Question

If \[ \frac{-x^2 + 6x + 1}{(x - 1)^2 (x^2 + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{Cx - 3}{x^2 + 2}, \] then \(A + B + C =\) ?

• A. 7
• B. 5
• C. 3
• D. \(\mathbf{2}\)

Hint:

To find constants in partial fractions, use suitable values of \(x\) to eliminate terms and solve systematically.

Answer 1

The Correct Option is D

We are given a partial fraction decomposition: \[ \frac{-x^2 + 6x + 1}{(x - 1)^2 (x^2 + 2)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{Cx - 3}{x^2 + 2} \] Multiply both sides by the common denominator \((x - 1)^2(x^2 + 2)\): \[ - x^2 + 6x + 1 = A(x - 1)(x^2 + 2) + B(x^2 + 2) + (Cx - 3)(x - 1)^2 \] We now expand both sides and compare coefficients. ### Step 1: Choose values to simplify **Let \(x = 1\):** LHS: \(-1^2 + 6(1) + 1 = -1 + 6 + 1 = 6\) RHS: \[ A(0)(1^2 + 2) + B(1^2 + 2) + (C(1) - 3)(0)^2 = 0 + 3B + 0 \Rightarrow 6 = 3B \Rightarrow B = 2 \] --- **Let \(x = 0\):** LHS: \(0 + 0 + 1 = 1\) RHS: \[ A(-1)(0^2 + 2) + B(0^2 + 2) + (-3)(-1)^2 = -2A + 2B - 3 \Rightarrow 1 = -2A + 4 - 3 = -2A + 1 \Rightarrow 0 = -2A \Rightarrow A = 0 \] --- **Let \(x = -1\):** LHS: \(-1 + (-6) + 1 = -6\) RHS: \[ A(-2)(3) + B(3) + (-C - 3)(-2)^2 = -6A + 3B + 4(-C - 3) \Rightarrow A = 0, B = 2 \Rightarrow -6 = 0 + 6 - 4(C + 3) \Rightarrow -6 = 6 - 4C - 12 \Rightarrow -6 = -4C - 6 \Rightarrow C = 0 \] --- ### Final Values: \[ A = 0,\quad B = 2,\quad C = 0 \Rightarrow A + B + C = 2 \] \[ \boxed{A + B + C = 2} \]