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: Write the partial fraction decomposition.
We are given $\dfrac{2x^2+1}{x^3-1} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+x+1}$, since $x^3 - 1 = (x-1)(x^2+x+1)$.
Step 2: Clear the denominator by multiplying both sides by $(x-1)(x^2+x+1)$.
$2x^2+1 = A(x^2+x+1) + (Bx+C)(x-1)$.
Step 3: Find $A$ by substituting $x=1$.
$2(1)+1 = A(1+1+1) \Rightarrow 3 = 3A \Rightarrow A = 1$.
Step 4: Expand and compare coefficients to find $B$ and $C$.
$2x^2+1 = (x^2+x+1) + (Bx+C)(x-1) = x^2+x+1 + Bx^2-Bx+Cx-C$.
Grouping: $(1+B)x^2 + (1-B+C)x + (1-C)$.
Matching with $2x^2 + 0 \cdot x + 1$:
$1+B=2 \Rightarrow B=1$; $1-B+C=0 \Rightarrow C=0$; $1-C=1$ ✓.
Step 5: Compute $7A + 2B + C$.
$7A + 2B + C = 7(1) + 2(1) + 0 = 7 + 2 + 0 = 9$.
Step 6: State the final answer.
With $A=1$, $B=1$, $C=0$, the value of $7A+2B+C = 9$. \[ \boxed{9} \]