Question

If
\[ \frac{x^4+24x^2+28}{(x^2+1)^3} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{Ex+F}{(x^2+1)^3}, \] then the value of \(A+B+C+D+E+F\) is

• A. \(21\)
• B. \(22\)
• C. \(28\)
• D. \(29\)

Hint:

In partial fractions, after multiplying by the denominator, compare coefficients of like powers of \(x\) to find unknown constants.

Answer 1

The Correct Option is C

Step 1: Multiply both sides by \((x^2+1)^3\).
\[ x^4+24x^2+28 = (Ax+B)(x^2+1)^2+(Cx+D)(x^2+1)+(Ex+F) \]

Step 2: Observe the nature of the numerator.
The left side contains only even powers of \(x\).
So, the coefficients of odd powers must be zero.
Hence,
\[ A=0,\quad C=0,\quad E=0 \]

Step 3: Substitute \(A=C=E=0\).
Now,
\[ x^4+24x^2+28 = B(x^2+1)^2+D(x^2+1)+F \]
Expanding,
\[ x^4+24x^2+28 = B(x^4+2x^2+1)+D(x^2+1)+F \]
\[ = Bx^4+(2B+D)x^2+(B+D+F) \]

Step 4: Compare coefficients.
Comparing coefficients of \(x^4\),
\[ B=1 \]
Comparing coefficients of \(x^2\),
\[ 2B+D=24 \]
\[ 2(1)+D=24 \]
\[ D=22 \]
Comparing constant terms,
\[ B+D+F=28 \]
\[ 1+22+F=28 \]
\[ F=5 \]

Step 5: Find the required sum.
\[ A+B+C+D+E+F=0+1+0+22+0+5 \]
\[ =28 \]

Step 6: Final conclusion.
Hence,
\[ \boxed{28} \]