Question:
If
\[
\frac{3x^2+x+2}
{(3x^2+x+4)(3x^2+x+1)}
=
\frac{Ax+B}{3x^2+x+4}
+
\frac{Cx+D}{3x^2+x+1},
\]
then \((A+B)+(C+D)\) is:
If
\[
\frac{3x^2+x+2}
{(3x^2+x+4)(3x^2+x+1)}
=
\frac{Ax+B}{3x^2+x+4}
+
\frac{Cx+D}{3x^2+x+1},
\]
then \((A+B)+(C+D)\) is:
Show Hint
When the same quadratic expression repeats several times, substitute it by a single variable before attempting partial fractions.
Updated On: Jun 9, 2026
- \( \frac13 \)
- \( \frac23 \)
- \(1\)
- \( \frac32 \)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
The same quadratic expression
\[
3x^2+x
\]
appears repeatedly. Introducing a substitution simplifies the algebra considerably.
Step 1: Substitute \(u=3x^2+x\). Then \[ \frac{3x^2+x+2} {(3x^2+x+4)(3x^2+x+1)} = \frac{u+2}{(u+4)(u+1)}. \] Let \[ \frac{u+2}{(u+4)(u+1)} = \frac{K_1}{u+4} + \frac{K_2}{u+1}. \]
Step 2: Find \(K_1\) and \(K_2\). Multiplying throughout by \[ (u+4)(u+1), \] we get \[ u+2 = K_1(u+1)+K_2(u+4). \] Using \(u=-4\), \[ -2=-3K_1, \] \[ K_1=\frac23. \] Using \(u=-1\), \[ 1=3K_2, \] \[ K_2=\frac13. \]
Step 3: Compare with the required form. Thus \[ \frac{u+2}{(u+4)(u+1)} = \frac{2/3}{u+4} + \frac{1/3}{u+1}. \] Replacing \(u=3x^2+x\), \[ A=0,\quad B=\frac23, \] \[ C=0,\quad D=\frac13. \] Hence \[ (A+B)+(C+D) = \frac23+\frac13 = 1. \] Therefore, \[ \boxed{1}. \]
Step 1: Substitute \(u=3x^2+x\). Then \[ \frac{3x^2+x+2} {(3x^2+x+4)(3x^2+x+1)} = \frac{u+2}{(u+4)(u+1)}. \] Let \[ \frac{u+2}{(u+4)(u+1)} = \frac{K_1}{u+4} + \frac{K_2}{u+1}. \]
Step 2: Find \(K_1\) and \(K_2\). Multiplying throughout by \[ (u+4)(u+1), \] we get \[ u+2 = K_1(u+1)+K_2(u+4). \] Using \(u=-4\), \[ -2=-3K_1, \] \[ K_1=\frac23. \] Using \(u=-1\), \[ 1=3K_2, \] \[ K_2=\frac13. \]
Step 3: Compare with the required form. Thus \[ \frac{u+2}{(u+4)(u+1)} = \frac{2/3}{u+4} + \frac{1/3}{u+1}. \] Replacing \(u=3x^2+x\), \[ A=0,\quad B=\frac23, \] \[ C=0,\quad D=\frac13. \] Hence \[ (A+B)+(C+D) = \frac23+\frac13 = 1. \] Therefore, \[ \boxed{1}. \]
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