AP EAMCET · Maths · Basic of Mathematics
If \(\frac{x^2-3 x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-3}+\frac{B}{(x-1)(x-2)}+\frac{C}{(x-1)(x-2)(x-3)}\) then \(\mathrm{B}=\)
- A 0
- B 1
- C -1
- D 0
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(x^2-3x+1 = A(x-1)(x-2) + B(x-3) + C\) Set \(x=1\): \( (1)^2-3(1)+1 = B(1-3)+C \) \(-1 = -2B+C \quad (1)\) Set \(x=2\): \( (2)^2-3(2)+1 = B(2-3)+C \) \(-1 = -B+C \quad (2)\) Subtract (2) from (1): \( (-1)-(-1) = (-2B+C)-(-B+C) \) \(0 = -2B+C+B-C \) \(0 = -B \) \(B = 0\)
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