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# For any two-real numbers, an operation defined by $a * b{\text{ }} = \;1 + ab$ is  $\left( A \right)$. 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  Verified
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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.