117199
If every element of a group \(G\) is its own inverse, then \(\mathbf{G}\) is
1 finite
2 infinite
3 not abelian
4 abelian
Explanation:
D According to question, \(\mathrm{a}=\mathrm{a}^{-1} \forall \mathrm{a} \in \mathrm{G}\) Then, for any \(a, b \in G\) \(\mathrm{ab} \in \mathrm{G}\) \(\Rightarrow \quad(a b)^{-1}=a b\) \(\Rightarrow \quad b^{-1} \mathrm{a}^{-1}=\mathrm{ab}\) \(\Rightarrow \quad \mathrm{ba}=\mathrm{ab}\left(\because \mathrm{b}^{-1}=\mathrm{b}, \mathrm{a}^{-1}=\mathrm{a}\right)\) Hence, \(\mathrm{G}\) is an abelian group.
Manipal UGET-2017
Sets, Relation and Function
117200
Let \(f: \mathbf{R} \rightarrow \mathbf{R}, \mathbf{g}: \mathbf{R} \rightarrow \mathbf{R}\), be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\) the function \((\text { fog })^{-1}(x)\) is equal to
1 \(\left(\frac{x+7}{2}\right)^{1 / 3}\)
2 \(\left(x-\frac{7}{2}\right)^{1 / 3}\)
3 \(\left(\frac{x-2}{7}\right)^{1 / 3}\)
4 \(\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}\)
Explanation:
D Given that, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=2 \mathrm{x}-3\) \(\mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}(\mathrm{x})=\mathrm{x}^3+5\) Now, \(\text { fog }=f(g(x))\) \(=2 g(x)-3=2\left(x^3+5\right)-3=2 x^3+10-3=2 x^3+7\) Now, we have to find (fog) \()^{-1}(\mathrm{x})\) Let, \(\Rightarrow \quad (\text { fog })^{-1}(x)=y=2 x^3+7\) \(y=2 x^3+7\) \(y-7=2 x^3\) \(x=\left[\frac{y-7}{2}\right]^{1 / 3}\)So, \(\quad(f \circ g)^{-1}(x)=\left[\frac{x-7}{2}\right]^{1 / 3}\)
Rajasthan PET-2011
Sets, Relation and Function
117201
If \(f: R \rightarrow R\) is defined as \(f(x)=(1-x)^{1 / 3}\) then \(f^{-1}(x)\) is
1 \((1-x)^{-1 / 3}\)
2 \((1-x)^3\)
3 \(1-x^3\)
4 \(1-x^{1 / 3}\)
Explanation:
C Given that, Let, \(\mathrm{f}(\mathrm{x})=(1-\mathrm{x})^{1 / 3}\) \(y=f(x)=(1-x)^{1 / 3}\) \(y^3=(1-x)\) \(x=f^{-1}(y)=1-y^3\) So, \(\mathrm{f}^{-1}(\mathrm{x})=1-\mathrm{x}^3\)
Rajasthan PET-2009
Sets, Relation and Function
117203
If \(\mathbf{f :} \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}\) to be defined by : \(f(x)=\frac{3 x+1}{5 x-3}\), then
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Sets, Relation and Function
117199
If every element of a group \(G\) is its own inverse, then \(\mathbf{G}\) is
1 finite
2 infinite
3 not abelian
4 abelian
Explanation:
D According to question, \(\mathrm{a}=\mathrm{a}^{-1} \forall \mathrm{a} \in \mathrm{G}\) Then, for any \(a, b \in G\) \(\mathrm{ab} \in \mathrm{G}\) \(\Rightarrow \quad(a b)^{-1}=a b\) \(\Rightarrow \quad b^{-1} \mathrm{a}^{-1}=\mathrm{ab}\) \(\Rightarrow \quad \mathrm{ba}=\mathrm{ab}\left(\because \mathrm{b}^{-1}=\mathrm{b}, \mathrm{a}^{-1}=\mathrm{a}\right)\) Hence, \(\mathrm{G}\) is an abelian group.
Manipal UGET-2017
Sets, Relation and Function
117200
Let \(f: \mathbf{R} \rightarrow \mathbf{R}, \mathbf{g}: \mathbf{R} \rightarrow \mathbf{R}\), be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\) the function \((\text { fog })^{-1}(x)\) is equal to
1 \(\left(\frac{x+7}{2}\right)^{1 / 3}\)
2 \(\left(x-\frac{7}{2}\right)^{1 / 3}\)
3 \(\left(\frac{x-2}{7}\right)^{1 / 3}\)
4 \(\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}\)
Explanation:
D Given that, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=2 \mathrm{x}-3\) \(\mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}(\mathrm{x})=\mathrm{x}^3+5\) Now, \(\text { fog }=f(g(x))\) \(=2 g(x)-3=2\left(x^3+5\right)-3=2 x^3+10-3=2 x^3+7\) Now, we have to find (fog) \()^{-1}(\mathrm{x})\) Let, \(\Rightarrow \quad (\text { fog })^{-1}(x)=y=2 x^3+7\) \(y=2 x^3+7\) \(y-7=2 x^3\) \(x=\left[\frac{y-7}{2}\right]^{1 / 3}\)So, \(\quad(f \circ g)^{-1}(x)=\left[\frac{x-7}{2}\right]^{1 / 3}\)
Rajasthan PET-2011
Sets, Relation and Function
117201
If \(f: R \rightarrow R\) is defined as \(f(x)=(1-x)^{1 / 3}\) then \(f^{-1}(x)\) is
1 \((1-x)^{-1 / 3}\)
2 \((1-x)^3\)
3 \(1-x^3\)
4 \(1-x^{1 / 3}\)
Explanation:
C Given that, Let, \(\mathrm{f}(\mathrm{x})=(1-\mathrm{x})^{1 / 3}\) \(y=f(x)=(1-x)^{1 / 3}\) \(y^3=(1-x)\) \(x=f^{-1}(y)=1-y^3\) So, \(\mathrm{f}^{-1}(\mathrm{x})=1-\mathrm{x}^3\)
Rajasthan PET-2009
Sets, Relation and Function
117203
If \(\mathbf{f :} \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}\) to be defined by : \(f(x)=\frac{3 x+1}{5 x-3}\), then
117199
If every element of a group \(G\) is its own inverse, then \(\mathbf{G}\) is
1 finite
2 infinite
3 not abelian
4 abelian
Explanation:
D According to question, \(\mathrm{a}=\mathrm{a}^{-1} \forall \mathrm{a} \in \mathrm{G}\) Then, for any \(a, b \in G\) \(\mathrm{ab} \in \mathrm{G}\) \(\Rightarrow \quad(a b)^{-1}=a b\) \(\Rightarrow \quad b^{-1} \mathrm{a}^{-1}=\mathrm{ab}\) \(\Rightarrow \quad \mathrm{ba}=\mathrm{ab}\left(\because \mathrm{b}^{-1}=\mathrm{b}, \mathrm{a}^{-1}=\mathrm{a}\right)\) Hence, \(\mathrm{G}\) is an abelian group.
Manipal UGET-2017
Sets, Relation and Function
117200
Let \(f: \mathbf{R} \rightarrow \mathbf{R}, \mathbf{g}: \mathbf{R} \rightarrow \mathbf{R}\), be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\) the function \((\text { fog })^{-1}(x)\) is equal to
1 \(\left(\frac{x+7}{2}\right)^{1 / 3}\)
2 \(\left(x-\frac{7}{2}\right)^{1 / 3}\)
3 \(\left(\frac{x-2}{7}\right)^{1 / 3}\)
4 \(\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}\)
Explanation:
D Given that, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=2 \mathrm{x}-3\) \(\mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}(\mathrm{x})=\mathrm{x}^3+5\) Now, \(\text { fog }=f(g(x))\) \(=2 g(x)-3=2\left(x^3+5\right)-3=2 x^3+10-3=2 x^3+7\) Now, we have to find (fog) \()^{-1}(\mathrm{x})\) Let, \(\Rightarrow \quad (\text { fog })^{-1}(x)=y=2 x^3+7\) \(y=2 x^3+7\) \(y-7=2 x^3\) \(x=\left[\frac{y-7}{2}\right]^{1 / 3}\)So, \(\quad(f \circ g)^{-1}(x)=\left[\frac{x-7}{2}\right]^{1 / 3}\)
Rajasthan PET-2011
Sets, Relation and Function
117201
If \(f: R \rightarrow R\) is defined as \(f(x)=(1-x)^{1 / 3}\) then \(f^{-1}(x)\) is
1 \((1-x)^{-1 / 3}\)
2 \((1-x)^3\)
3 \(1-x^3\)
4 \(1-x^{1 / 3}\)
Explanation:
C Given that, Let, \(\mathrm{f}(\mathrm{x})=(1-\mathrm{x})^{1 / 3}\) \(y=f(x)=(1-x)^{1 / 3}\) \(y^3=(1-x)\) \(x=f^{-1}(y)=1-y^3\) So, \(\mathrm{f}^{-1}(\mathrm{x})=1-\mathrm{x}^3\)
Rajasthan PET-2009
Sets, Relation and Function
117203
If \(\mathbf{f :} \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}\) to be defined by : \(f(x)=\frac{3 x+1}{5 x-3}\), then
117199
If every element of a group \(G\) is its own inverse, then \(\mathbf{G}\) is
1 finite
2 infinite
3 not abelian
4 abelian
Explanation:
D According to question, \(\mathrm{a}=\mathrm{a}^{-1} \forall \mathrm{a} \in \mathrm{G}\) Then, for any \(a, b \in G\) \(\mathrm{ab} \in \mathrm{G}\) \(\Rightarrow \quad(a b)^{-1}=a b\) \(\Rightarrow \quad b^{-1} \mathrm{a}^{-1}=\mathrm{ab}\) \(\Rightarrow \quad \mathrm{ba}=\mathrm{ab}\left(\because \mathrm{b}^{-1}=\mathrm{b}, \mathrm{a}^{-1}=\mathrm{a}\right)\) Hence, \(\mathrm{G}\) is an abelian group.
Manipal UGET-2017
Sets, Relation and Function
117200
Let \(f: \mathbf{R} \rightarrow \mathbf{R}, \mathbf{g}: \mathbf{R} \rightarrow \mathbf{R}\), be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\) the function \((\text { fog })^{-1}(x)\) is equal to
1 \(\left(\frac{x+7}{2}\right)^{1 / 3}\)
2 \(\left(x-\frac{7}{2}\right)^{1 / 3}\)
3 \(\left(\frac{x-2}{7}\right)^{1 / 3}\)
4 \(\left(\frac{\mathrm{x}-7}{2}\right)^{1 / 3}\)
Explanation:
D Given that, \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=2 \mathrm{x}-3\) \(\mathrm{~g}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}(\mathrm{x})=\mathrm{x}^3+5\) Now, \(\text { fog }=f(g(x))\) \(=2 g(x)-3=2\left(x^3+5\right)-3=2 x^3+10-3=2 x^3+7\) Now, we have to find (fog) \()^{-1}(\mathrm{x})\) Let, \(\Rightarrow \quad (\text { fog })^{-1}(x)=y=2 x^3+7\) \(y=2 x^3+7\) \(y-7=2 x^3\) \(x=\left[\frac{y-7}{2}\right]^{1 / 3}\)So, \(\quad(f \circ g)^{-1}(x)=\left[\frac{x-7}{2}\right]^{1 / 3}\)
Rajasthan PET-2011
Sets, Relation and Function
117201
If \(f: R \rightarrow R\) is defined as \(f(x)=(1-x)^{1 / 3}\) then \(f^{-1}(x)\) is
1 \((1-x)^{-1 / 3}\)
2 \((1-x)^3\)
3 \(1-x^3\)
4 \(1-x^{1 / 3}\)
Explanation:
C Given that, Let, \(\mathrm{f}(\mathrm{x})=(1-\mathrm{x})^{1 / 3}\) \(y=f(x)=(1-x)^{1 / 3}\) \(y^3=(1-x)\) \(x=f^{-1}(y)=1-y^3\) So, \(\mathrm{f}^{-1}(\mathrm{x})=1-\mathrm{x}^3\)
Rajasthan PET-2009
Sets, Relation and Function
117203
If \(\mathbf{f :} \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}\) to be defined by : \(f(x)=\frac{3 x+1}{5 x-3}\), then