117156
If the set of integers with the operation defined by \(a * b=a+b-1\) forms a group, what is the inverse of \(a\) ?
1 \(-\mathrm{a}\)
2 \(2 \mathrm{a}\)
3 \(2-\mathrm{a}\)
4 \(1-\mathrm{a}\)
Explanation:
Exp: (C) : Let the Algebraic structure \((\mathrm{G}, *)\) form a group where * is the Binary operation such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Let e be the identity element such that a.e \(=\mathrm{a}=\) e.a. or \(\mathrm{a}+\mathrm{e}-1=\mathrm{a}\) or \(\mathrm{e}=1\) is the identity element by using the property of inverse, we have- element \(b\) is said to be the inverse of a where \(a, b \in G\). such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{e}\) i.e. \(a+b-1=1\) or \(\mathrm{b}=2-\mathrm{a}\) Hence inverse of element \(a\) is given by \((2-a)\).
[SCRA-2012]
Sets, Relation and Function
117171
Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\). the function \((f \circ g)^{-1}(x)\) is equal to
117156
If the set of integers with the operation defined by \(a * b=a+b-1\) forms a group, what is the inverse of \(a\) ?
1 \(-\mathrm{a}\)
2 \(2 \mathrm{a}\)
3 \(2-\mathrm{a}\)
4 \(1-\mathrm{a}\)
Explanation:
Exp: (C) : Let the Algebraic structure \((\mathrm{G}, *)\) form a group where * is the Binary operation such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Let e be the identity element such that a.e \(=\mathrm{a}=\) e.a. or \(\mathrm{a}+\mathrm{e}-1=\mathrm{a}\) or \(\mathrm{e}=1\) is the identity element by using the property of inverse, we have- element \(b\) is said to be the inverse of a where \(a, b \in G\). such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{e}\) i.e. \(a+b-1=1\) or \(\mathrm{b}=2-\mathrm{a}\) Hence inverse of element \(a\) is given by \((2-a)\).
[SCRA-2012]
Sets, Relation and Function
117171
Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\). the function \((f \circ g)^{-1}(x)\) is equal to
117156
If the set of integers with the operation defined by \(a * b=a+b-1\) forms a group, what is the inverse of \(a\) ?
1 \(-\mathrm{a}\)
2 \(2 \mathrm{a}\)
3 \(2-\mathrm{a}\)
4 \(1-\mathrm{a}\)
Explanation:
Exp: (C) : Let the Algebraic structure \((\mathrm{G}, *)\) form a group where * is the Binary operation such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Let e be the identity element such that a.e \(=\mathrm{a}=\) e.a. or \(\mathrm{a}+\mathrm{e}-1=\mathrm{a}\) or \(\mathrm{e}=1\) is the identity element by using the property of inverse, we have- element \(b\) is said to be the inverse of a where \(a, b \in G\). such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{e}\) i.e. \(a+b-1=1\) or \(\mathrm{b}=2-\mathrm{a}\) Hence inverse of element \(a\) is given by \((2-a)\).
[SCRA-2012]
Sets, Relation and Function
117171
Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\). the function \((f \circ g)^{-1}(x)\) is equal to
117156
If the set of integers with the operation defined by \(a * b=a+b-1\) forms a group, what is the inverse of \(a\) ?
1 \(-\mathrm{a}\)
2 \(2 \mathrm{a}\)
3 \(2-\mathrm{a}\)
4 \(1-\mathrm{a}\)
Explanation:
Exp: (C) : Let the Algebraic structure \((\mathrm{G}, *)\) form a group where * is the Binary operation such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-1\) Let e be the identity element such that a.e \(=\mathrm{a}=\) e.a. or \(\mathrm{a}+\mathrm{e}-1=\mathrm{a}\) or \(\mathrm{e}=1\) is the identity element by using the property of inverse, we have- element \(b\) is said to be the inverse of a where \(a, b \in G\). such that - \(\mathrm{a} * \mathrm{~b}=\mathrm{e}\) i.e. \(a+b-1=1\) or \(\mathrm{b}=2-\mathrm{a}\) Hence inverse of element \(a\) is given by \((2-a)\).
[SCRA-2012]
Sets, Relation and Function
117171
Let \(f: R \rightarrow R, g: R \rightarrow R\) be two functions such that \(f(x)=2 x-3, g(x)=x^3+5\). the function \((f \circ g)^{-1}(x)\) is equal to
