Question

If \[ \frac{x^2+1}{x^4+4} = \frac{Ax+B}{x^2-2x+2} + \frac{Cx+D}{x^2+2x+2}, \] then \(3A+2B+3C=\)

• A. \(-D\)
• B. \(D\)
• C. \(2D\)
• D. \(-2D\)

Hint:

In partial fractions, after multiplying by the denominator, compare coefficients of equal powers of \(x\). This gives equations for unknown constants.

Answer 1

The Correct Option is C

Step 1: Factorize the denominator.
We know that \[ x^4+4=(x^2-2x+2)(x^2+2x+2). \] So the given partial fraction form is valid.

Step 2: Multiply both sides by \(x^4+4\).
\[ x^2+1=(Ax+B)(x^2+2x+2)+(Cx+D)(x^2-2x+2). \]

Step 3: Expand the right-hand side.
\[ (Ax+B)(x^2+2x+2) = Ax^3+(2A+B)x^2+(2A+2B)x+2B. \] Also, \[ (Cx+D)(x^2-2x+2) = Cx^3+(-2C+D)x^2+(2C-2D)x+2D. \] Adding both expressions, \[ x^2+1 = (A+C)x^3+(2A+B-2C+D)x^2+(2A+2B+2C-2D)x+(2B+2D). \]

Step 4: Compare coefficients.
Comparing with \[ x^2+1, \] we get \[ A+C=0, \] \[ 2A+B-2C+D=1, \] \[ 2A+2B+2C-2D=0, \] \[ 2B+2D=1. \] From \[ A+C=0, \] we get \[ C=-A. \] From \[ 2A+2B+2C-2D=0, \] we get \[ A+B+C-D=0. \] Using \(C=-A\), \[ B-D=0. \] Hence, \[ B=D. \] Now from \[ 2B+2D=1, \] we get \[ B+D=\frac{1}{2}. \] Since \(B=D\), \[ 2D=\frac{1}{2} \] \[ D=\frac{1}{4}. \] Thus, \[ B=\frac{1}{4}. \] Using \[ 2A+B-2C+D=1, \] and \(C=-A\), \[ 2A+B+2A+D=1. \] \[ 4A+B+D=1. \] \[ 4A+\frac{1}{4}+\frac{1}{4}=1. \] \[ 4A+\frac{1}{2}=1. \] \[ 4A=\frac{1}{2}. \] \[ A=\frac{1}{8}. \] Therefore, \[ C=-\frac{1}{8}. \]

Step 5: Find \(3A+2B+3C\).
\[ 3A+2B+3C = 3\left(\frac{1}{8}\right)+2\left(\frac{1}{4}\right)+3\left(-\frac{1}{8}\right). \] \[ = \frac{3}{8}+\frac{1}{2}-\frac{3}{8}. \] \[ = \frac{1}{2}. \] Since \[ D=\frac{1}{4}, \] we get \[ \frac{1}{2}=2D. \]

Step 6: Final conclusion.
Therefore, \[ \boxed{2D} \]