrm{n}}=\frac{\mathrm{x}_2-\mathrm{m}}{\mathrm{x}_2-\mathrm{n}} \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is one-one. Let, \(\lambda \in \mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=\lambda\) \(\therefore \frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}=\lambda \therefore \mathrm{x}=\frac{\mathrm{m}-\mathrm{n} \lambda}{1-\lambda}\) \(\mathrm{X}\) is not defined for \(\lambda=1\), also \(\mathrm{x}\) is not real. \(\therefore \mathrm{f}(\mathrm{x})\) is not an onto function. If a function is not onto, then it is an into function. Hence, \(f\) is one-one into function.
COMEDK 2020
Sets, Relation and Function
117118
The function \(f:[0, \infty) \rightarrow \mathbf{R}\) given by \(f(x)=\frac{\mathbf{x}}{\mathbf{x}+\mathbf{1}}\) is
1 one-one and onto
2 one-one but not onto
3 onto but not one-one
4 Neither one-one nor onto
Explanation:
Exp: (B) : Given, The function \(\mathrm{f}:[0, \infty) \rightarrow R\) given by \(f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) So, \(\mathrm{f}(\mathrm{x})\) is not contains the value 1 . Now, then, \(x \in[0, \infty)\), it is a one-one function but not onto function. Because every point in the co-domain is not a value of \(\mathrm{f}(\mathrm{x})\). Hence, function \(\mathrm{f}\) is one-one but not onto
117114
The number of elements in the set \(S=\left\{x \in R: 2 \cos \left(\frac{x^2+6}{6}\right)=4^x+4^{-x}\right\}\) is :
1 1
2 3
3 0
4 Infinite
Explanation:
Exp: (A) : \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=4^{\mathrm{x}}+4^{-\mathrm{x}}\) L.H.S \(\leq 2 . \&\) R.H.S. \(=2\) \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=2 \quad 4^{\mathrm{x}}+4^{-\mathrm{x}}=2\) It is possible only \(x=0\) Hence, it gives only one solution.
[JEE Main-29.07.2022
Sets, Relation and Function
117115
Let \(f(x)\) be a function such that \(f(x+y)=f(x)\). \(f(y)\) for all \(x, y \in N\). If \(f(1)=3\) and \(\sum_{k=1}^n f(k)=\) 3279 , then the value of \(n\) is
1 8
2 6
3 7
4 9
Explanation:
Exp: (C) : Given \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{n}\) Given, \(f(1)=3\) \(f(2)=f(1) . f(1)=3 \times 3=3^2\) \(f(3)=f(2) . f(1)=3^2 \times 3=3^3\) \(\therefore \quad \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots . .+\mathrm{f}(\mathrm{n})=3279\) L. H. S. is in G.P. which is \(3+3^2+3^3+\ldots \ldots f(n)=3279\) \(\Rightarrow\) Here, \(\mathrm{a}=3\) \(\mathrm{r}=3\) \(\frac{\left(\mathrm{r}^{\mathrm{n}}-1\right)}{(\mathrm{r}-1)}=3 \cdot \frac{\left(3^{\mathrm{n}}-1\right)}{(3-1)}=3279\) \(\therefore \quad \frac{3^{\mathrm{n}}-1}{2}=1093\) \(\therefore \quad 3^{\mathrm{n}}-1=2186\) \(3^{\mathrm{n}}=2187\) \(\therefore \quad 3^{\mathrm{n}}=3^7\) Hence, \(\mathrm{n}=7\)
[JEE Main-24.01.2023
Sets, Relation and Function
117116
If \(f: R \rightarrow A\) defined by \(f(x)=\frac{1}{x^2+2 x+2} \forall x \in R\) is surjective, then \(\mathbf{A}=\)
117117
Let \(\mathbf{f}: \mathbf{R}-\{\mathbf{n}\} \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\frac{x-m}{x-n}\) such that \(m \neq n\), then
1 \(f\) is one one into function
2 \(f\) is one one onto function
3 \(f\) is many one into function
4 \(f\) is many one onto function
Explanation:
Exp: (A) : Given \(f(x)=\frac{x-m}{x-n}\) where \(m \neq n\), \(\forall \mathrm{x} \in \mathrm{R}-\{\mathrm{n}\}\). Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) \(\therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1-\mathrm{m}}{\mathrm{x}_1-\mathrm{n}}=\frac{\mathrm{x}_2-\mathrm{m}}{\mathrm{x}_2-\mathrm{n}} \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is one-one. Let, \(\lambda \in \mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=\lambda\) \(\therefore \frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}=\lambda \therefore \mathrm{x}=\frac{\mathrm{m}-\mathrm{n} \lambda}{1-\lambda}\) \(\mathrm{X}\) is not defined for \(\lambda=1\), also \(\mathrm{x}\) is not real. \(\therefore \mathrm{f}(\mathrm{x})\) is not an onto function. If a function is not onto, then it is an into function. Hence, \(f\) is one-one into function.
COMEDK 2020
Sets, Relation and Function
117118
The function \(f:[0, \infty) \rightarrow \mathbf{R}\) given by \(f(x)=\frac{\mathbf{x}}{\mathbf{x}+\mathbf{1}}\) is
1 one-one and onto
2 one-one but not onto
3 onto but not one-one
4 Neither one-one nor onto
Explanation:
Exp: (B) : Given, The function \(\mathrm{f}:[0, \infty) \rightarrow R\) given by \(f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) So, \(\mathrm{f}(\mathrm{x})\) is not contains the value 1 . Now, then, \(x \in[0, \infty)\), it is a one-one function but not onto function. Because every point in the co-domain is not a value of \(\mathrm{f}(\mathrm{x})\). Hence, function \(\mathrm{f}\) is one-one but not onto
117114
The number of elements in the set \(S=\left\{x \in R: 2 \cos \left(\frac{x^2+6}{6}\right)=4^x+4^{-x}\right\}\) is :
1 1
2 3
3 0
4 Infinite
Explanation:
Exp: (A) : \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=4^{\mathrm{x}}+4^{-\mathrm{x}}\) L.H.S \(\leq 2 . \&\) R.H.S. \(=2\) \(2 \cos \left(\frac{\mathrm{x}^2+\mathrm{x}}{6}\right)=2 \quad 4^{\mathrm{x}}+4^{-\mathrm{x}}=2\) It is possible only \(x=0\) Hence, it gives only one solution.
[JEE Main-29.07.2022
Sets, Relation and Function
117115
Let \(f(x)\) be a function such that \(f(x+y)=f(x)\). \(f(y)\) for all \(x, y \in N\). If \(f(1)=3\) and \(\sum_{k=1}^n f(k)=\) 3279 , then the value of \(n\) is
1 8
2 6
3 7
4 9
Explanation:
Exp: (C) : Given \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{n}\) Given, \(f(1)=3\) \(f(2)=f(1) . f(1)=3 \times 3=3^2\) \(f(3)=f(2) . f(1)=3^2 \times 3=3^3\) \(\therefore \quad \mathrm{f}(1)+\mathrm{f}(2)+\mathrm{f}(3)+\ldots . .+\mathrm{f}(\mathrm{n})=3279\) L. H. S. is in G.P. which is \(3+3^2+3^3+\ldots \ldots f(n)=3279\) \(\Rightarrow\) Here, \(\mathrm{a}=3\) \(\mathrm{r}=3\) \(\frac{\left(\mathrm{r}^{\mathrm{n}}-1\right)}{(\mathrm{r}-1)}=3 \cdot \frac{\left(3^{\mathrm{n}}-1\right)}{(3-1)}=3279\) \(\therefore \quad \frac{3^{\mathrm{n}}-1}{2}=1093\) \(\therefore \quad 3^{\mathrm{n}}-1=2186\) \(3^{\mathrm{n}}=2187\) \(\therefore \quad 3^{\mathrm{n}}=3^7\) Hence, \(\mathrm{n}=7\)
[JEE Main-24.01.2023
Sets, Relation and Function
117116
If \(f: R \rightarrow A\) defined by \(f(x)=\frac{1}{x^2+2 x+2} \forall x \in R\) is surjective, then \(\mathbf{A}=\)
117117
Let \(\mathbf{f}: \mathbf{R}-\{\mathbf{n}\} \rightarrow \mathbf{R}\) be a function defined by \(f(x)=\frac{x-m}{x-n}\) such that \(m \neq n\), then
1 \(f\) is one one into function
2 \(f\) is one one onto function
3 \(f\) is many one into function
4 \(f\) is many one onto function
Explanation:
Exp: (A) : Given \(f(x)=\frac{x-m}{x-n}\) where \(m \neq n\), \(\forall \mathrm{x} \in \mathrm{R}-\{\mathrm{n}\}\). Let \(\mathrm{x}_1 \mathrm{x}_2 \in \mathrm{R}\) \(\therefore \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) \Rightarrow \frac{\mathrm{x}_1-\mathrm{m}}{\mathrm{x}_1-\mathrm{n}}=\frac{\mathrm{x}_2-\mathrm{m}}{\mathrm{x}_2-\mathrm{n}} \Rightarrow \mathrm{x}_1=\mathrm{x}_2\) \(\therefore \mathrm{f}\) is one-one. Let, \(\lambda \in \mathrm{R}\) such that \(\mathrm{f}(\mathrm{x})=\lambda\) \(\therefore \frac{\mathrm{x}-\mathrm{m}}{\mathrm{x}-\mathrm{n}}=\lambda \therefore \mathrm{x}=\frac{\mathrm{m}-\mathrm{n} \lambda}{1-\lambda}\) \(\mathrm{X}\) is not defined for \(\lambda=1\), also \(\mathrm{x}\) is not real. \(\therefore \mathrm{f}(\mathrm{x})\) is not an onto function. If a function is not onto, then it is an into function. Hence, \(f\) is one-one into function.
COMEDK 2020
Sets, Relation and Function
117118
The function \(f:[0, \infty) \rightarrow \mathbf{R}\) given by \(f(x)=\frac{\mathbf{x}}{\mathbf{x}+\mathbf{1}}\) is
1 one-one and onto
2 one-one but not onto
3 onto but not one-one
4 Neither one-one nor onto
Explanation:
Exp: (B) : Given, The function \(\mathrm{f}:[0, \infty) \rightarrow R\) given by \(f(x)=\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\) So, \(\mathrm{f}(\mathrm{x})\) is not contains the value 1 . Now, then, \(x \in[0, \infty)\), it is a one-one function but not onto function. Because every point in the co-domain is not a value of \(\mathrm{f}(\mathrm{x})\). Hence, function \(\mathrm{f}\) is one-one but not onto
