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. a. 413
B. b. 134
C. c, 431
D. d, 143

Answer

Let's solve the question step by step.

### Step 1: Solve for A
We start with the equation given in option A:
\[ \frac{3x}{(x-a)(x-b)} = \frac{2}{x-a} + \frac{1}{x-b} \]
To combine the right-hand side, we take the LCM:
\[ \frac{2(x-b) + 1(x-a)}{(x-a)(x-b)} = \frac{2x - 2b + x - a}{(x-a)(x-b)} = \frac{3x - (2b + a)}{(x-a)(x-b)} \]
Setting the numerators equal gives:
\[ 3x = 3x - (2b + a) \]
This simplifies to:
\[ 0 = - (2b + a) \implies 2b + a = 0 \implies a = -2b \]
Thus, the ratio \( \frac{a}{b} = \frac{-2b}{b} = -2:1 \).

### Step 2: Solve for B
Next, we consider option B:
\[ \frac{x+4}{(x^2-4)(x+1)} = \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x+1} \]
The LCM of the right-hand side is \((x-2)(x+2)(x+1)\). We rewrite the equation:
\[ x + 4 = A(x+2)(x+1) + B(x-2)(x+1) + C(x-2)(x+2) \]
To find \(A\), \(B\), and \(C\), we substitute convenient values for \(x\).

1. Let \(x = 2\):
\[ 2 + 4 = A(4)(3) \implies 6 = 12A \implies A = \frac{1}{2} \]

2. Let \(x = -2\):
\[ -2 + 4 = B(-4)(-1) \implies 2 = 4B \implies B = \frac{1}{2} \]

3. Let \(x = -1\):
\[ -1 + 4 = C(-3)(1) \implies 3 = -3C \implies C = -1 \]

Now, we find \(A + B + C\):
\[ A + B + C = \frac{1}{2} + \frac{1}{2} - 1 = 0 \]

### Step 3: Solve for C
Now we solve option C:
\[ \frac{2x+1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1} \]
The LCM is \((x-1)(x^2+1)\):
\[ 2x + 1 = A(x^2 + 1) + (Bx + C)(x - 1) \]
Expanding the right-hand side:
\[ 2x + 1 = Ax^2 + A + Bx^2 - Bx + Cx - C \]
Combining like terms:
\[ 2x + 1 = (A + B)x^2 + (C - B)x + A - C \]
Setting coefficients equal gives:
1. \(A + B = 0\)
2. \(C - B = 2\)
3. \(A - C = 1\)

From \(A + B = 0\), we have \(B = -A\). Substituting into the second equation:
\[ C + A = 2 \implies C = 2 - A \]
Substituting into the third equation:
\[ A - (2 - A) = 1 \implies 2A - 2 = 1 \implies 2A = 3 \implies A = \frac{3}{2} \]
Thus, \(B = -\frac{3}{2}\) and \(C = 2 - \frac{3}{2} = \frac{1}{2}\).

### Final Answers
- For A: \(a:b = -2:1\) (matches with 4)
- For B: \(A + B + C = 0\) (matches with 1)
- For C: \(C = \frac{1}{2}\) (matches with 3)