Question:
If $\frac{3x + 4}{(x-1)(x-2)^2} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$, then $A + B + C =$
If $\frac{3x + 4}{(x-1)(x-2)^2} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$, then $A + B + C =$
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To find the sum of coefficients in a partial fraction identity of this type, you can sometimes substitute strategic values, but here, the direct determination of $A, B, C$ is extremely fast.
Updated On: May 31, 2026
- 10
- 20
- 17
- 14
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The Correct Option is A
Solution and Explanation
Step 1: Concept
We decompose a rational function into partial fractions. We can solve for the coefficients $A$, $B$, and $C$ by multiplying through by the common denominator and substituting specific values of $x$.
Step 2: Meaning
The given equation is: \[ 3x + 4 = A(x-2)^2 + B(x-1)(x-2) + C(x-1) \]
Step 3: Analysis
To find $A$, substitute $x = 1$: \[ 3(1) + 4 = A(1-2)^2 \implies 7 = A(1) \implies A = 7 \] To find $C$, substitute $x = 2$: \[ 3(2) + 4 = C(2-1) \implies 10 = C(1) \implies C = 10 \] To find $B$, compare the coefficient of $x^2$ on both sides: \[ 0 = A + B \implies B = -A = -7 \] Now, calculate $A + B + C$: \[ A + B + C = 7 + (-7) + 10 = 10 \]
Step 4: Conclusion
The sum of the coefficients $A + B + C$ is equal to $10$. Final Answer: (A)
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