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
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 linear factors, substitute the roots of the denominator factors to find constants quickly.
Updated On: May 7, 2026
- \((1,-5,4)\)
- \((1,5,4)\)
- \((4,5,1)\)
- \((1,4,5)\)
Show Solution
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The Correct Option is A
Solution and Explanation
We are given
\[
\frac{3x-1}{(x-1)(x-2)(x-3)}
=
\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}.
\]
Multiplying both sides by \((x-1)(x-2)(x-3)\), we get
\[
3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2).
\]
Now use suitable values of \(x\) to find \(A\), \(B\), and \(C\).
Put
\[
x=1.
\]
Then,
\[
3(1)-1=A(1-2)(1-3).
\]
\[
2=A(-1)(-2).
\]
\[
2=2A.
\]
\[
A=1.
\]
Now put
\[
x=2.
\]
Then,
\[
3(2)-1=B(2-1)(2-3).
\]
\[
6-1=B(1)(-1).
\]
\[
5=-B.
\]
\[
B=-5.
\]
Now put
\[
x=3.
\]
Then,
\[
3(3)-1=C(3-1)(3-2).
\]
\[
9-1=C(2)(1).
\]
\[
8=2C.
\]
\[
C=4.
\]
Therefore,
\[
(A,B,C)=(1,-5,4).
\]
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