If `(1)/((1+2x)(1-x^(2)))=(A)/(1+2x)+(B)/(1+x)+(C)/(1-x)` then assending order of A, B, C.

Options

A. A, B, C
B. B, C, A
C. C, A, B
D. B, A, C

Answer

To solve the equation

\[ \frac{1}{(1 + 2x)(1 - x^2)} = \frac{A}{1 + 2x} + \frac{B}{1 + x} + \frac{C}{1 - x} \]

we will first express the left-hand side in terms of partial fractions. Let's break down the steps:

### Step 1: Rewrite the denominator
The denominator can be factored as follows:

\[ 1 - x^2 = (1 - x)(1 + x) \]

Thus, we can rewrite the equation as:

\[ \frac{1}{(1 + 2x)(1 - x)(1 + x)} = \frac{A}{1 + 2x} + \frac{B}{1 + x} + \frac{C}{1 - x} \]

### Step 2: Find a common denominator
The common denominator for the right-hand side is:

\[ (1 + 2x)(1 + x)(1 - x) \]

Now, we can express the right-hand side as:

\[ \frac{A(1 + x)(1 - x) + B(1 + 2x)(1 - x) + C(1 + 2x)(1 + x)}{(1 + 2x)(1 + x)(1 - x)} \]

### Step 3: Set the numerators equal
Since the denominators are equal, we can set the numerators equal to each other:

\[ 1 = A(1 + x)(1 - x) + B(1 + 2x)(1 - x) + C(1 + 2x)(1 + x) \]

### Step 4: Expand the right-hand side
Now, we will expand the right-hand side:

1. For \(A(1 + x)(1 - x)\):
\[ A(1 - x^2) \]

2. For \(B(1 + 2x)(1 - x)\):
\[ B(1 - x + 2x - 2x^2) = B(1 + x - 2x^2) \]

3. For \(C(1 + 2x)(1 + x)\):
\[ C(1 + x + 2x + 2x^2) = C(1 + 3x + 2x^2) \]

Combining these, we have:

\[ 1 = A(1 - x^2) + B(1 + x - 2x^2) + C(1 + 3x + 2x^2) \]

### Step 5: Collect like terms
Now, we can collect like terms:

\[ 1 = (A + B + C) + (B + 3C)x + (-A - 2B + 2C)x^2 \]

### Step 6: Set up equations
Now, we can set up the equations by comparing coefficients:

1. Constant term: \(A + B + C = 1\)
2. Coefficient of \(x\): \(B + 3C = 0\)
3. Coefficient of \(x^2\): \(-A - 2B + 2C = 0\)

### Step 7: Solve the system of equations
From the second equation \(B + 3C = 0\), we can express \(B\) in terms of \(C\):

\[ B = -3C \]

Substituting \(B\) into the first equation:

\[ A - 3C + C = 1 \implies A - 2C = 1 \implies A = 1 + 2C \]

Now substitute \(B\) and \(A\) into the third equation:

\[ -(1 + 2C) - 2(-3C) + 2C = 0 \]
\[ -1 - 2C + 6C + 2C = 0 \implies -1 + 6C = 0 \implies 6C = 1 \implies C = \frac{1}{6} \]

Now substituting \(C\) back to find \(B\):

\[ B = -3C = -3 \times \frac{1}{6} = -\frac{1}{2} \]

And substituting \(C\) into the equation for \(A\):

\[ A = 1 + 2C = 1 + 2 \times \frac{1}{6} = 1 + \frac{1}{3} = \frac{4}{3} \]

### Step 8: Summary of values
Thus, we have:

- \(A = \frac{4}{3}\)
- \(B = -\frac{1}{2}\)
- \(C = \frac{1}{6}\)

### Step 9: Ascending order of A, B, C
Now we can arrange \(A\), \(B\), and \(C\) in ascending order:

- \(B = -\frac{1}{2}\)
- \(C = \frac{1}{6}\)
- \(A = \frac{4}{3}\)

Thus, the ascending order is \(B < C < A\).

### Final Answer
The ascending order of \(A\), \(B\), and \(C\) is:

\[ B, C, A \]