Question

For any two-real numbers, an operation defined by $$a\ast b=1+ab$$ 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

Answer 1

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

We have been given the operator $$\ast$$ such that:

$$a\ast b=1+ab\text{ (1) ; }a,b\in \text{R}$$
Since $$(1+ab)$$also belongs to $$R$$(Real Numbers Set),
Operator $$\ast$$ satisfies closure property
$$a\ast b$$ is a binary operation.

For binary operation to be commutative, we would have the following condition:
$$a\ast b=b\ast a\text{(2)}$$

We need to check condition (2) for operation (1)
$$\begin{gathered} a\ast b=1+ab \\ b\ast a=1+ba \end{gathered}$$

Since multiplication operator is commutative, we have
$$\begin{gathered} ab=ba \\ \Rightarrow a\ast b=1+ab=1+ba=b\ast a \end{gathered}$$

Hence condition (2) is satisfied.
Therefore, operation (1) is commutative.
For binary operation to be associative, we would have the following condition:
$$a\ast \left(b\ast c\right)=\left(a\ast b\right)\ast c\text{ (3)}$$
We need to check for condition (3) for operator (1)
$$\begin{gathered} a\ast \left(b\ast c\right)=a\ast \left(1+bc\right)=1+a(1+bc)=1+a+abc \\ \left(a\ast b\right)\ast c=\left(1+ab\right)\ast c=1+\left(1+ab\right)c=1+c+abc \end{gathered}$$
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.