Question

If \( \frac{2x^3 + 1}{2x^2 - x - 6} = ax + b + \frac{A}{px - 2} + \frac{B}{2x + q} \), then \( 51apB = \):

• A. \( 23bqA \)
• B. \( 69bqA \)
• C. \( 7bqA \)
• D. \( 17bqA \) \bigskip

Hint:

When handling complex rational expressions, match the numerators and solve for unknowns by equating the coefficients of corresponding terms.

Answer 1

The Correct Option is A

We are given the equation: \[ \frac{2x^3 + 1}{2x^2 - x - 6} = ax + b + \frac{A}{px - 2} + \frac{B}{2x + q} \] and need to determine the value of \( 51apB \) in terms of \( 23bqA \). 

--- Step 1: Factorize the Denominator Factorizing \( 2x^2 - x - 6 \): \[ 2x^2 - x - 6 = (px - 2)(2x + q) \] Comparing coefficients: \[ p \cdot 2 = 2, \quad -2q + 2p = -1, \quad -2q = -6 \] Solving: \[ q = 3, \quad p = 1 \] Thus, the denominator factors as: \[ ( x - 2)(2x + 3) \] 

--- Step 2: Partial Fraction Decomposition The given expression can be rewritten as: \[ \frac{2x^3 + 1}{(x - 2)(2x + 3)} = ax + b + \frac{A}{x - 2} + \frac{B}{2x + 3} \] Multiplying both sides by \( (x - 2)(2x + 3) \): \[ 2x^3 + 1 = (ax + b)(x - 2)(2x + 3) + A(2x + 3) + B(x - 2) \] Expanding: \[ (ax + b)(2x^2 - x - 6) + A(2x + 3) + B(x - 2) = 2x^3 + 1 \] Expanding further: \[ (2ax^3 - ax^2 - 6ax + 2bx^2 - bx - 6b) + (2Ax + 3A) + (Bx - 2B) = 2x^3 + 1 \] Grouping terms: \[ 2ax^3 + (-a + 2b)x^2 + (-6a - b + 2A + B)x + (-6b + 3A - 2B) = 2x^3 + 1 \] Comparing coefficients: 1. For \( x^3 \): \[ 2a = 2 \Rightarrow a = 1 \] 2. For \( x^2 \): \[ -a + 2b = 0 \Rightarrow -1 + 2b = 0 \Rightarrow b = \frac{1}{2} \] 3. For \( x \): \[ -6a - b + 2A + B = 0 \] Substituting \( a = 1, b = \frac{1}{2} \): \[ -6(1) - \frac{1}{2} + 2A + B = 0 \] \[ -6 - \frac{1}{2} + 2A + B = 0 \] \[ 2A + B = \frac{13}{2} \] 4. Constant term: \[ -6b + 3A - 2B = 1 \] Substituting \( b = \frac{1}{2} \): \[ -6\left(\frac{1}{2}\right) + 3A - 2B = 1 \] \[ -3 + 3A - 2B = 1 \] \[ 3A - 2B = 4 \] Solving the system: 1. \( 2A + B = \frac{13}{2} \) 2. \( 3A - 2B = 4 \) Multiplying the first equation by 2: \[ 4A + 2B = 13 \] Adding both equations: \[ (4A + 2B) + (3A - 2B) = 13 + 4 \] \[ 7A = 17 \Rightarrow A = \frac{17}{7} \] Substituting into \( 2A + B = \frac{13}{2} \): \[ 2\left(\frac{17}{7}\right) + B = \frac{13}{2} \] \[ \frac{34}{7} + B = \frac{13}{2} \] \[ B = \frac{13}{2} - \frac{34}{7} \] \[ B = \frac{91 - 68}{14} = \frac{23}{14} \] 

--- Step 3: Compute \( 51apB \) \[ 51apB = 51 \times 1 \times 1 \times \frac{23}{14} = \frac{51 \times 23}{14} = \frac{1173}{14} \] Since \( 23bqA = 23 \times \frac{1}{2} \times 3 \times \frac{17}{7} = \frac{23 \times 51}{14} = \frac{1173}{14} \), we conclude: \[ 51apB = 23bqA \] 

Final Answer: \(\boxed{23bqA}\)