117131
Let \(f, g: N \rightarrow N\), such that \(f(n+1)=f(n)+\) \(\mathbf{f}(1) \forall \mathbf{n} \in \mathbf{N}\) and \(g\) be any arbitrary function. Which of the following statements is not true?
1 if fog is one-one, then \(g\) is one - one
2 if \(\mathrm{f}\) is onto, then \(\mathrm{f}(\mathrm{n})=\mathrm{n}, \forall \mathrm{n} \in \mathrm{N}\).
3 \(f\) is one-one
4 if \(g\) is onto, then fog is one-one
Explanation:
Exp: (D) : Given, \(\mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+\mathrm{f}(1)\) \(\mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=\mathrm{f}(1)\) Since, Above terms are in A.P. with common difference \(=\mathrm{f}(1)\) General term \(T_n=f(1)+(n-1) f(1)=n f(1)\) \(\mathrm{f}(\mathrm{n})=\mathrm{nf}(1)\) For fog to be one-one, \(g\) must be one-one. For \(f\) to be onto, \(f(n)\) should take all the values of natural numbers. \(\mathrm{f}(\mathrm{n})=\mathrm{n}\) If \(g\) is many one, then fog is many one. So, if \(\mathrm{g}\) is onto then fog is one-one is incorrect.
[JEE Main 25.02.2021 Shift-I]
Sets, Relation and Function
117132
If \(R\) denotes the set of all real numbers then the function \(f: R \rightarrow R\) defined by \(f(x)=|x|\) is
1 injective and surjective
2 surjective
3 injective.
4 neither injective nor surjective.
Explanation:
Exp: (D) : Given that, \(f(x)=|x| \geq 0\) for all \(x\) So the range of is \([0, \infty) \neq\) co domain. Hence, \(\mathrm{f}\) is not onto function (nor injective function). Also \(\mathrm{f}(-1)=\mathrm{f}(1)=1\) \(\mathrm{f}\) is many one function( non surjective function).
[MHT CET-2022]
Sets, Relation and Function
117133
In the function \(\mathbf{f}: \mathbf{R}-\{-1,1\} \rightarrow \mathbf{A}\) defined by \(f(x) f(x)=\frac{x^2}{1-x^2}\) is surjective, then a is equal to
1 \(\mathrm{R}-\{-1\}\)
2 \([0, \infty)\)
3 \(\mathrm{R}-[-1,0)\)
4 \(\mathrm{R}-(-1,0)\)
Explanation:
Exp: (C) : Given function, \(\text { f: } \mathrm{R}-\{-1,1\}\) \(f(x)=\frac{x^2}{1-x^2}\) is surjective \(\because \mathrm{f}\) is surjective, so every element in co-domain must have a pre-image in domain. \(\text { Let, } y=\frac{x^2}{1-x^2}\) \(y\left(1-x^2\right)=x^2\) \(y-y^2=x^2\) \(y=x^2+y x^2\) \(y=x^2(1+y)\) \(x^2=\frac{y}{1+y}\) \(x=\sqrt{\frac{y}{1+y}}\) For \(\mathrm{x}\) to defined. \(1+y \neq 0\) \(y \neq-1\) and \(\frac{\mathrm{y}}{1+\mathrm{y}} \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-1) \cup[0, \infty]\) From equation (i) and (ii) we get \(y \in R-[-1,0)\)
[MHT CET-2022]
Sets, Relation and Function
117135
Let \(f: R \rightarrow R\) be defined by \(f(x)=\cos x\). Then
1 \(f\) is one - one and odd
2 \(f\) is odd but not one - one
3 \(f\) is even and onto
4 \(f\) is one- one and even
5 \(f\) is even but not onto
Explanation:
E Given, \(f(x)=\cos x\) For even function - \(f(-x)=f(x)\) \(\cos (-x)=\cos x\) Hence, \(f(x)\) is even function . Let, \(y=\cos x\) \(x=\cos ^{-1} y\) \(\cos ^{-1} \mathrm{y}\) does not defined in real number because \(\cos ^{-1} \mathrm{y}\) is defined in range \(\in[-1,1]\). Hence, \(f(x)\) is not onto function . Therefore, \(f(x)\) is even function but not onto function.
[Kerala CEE-2022]
Sets, Relation and Function
117061
Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then \(f\) 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:
C Given, \(f(x)=\max \{(x+1),(1-x), 2\}\) \(f(x)=\left\{\begin{array}{ccc}1-x & , & x \leq-1 \\ 2, & , & -1\lt x\lt 1 \\ x+1 & , & x \geq 1\end{array}\right.\) Neither one-one nor onto.
117131
Let \(f, g: N \rightarrow N\), such that \(f(n+1)=f(n)+\) \(\mathbf{f}(1) \forall \mathbf{n} \in \mathbf{N}\) and \(g\) be any arbitrary function. Which of the following statements is not true?
1 if fog is one-one, then \(g\) is one - one
2 if \(\mathrm{f}\) is onto, then \(\mathrm{f}(\mathrm{n})=\mathrm{n}, \forall \mathrm{n} \in \mathrm{N}\).
3 \(f\) is one-one
4 if \(g\) is onto, then fog is one-one
Explanation:
Exp: (D) : Given, \(\mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+\mathrm{f}(1)\) \(\mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=\mathrm{f}(1)\) Since, Above terms are in A.P. with common difference \(=\mathrm{f}(1)\) General term \(T_n=f(1)+(n-1) f(1)=n f(1)\) \(\mathrm{f}(\mathrm{n})=\mathrm{nf}(1)\) For fog to be one-one, \(g\) must be one-one. For \(f\) to be onto, \(f(n)\) should take all the values of natural numbers. \(\mathrm{f}(\mathrm{n})=\mathrm{n}\) If \(g\) is many one, then fog is many one. So, if \(\mathrm{g}\) is onto then fog is one-one is incorrect.
[JEE Main 25.02.2021 Shift-I]
Sets, Relation and Function
117132
If \(R\) denotes the set of all real numbers then the function \(f: R \rightarrow R\) defined by \(f(x)=|x|\) is
1 injective and surjective
2 surjective
3 injective.
4 neither injective nor surjective.
Explanation:
Exp: (D) : Given that, \(f(x)=|x| \geq 0\) for all \(x\) So the range of is \([0, \infty) \neq\) co domain. Hence, \(\mathrm{f}\) is not onto function (nor injective function). Also \(\mathrm{f}(-1)=\mathrm{f}(1)=1\) \(\mathrm{f}\) is many one function( non surjective function).
[MHT CET-2022]
Sets, Relation and Function
117133
In the function \(\mathbf{f}: \mathbf{R}-\{-1,1\} \rightarrow \mathbf{A}\) defined by \(f(x) f(x)=\frac{x^2}{1-x^2}\) is surjective, then a is equal to
1 \(\mathrm{R}-\{-1\}\)
2 \([0, \infty)\)
3 \(\mathrm{R}-[-1,0)\)
4 \(\mathrm{R}-(-1,0)\)
Explanation:
Exp: (C) : Given function, \(\text { f: } \mathrm{R}-\{-1,1\}\) \(f(x)=\frac{x^2}{1-x^2}\) is surjective \(\because \mathrm{f}\) is surjective, so every element in co-domain must have a pre-image in domain. \(\text { Let, } y=\frac{x^2}{1-x^2}\) \(y\left(1-x^2\right)=x^2\) \(y-y^2=x^2\) \(y=x^2+y x^2\) \(y=x^2(1+y)\) \(x^2=\frac{y}{1+y}\) \(x=\sqrt{\frac{y}{1+y}}\) For \(\mathrm{x}\) to defined. \(1+y \neq 0\) \(y \neq-1\) and \(\frac{\mathrm{y}}{1+\mathrm{y}} \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-1) \cup[0, \infty]\) From equation (i) and (ii) we get \(y \in R-[-1,0)\)
[MHT CET-2022]
Sets, Relation and Function
117135
Let \(f: R \rightarrow R\) be defined by \(f(x)=\cos x\). Then
1 \(f\) is one - one and odd
2 \(f\) is odd but not one - one
3 \(f\) is even and onto
4 \(f\) is one- one and even
5 \(f\) is even but not onto
Explanation:
E Given, \(f(x)=\cos x\) For even function - \(f(-x)=f(x)\) \(\cos (-x)=\cos x\) Hence, \(f(x)\) is even function . Let, \(y=\cos x\) \(x=\cos ^{-1} y\) \(\cos ^{-1} \mathrm{y}\) does not defined in real number because \(\cos ^{-1} \mathrm{y}\) is defined in range \(\in[-1,1]\). Hence, \(f(x)\) is not onto function . Therefore, \(f(x)\) is even function but not onto function.
[Kerala CEE-2022]
Sets, Relation and Function
117061
Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then \(f\) 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:
C Given, \(f(x)=\max \{(x+1),(1-x), 2\}\) \(f(x)=\left\{\begin{array}{ccc}1-x & , & x \leq-1 \\ 2, & , & -1\lt x\lt 1 \\ x+1 & , & x \geq 1\end{array}\right.\) Neither one-one nor onto.
117131
Let \(f, g: N \rightarrow N\), such that \(f(n+1)=f(n)+\) \(\mathbf{f}(1) \forall \mathbf{n} \in \mathbf{N}\) and \(g\) be any arbitrary function. Which of the following statements is not true?
1 if fog is one-one, then \(g\) is one - one
2 if \(\mathrm{f}\) is onto, then \(\mathrm{f}(\mathrm{n})=\mathrm{n}, \forall \mathrm{n} \in \mathrm{N}\).
3 \(f\) is one-one
4 if \(g\) is onto, then fog is one-one
Explanation:
Exp: (D) : Given, \(\mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+\mathrm{f}(1)\) \(\mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=\mathrm{f}(1)\) Since, Above terms are in A.P. with common difference \(=\mathrm{f}(1)\) General term \(T_n=f(1)+(n-1) f(1)=n f(1)\) \(\mathrm{f}(\mathrm{n})=\mathrm{nf}(1)\) For fog to be one-one, \(g\) must be one-one. For \(f\) to be onto, \(f(n)\) should take all the values of natural numbers. \(\mathrm{f}(\mathrm{n})=\mathrm{n}\) If \(g\) is many one, then fog is many one. So, if \(\mathrm{g}\) is onto then fog is one-one is incorrect.
[JEE Main 25.02.2021 Shift-I]
Sets, Relation and Function
117132
If \(R\) denotes the set of all real numbers then the function \(f: R \rightarrow R\) defined by \(f(x)=|x|\) is
1 injective and surjective
2 surjective
3 injective.
4 neither injective nor surjective.
Explanation:
Exp: (D) : Given that, \(f(x)=|x| \geq 0\) for all \(x\) So the range of is \([0, \infty) \neq\) co domain. Hence, \(\mathrm{f}\) is not onto function (nor injective function). Also \(\mathrm{f}(-1)=\mathrm{f}(1)=1\) \(\mathrm{f}\) is many one function( non surjective function).
[MHT CET-2022]
Sets, Relation and Function
117133
In the function \(\mathbf{f}: \mathbf{R}-\{-1,1\} \rightarrow \mathbf{A}\) defined by \(f(x) f(x)=\frac{x^2}{1-x^2}\) is surjective, then a is equal to
1 \(\mathrm{R}-\{-1\}\)
2 \([0, \infty)\)
3 \(\mathrm{R}-[-1,0)\)
4 \(\mathrm{R}-(-1,0)\)
Explanation:
Exp: (C) : Given function, \(\text { f: } \mathrm{R}-\{-1,1\}\) \(f(x)=\frac{x^2}{1-x^2}\) is surjective \(\because \mathrm{f}\) is surjective, so every element in co-domain must have a pre-image in domain. \(\text { Let, } y=\frac{x^2}{1-x^2}\) \(y\left(1-x^2\right)=x^2\) \(y-y^2=x^2\) \(y=x^2+y x^2\) \(y=x^2(1+y)\) \(x^2=\frac{y}{1+y}\) \(x=\sqrt{\frac{y}{1+y}}\) For \(\mathrm{x}\) to defined. \(1+y \neq 0\) \(y \neq-1\) and \(\frac{\mathrm{y}}{1+\mathrm{y}} \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-1) \cup[0, \infty]\) From equation (i) and (ii) we get \(y \in R-[-1,0)\)
[MHT CET-2022]
Sets, Relation and Function
117135
Let \(f: R \rightarrow R\) be defined by \(f(x)=\cos x\). Then
1 \(f\) is one - one and odd
2 \(f\) is odd but not one - one
3 \(f\) is even and onto
4 \(f\) is one- one and even
5 \(f\) is even but not onto
Explanation:
E Given, \(f(x)=\cos x\) For even function - \(f(-x)=f(x)\) \(\cos (-x)=\cos x\) Hence, \(f(x)\) is even function . Let, \(y=\cos x\) \(x=\cos ^{-1} y\) \(\cos ^{-1} \mathrm{y}\) does not defined in real number because \(\cos ^{-1} \mathrm{y}\) is defined in range \(\in[-1,1]\). Hence, \(f(x)\) is not onto function . Therefore, \(f(x)\) is even function but not onto function.
[Kerala CEE-2022]
Sets, Relation and Function
117061
Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then \(f\) 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:
C Given, \(f(x)=\max \{(x+1),(1-x), 2\}\) \(f(x)=\left\{\begin{array}{ccc}1-x & , & x \leq-1 \\ 2, & , & -1\lt x\lt 1 \\ x+1 & , & x \geq 1\end{array}\right.\) Neither one-one nor onto.
117131
Let \(f, g: N \rightarrow N\), such that \(f(n+1)=f(n)+\) \(\mathbf{f}(1) \forall \mathbf{n} \in \mathbf{N}\) and \(g\) be any arbitrary function. Which of the following statements is not true?
1 if fog is one-one, then \(g\) is one - one
2 if \(\mathrm{f}\) is onto, then \(\mathrm{f}(\mathrm{n})=\mathrm{n}, \forall \mathrm{n} \in \mathrm{N}\).
3 \(f\) is one-one
4 if \(g\) is onto, then fog is one-one
Explanation:
Exp: (D) : Given, \(\mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+\mathrm{f}(1)\) \(\mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=\mathrm{f}(1)\) Since, Above terms are in A.P. with common difference \(=\mathrm{f}(1)\) General term \(T_n=f(1)+(n-1) f(1)=n f(1)\) \(\mathrm{f}(\mathrm{n})=\mathrm{nf}(1)\) For fog to be one-one, \(g\) must be one-one. For \(f\) to be onto, \(f(n)\) should take all the values of natural numbers. \(\mathrm{f}(\mathrm{n})=\mathrm{n}\) If \(g\) is many one, then fog is many one. So, if \(\mathrm{g}\) is onto then fog is one-one is incorrect.
[JEE Main 25.02.2021 Shift-I]
Sets, Relation and Function
117132
If \(R\) denotes the set of all real numbers then the function \(f: R \rightarrow R\) defined by \(f(x)=|x|\) is
1 injective and surjective
2 surjective
3 injective.
4 neither injective nor surjective.
Explanation:
Exp: (D) : Given that, \(f(x)=|x| \geq 0\) for all \(x\) So the range of is \([0, \infty) \neq\) co domain. Hence, \(\mathrm{f}\) is not onto function (nor injective function). Also \(\mathrm{f}(-1)=\mathrm{f}(1)=1\) \(\mathrm{f}\) is many one function( non surjective function).
[MHT CET-2022]
Sets, Relation and Function
117133
In the function \(\mathbf{f}: \mathbf{R}-\{-1,1\} \rightarrow \mathbf{A}\) defined by \(f(x) f(x)=\frac{x^2}{1-x^2}\) is surjective, then a is equal to
1 \(\mathrm{R}-\{-1\}\)
2 \([0, \infty)\)
3 \(\mathrm{R}-[-1,0)\)
4 \(\mathrm{R}-(-1,0)\)
Explanation:
Exp: (C) : Given function, \(\text { f: } \mathrm{R}-\{-1,1\}\) \(f(x)=\frac{x^2}{1-x^2}\) is surjective \(\because \mathrm{f}\) is surjective, so every element in co-domain must have a pre-image in domain. \(\text { Let, } y=\frac{x^2}{1-x^2}\) \(y\left(1-x^2\right)=x^2\) \(y-y^2=x^2\) \(y=x^2+y x^2\) \(y=x^2(1+y)\) \(x^2=\frac{y}{1+y}\) \(x=\sqrt{\frac{y}{1+y}}\) For \(\mathrm{x}\) to defined. \(1+y \neq 0\) \(y \neq-1\) and \(\frac{\mathrm{y}}{1+\mathrm{y}} \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-1) \cup[0, \infty]\) From equation (i) and (ii) we get \(y \in R-[-1,0)\)
[MHT CET-2022]
Sets, Relation and Function
117135
Let \(f: R \rightarrow R\) be defined by \(f(x)=\cos x\). Then
1 \(f\) is one - one and odd
2 \(f\) is odd but not one - one
3 \(f\) is even and onto
4 \(f\) is one- one and even
5 \(f\) is even but not onto
Explanation:
E Given, \(f(x)=\cos x\) For even function - \(f(-x)=f(x)\) \(\cos (-x)=\cos x\) Hence, \(f(x)\) is even function . Let, \(y=\cos x\) \(x=\cos ^{-1} y\) \(\cos ^{-1} \mathrm{y}\) does not defined in real number because \(\cos ^{-1} \mathrm{y}\) is defined in range \(\in[-1,1]\). Hence, \(f(x)\) is not onto function . Therefore, \(f(x)\) is even function but not onto function.
[Kerala CEE-2022]
Sets, Relation and Function
117061
Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then \(f\) 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:
C Given, \(f(x)=\max \{(x+1),(1-x), 2\}\) \(f(x)=\left\{\begin{array}{ccc}1-x & , & x \leq-1 \\ 2, & , & -1\lt x\lt 1 \\ x+1 & , & x \geq 1\end{array}\right.\) Neither one-one nor onto.
117131
Let \(f, g: N \rightarrow N\), such that \(f(n+1)=f(n)+\) \(\mathbf{f}(1) \forall \mathbf{n} \in \mathbf{N}\) and \(g\) be any arbitrary function. Which of the following statements is not true?
1 if fog is one-one, then \(g\) is one - one
2 if \(\mathrm{f}\) is onto, then \(\mathrm{f}(\mathrm{n})=\mathrm{n}, \forall \mathrm{n} \in \mathrm{N}\).
3 \(f\) is one-one
4 if \(g\) is onto, then fog is one-one
Explanation:
Exp: (D) : Given, \(\mathrm{f}(\mathrm{n}+1)=\mathrm{f}(\mathrm{n})+\mathrm{f}(1)\) \(\mathrm{f}(\mathrm{n}+1)-\mathrm{f}(\mathrm{n})=\mathrm{f}(1)\) Since, Above terms are in A.P. with common difference \(=\mathrm{f}(1)\) General term \(T_n=f(1)+(n-1) f(1)=n f(1)\) \(\mathrm{f}(\mathrm{n})=\mathrm{nf}(1)\) For fog to be one-one, \(g\) must be one-one. For \(f\) to be onto, \(f(n)\) should take all the values of natural numbers. \(\mathrm{f}(\mathrm{n})=\mathrm{n}\) If \(g\) is many one, then fog is many one. So, if \(\mathrm{g}\) is onto then fog is one-one is incorrect.
[JEE Main 25.02.2021 Shift-I]
Sets, Relation and Function
117132
If \(R\) denotes the set of all real numbers then the function \(f: R \rightarrow R\) defined by \(f(x)=|x|\) is
1 injective and surjective
2 surjective
3 injective.
4 neither injective nor surjective.
Explanation:
Exp: (D) : Given that, \(f(x)=|x| \geq 0\) for all \(x\) So the range of is \([0, \infty) \neq\) co domain. Hence, \(\mathrm{f}\) is not onto function (nor injective function). Also \(\mathrm{f}(-1)=\mathrm{f}(1)=1\) \(\mathrm{f}\) is many one function( non surjective function).
[MHT CET-2022]
Sets, Relation and Function
117133
In the function \(\mathbf{f}: \mathbf{R}-\{-1,1\} \rightarrow \mathbf{A}\) defined by \(f(x) f(x)=\frac{x^2}{1-x^2}\) is surjective, then a is equal to
1 \(\mathrm{R}-\{-1\}\)
2 \([0, \infty)\)
3 \(\mathrm{R}-[-1,0)\)
4 \(\mathrm{R}-(-1,0)\)
Explanation:
Exp: (C) : Given function, \(\text { f: } \mathrm{R}-\{-1,1\}\) \(f(x)=\frac{x^2}{1-x^2}\) is surjective \(\because \mathrm{f}\) is surjective, so every element in co-domain must have a pre-image in domain. \(\text { Let, } y=\frac{x^2}{1-x^2}\) \(y\left(1-x^2\right)=x^2\) \(y-y^2=x^2\) \(y=x^2+y x^2\) \(y=x^2(1+y)\) \(x^2=\frac{y}{1+y}\) \(x=\sqrt{\frac{y}{1+y}}\) For \(\mathrm{x}\) to defined. \(1+y \neq 0\) \(y \neq-1\) and \(\frac{\mathrm{y}}{1+\mathrm{y}} \geq 0\) \(\Rightarrow \mathrm{y} \in(-\infty,-1) \cup[0, \infty]\) From equation (i) and (ii) we get \(y \in R-[-1,0)\)
[MHT CET-2022]
Sets, Relation and Function
117135
Let \(f: R \rightarrow R\) be defined by \(f(x)=\cos x\). Then
1 \(f\) is one - one and odd
2 \(f\) is odd but not one - one
3 \(f\) is even and onto
4 \(f\) is one- one and even
5 \(f\) is even but not onto
Explanation:
E Given, \(f(x)=\cos x\) For even function - \(f(-x)=f(x)\) \(\cos (-x)=\cos x\) Hence, \(f(x)\) is even function . Let, \(y=\cos x\) \(x=\cos ^{-1} y\) \(\cos ^{-1} \mathrm{y}\) does not defined in real number because \(\cos ^{-1} \mathrm{y}\) is defined in range \(\in[-1,1]\). Hence, \(f(x)\) is not onto function . Therefore, \(f(x)\) is even function but not onto function.
[Kerala CEE-2022]
Sets, Relation and Function
117061
Define \(f: R \rightarrow R\) by \(f(x)=\max \{x+1,1-x, 2\}\). Then \(f\) 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:
C Given, \(f(x)=\max \{(x+1),(1-x), 2\}\) \(f(x)=\left\{\begin{array}{ccc}1-x & , & x \leq-1 \\ 2, & , & -1\lt x\lt 1 \\ x+1 & , & x \geq 1\end{array}\right.\) Neither one-one nor onto.
