Question

If \( \frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1} \), then \( 3A+2B-C = \)

• A. \( \frac{8}{5} \)
• B. \( \frac{16}{5} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{19}{5} \)

Hint:

Comparing the coefficients of the highest power of \( x \) (here \( x^2 \)) and the constant term is usually the fastest way to find remaining coefficients after using the Heaviside cover-up method.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We find the constants \( A, B, C \) using Partial Fraction Decomposition.
Step 2: Computation Steps:
Equation: \[ \frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1} \] Finding A: Using the "Cover-up Method", multiply by \( (x+2) \) and set \( x = -2 \). \[ A = \left. \frac{x^2-3}{x^2+1} \right|_{x=-2} = \frac{(-2)^2 - 3}{(-2)^2 + 1} = \frac{4-3}{4+1} = \frac{1}{5} \] Finding B and C: Compare coefficients or substitute values. Multiply the original equation by \( (x+2)(x^2+1) \): \[ x^2 - 3 = A(x^2+1) + (Bx+C)(x+2) \] Substitute \( A = 1/5 \): \[ x^2 - 3 = \frac{1}{5}(x^2+1) + (Bx+C)(x+2) \] Compare coefficient of \( x^2 \): \[ 1 = \frac{1}{5} + B \implies B = 1 - \frac{1}{5} = \frac{4}{5} \] Compare constant terms (put \( x=0 \)): \[ -3 = \frac{1}{5}(1) + C(2) \] \[ -3 - \frac{1}{5} = 2C \implies -\frac{16}{5} = 2C \implies C = -\frac{8}{5} \] Calculation Process Required Value: \[ 3A + 2B - C = 3\left(\frac{1}{5}\right) + 2\left(\frac{4}{5}\right) - \left(-\frac{8}{5}\right) \] \[ = \frac{3}{5} + \frac{8}{5} + \frac{8}{5} = \frac{19}{5} \]
Step 4: Required Answer:
The value is \( 19/5 \).