Statement-I :
If `(3x+4)/((x+1)^(2)(x-1))=(A)/(x-1)+(B)/(x+1)+(C)/((x+1)^(2))` then `A=7//4`
Statement-II :
If `(px+q)/((2x-3)^(2))=(1)/(2x-3)+(3)/((2x-3)^(2))` then `p=2, q=3`. Which of the above statements is true

Options

A. only I is true
B. only II is true
C. both I & II are true
D. neither I nor II true

Answer

To solve the problem, we need to analyze both statements separately and determine their validity.

### Statement I:
We need to verify if:
\[ \frac{3x + 4}{(x + 1)^2 (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \]
is true for \( A = \frac{7}{4} \).

**Step 1: Set up the equation**
We start with the left-hand side (LHS):
\[ \frac{3x + 4}{(x + 1)^2 (x - 1)} \]
And the right-hand side (RHS):
\[ \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2} \]

**Step 2: Find a common denominator for the RHS**
The common denominator is \((x - 1)(x + 1)^2\). Thus, we rewrite the RHS:
\[ \frac{A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)}{(x - 1)(x + 1)^2} \]

**Step 3: Equate the numerators**
Now we equate the numerators:
\[ 3x + 4 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) \]

**Step 4: Expand the right-hand side**
Expanding each term:
- \(A(x + 1)^2 = A(x^2 + 2x + 1)\)
- \(B(x - 1)(x + 1) = B(x^2 - 1)\)
- \(C(x - 1) = Cx - C\)

Combining these gives:
\[ Ax^2 + 2Ax + A + Bx^2 - B + Cx - C \]
Combining like terms:
\[ (A + B)x^2 + (2A + C)x + (A - B - C) \]

**Step 5: Set coefficients equal**
Now we compare coefficients with \(3x + 4\):
1. Coefficient of \(x^2\): \(A + B = 0\)
2. Coefficient of \(x\): \(2A + C = 3\)
3. Constant term: \(A - B - C = 4\)

**Step 6: Solve the equations**
From \(A + B = 0\), we have \(B = -A\).
Substituting \(B\) in the other equations:
- \(2A + C = 3\)
- \(A - (-A) - C = 4 \Rightarrow 2A - C = 4\)

Now we can solve these two equations:
1. \(2A + C = 3\)
2. \(2A - C = 4\)

Adding these equations:
\[ 4A = 7 \Rightarrow A = \frac{7}{4} \]

Substituting \(A\) back into \(2A + C = 3\):
\[ 2 \cdot \frac{7}{4} + C = 3 \Rightarrow \frac{14}{4} + C = 3 \Rightarrow C = 3 - \frac{14}{4} = 3 - \frac{7}{2} = \frac{6}{2} - \frac{7}{2} = -\frac{1}{2} \]

Now substituting \(A\) back to find \(B\):
\[ B = -A = -\frac{7}{4} \]

**Conclusion for Statement I:**
The value of \(A\) is indeed \(\frac{7}{4}\), so Statement I is true.

---

### Statement II:
We need to verify if:
\[ \frac{px + q}{(2x - 3)^2} = \frac{1}{2x - 3} + \frac{3}{(2x - 3)^2} \]
is true for \(p = 2\) and \(q = 3\).

**Step 1: Set up the equation**
The LHS is:
\[ \frac{px + q}{(2x - 3)^2} \]
The RHS is:
\[ \frac{1}{2x - 3} + \frac{3}{(2x - 3)^2} \]

**Step 2: Find a common denominator for the RHS**
The common denominator is \((2x - 3)^2\):
\[ \frac{(2x - 3) + 3}{(2x - 3)^2} = \frac{2x}{(2x - 3)^2} \]

**Step 3: Equate the numerators**
Now we equate the numerators:
\[ px + q = 2x \]

**Step 4: Substitute values**
Substituting \(p = 2\) and \(q = 3\):
\[ 2x + 3 \neq 2x \]

**Conclusion for Statement II:**
The equation does not hold true, hence Statement II is false.

### Final Answer:
- Statement I is true.
- Statement II is false.