Question

If \[ \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}, \] then the values of \((A,B,C)\) are

• A. \((1,-5,4)\)
• B. \((1,5,4)\)
• C. \((4,5,1)\)
• D. \((1,4,5)\)

Hint:

For partial fractions with factors \((x-a)\), substitute \(x=a\) to quickly find the corresponding constant.

Answer 1

The Correct Option is A

Concept:
For partial fractions with distinct linear factors, use the cover-up method or substitution method.

Step 1: Given: \[ \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \]

Step 2: Multiply both sides by \((x-1)(x-2)(x-3)\): \[ 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \]

Step 3: Put \(x=1\): \[ 3(1)-1=A(1-2)(1-3) \] \[ 2=A(-1)(-2) \] \[ 2=2A \] \[ A=1 \]

Step 4: Put \(x=2\): \[ 3(2)-1=B(2-1)(2-3) \] \[ 5=B(1)(-1) \] \[ B=-5 \]

Step 5: Put \(x=3\): \[ 3(3)-1=C(3-1)(3-2) \] \[ 8=C(2)(1) \] \[ C=4 \]

Step 6: Therefore: \[ (A,B,C)=(1,-5,4) \] \[ \boxed{(1,-5,4)} \]