Question:
If we resolve the rational fraction
\[
\frac{1}{(1-2x)^2(1-3x)}
\]
into partial fractions of the form
\[
\frac{A}{1-3x}+\frac{B}{1-2x}+\frac{C}{(1-2x)^2},
\]
then what is the \(\min\{A,B,C\}=\)
If we resolve the rational fraction
\[
\frac{1}{(1-2x)^2(1-3x)}
\]
into partial fractions of the form
\[
\frac{A}{1-3x}+\frac{B}{1-2x}+\frac{C}{(1-2x)^2},
\]
then what is the \(\min\{A,B,C\}=\)
Show Hint
In partial fractions, after multiplying by the common denominator, substitute the roots of denominator factors to quickly find constants.
Updated On: Jun 25, 2026
- \(1\)
- \(9\)
- \(-2\)
- \(-6\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the given partial fraction form.
We have \[ \frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x}+\frac{B}{1-2x}+\frac{C}{(1-2x)^2} \]
Step 2: Multiply both sides by \((1-2x)^2(1-3x)\).
\[ 1=A(1-2x)^2+B(1-2x)(1-3x)+C(1-3x) \]
Step 3: Find \(A\).
Put \[ x=\frac{1}{3} \] Then, \[ 1-3x=0 \] So, \[ 1=A\left(1-\frac{2}{3}\right)^2 \] \[ 1=A\left(\frac{1}{3}\right)^2 \] \[ 1=\frac{A}{9} \] Hence, \[ A=9 \]
Step 4: Find \(C\).
Put \[ x=\frac{1}{2} \] Then, \[ 1-2x=0 \] So, \[ 1=C\left(1-\frac{3}{2}\right) \] \[ 1=C\left(-\frac{1}{2}\right) \] Hence, \[ C=-2 \]
Step 5: Find \(B\).
Now compare the coefficient of \(x^2\) in \[ 1=A(1-2x)^2+B(1-2x)(1-3x)+C(1-3x) \] The coefficient of \(x^2\) from \(A(1-2x)^2\) is \[ 4A \] The coefficient of \(x^2\) from \(B(1-2x)(1-3x)\) is \[ 6B \] The term \(C(1-3x)\) has no \(x^2\) term.
Since the left side has no \(x^2\) term, \[ 4A+6B=0 \] Substitute \[ A=9 \] \[ 4(9)+6B=0 \] \[ 36+6B=0 \] \[ 6B=-36 \] \[ B=-6 \]
Step 6: Find the minimum value.
We have \[ A=9,\quad B=-6,\quad C=-2 \] Therefore, \[ \min\{A,B,C\}=-6 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{-6} \]
We have \[ \frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x}+\frac{B}{1-2x}+\frac{C}{(1-2x)^2} \]
Step 2: Multiply both sides by \((1-2x)^2(1-3x)\).
\[ 1=A(1-2x)^2+B(1-2x)(1-3x)+C(1-3x) \]
Step 3: Find \(A\).
Put \[ x=\frac{1}{3} \] Then, \[ 1-3x=0 \] So, \[ 1=A\left(1-\frac{2}{3}\right)^2 \] \[ 1=A\left(\frac{1}{3}\right)^2 \] \[ 1=\frac{A}{9} \] Hence, \[ A=9 \]
Step 4: Find \(C\).
Put \[ x=\frac{1}{2} \] Then, \[ 1-2x=0 \] So, \[ 1=C\left(1-\frac{3}{2}\right) \] \[ 1=C\left(-\frac{1}{2}\right) \] Hence, \[ C=-2 \]
Step 5: Find \(B\).
Now compare the coefficient of \(x^2\) in \[ 1=A(1-2x)^2+B(1-2x)(1-3x)+C(1-3x) \] The coefficient of \(x^2\) from \(A(1-2x)^2\) is \[ 4A \] The coefficient of \(x^2\) from \(B(1-2x)(1-3x)\) is \[ 6B \] The term \(C(1-3x)\) has no \(x^2\) term.
Since the left side has no \(x^2\) term, \[ 4A+6B=0 \] Substitute \[ A=9 \] \[ 4(9)+6B=0 \] \[ 36+6B=0 \] \[ 6B=-36 \] \[ B=-6 \]
Step 6: Find the minimum value.
We have \[ A=9,\quad B=-6,\quad C=-2 \] Therefore, \[ \min\{A,B,C\}=-6 \]
Step 7: Final conclusion.
Therefore, \[ \boxed{-6} \]
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