For the binary operation * defined on `R-{1}` by the rule `a`*`b=a+b+a b` for all `a ,\ b in R-{1}`, the inverse of `a` is

Options

A. `a`
B. `-a/(a+1)`
C. `1/a`
D. `a^2`

Explanation

To find the inverse of the binary operation defined on \( R - \{1\} \) by the rule \( a * b = a + b + ab \), we will follow these steps:

### Step 1: Identify the Identity Element
The identity element \( e \) for the operation \( * \) must satisfy the condition:
\[ a * e = a \]
for all \( a \in R - \{1\} \).

Substituting \( b \) with \( e \) in the operation:
\[ a * e = a + e + ae \]
Setting this equal to \( a \):
\[ a + e + ae = a \]

### Step 2: Simplify the Equation
Subtract \( a \) from both sides:
\[ e + ae = 0 \]
Factoring out \( e \):
\[ e(1 + a) = 0 \]

### Step 3: Solve for the Identity Element
Since \( a \) can be any number in \( R - \{1\} \), \( 1 + a \) is never zero. Therefore, we must have:
\[ e = 0 \]
Thus, the identity element \( e \) is \( 0 \).

### Step 4: Find the Inverse
The inverse \( x \) of \( a \) must satisfy:
\[ a * x = e \]
Substituting \( e = 0 \):
\[ a * x = 0 \]
Using the operation definition:
\[ a + x + ax = 0 \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ x + ax = -a \]
Factoring out \( x \):
\[ x(1 + a) = -a \]

### Step 6: Solve for the Inverse
Dividing both sides by \( 1 + a \) (which is non-zero since \( a \neq -1 \)):
\[ x = \frac{-a}{1 + a} \]

### Conclusion
The inverse of \( a \) under the operation \( * \) is:
\[ \boxed{\frac{-a}{1 + a}} \]