If the binomial expansion of `(a +b x)^-2` is `1/4-3x+.......,` then `(a, b) =`

Options

A. `(2,12)`
B. `(2,8)`
C. `(-2, -12)`
D. none of these

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) in the binomial expansion of \( (a + bx)^{-2} \) given that it is equal to \( \frac{1}{4} - 3x + \ldots \).

### Step 1: Understand the Binomial Expansion
The binomial expansion for \( (1 + x)^{-n} \) can be expressed as:
\[ (1 + x)^{-n} = 1 - nx + \frac{n(n+1)}{2!}x^2 - \ldots \]
For our case, we have \( (a + bx)^{-2} \). We can factor out \( a^{-2} \):
\[ (a + bx)^{-2} = a^{-2} \left(1 + \frac{bx}{a}\right)^{-2} \]

### Step 2: Apply the Binomial Expansion
Using the binomial expansion for \( (1 + u)^{-2} \) where \( u = \frac{bx}{a} \):
\[ (1 + u)^{-2} = 1 - 2u + \frac{2(3)}{2!}u^2 - \ldots \]
Substituting \( u = \frac{bx}{a} \):
\[ (1 + \frac{bx}{a})^{-2} = 1 - 2\left(\frac{bx}{a}\right) + 3\left(\frac{bx}{a}\right)^2 - \ldots \]

### Step 3: Combine the Results
Now, substituting back into our expression:
\[ (a + bx)^{-2} = a^{-2} \left(1 - 2\left(\frac{bx}{a}\right) + 3\left(\frac{bx}{a}\right)^2 - \ldots\right) \]
This gives:
\[ = a^{-2} - \frac{2b}{a^3}x + \frac{3b^2}{a^4}x^2 - \ldots \]

### Step 4: Compare Coefficients
We know from the problem statement that:
\[ \frac{1}{4} - 3x + \ldots \]
Thus, we can equate coefficients:
1. For the constant term:
\[ \frac{1}{a^2} = \frac{1}{4} \implies a^2 = 4 \implies a = \pm 2 \]
2. For the coefficient of \( x \):
\[ -\frac{2b}{a^3} = -3 \implies \frac{2b}{a^3} = 3 \implies 2b = 3a^3 \implies b = \frac{3a^3}{2} \]

### Step 5: Calculate Values of \( b \)
Now substituting the values of \( a \):
1. If \( a = 2 \):
\[ b = \frac{3(2^3)}{2} = \frac{3 \cdot 8}{2} = 12 \]
2. If \( a = -2 \):
\[ b = \frac{3(-2)^3}{2} = \frac{3(-8)}{2} = -12 \]

### Final Result
Thus, the pairs \( (a, b) \) are:
\[ (2, 12) \quad \text{and} \quad (-2, -12) \]