The values of a and b for which the map `f: R to R`, given by f(x)=ax+b`(a,b in R)` is a bijection with fof as indentity function, are

Options

A. `a=1, bin R`
B. `(a=1, b=0) or, (a=-1, b in R)`
C. `a=pm1, b in R`
D. `a= pm 1, b=0`

Explanation

To determine the values of \( a \) and \( b \) for which the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = ax + b \) is a bijection and \( f(f(x)) \) is the identity function, we can follow these steps:

### Step 1: Understand the condition for bijection
A function is a bijection if it is both injective (one-to-one) and surjective (onto). For the function \( f(x) = ax + b \) to be injective, the coefficient \( a \) must not be zero. Thus, we have:
\[ a \neq 0 \]

### Step 2: Find the expression for \( f(f(x)) \)
We need to compute \( f(f(x)) \):
\[ f(f(x)) = f(ax + b) = a(ax + b) + b = a^2x + ab + b \]

### Step 3: Set \( f(f(x)) \) equal to \( x \)
Since we want \( f(f(x)) \) to be the identity function, we set:
\[ f(f(x)) = x \]
This gives us the equation:
\[ a^2x + ab + b = x \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ (a^2 - 1)x + (ab + b) = 0 \]
For this equation to hold for all \( x \), both coefficients must be zero:
1. \( a^2 - 1 = 0 \)
2. \( ab + b = 0 \)

### Step 5: Solve the first equation
From \( a^2 - 1 = 0 \), we can factor it as:
\[ (a - 1)(a + 1) = 0 \]
Thus, we find:
\[ a = 1 \quad \text{or} \quad a = -1 \]

### Step 6: Solve the second equation
Now, we analyze the second equation \( ab + b = 0 \):
\[ b(a + 1) = 0 \]
This gives us two cases:
1. \( b = 0 \)
2. \( a + 1 = 0 \) which implies \( a = -1 \)

### Step 7: Determine values of \( b \) for each case of \( a \)
- **Case 1**: If \( a = 1 \):
- From \( b(a + 1) = 0 \), we have \( b(1 + 1) = 0 \) which leads to \( b = 0 \).

- **Case 2**: If \( a = -1 \):
- From \( b(-1 + 1) = 0 \), this gives \( b(0) = 0 \), meaning \( b \) can be any real number.

### Conclusion
The values of \( a \) and \( b \) for which the function \( f(x) = ax + b \) is a bijection and \( f(f(x)) \) is the identity function are:
1. \( (a, b) = (1, 0) \)
2. \( (a, b) = (-1, b) \) where \( b \in \mathbb{R} \)