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$\left( A \right)$. Commutative but not associative

$\left( B \right)$. Associative but not commutative

$\left( C \right)$. Neither commutative nor associative

$\left( D \right)$. Both commutative and associative

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Hint: Use commutative and associative property for the given operation.

We have been given the operator \[ * \] such that:

\[{\text{ }}a * b = 1 + ab{\text{ (1) ; }}a,b{\text{ }} \in {\text{R}}\]

Since \[(1 + ab){\text{ }}\]also belongs to \[R{\text{ }}\](Real Numbers Set),

Operator \[ * \] satisfies closure property

\[a * b\] is a binary operation.

For binary operation to be commutative, we would have the following condition:

\[a * b = b * a {\text{(2)}}\]

We need to check condition (2) for operation (1)

\[

a * b = 1 + ab \\

b * a = 1 + ba \\

\]

Since multiplication operator is commutative, we have

\[

ab = ba \\

\Rightarrow a * b = 1 + ab = 1 + ba = b * a \\

\]

Hence condition (2) is satisfied.

Therefore, operation (1) is commutative.

For binary operation to be associative, we would have the following condition:

\[a * \left( {b * c} \right) = \left( {a * b} \right) * c{\text{ (3)}}\]

We need to check for condition (3) for operator (1)

\[

a * \left( {b * c} \right) = a * \left( {1 + bc} \right) = 1 + a(1 + bc) = 1 + a + abc \\

\left( {a * b} \right) * c = \left( {1 + ab} \right) * c = 1 + \left( {1 + ab} \right)c = 1 + c + abc \\

\]

Since \[1 + a + abc \ne 1 + c + abc\], condition (3) is not satisfied.

Therefore, operation (1) is not associative.

Hence the correct option is $\left( A \right)$. Commutative but not associative.

Note: Always try to remember the basic conditions for associativity and commutativity. Commutativity does not imply associativity.

We have been given the operator \[ * \] such that:

\[{\text{ }}a * b = 1 + ab{\text{ (1) ; }}a,b{\text{ }} \in {\text{R}}\]

Since \[(1 + ab){\text{ }}\]also belongs to \[R{\text{ }}\](Real Numbers Set),

Operator \[ * \] satisfies closure property

\[a * b\] is a binary operation.

For binary operation to be commutative, we would have the following condition:

\[a * b = b * a {\text{(2)}}\]

We need to check condition (2) for operation (1)

\[

a * b = 1 + ab \\

b * a = 1 + ba \\

\]

Since multiplication operator is commutative, we have

\[

ab = ba \\

\Rightarrow a * b = 1 + ab = 1 + ba = b * a \\

\]

Hence condition (2) is satisfied.

Therefore, operation (1) is commutative.

For binary operation to be associative, we would have the following condition:

\[a * \left( {b * c} \right) = \left( {a * b} \right) * c{\text{ (3)}}\]

We need to check for condition (3) for operator (1)

\[

a * \left( {b * c} \right) = a * \left( {1 + bc} \right) = 1 + a(1 + bc) = 1 + a + abc \\

\left( {a * b} \right) * c = \left( {1 + ab} \right) * c = 1 + \left( {1 + ab} \right)c = 1 + c + abc \\

\]

Since \[1 + a + abc \ne 1 + c + abc\], condition (3) is not satisfied.

Therefore, operation (1) is not associative.

Hence the correct option is $\left( A \right)$. Commutative but not associative.

Note: Always try to remember the basic conditions for associativity and commutativity. Commutativity does not imply associativity.