Observe the following Lists
`ul("List - I")`
A) If `(3x)/((x-a)(x-b))=(2)/(x-a)+(1)/(x-b)` then `a:b` is
B) If `(x+4)/((x^(2)-4)(x+1))=(A)/(x-2)+(B)/(x+2)+(C)/(x+1)`
then `A+B+C` is
C) If `(2x+1)/((x-1)(x^(2)+1))=(A)/(x-1)+(Bx+C)/(x^(2)+1)` then `C=`
`ul("List - II")`
1) Slope of `x` -axis
2) `sin. (3 pi)/(2)`
3) `{:(" "Lt),(x rarr 0):} (Tan x-Sinx)/(x^(3))`
4) Slope of the line `6x+3y-7=0`
The correct match is

Options

A. 4 1 3
B. 4 1 3
C. 4 1 2
D. 2 3 1

Answer

To solve the given problem, we will analyze each part of the question step by step.

### Part A: Finding the ratio \( a:b \)

We start with the equation:
\[ \frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b} \]

1. **Combine the right-hand side**:
\[ \frac{2}{x-a} + \frac{1}{x-b} = \frac{2(x-b) + 1(x-a)}{(x-a)(x-b)} = \frac{2x - 2b + x - a}{(x-a)(x-b)} = \frac{3x - (a + 2b)}{(x-a)(x-b)} \]

2. **Set the numerators equal**:
\[ 3x = 3x - (a + 2b) \]
This implies:
\[ a + 2b = 0 \quad \Rightarrow \quad a = -2b \]

3. **Find the ratio \( a:b \)**:
\[ \frac{a}{b} = \frac{-2b}{b} = -2 \quad \Rightarrow \quad a:b = -2:1 \]

### Part B: Finding \( A + B + C \)

We have:
\[ \frac{x+4}{(x^2-4)(x+1)} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x+1} \]

1. **Factor the denominator**:
\[ x^2 - 4 = (x-2)(x+2) \]
Thus, the equation becomes:
\[ \frac{x+4}{(x-2)(x+2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x+1} \]

2. **Combine the right-hand side**:
\[ \frac{A(x+2)(x+1) + B(x-2)(x+1) + C(x-2)(x+2)}{(x-2)(x+2)(x+1)} \]

3. **Set the numerators equal**:
\[ x + 4 = A(x^2 + 3x + 2) + B(x^2 - x - 2) + C(x^2 - 4) \]

4. **Equate coefficients**:
- Coefficient of \( x^2 \): \( A + B + C = 0 \)
- Coefficient of \( x \): \( 3A - B = 1 \)
- Constant term: \( 2A - 2B - 4C = 4 \)

5. **Solve the equations**:
From \( A + B + C = 0 \), we can express \( C = -A - B \).
Substitute \( C \) into the other equations and solve for \( A \), \( B \), and \( C \).

6. **Find \( A + B + C \)**:
Since \( A + B + C = 0 \), we conclude:
\[ A + B + C = 0 \]

### Part C: Finding \( C \)

We have:
\[ \frac{2x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \]

1. **Combine the right-hand side**:
\[ \frac{A(x^2+1) + (Bx+C)(x-1)}{(x-1)(x^2+1)} \]

2. **Set the numerators equal**:
\[ 2x + 1 = A(x^2 + 1) + (Bx^2 + Cx - Bx - C) \]

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

4. **Solve the equations**:
From \( A + B = 0 \), we have \( B = -A \).
Substitute \( B \) into the other equations and solve for \( A \), \( B \), and \( C \).

5. **Find \( C \)**:
After solving, we find \( C = \frac{1}{2} \).

### Final Results

- For part A, \( a:b = -2:1 \)
- For part B, \( A + B + C = 0 \)
- For part C, \( C = \frac{1}{2} \)