117177
If the operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers ' \(a\) ' and ' \(b\) ', then \((2 \oplus 3)\) ๑ \(\mathbf{4}=\)
1 182
2 185
3 181
4 184
Explanation:
Exp: (B) : Given, The operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers 'a' and 'b'. Then, \((2 \oplus 3) \oplus 4=\left(2^2+3^2\right) \oplus 4\) \(=(4+9) \oplus 4\) \(=13 \oplus 4\) \(=13^2+4^2\) \(=169+16\)So, \((2 \oplus 3) \oplus 4=185\)
[Karnataka CET-2015]
Sets, Relation and Function
117178
Let * be a binary operation defined on \(\mathbf{R}\) by \(\mathbf{a}^*\) \(\mathbf{b}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{4}} \forall \mathbf{a}, \mathbf{b} \in \mathbf{R}\) then the operation * is
1 Commutative and Associative
2 Commutative but not Associative
3 Associative but not Commutative
4 Neither Associative nor Commutative
Explanation:
Exp: (B) : Given, Let, \(*\) be a binary operation defined on R by \(a * b=\) \(\frac{\mathrm{a}+\mathrm{b}}{4} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\). Then, we know that, a binary operation \(*\) on \(\mathrm{R}\) is commutative if - \(\mathrm{a} * \mathrm{~b} =\mathrm{b} * \mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4} \quad(\therefore \text { both side are equal })\) Which is true. And, \(A\) binary operation * on \(\mathrm{R}\) is associative if, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c}), \forall \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}\) \(\Rightarrow \quad \left(\frac{\mathrm{a}+\mathrm{b}}{4}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}+\mathrm{c}}{4}\right)\) \(\Rightarrow \quad \frac{\frac{\mathrm{a}+\mathrm{b}}{4}+\mathrm{c}}{4}=\frac{\mathrm{a}+\frac{\mathrm{b}+\mathrm{c}}{4}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}+4 \mathrm{c}}{16}=\frac{4 \mathrm{a}+\mathrm{b}+\mathrm{c}}{16}(\therefore \text { both side are not equal })\) Which is not correct. So, the operation * is commutative but not associative.
[Karnataka CET 2016]
Sets, Relation and Function
117179
Binary operation * on \(R-\{-1\}\) defined by \(a{ }^* \mathbf{b}\) \(\frac{\mathbf{a}}{\mathrm{b}+\mathbf{1}}\) is
1 * is associative and commutative
2 * is neither associative nor commutative
3 * is commutative but not associative
4 * is associative but not commutative
Explanation:
Exp: (B) : Given, Binary operation * on \(\mathrm{R}-\{-1\}\) defined by \(\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}}{\mathrm{b}+1}\) Then, we know that, a binary operation * on \(\mathrm{R}-\{-1\}\) is commutative if - \(\Rightarrow \frac{a}{b+1} \neq \frac{b}{a+1}\) Which is not true. And, A binary operation * on \(\mathrm{R}-\{-1\}\) is associative it \((\mathrm{a} * \mathrm{~b}) *(\mathrm{c})=(\mathrm{a}) *(\mathrm{~b} * \mathrm{c})\) \(\left(\frac{\mathrm{a}}{\mathrm{b}+1}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}}{\mathrm{c}+1}\right)\) \(\frac{\frac{\mathrm{a}}{\mathrm{b}+1}}{\mathrm{c}+1}=\frac{\mathrm{a}}{\frac{\mathrm{b}}{\mathrm{c}+1}+1}\) \(\frac{\mathrm{a}}{(\mathrm{b}+1)(\mathrm{c}+1)} \neq \frac{\mathrm{a}(\mathrm{c}+1)}{\mathrm{b}+\mathrm{c}+1}\) Which is not true. So, \({ }^*\) is neither associative nor commutative.
[Karnataka CET 2017]
Sets, Relation and Function
117180
In the group (Z, \(\left.{ }^*\right)\), if \(\mathbf{a}{ }^* \mathbf{b}=\mathbf{a}+\mathbf{b}-\mathbf{n} \forall \mathbf{a}, \mathbf{b} \in \mathbf{Z}\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
1 \(\mathrm{n}\)
2 \(-n\)
3 \(-3 n\)
4 \(3 \mathrm{n}\)
Explanation:
Exp: (D) : Given, In this the group \((\mathrm{Z}, *)\), If \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{n} \quad \forall \mathrm{a}, \mathrm{b} \in \mathrm{Z}\) Where, \(\mathrm{n}\) is a fixed integer. In this question, we can see that the identity element of the group is \(\mathrm{n}\). Then, \(\quad \mathrm{a} * \mathrm{n}=\mathrm{a} \quad \forall \mathrm{a} \in \mathrm{Z}\) So, let the inverse of \((-n)\) be \(Z\). Then, \(-\mathrm{n} * \mathrm{Z}=\) identity \(=\mathrm{n}\) \(-n+Z-n=n\) \(Z-2 n=n\) \(Z=3 n\)
[Karnataka CET 2013]
Sets, Relation and Function
117181
For any two real numbers, an operation * defined by \(a * b=1+a b\) is
1 neither commutative nor associative
2 commutative but not associative
3 both commutative and associative
4 associative but not commutative.
Explanation:
Exp: (B) : Given, For any two real number, an operation * defined by a * \(\mathrm{b}=1+\mathrm{ab}\). Then, A binary operation \(* \mathrm{R}\) is commutative if - \(a * b=b * a\) \(1+a b=1+b a\) \(1+a b=1+a b\) Which is true. And, A binary operation * on \(\mathrm{R}\) is associative if - \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})\) \((1+\mathrm{ab}) * \mathrm{c}=\mathrm{a} *(1+\mathrm{bc})\) \(1+(1+\mathrm{ab}) \mathrm{c}=1+\mathrm{a}(1+\mathrm{bc})\) \(1+\mathrm{c}+\mathrm{abc} \neq 1+\mathrm{a}+\mathrm{abc}\) Which is not true. So, \(*\) commutative but not associative.
117177
If the operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers ' \(a\) ' and ' \(b\) ', then \((2 \oplus 3)\) ๑ \(\mathbf{4}=\)
1 182
2 185
3 181
4 184
Explanation:
Exp: (B) : Given, The operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers 'a' and 'b'. Then, \((2 \oplus 3) \oplus 4=\left(2^2+3^2\right) \oplus 4\) \(=(4+9) \oplus 4\) \(=13 \oplus 4\) \(=13^2+4^2\) \(=169+16\)So, \((2 \oplus 3) \oplus 4=185\)
[Karnataka CET-2015]
Sets, Relation and Function
117178
Let * be a binary operation defined on \(\mathbf{R}\) by \(\mathbf{a}^*\) \(\mathbf{b}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{4}} \forall \mathbf{a}, \mathbf{b} \in \mathbf{R}\) then the operation * is
1 Commutative and Associative
2 Commutative but not Associative
3 Associative but not Commutative
4 Neither Associative nor Commutative
Explanation:
Exp: (B) : Given, Let, \(*\) be a binary operation defined on R by \(a * b=\) \(\frac{\mathrm{a}+\mathrm{b}}{4} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\). Then, we know that, a binary operation \(*\) on \(\mathrm{R}\) is commutative if - \(\mathrm{a} * \mathrm{~b} =\mathrm{b} * \mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4} \quad(\therefore \text { both side are equal })\) Which is true. And, \(A\) binary operation * on \(\mathrm{R}\) is associative if, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c}), \forall \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}\) \(\Rightarrow \quad \left(\frac{\mathrm{a}+\mathrm{b}}{4}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}+\mathrm{c}}{4}\right)\) \(\Rightarrow \quad \frac{\frac{\mathrm{a}+\mathrm{b}}{4}+\mathrm{c}}{4}=\frac{\mathrm{a}+\frac{\mathrm{b}+\mathrm{c}}{4}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}+4 \mathrm{c}}{16}=\frac{4 \mathrm{a}+\mathrm{b}+\mathrm{c}}{16}(\therefore \text { both side are not equal })\) Which is not correct. So, the operation * is commutative but not associative.
[Karnataka CET 2016]
Sets, Relation and Function
117179
Binary operation * on \(R-\{-1\}\) defined by \(a{ }^* \mathbf{b}\) \(\frac{\mathbf{a}}{\mathrm{b}+\mathbf{1}}\) is
1 * is associative and commutative
2 * is neither associative nor commutative
3 * is commutative but not associative
4 * is associative but not commutative
Explanation:
Exp: (B) : Given, Binary operation * on \(\mathrm{R}-\{-1\}\) defined by \(\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}}{\mathrm{b}+1}\) Then, we know that, a binary operation * on \(\mathrm{R}-\{-1\}\) is commutative if - \(\Rightarrow \frac{a}{b+1} \neq \frac{b}{a+1}\) Which is not true. And, A binary operation * on \(\mathrm{R}-\{-1\}\) is associative it \((\mathrm{a} * \mathrm{~b}) *(\mathrm{c})=(\mathrm{a}) *(\mathrm{~b} * \mathrm{c})\) \(\left(\frac{\mathrm{a}}{\mathrm{b}+1}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}}{\mathrm{c}+1}\right)\) \(\frac{\frac{\mathrm{a}}{\mathrm{b}+1}}{\mathrm{c}+1}=\frac{\mathrm{a}}{\frac{\mathrm{b}}{\mathrm{c}+1}+1}\) \(\frac{\mathrm{a}}{(\mathrm{b}+1)(\mathrm{c}+1)} \neq \frac{\mathrm{a}(\mathrm{c}+1)}{\mathrm{b}+\mathrm{c}+1}\) Which is not true. So, \({ }^*\) is neither associative nor commutative.
[Karnataka CET 2017]
Sets, Relation and Function
117180
In the group (Z, \(\left.{ }^*\right)\), if \(\mathbf{a}{ }^* \mathbf{b}=\mathbf{a}+\mathbf{b}-\mathbf{n} \forall \mathbf{a}, \mathbf{b} \in \mathbf{Z}\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
1 \(\mathrm{n}\)
2 \(-n\)
3 \(-3 n\)
4 \(3 \mathrm{n}\)
Explanation:
Exp: (D) : Given, In this the group \((\mathrm{Z}, *)\), If \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{n} \quad \forall \mathrm{a}, \mathrm{b} \in \mathrm{Z}\) Where, \(\mathrm{n}\) is a fixed integer. In this question, we can see that the identity element of the group is \(\mathrm{n}\). Then, \(\quad \mathrm{a} * \mathrm{n}=\mathrm{a} \quad \forall \mathrm{a} \in \mathrm{Z}\) So, let the inverse of \((-n)\) be \(Z\). Then, \(-\mathrm{n} * \mathrm{Z}=\) identity \(=\mathrm{n}\) \(-n+Z-n=n\) \(Z-2 n=n\) \(Z=3 n\)
[Karnataka CET 2013]
Sets, Relation and Function
117181
For any two real numbers, an operation * defined by \(a * b=1+a b\) is
1 neither commutative nor associative
2 commutative but not associative
3 both commutative and associative
4 associative but not commutative.
Explanation:
Exp: (B) : Given, For any two real number, an operation * defined by a * \(\mathrm{b}=1+\mathrm{ab}\). Then, A binary operation \(* \mathrm{R}\) is commutative if - \(a * b=b * a\) \(1+a b=1+b a\) \(1+a b=1+a b\) Which is true. And, A binary operation * on \(\mathrm{R}\) is associative if - \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})\) \((1+\mathrm{ab}) * \mathrm{c}=\mathrm{a} *(1+\mathrm{bc})\) \(1+(1+\mathrm{ab}) \mathrm{c}=1+\mathrm{a}(1+\mathrm{bc})\) \(1+\mathrm{c}+\mathrm{abc} \neq 1+\mathrm{a}+\mathrm{abc}\) Which is not true. So, \(*\) commutative but not associative.
117177
If the operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers ' \(a\) ' and ' \(b\) ', then \((2 \oplus 3)\) ๑ \(\mathbf{4}=\)
1 182
2 185
3 181
4 184
Explanation:
Exp: (B) : Given, The operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers 'a' and 'b'. Then, \((2 \oplus 3) \oplus 4=\left(2^2+3^2\right) \oplus 4\) \(=(4+9) \oplus 4\) \(=13 \oplus 4\) \(=13^2+4^2\) \(=169+16\)So, \((2 \oplus 3) \oplus 4=185\)
[Karnataka CET-2015]
Sets, Relation and Function
117178
Let * be a binary operation defined on \(\mathbf{R}\) by \(\mathbf{a}^*\) \(\mathbf{b}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{4}} \forall \mathbf{a}, \mathbf{b} \in \mathbf{R}\) then the operation * is
1 Commutative and Associative
2 Commutative but not Associative
3 Associative but not Commutative
4 Neither Associative nor Commutative
Explanation:
Exp: (B) : Given, Let, \(*\) be a binary operation defined on R by \(a * b=\) \(\frac{\mathrm{a}+\mathrm{b}}{4} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\). Then, we know that, a binary operation \(*\) on \(\mathrm{R}\) is commutative if - \(\mathrm{a} * \mathrm{~b} =\mathrm{b} * \mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4} \quad(\therefore \text { both side are equal })\) Which is true. And, \(A\) binary operation * on \(\mathrm{R}\) is associative if, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c}), \forall \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}\) \(\Rightarrow \quad \left(\frac{\mathrm{a}+\mathrm{b}}{4}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}+\mathrm{c}}{4}\right)\) \(\Rightarrow \quad \frac{\frac{\mathrm{a}+\mathrm{b}}{4}+\mathrm{c}}{4}=\frac{\mathrm{a}+\frac{\mathrm{b}+\mathrm{c}}{4}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}+4 \mathrm{c}}{16}=\frac{4 \mathrm{a}+\mathrm{b}+\mathrm{c}}{16}(\therefore \text { both side are not equal })\) Which is not correct. So, the operation * is commutative but not associative.
[Karnataka CET 2016]
Sets, Relation and Function
117179
Binary operation * on \(R-\{-1\}\) defined by \(a{ }^* \mathbf{b}\) \(\frac{\mathbf{a}}{\mathrm{b}+\mathbf{1}}\) is
1 * is associative and commutative
2 * is neither associative nor commutative
3 * is commutative but not associative
4 * is associative but not commutative
Explanation:
Exp: (B) : Given, Binary operation * on \(\mathrm{R}-\{-1\}\) defined by \(\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}}{\mathrm{b}+1}\) Then, we know that, a binary operation * on \(\mathrm{R}-\{-1\}\) is commutative if - \(\Rightarrow \frac{a}{b+1} \neq \frac{b}{a+1}\) Which is not true. And, A binary operation * on \(\mathrm{R}-\{-1\}\) is associative it \((\mathrm{a} * \mathrm{~b}) *(\mathrm{c})=(\mathrm{a}) *(\mathrm{~b} * \mathrm{c})\) \(\left(\frac{\mathrm{a}}{\mathrm{b}+1}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}}{\mathrm{c}+1}\right)\) \(\frac{\frac{\mathrm{a}}{\mathrm{b}+1}}{\mathrm{c}+1}=\frac{\mathrm{a}}{\frac{\mathrm{b}}{\mathrm{c}+1}+1}\) \(\frac{\mathrm{a}}{(\mathrm{b}+1)(\mathrm{c}+1)} \neq \frac{\mathrm{a}(\mathrm{c}+1)}{\mathrm{b}+\mathrm{c}+1}\) Which is not true. So, \({ }^*\) is neither associative nor commutative.
[Karnataka CET 2017]
Sets, Relation and Function
117180
In the group (Z, \(\left.{ }^*\right)\), if \(\mathbf{a}{ }^* \mathbf{b}=\mathbf{a}+\mathbf{b}-\mathbf{n} \forall \mathbf{a}, \mathbf{b} \in \mathbf{Z}\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
1 \(\mathrm{n}\)
2 \(-n\)
3 \(-3 n\)
4 \(3 \mathrm{n}\)
Explanation:
Exp: (D) : Given, In this the group \((\mathrm{Z}, *)\), If \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{n} \quad \forall \mathrm{a}, \mathrm{b} \in \mathrm{Z}\) Where, \(\mathrm{n}\) is a fixed integer. In this question, we can see that the identity element of the group is \(\mathrm{n}\). Then, \(\quad \mathrm{a} * \mathrm{n}=\mathrm{a} \quad \forall \mathrm{a} \in \mathrm{Z}\) So, let the inverse of \((-n)\) be \(Z\). Then, \(-\mathrm{n} * \mathrm{Z}=\) identity \(=\mathrm{n}\) \(-n+Z-n=n\) \(Z-2 n=n\) \(Z=3 n\)
[Karnataka CET 2013]
Sets, Relation and Function
117181
For any two real numbers, an operation * defined by \(a * b=1+a b\) is
1 neither commutative nor associative
2 commutative but not associative
3 both commutative and associative
4 associative but not commutative.
Explanation:
Exp: (B) : Given, For any two real number, an operation * defined by a * \(\mathrm{b}=1+\mathrm{ab}\). Then, A binary operation \(* \mathrm{R}\) is commutative if - \(a * b=b * a\) \(1+a b=1+b a\) \(1+a b=1+a b\) Which is true. And, A binary operation * on \(\mathrm{R}\) is associative if - \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})\) \((1+\mathrm{ab}) * \mathrm{c}=\mathrm{a} *(1+\mathrm{bc})\) \(1+(1+\mathrm{ab}) \mathrm{c}=1+\mathrm{a}(1+\mathrm{bc})\) \(1+\mathrm{c}+\mathrm{abc} \neq 1+\mathrm{a}+\mathrm{abc}\) Which is not true. So, \(*\) commutative but not associative.
117177
If the operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers ' \(a\) ' and ' \(b\) ', then \((2 \oplus 3)\) ๑ \(\mathbf{4}=\)
1 182
2 185
3 181
4 184
Explanation:
Exp: (B) : Given, The operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers 'a' and 'b'. Then, \((2 \oplus 3) \oplus 4=\left(2^2+3^2\right) \oplus 4\) \(=(4+9) \oplus 4\) \(=13 \oplus 4\) \(=13^2+4^2\) \(=169+16\)So, \((2 \oplus 3) \oplus 4=185\)
[Karnataka CET-2015]
Sets, Relation and Function
117178
Let * be a binary operation defined on \(\mathbf{R}\) by \(\mathbf{a}^*\) \(\mathbf{b}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{4}} \forall \mathbf{a}, \mathbf{b} \in \mathbf{R}\) then the operation * is
1 Commutative and Associative
2 Commutative but not Associative
3 Associative but not Commutative
4 Neither Associative nor Commutative
Explanation:
Exp: (B) : Given, Let, \(*\) be a binary operation defined on R by \(a * b=\) \(\frac{\mathrm{a}+\mathrm{b}}{4} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\). Then, we know that, a binary operation \(*\) on \(\mathrm{R}\) is commutative if - \(\mathrm{a} * \mathrm{~b} =\mathrm{b} * \mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4} \quad(\therefore \text { both side are equal })\) Which is true. And, \(A\) binary operation * on \(\mathrm{R}\) is associative if, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c}), \forall \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}\) \(\Rightarrow \quad \left(\frac{\mathrm{a}+\mathrm{b}}{4}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}+\mathrm{c}}{4}\right)\) \(\Rightarrow \quad \frac{\frac{\mathrm{a}+\mathrm{b}}{4}+\mathrm{c}}{4}=\frac{\mathrm{a}+\frac{\mathrm{b}+\mathrm{c}}{4}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}+4 \mathrm{c}}{16}=\frac{4 \mathrm{a}+\mathrm{b}+\mathrm{c}}{16}(\therefore \text { both side are not equal })\) Which is not correct. So, the operation * is commutative but not associative.
[Karnataka CET 2016]
Sets, Relation and Function
117179
Binary operation * on \(R-\{-1\}\) defined by \(a{ }^* \mathbf{b}\) \(\frac{\mathbf{a}}{\mathrm{b}+\mathbf{1}}\) is
1 * is associative and commutative
2 * is neither associative nor commutative
3 * is commutative but not associative
4 * is associative but not commutative
Explanation:
Exp: (B) : Given, Binary operation * on \(\mathrm{R}-\{-1\}\) defined by \(\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}}{\mathrm{b}+1}\) Then, we know that, a binary operation * on \(\mathrm{R}-\{-1\}\) is commutative if - \(\Rightarrow \frac{a}{b+1} \neq \frac{b}{a+1}\) Which is not true. And, A binary operation * on \(\mathrm{R}-\{-1\}\) is associative it \((\mathrm{a} * \mathrm{~b}) *(\mathrm{c})=(\mathrm{a}) *(\mathrm{~b} * \mathrm{c})\) \(\left(\frac{\mathrm{a}}{\mathrm{b}+1}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}}{\mathrm{c}+1}\right)\) \(\frac{\frac{\mathrm{a}}{\mathrm{b}+1}}{\mathrm{c}+1}=\frac{\mathrm{a}}{\frac{\mathrm{b}}{\mathrm{c}+1}+1}\) \(\frac{\mathrm{a}}{(\mathrm{b}+1)(\mathrm{c}+1)} \neq \frac{\mathrm{a}(\mathrm{c}+1)}{\mathrm{b}+\mathrm{c}+1}\) Which is not true. So, \({ }^*\) is neither associative nor commutative.
[Karnataka CET 2017]
Sets, Relation and Function
117180
In the group (Z, \(\left.{ }^*\right)\), if \(\mathbf{a}{ }^* \mathbf{b}=\mathbf{a}+\mathbf{b}-\mathbf{n} \forall \mathbf{a}, \mathbf{b} \in \mathbf{Z}\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
1 \(\mathrm{n}\)
2 \(-n\)
3 \(-3 n\)
4 \(3 \mathrm{n}\)
Explanation:
Exp: (D) : Given, In this the group \((\mathrm{Z}, *)\), If \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{n} \quad \forall \mathrm{a}, \mathrm{b} \in \mathrm{Z}\) Where, \(\mathrm{n}\) is a fixed integer. In this question, we can see that the identity element of the group is \(\mathrm{n}\). Then, \(\quad \mathrm{a} * \mathrm{n}=\mathrm{a} \quad \forall \mathrm{a} \in \mathrm{Z}\) So, let the inverse of \((-n)\) be \(Z\). Then, \(-\mathrm{n} * \mathrm{Z}=\) identity \(=\mathrm{n}\) \(-n+Z-n=n\) \(Z-2 n=n\) \(Z=3 n\)
[Karnataka CET 2013]
Sets, Relation and Function
117181
For any two real numbers, an operation * defined by \(a * b=1+a b\) is
1 neither commutative nor associative
2 commutative but not associative
3 both commutative and associative
4 associative but not commutative.
Explanation:
Exp: (B) : Given, For any two real number, an operation * defined by a * \(\mathrm{b}=1+\mathrm{ab}\). Then, A binary operation \(* \mathrm{R}\) is commutative if - \(a * b=b * a\) \(1+a b=1+b a\) \(1+a b=1+a b\) Which is true. And, A binary operation * on \(\mathrm{R}\) is associative if - \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})\) \((1+\mathrm{ab}) * \mathrm{c}=\mathrm{a} *(1+\mathrm{bc})\) \(1+(1+\mathrm{ab}) \mathrm{c}=1+\mathrm{a}(1+\mathrm{bc})\) \(1+\mathrm{c}+\mathrm{abc} \neq 1+\mathrm{a}+\mathrm{abc}\) Which is not true. So, \(*\) commutative but not associative.
117177
If the operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers ' \(a\) ' and ' \(b\) ', then \((2 \oplus 3)\) ๑ \(\mathbf{4}=\)
1 182
2 185
3 181
4 184
Explanation:
Exp: (B) : Given, The operation \(\oplus\) is defined by \(a \oplus b=a^2+b^2\) for all real numbers 'a' and 'b'. Then, \((2 \oplus 3) \oplus 4=\left(2^2+3^2\right) \oplus 4\) \(=(4+9) \oplus 4\) \(=13 \oplus 4\) \(=13^2+4^2\) \(=169+16\)So, \((2 \oplus 3) \oplus 4=185\)
[Karnataka CET-2015]
Sets, Relation and Function
117178
Let * be a binary operation defined on \(\mathbf{R}\) by \(\mathbf{a}^*\) \(\mathbf{b}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{4}} \forall \mathbf{a}, \mathbf{b} \in \mathbf{R}\) then the operation * is
1 Commutative and Associative
2 Commutative but not Associative
3 Associative but not Commutative
4 Neither Associative nor Commutative
Explanation:
Exp: (B) : Given, Let, \(*\) be a binary operation defined on R by \(a * b=\) \(\frac{\mathrm{a}+\mathrm{b}}{4} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\). Then, we know that, a binary operation \(*\) on \(\mathrm{R}\) is commutative if - \(\mathrm{a} * \mathrm{~b} =\mathrm{b} * \mathrm{a}, \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}}{4} =\frac{\mathrm{b}+\mathrm{a}}{4} \quad(\therefore \text { both side are equal })\) Which is true. And, \(A\) binary operation * on \(\mathrm{R}\) is associative if, \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c}), \forall \mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{R}\) \(\Rightarrow \quad \left(\frac{\mathrm{a}+\mathrm{b}}{4}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}+\mathrm{c}}{4}\right)\) \(\Rightarrow \quad \frac{\frac{\mathrm{a}+\mathrm{b}}{4}+\mathrm{c}}{4}=\frac{\mathrm{a}+\frac{\mathrm{b}+\mathrm{c}}{4}}{4}\) \(\Rightarrow \quad \frac{\mathrm{a}+\mathrm{b}+4 \mathrm{c}}{16}=\frac{4 \mathrm{a}+\mathrm{b}+\mathrm{c}}{16}(\therefore \text { both side are not equal })\) Which is not correct. So, the operation * is commutative but not associative.
[Karnataka CET 2016]
Sets, Relation and Function
117179
Binary operation * on \(R-\{-1\}\) defined by \(a{ }^* \mathbf{b}\) \(\frac{\mathbf{a}}{\mathrm{b}+\mathbf{1}}\) is
1 * is associative and commutative
2 * is neither associative nor commutative
3 * is commutative but not associative
4 * is associative but not commutative
Explanation:
Exp: (B) : Given, Binary operation * on \(\mathrm{R}-\{-1\}\) defined by \(\mathrm{a} * \mathrm{~b}=\frac{\mathrm{a}}{\mathrm{b}+1}\) Then, we know that, a binary operation * on \(\mathrm{R}-\{-1\}\) is commutative if - \(\Rightarrow \frac{a}{b+1} \neq \frac{b}{a+1}\) Which is not true. And, A binary operation * on \(\mathrm{R}-\{-1\}\) is associative it \((\mathrm{a} * \mathrm{~b}) *(\mathrm{c})=(\mathrm{a}) *(\mathrm{~b} * \mathrm{c})\) \(\left(\frac{\mathrm{a}}{\mathrm{b}+1}\right) * \mathrm{c}=\mathrm{a} *\left(\frac{\mathrm{b}}{\mathrm{c}+1}\right)\) \(\frac{\frac{\mathrm{a}}{\mathrm{b}+1}}{\mathrm{c}+1}=\frac{\mathrm{a}}{\frac{\mathrm{b}}{\mathrm{c}+1}+1}\) \(\frac{\mathrm{a}}{(\mathrm{b}+1)(\mathrm{c}+1)} \neq \frac{\mathrm{a}(\mathrm{c}+1)}{\mathrm{b}+\mathrm{c}+1}\) Which is not true. So, \({ }^*\) is neither associative nor commutative.
[Karnataka CET 2017]
Sets, Relation and Function
117180
In the group (Z, \(\left.{ }^*\right)\), if \(\mathbf{a}{ }^* \mathbf{b}=\mathbf{a}+\mathbf{b}-\mathbf{n} \forall \mathbf{a}, \mathbf{b} \in \mathbf{Z}\) where \(n\) is a fixed integer, then the inverse of \((-n)\) is
1 \(\mathrm{n}\)
2 \(-n\)
3 \(-3 n\)
4 \(3 \mathrm{n}\)
Explanation:
Exp: (D) : Given, In this the group \((\mathrm{Z}, *)\), If \(\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{n} \quad \forall \mathrm{a}, \mathrm{b} \in \mathrm{Z}\) Where, \(\mathrm{n}\) is a fixed integer. In this question, we can see that the identity element of the group is \(\mathrm{n}\). Then, \(\quad \mathrm{a} * \mathrm{n}=\mathrm{a} \quad \forall \mathrm{a} \in \mathrm{Z}\) So, let the inverse of \((-n)\) be \(Z\). Then, \(-\mathrm{n} * \mathrm{Z}=\) identity \(=\mathrm{n}\) \(-n+Z-n=n\) \(Z-2 n=n\) \(Z=3 n\)
[Karnataka CET 2013]
Sets, Relation and Function
117181
For any two real numbers, an operation * defined by \(a * b=1+a b\) is
1 neither commutative nor associative
2 commutative but not associative
3 both commutative and associative
4 associative but not commutative.
Explanation:
Exp: (B) : Given, For any two real number, an operation * defined by a * \(\mathrm{b}=1+\mathrm{ab}\). Then, A binary operation \(* \mathrm{R}\) is commutative if - \(a * b=b * a\) \(1+a b=1+b a\) \(1+a b=1+a b\) Which is true. And, A binary operation * on \(\mathrm{R}\) is associative if - \((\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a} *(\mathrm{~b} * \mathrm{c})\) \((1+\mathrm{ab}) * \mathrm{c}=\mathrm{a} *(1+\mathrm{bc})\) \(1+(1+\mathrm{ab}) \mathrm{c}=1+\mathrm{a}(1+\mathrm{bc})\) \(1+\mathrm{c}+\mathrm{abc} \neq 1+\mathrm{a}+\mathrm{abc}\) Which is not true. So, \(*\) commutative but not associative.
