117106
Let a function \(f:(0, \infty) \rightarrow(0, \infty)\) be defined by \(f(x)=\left|1-\frac{1}{x}\right|\). Then, \(f\) is
1 injective only.
2 both injective as well as surjective.
3 not injective but it is surjective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given, \(f(x)=\left|1-\frac{1}{x}\right|\) \(f(x)= \begin{cases}1-\frac{1}{x} & x \in(1, \infty) \\ \frac{1}{x}-1 & x \in(0,1)\end{cases}\) \(f^{\prime}(x)= \begin{cases}\frac{1}{x^2} x \in(1, \infty) \\ -\frac{1}{x^2} x \in(0,1)\end{cases}\) This shows \(\mathrm{f}(\mathrm{x})\) is not injective. Since range of the function is equal to codomain function is surjective.
[JEE Main 11.01.2019 Shift - II]
Sets, Relation and Function
117107
The function \(f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\) is
1 invertible.
2 injective but not surjective.
3 surjective but nor injective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given function, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1-\mathrm{x}^2}\) On differentiating the given function with respect to \(x\) we get - \(\frac{d}{d x}[f(x)]=\frac{d}{d x}\left(\frac{x}{1+x^2}\right)\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \frac{d}{d x}(x)-x \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\) \({\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d \mu}{d x}-u \frac{d v}{d x}}{v^2}\right]}\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}\) \(f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2}\) Now for increasing intervals \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right)}>0 \quad\left[\because\left(1+\mathrm{x}^2\right)^2 \geq 0\right]\) \(1-\mathrm{x}^2>0\) \((1-\mathrm{x})(1+\mathrm{x})>0\) \((\mathrm{x}-1)(\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in \cdot(-1,1)\) Thus, the given function is increasing in nature in \((-1,1)\) and decreasing in nature in \(\mathrm{R}-(-1,1)\). Hence, the given function is not an injective function Now, Let \(y=\frac{x}{1+x^2}\) \(y\left(1+x^2\right)=x\) \(y x^2-x+y=0\) For real solutions, \(\mathrm{D} \geq 0\) \(1-4 y^2 \geq 0\) \(4 y^2-1 \leq 0\) \(4\left(y^2-\frac{1}{4}\right) \leq 0\) \(y^2-\left(\frac{1}{2}\right)^2 \leq 0\) \(\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \leq 0\) \(y \in\left[\frac{-1}{2}, \frac{1}{2}\right]\) The range of the given function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\) Hence, \(\mathrm{f}(\mathrm{x})\) is surjective but not injective
[JEE Main-2017]
Sets, Relation and Function
117108
Let \(x\) denote the total number of one-one functions from a set \(A\) with 3 elements to a set \(B\) with 5 elements and \(y\) denote the total number of one-one functions from the set \(A\) to the set \(\mathbf{A} \times \mathbf{B}\). Then,
1 \(2 y=91 x\)
2 \(2 y=273 x\)
3 \(y=91 x\)
4 \(y=273 x\)
Explanation:
Exp: (C) : Given, Total number of one-one function from a set \(A\) with 3 elements to a set B with 5 elements and y denoting total number of one one function. As we know that, no of one-one function \({ }^q \mathrm{C}_{\mathrm{p}} \times \mathrm{p}\) ! \(\mathrm{x}={ }^5 \mathrm{C}_3 \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 ! 3 !} \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 !}\) \(\mathrm{x}=5 \times 4 \times 3=60\) The number of one-one function \(y={ }^{15} C_3 \times 3 !\) \(y=\frac{15 !}{12 ! 3 !} \times 3 !\) \(y=\frac{15 !}{12 !}\) \(y=15 \times 14 \times 13=2730\) Therefore, \(\frac{x}{y}=\frac{60}{2730}\) \(273 x=6 y\) \(2 y=91 x\)
[JEE Main 25.02.2021 Shift -II]
Sets, Relation and Function
117109
If \(f(x)=x^2-2 x+4\), then the set of values of \(x\) satisfying \(f(x-1)=f(x+1)\) is
1 \(\{-1\}\)
2 \(\{-1,1\}\)
3 \(\{1\}\)
4 \(\{1,2\}\)
Explanation:
Exp: (C) : Given that, \(f(x)=x^2-2 x+4\) According to question, \(f(x-1)=f(x+1)\) \((x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4\) \(x^2+1-2 x-2 x+2+4=x^2+1+2 x-2 x-2+4\) \(4 x=4\) \(x=1\)So, the only solution is \(\{1\}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
117106
Let a function \(f:(0, \infty) \rightarrow(0, \infty)\) be defined by \(f(x)=\left|1-\frac{1}{x}\right|\). Then, \(f\) is
1 injective only.
2 both injective as well as surjective.
3 not injective but it is surjective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given, \(f(x)=\left|1-\frac{1}{x}\right|\) \(f(x)= \begin{cases}1-\frac{1}{x} & x \in(1, \infty) \\ \frac{1}{x}-1 & x \in(0,1)\end{cases}\) \(f^{\prime}(x)= \begin{cases}\frac{1}{x^2} x \in(1, \infty) \\ -\frac{1}{x^2} x \in(0,1)\end{cases}\) This shows \(\mathrm{f}(\mathrm{x})\) is not injective. Since range of the function is equal to codomain function is surjective.
[JEE Main 11.01.2019 Shift - II]
Sets, Relation and Function
117107
The function \(f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\) is
1 invertible.
2 injective but not surjective.
3 surjective but nor injective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given function, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1-\mathrm{x}^2}\) On differentiating the given function with respect to \(x\) we get - \(\frac{d}{d x}[f(x)]=\frac{d}{d x}\left(\frac{x}{1+x^2}\right)\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \frac{d}{d x}(x)-x \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\) \({\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d \mu}{d x}-u \frac{d v}{d x}}{v^2}\right]}\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}\) \(f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2}\) Now for increasing intervals \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right)}>0 \quad\left[\because\left(1+\mathrm{x}^2\right)^2 \geq 0\right]\) \(1-\mathrm{x}^2>0\) \((1-\mathrm{x})(1+\mathrm{x})>0\) \((\mathrm{x}-1)(\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in \cdot(-1,1)\) Thus, the given function is increasing in nature in \((-1,1)\) and decreasing in nature in \(\mathrm{R}-(-1,1)\). Hence, the given function is not an injective function Now, Let \(y=\frac{x}{1+x^2}\) \(y\left(1+x^2\right)=x\) \(y x^2-x+y=0\) For real solutions, \(\mathrm{D} \geq 0\) \(1-4 y^2 \geq 0\) \(4 y^2-1 \leq 0\) \(4\left(y^2-\frac{1}{4}\right) \leq 0\) \(y^2-\left(\frac{1}{2}\right)^2 \leq 0\) \(\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \leq 0\) \(y \in\left[\frac{-1}{2}, \frac{1}{2}\right]\) The range of the given function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\) Hence, \(\mathrm{f}(\mathrm{x})\) is surjective but not injective
[JEE Main-2017]
Sets, Relation and Function
117108
Let \(x\) denote the total number of one-one functions from a set \(A\) with 3 elements to a set \(B\) with 5 elements and \(y\) denote the total number of one-one functions from the set \(A\) to the set \(\mathbf{A} \times \mathbf{B}\). Then,
1 \(2 y=91 x\)
2 \(2 y=273 x\)
3 \(y=91 x\)
4 \(y=273 x\)
Explanation:
Exp: (C) : Given, Total number of one-one function from a set \(A\) with 3 elements to a set B with 5 elements and y denoting total number of one one function. As we know that, no of one-one function \({ }^q \mathrm{C}_{\mathrm{p}} \times \mathrm{p}\) ! \(\mathrm{x}={ }^5 \mathrm{C}_3 \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 ! 3 !} \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 !}\) \(\mathrm{x}=5 \times 4 \times 3=60\) The number of one-one function \(y={ }^{15} C_3 \times 3 !\) \(y=\frac{15 !}{12 ! 3 !} \times 3 !\) \(y=\frac{15 !}{12 !}\) \(y=15 \times 14 \times 13=2730\) Therefore, \(\frac{x}{y}=\frac{60}{2730}\) \(273 x=6 y\) \(2 y=91 x\)
[JEE Main 25.02.2021 Shift -II]
Sets, Relation and Function
117109
If \(f(x)=x^2-2 x+4\), then the set of values of \(x\) satisfying \(f(x-1)=f(x+1)\) is
1 \(\{-1\}\)
2 \(\{-1,1\}\)
3 \(\{1\}\)
4 \(\{1,2\}\)
Explanation:
Exp: (C) : Given that, \(f(x)=x^2-2 x+4\) According to question, \(f(x-1)=f(x+1)\) \((x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4\) \(x^2+1-2 x-2 x+2+4=x^2+1+2 x-2 x-2+4\) \(4 x=4\) \(x=1\)So, the only solution is \(\{1\}\).
117106
Let a function \(f:(0, \infty) \rightarrow(0, \infty)\) be defined by \(f(x)=\left|1-\frac{1}{x}\right|\). Then, \(f\) is
1 injective only.
2 both injective as well as surjective.
3 not injective but it is surjective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given, \(f(x)=\left|1-\frac{1}{x}\right|\) \(f(x)= \begin{cases}1-\frac{1}{x} & x \in(1, \infty) \\ \frac{1}{x}-1 & x \in(0,1)\end{cases}\) \(f^{\prime}(x)= \begin{cases}\frac{1}{x^2} x \in(1, \infty) \\ -\frac{1}{x^2} x \in(0,1)\end{cases}\) This shows \(\mathrm{f}(\mathrm{x})\) is not injective. Since range of the function is equal to codomain function is surjective.
[JEE Main 11.01.2019 Shift - II]
Sets, Relation and Function
117107
The function \(f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\) is
1 invertible.
2 injective but not surjective.
3 surjective but nor injective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given function, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1-\mathrm{x}^2}\) On differentiating the given function with respect to \(x\) we get - \(\frac{d}{d x}[f(x)]=\frac{d}{d x}\left(\frac{x}{1+x^2}\right)\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \frac{d}{d x}(x)-x \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\) \({\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d \mu}{d x}-u \frac{d v}{d x}}{v^2}\right]}\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}\) \(f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2}\) Now for increasing intervals \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right)}>0 \quad\left[\because\left(1+\mathrm{x}^2\right)^2 \geq 0\right]\) \(1-\mathrm{x}^2>0\) \((1-\mathrm{x})(1+\mathrm{x})>0\) \((\mathrm{x}-1)(\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in \cdot(-1,1)\) Thus, the given function is increasing in nature in \((-1,1)\) and decreasing in nature in \(\mathrm{R}-(-1,1)\). Hence, the given function is not an injective function Now, Let \(y=\frac{x}{1+x^2}\) \(y\left(1+x^2\right)=x\) \(y x^2-x+y=0\) For real solutions, \(\mathrm{D} \geq 0\) \(1-4 y^2 \geq 0\) \(4 y^2-1 \leq 0\) \(4\left(y^2-\frac{1}{4}\right) \leq 0\) \(y^2-\left(\frac{1}{2}\right)^2 \leq 0\) \(\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \leq 0\) \(y \in\left[\frac{-1}{2}, \frac{1}{2}\right]\) The range of the given function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\) Hence, \(\mathrm{f}(\mathrm{x})\) is surjective but not injective
[JEE Main-2017]
Sets, Relation and Function
117108
Let \(x\) denote the total number of one-one functions from a set \(A\) with 3 elements to a set \(B\) with 5 elements and \(y\) denote the total number of one-one functions from the set \(A\) to the set \(\mathbf{A} \times \mathbf{B}\). Then,
1 \(2 y=91 x\)
2 \(2 y=273 x\)
3 \(y=91 x\)
4 \(y=273 x\)
Explanation:
Exp: (C) : Given, Total number of one-one function from a set \(A\) with 3 elements to a set B with 5 elements and y denoting total number of one one function. As we know that, no of one-one function \({ }^q \mathrm{C}_{\mathrm{p}} \times \mathrm{p}\) ! \(\mathrm{x}={ }^5 \mathrm{C}_3 \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 ! 3 !} \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 !}\) \(\mathrm{x}=5 \times 4 \times 3=60\) The number of one-one function \(y={ }^{15} C_3 \times 3 !\) \(y=\frac{15 !}{12 ! 3 !} \times 3 !\) \(y=\frac{15 !}{12 !}\) \(y=15 \times 14 \times 13=2730\) Therefore, \(\frac{x}{y}=\frac{60}{2730}\) \(273 x=6 y\) \(2 y=91 x\)
[JEE Main 25.02.2021 Shift -II]
Sets, Relation and Function
117109
If \(f(x)=x^2-2 x+4\), then the set of values of \(x\) satisfying \(f(x-1)=f(x+1)\) is
1 \(\{-1\}\)
2 \(\{-1,1\}\)
3 \(\{1\}\)
4 \(\{1,2\}\)
Explanation:
Exp: (C) : Given that, \(f(x)=x^2-2 x+4\) According to question, \(f(x-1)=f(x+1)\) \((x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4\) \(x^2+1-2 x-2 x+2+4=x^2+1+2 x-2 x-2+4\) \(4 x=4\) \(x=1\)So, the only solution is \(\{1\}\).
117106
Let a function \(f:(0, \infty) \rightarrow(0, \infty)\) be defined by \(f(x)=\left|1-\frac{1}{x}\right|\). Then, \(f\) is
1 injective only.
2 both injective as well as surjective.
3 not injective but it is surjective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given, \(f(x)=\left|1-\frac{1}{x}\right|\) \(f(x)= \begin{cases}1-\frac{1}{x} & x \in(1, \infty) \\ \frac{1}{x}-1 & x \in(0,1)\end{cases}\) \(f^{\prime}(x)= \begin{cases}\frac{1}{x^2} x \in(1, \infty) \\ -\frac{1}{x^2} x \in(0,1)\end{cases}\) This shows \(\mathrm{f}(\mathrm{x})\) is not injective. Since range of the function is equal to codomain function is surjective.
[JEE Main 11.01.2019 Shift - II]
Sets, Relation and Function
117107
The function \(f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]\) defined as \(f(x)=\frac{x}{1+x^2}\) is
1 invertible.
2 injective but not surjective.
3 surjective but nor injective.
4 neither injective nor surjective.
Explanation:
Exp: (C) : Given function, \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1-\mathrm{x}^2}\) On differentiating the given function with respect to \(x\) we get - \(\frac{d}{d x}[f(x)]=\frac{d}{d x}\left(\frac{x}{1+x^2}\right)\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \frac{d}{d x}(x)-x \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\) \({\left[\because \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d \mu}{d x}-u \frac{d v}{d x}}{v^2}\right]}\) \(f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}\) \(f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2}\) Now for increasing intervals \(\mathrm{f}^{\prime}(\mathrm{x})>0\) \(\frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right)}>0 \quad\left[\because\left(1+\mathrm{x}^2\right)^2 \geq 0\right]\) \(1-\mathrm{x}^2>0\) \((1-\mathrm{x})(1+\mathrm{x})>0\) \((\mathrm{x}-1)(\mathrm{x}+1)\lt 0\) \(\mathrm{x} \in \cdot(-1,1)\) Thus, the given function is increasing in nature in \((-1,1)\) and decreasing in nature in \(\mathrm{R}-(-1,1)\). Hence, the given function is not an injective function Now, Let \(y=\frac{x}{1+x^2}\) \(y\left(1+x^2\right)=x\) \(y x^2-x+y=0\) For real solutions, \(\mathrm{D} \geq 0\) \(1-4 y^2 \geq 0\) \(4 y^2-1 \leq 0\) \(4\left(y^2-\frac{1}{4}\right) \leq 0\) \(y^2-\left(\frac{1}{2}\right)^2 \leq 0\) \(\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right) \leq 0\) \(y \in\left[\frac{-1}{2}, \frac{1}{2}\right]\) The range of the given function is \(\left[\frac{-1}{2}, \frac{1}{2}\right]\) Hence, \(\mathrm{f}(\mathrm{x})\) is surjective but not injective
[JEE Main-2017]
Sets, Relation and Function
117108
Let \(x\) denote the total number of one-one functions from a set \(A\) with 3 elements to a set \(B\) with 5 elements and \(y\) denote the total number of one-one functions from the set \(A\) to the set \(\mathbf{A} \times \mathbf{B}\). Then,
1 \(2 y=91 x\)
2 \(2 y=273 x\)
3 \(y=91 x\)
4 \(y=273 x\)
Explanation:
Exp: (C) : Given, Total number of one-one function from a set \(A\) with 3 elements to a set B with 5 elements and y denoting total number of one one function. As we know that, no of one-one function \({ }^q \mathrm{C}_{\mathrm{p}} \times \mathrm{p}\) ! \(\mathrm{x}={ }^5 \mathrm{C}_3 \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 ! 3 !} \times 3 !\) \(\mathrm{x}=\frac{5 !}{2 !}\) \(\mathrm{x}=5 \times 4 \times 3=60\) The number of one-one function \(y={ }^{15} C_3 \times 3 !\) \(y=\frac{15 !}{12 ! 3 !} \times 3 !\) \(y=\frac{15 !}{12 !}\) \(y=15 \times 14 \times 13=2730\) Therefore, \(\frac{x}{y}=\frac{60}{2730}\) \(273 x=6 y\) \(2 y=91 x\)
[JEE Main 25.02.2021 Shift -II]
Sets, Relation and Function
117109
If \(f(x)=x^2-2 x+4\), then the set of values of \(x\) satisfying \(f(x-1)=f(x+1)\) is
1 \(\{-1\}\)
2 \(\{-1,1\}\)
3 \(\{1\}\)
4 \(\{1,2\}\)
Explanation:
Exp: (C) : Given that, \(f(x)=x^2-2 x+4\) According to question, \(f(x-1)=f(x+1)\) \((x-1)^2-2(x-1)+4=(x+1)^2-2(x+1)+4\) \(x^2+1-2 x-2 x+2+4=x^2+1+2 x-2 x-2+4\) \(4 x=4\) \(x=1\)So, the only solution is \(\{1\}\).