Question

If \[ \frac{2x^2+1}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}, \] then \[ 7A+2B+C= \]

• A. \(8\)
• B. \(9\)
• C. \(10\)
• D. \(11\)

Hint:

For partial fractions, first factorize the denominator completely, then take LCM and compare coefficients of like powers of \(x\).

Answer 1

The Correct Option is B

Step 1: Factorize the denominator.
We know that \[ x^3-1=(x-1)(x^2+x+1) \] So, \[ \frac{2x^2+1}{x^3-1} = \frac{2x^2+1}{(x-1)(x^2+x+1)} \]

Step 2: Write the given partial fraction form.
Given, \[ \frac{2x^2+1}{(x-1)(x^2+x+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1} \] Taking LCM on the right side, \[ \frac{2x^2+1}{(x-1)(x^2+x+1)} = \frac{A(x^2+x+1)+(Bx+C)(x-1)}{(x-1)(x^2+x+1)} \]

Step 3: Equate the numerators.
Therefore, \[ 2x^2+1=A(x^2+x+1)+(Bx+C)(x-1) \] Expanding, \[ 2x^2+1=Ax^2+Ax+A+Bx^2-Bx+Cx-C \] \[ 2x^2+1=(A+B)x^2+(A-B+C)x+(A-C) \]

Step 4: Compare coefficients.
Comparing coefficients of \(x^2\), \(x\), and constant terms, we get \[ A+B=2 \] \[ A-B+C=0 \] \[ A-C=1 \]

Step 5: Solve for \(A\), \(B\), and \(C\).
From \[ A-C=1, \] we get \[ C=A-1 \] Substitute this in \[ A-B+C=0 \] \[ A-B+A-1=0 \] \[ 2A-B=1 \] So, \[ B=2A-1 \] Now use \[ A+B=2 \] \[ A+2A-1=2 \] \[ 3A=3 \] \[ A=1 \] Thus, \[ B=2(1)-1=1 \] and \[ C=1-1=0 \]

Step 6: Find the required value.
Now, \[ 7A+2B+C=7(1)+2(1)+0 \] \[ =7+2 \] \[ =9 \] Therefore, \[ \boxed{9} \]