QBManager

Sets, Relation and Function

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 $4 =$

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, $(2 \oplus 3) \oplus 4 = (2^{2} + 3^{2}) \oplus 4$
$= (4 + 9) \oplus 4$
$= 13 \oplus 4$
$= 13^{2} + 4^{2}$
$= 169 + 16$ So, $(2 \oplus 3) \oplus 4 = 185$

[Karnataka CET-2015]

Sets, Relation and Function

117178 Let * be a binary operation defined on $R$ by $a^{*}$ $b = \frac{a + b}{4} \forall a, b \in R$ 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 $a * b =$ $\frac{a + b}{4} \forall a, b \in R$ .
Then, we know that, a binary operation $*$ on $R$ is commutative if -
$a * b = b * a, \forall a, b \in R$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4}$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4} (∴ both side are equal)$
Which is true.
And, $A$ binary operation * on $R$ is associative if,
$(a * b) * c = a * (b * c), \forall a, b, c \in R$
$\Rightarrow (\frac{a + b}{4}) * c = a * (\frac{b + c}{4})$
$\Rightarrow \frac{\frac{a + b}{4} + c}{4} = \frac{a + \frac{b + c}{4}}{4}$
$\Rightarrow \frac{a + b + 4 c}{16} = \frac{4 a + b + c}{16} (∴ both side are not equal)$
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 $a^{*} b$ $\frac{a}{b + 1}$ 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 $R - {- 1}$ defined by $a * b = \frac{a}{b + 1}$
Then, we know that, a binary operation * on $R - {- 1}$ is commutative if -
$\Rightarrow \frac{a}{b + 1} \neq \frac{b}{a + 1}$
Which is not true.
And, A binary operation * on $R - {- 1}$ is associative it
$(a * b) * (c) = (a) * (b * c)$
$(\frac{a}{b + 1}) * c = a * (\frac{b}{c + 1})$
$\frac{\frac{a}{b + 1}}{c + 1} = \frac{a}{\frac{b}{c + 1} + 1}$
$\frac{a}{(b + 1) (c + 1)} \neq \frac{a (c + 1)}{b + c + 1}$
Which is not true.
So, $^{*}$ is neither associative nor commutative.

[Karnataka CET 2017]

Sets, Relation and Function

117180 In the group (Z, $^{*})$ , if $a^{*} b = a + b - n \forall a, b \in Z$ where $n$ is a fixed integer, then the inverse of $(- n)$ is

1

n

2

- n

3

- 3 n

4

3 n

Explanation:

Exp: (D) : Given,
In this the group $(Z, *)$ ,
If $a * b = a + b - n \forall a, b \in Z$
Where, $n$ is a fixed integer.
In this question, we can see that the identity element of the group is $n$ .
Then, $a * n = a \forall a \in Z$
So, let the inverse of $(- n)$ be $Z$ .
Then, $- n * Z =$ identity $= n$
$- n + Z - n = n$
$Z - 2 n = n$
$Z = 3 n$

[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 * $b = 1 + ab$ .
Then, A binary operation $* R$ 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 $R$ is associative if -
$(a * b) * c = a * (b * c)$
$(1 + ab) * c = a * (1 + bc)$
$1 + (1 + ab) c = 1 + a (1 + bc)$
$1 + c + abc \neq 1 + a + abc$
Which is not true.
So, $*$ commutative but not associative.

[Karnataka CET 2014]

Sets, Relation and Function

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 $4 =$

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, $(2 \oplus 3) \oplus 4 = (2^{2} + 3^{2}) \oplus 4$
$= (4 + 9) \oplus 4$
$= 13 \oplus 4$
$= 13^{2} + 4^{2}$
$= 169 + 16$ So, $(2 \oplus 3) \oplus 4 = 185$

[Karnataka CET-2015]

Sets, Relation and Function

117178 Let * be a binary operation defined on $R$ by $a^{*}$ $b = \frac{a + b}{4} \forall a, b \in R$ 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 $a * b =$ $\frac{a + b}{4} \forall a, b \in R$ .
Then, we know that, a binary operation $*$ on $R$ is commutative if -
$a * b = b * a, \forall a, b \in R$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4}$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4} (∴ both side are equal)$
Which is true.
And, $A$ binary operation * on $R$ is associative if,
$(a * b) * c = a * (b * c), \forall a, b, c \in R$
$\Rightarrow (\frac{a + b}{4}) * c = a * (\frac{b + c}{4})$
$\Rightarrow \frac{\frac{a + b}{4} + c}{4} = \frac{a + \frac{b + c}{4}}{4}$
$\Rightarrow \frac{a + b + 4 c}{16} = \frac{4 a + b + c}{16} (∴ both side are not equal)$
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 $a^{*} b$ $\frac{a}{b + 1}$ 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 $R - {- 1}$ defined by $a * b = \frac{a}{b + 1}$
Then, we know that, a binary operation * on $R - {- 1}$ is commutative if -
$\Rightarrow \frac{a}{b + 1} \neq \frac{b}{a + 1}$
Which is not true.
And, A binary operation * on $R - {- 1}$ is associative it
$(a * b) * (c) = (a) * (b * c)$
$(\frac{a}{b + 1}) * c = a * (\frac{b}{c + 1})$
$\frac{\frac{a}{b + 1}}{c + 1} = \frac{a}{\frac{b}{c + 1} + 1}$
$\frac{a}{(b + 1) (c + 1)} \neq \frac{a (c + 1)}{b + c + 1}$
Which is not true.
So, $^{*}$ is neither associative nor commutative.

[Karnataka CET 2017]

Sets, Relation and Function

117180 In the group (Z, $^{*})$ , if $a^{*} b = a + b - n \forall a, b \in Z$ where $n$ is a fixed integer, then the inverse of $(- n)$ is

1

n

2

- n

3

- 3 n

4

3 n

Explanation:

Exp: (D) : Given,
In this the group $(Z, *)$ ,
If $a * b = a + b - n \forall a, b \in Z$
Where, $n$ is a fixed integer.
In this question, we can see that the identity element of the group is $n$ .
Then, $a * n = a \forall a \in Z$
So, let the inverse of $(- n)$ be $Z$ .
Then, $- n * Z =$ identity $= n$
$- n + Z - n = n$
$Z - 2 n = n$
$Z = 3 n$

[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 * $b = 1 + ab$ .
Then, A binary operation $* R$ 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 $R$ is associative if -
$(a * b) * c = a * (b * c)$
$(1 + ab) * c = a * (1 + bc)$
$1 + (1 + ab) c = 1 + a (1 + bc)$
$1 + c + abc \neq 1 + a + abc$
Which is not true.
So, $*$ commutative but not associative.

[Karnataka CET 2014]

Sets, Relation and Function

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 $4 =$

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, $(2 \oplus 3) \oplus 4 = (2^{2} + 3^{2}) \oplus 4$
$= (4 + 9) \oplus 4$
$= 13 \oplus 4$
$= 13^{2} + 4^{2}$
$= 169 + 16$ So, $(2 \oplus 3) \oplus 4 = 185$

[Karnataka CET-2015]

Sets, Relation and Function

117178 Let * be a binary operation defined on $R$ by $a^{*}$ $b = \frac{a + b}{4} \forall a, b \in R$ 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 $a * b =$ $\frac{a + b}{4} \forall a, b \in R$ .
Then, we know that, a binary operation $*$ on $R$ is commutative if -
$a * b = b * a, \forall a, b \in R$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4}$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4} (∴ both side are equal)$
Which is true.
And, $A$ binary operation * on $R$ is associative if,
$(a * b) * c = a * (b * c), \forall a, b, c \in R$
$\Rightarrow (\frac{a + b}{4}) * c = a * (\frac{b + c}{4})$
$\Rightarrow \frac{\frac{a + b}{4} + c}{4} = \frac{a + \frac{b + c}{4}}{4}$
$\Rightarrow \frac{a + b + 4 c}{16} = \frac{4 a + b + c}{16} (∴ both side are not equal)$
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 $a^{*} b$ $\frac{a}{b + 1}$ 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 $R - {- 1}$ defined by $a * b = \frac{a}{b + 1}$
Then, we know that, a binary operation * on $R - {- 1}$ is commutative if -
$\Rightarrow \frac{a}{b + 1} \neq \frac{b}{a + 1}$
Which is not true.
And, A binary operation * on $R - {- 1}$ is associative it
$(a * b) * (c) = (a) * (b * c)$
$(\frac{a}{b + 1}) * c = a * (\frac{b}{c + 1})$
$\frac{\frac{a}{b + 1}}{c + 1} = \frac{a}{\frac{b}{c + 1} + 1}$
$\frac{a}{(b + 1) (c + 1)} \neq \frac{a (c + 1)}{b + c + 1}$
Which is not true.
So, $^{*}$ is neither associative nor commutative.

[Karnataka CET 2017]

Sets, Relation and Function

117180 In the group (Z, $^{*})$ , if $a^{*} b = a + b - n \forall a, b \in Z$ where $n$ is a fixed integer, then the inverse of $(- n)$ is

1

n

2

- n

3

- 3 n

4

3 n

Explanation:

Exp: (D) : Given,
In this the group $(Z, *)$ ,
If $a * b = a + b - n \forall a, b \in Z$
Where, $n$ is a fixed integer.
In this question, we can see that the identity element of the group is $n$ .
Then, $a * n = a \forall a \in Z$
So, let the inverse of $(- n)$ be $Z$ .
Then, $- n * Z =$ identity $= n$
$- n + Z - n = n$
$Z - 2 n = n$
$Z = 3 n$

[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 * $b = 1 + ab$ .
Then, A binary operation $* R$ 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 $R$ is associative if -
$(a * b) * c = a * (b * c)$
$(1 + ab) * c = a * (1 + bc)$
$1 + (1 + ab) c = 1 + a (1 + bc)$
$1 + c + abc \neq 1 + a + abc$
Which is not true.
So, $*$ commutative but not associative.

[Karnataka CET 2014]

Sets, Relation and Function

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 $4 =$

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, $(2 \oplus 3) \oplus 4 = (2^{2} + 3^{2}) \oplus 4$
$= (4 + 9) \oplus 4$
$= 13 \oplus 4$
$= 13^{2} + 4^{2}$
$= 169 + 16$ So, $(2 \oplus 3) \oplus 4 = 185$

[Karnataka CET-2015]

Sets, Relation and Function

117178 Let * be a binary operation defined on $R$ by $a^{*}$ $b = \frac{a + b}{4} \forall a, b \in R$ 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 $a * b =$ $\frac{a + b}{4} \forall a, b \in R$ .
Then, we know that, a binary operation $*$ on $R$ is commutative if -
$a * b = b * a, \forall a, b \in R$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4}$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4} (∴ both side are equal)$
Which is true.
And, $A$ binary operation * on $R$ is associative if,
$(a * b) * c = a * (b * c), \forall a, b, c \in R$
$\Rightarrow (\frac{a + b}{4}) * c = a * (\frac{b + c}{4})$
$\Rightarrow \frac{\frac{a + b}{4} + c}{4} = \frac{a + \frac{b + c}{4}}{4}$
$\Rightarrow \frac{a + b + 4 c}{16} = \frac{4 a + b + c}{16} (∴ both side are not equal)$
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 $a^{*} b$ $\frac{a}{b + 1}$ 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 $R - {- 1}$ defined by $a * b = \frac{a}{b + 1}$
Then, we know that, a binary operation * on $R - {- 1}$ is commutative if -
$\Rightarrow \frac{a}{b + 1} \neq \frac{b}{a + 1}$
Which is not true.
And, A binary operation * on $R - {- 1}$ is associative it
$(a * b) * (c) = (a) * (b * c)$
$(\frac{a}{b + 1}) * c = a * (\frac{b}{c + 1})$
$\frac{\frac{a}{b + 1}}{c + 1} = \frac{a}{\frac{b}{c + 1} + 1}$
$\frac{a}{(b + 1) (c + 1)} \neq \frac{a (c + 1)}{b + c + 1}$
Which is not true.
So, $^{*}$ is neither associative nor commutative.

[Karnataka CET 2017]

Sets, Relation and Function

117180 In the group (Z, $^{*})$ , if $a^{*} b = a + b - n \forall a, b \in Z$ where $n$ is a fixed integer, then the inverse of $(- n)$ is

1

n

2

- n

3

- 3 n

4

3 n

Explanation:

Exp: (D) : Given,
In this the group $(Z, *)$ ,
If $a * b = a + b - n \forall a, b \in Z$
Where, $n$ is a fixed integer.
In this question, we can see that the identity element of the group is $n$ .
Then, $a * n = a \forall a \in Z$
So, let the inverse of $(- n)$ be $Z$ .
Then, $- n * Z =$ identity $= n$
$- n + Z - n = n$
$Z - 2 n = n$
$Z = 3 n$

[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 * $b = 1 + ab$ .
Then, A binary operation $* R$ 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 $R$ is associative if -
$(a * b) * c = a * (b * c)$
$(1 + ab) * c = a * (1 + bc)$
$1 + (1 + ab) c = 1 + a (1 + bc)$
$1 + c + abc \neq 1 + a + abc$
Which is not true.
So, $*$ commutative but not associative.

[Karnataka CET 2014]

Sets, Relation and Function

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 $4 =$

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, $(2 \oplus 3) \oplus 4 = (2^{2} + 3^{2}) \oplus 4$
$= (4 + 9) \oplus 4$
$= 13 \oplus 4$
$= 13^{2} + 4^{2}$
$= 169 + 16$ So, $(2 \oplus 3) \oplus 4 = 185$

[Karnataka CET-2015]

Sets, Relation and Function

117178 Let * be a binary operation defined on $R$ by $a^{*}$ $b = \frac{a + b}{4} \forall a, b \in R$ 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 $a * b =$ $\frac{a + b}{4} \forall a, b \in R$ .
Then, we know that, a binary operation $*$ on $R$ is commutative if -
$a * b = b * a, \forall a, b \in R$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4}$
$\Rightarrow \frac{a + b}{4} = \frac{b + a}{4} (∴ both side are equal)$
Which is true.
And, $A$ binary operation * on $R$ is associative if,
$(a * b) * c = a * (b * c), \forall a, b, c \in R$
$\Rightarrow (\frac{a + b}{4}) * c = a * (\frac{b + c}{4})$
$\Rightarrow \frac{\frac{a + b}{4} + c}{4} = \frac{a + \frac{b + c}{4}}{4}$
$\Rightarrow \frac{a + b + 4 c}{16} = \frac{4 a + b + c}{16} (∴ both side are not equal)$
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 $a^{*} b$ $\frac{a}{b + 1}$ 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 $R - {- 1}$ defined by $a * b = \frac{a}{b + 1}$
Then, we know that, a binary operation * on $R - {- 1}$ is commutative if -
$\Rightarrow \frac{a}{b + 1} \neq \frac{b}{a + 1}$
Which is not true.
And, A binary operation * on $R - {- 1}$ is associative it
$(a * b) * (c) = (a) * (b * c)$
$(\frac{a}{b + 1}) * c = a * (\frac{b}{c + 1})$
$\frac{\frac{a}{b + 1}}{c + 1} = \frac{a}{\frac{b}{c + 1} + 1}$
$\frac{a}{(b + 1) (c + 1)} \neq \frac{a (c + 1)}{b + c + 1}$
Which is not true.
So, $^{*}$ is neither associative nor commutative.

[Karnataka CET 2017]

Sets, Relation and Function

117180 In the group (Z, $^{*})$ , if $a^{*} b = a + b - n \forall a, b \in Z$ where $n$ is a fixed integer, then the inverse of $(- n)$ is

1

n

2

- n

3

- 3 n

4

3 n

Explanation:

Exp: (D) : Given,
In this the group $(Z, *)$ ,
If $a * b = a + b - n \forall a, b \in Z$
Where, $n$ is a fixed integer.
In this question, we can see that the identity element of the group is $n$ .
Then, $a * n = a \forall a \in Z$
So, let the inverse of $(- n)$ be $Z$ .
Then, $- n * Z =$ identity $= n$
$- n + Z - n = n$
$Z - 2 n = n$
$Z = 3 n$

[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 * $b = 1 + ab$ .
Then, A binary operation $* R$ 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 $R$ is associative if -
$(a * b) * c = a * (b * c)$
$(1 + ab) * c = a * (1 + bc)$
$1 + (1 + ab) c = 1 + a (1 + bc)$
$1 + c + abc \neq 1 + a + abc$
Which is not true.
So, $*$ commutative but not associative.

[Karnataka CET 2014]

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