Question:medium

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

Show Hint

For partial fractions with factors \((x-a)\), substitute \(x=a\) to quickly find the corresponding constant.
  • \((1,-5,4)\)
  • \((1,5,4)\)
  • \((4,5,1)\)
  • \((1,4,5)\)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
This is a partial fraction decomposition problem with three distinct linear factors in the denominator. The most efficient way to solve this is using the "cover-up" method.
Step 2: Key Formula or Approach:
By combining the fractions on the right side, we obtain the identity: \[ 3x-1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) \] We will find A, B, and C by substituting the roots of the denominator ($x=1, x=2, x=3$) into this identity.
Step 3: Detailed Explanation:
To find A (let x = 1): Substitute $x=1$ into the identity. The terms with B and C will become zero. \[ 3(1)-1 = A(1-2)(1-3) + B(0) + C(0) \] \[ 2 = A(-1)(-2) \] \[ 2 = 2A \implies A = 1 \] To find B (let x = 2): Substitute $x=2$ into the identity. The terms with A and C will become zero. \[ 3(2)-1 = A(0) + B(2-1)(2-3) + C(0) \] \[ 5 = B(1)(-1) \] \[ 5 = -B \implies B = -5 \] To find C (let x = 3): Substitute $x=3$ into the identity. The terms with A and B will become zero. \[ 3(3)-1 = A(0) + B(0) + C(3-1)(3-2) \] \[ 8 = C(2)(1) \] \[ 8 = 2C \implies C = 4 \] Step 4: Final Answer:
The values are A = 1, B = -5, and C = 4. This corresponds to the tuple (1, -5, 4). Therefore, option (A) is the correct answer.
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