Question

\( \frac{x^3}{(x^2+1)(x^2+3)} = \)

• A. \( \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3} \) for some A,B,C,D \(\in R \setminus \{0\}\) లు
  % Telugu: A,B,C,D లు శూన్యేతర వాస్తవ సంఖ్యలు
• B. \( \frac{Ax+B}{x^2+1} + \frac{Cx}{x^2+3} \) for some A,B,C \(\in R \setminus \{0\}\) లు
• C. \( \frac{Ax}{x^2+1} + \frac{Bx}{x^2+3} \) for some A,B \(\in R \setminus \{0\}\) లు
• D. \( 1 + \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3} \) for some A,B,C,D \(\in R\)

Hint:

For partial fractions of \(N(x)/D(x)\):
If deg(N) \(\ge\) deg(D) (improper), first perform polynomial long division to get Quotient + Proper Fraction.
If deg(N)<deg(D) (proper), proceed to factor denominator.
For an irreducible quadratic factor \(ax^2+bx+c\), the corresponding partial fraction term is \(\frac{Ax+B}{ax^2+bx+c}\).
If the rational function is odd, then in the partial fraction terms \(\frac{Ax+B}{x^2+k}\), B must be 0. If even, A must be 0.

Answer 1

The Correct Option is C

Let \(f(x) = \frac{x^3}{(x^2+1)(x^2+3)}\). The degree of the numerator is 3. The degree of the denominator is \(2+2=4\). Since the degree of the numerator (3) is less than the degree of the denominator (4), the fraction is proper. Polynomial long division is not needed first. The denominator has irreducible quadratic factors \(x^2+1\) and \(x^2+3\). The general form of partial fraction decomposition for this proper rational function is: \[ \frac{x^3}{(x^2+1)(x^2+3)} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3} \] where A, B, C, D are real constants to be determined. Now let's consider the options: % Option (a) \( \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3} \) for some A,B,C,D \(\in R \setminus \{0\}\). This is the general form, but the condition "non-zero" might not hold for all. % Option (b) \( \frac{Ax+B}{x^2+1} + \frac{Cx}{x^2+3} \). This implies D=0. % Option (c) \( \frac{Ax}{x^2+1} + \frac{Bx}{x^2+3} \). This implies B=0 and D=0 (using B for the second numerator constant). % Option (d) \( 1 + \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2+3} \). This implies the fraction was improper, with a quotient of 1 from polynomial division. This is incorrect as the given fraction is proper. The checkmark in the image is on option (d). This indicates that the problem might have intended an improper fraction. If the numerator was, for example, \(x^4\) or higher, then polynomial division would yield a quotient plus a proper fraction. However, as written, \(f(x) = \frac{x^3}{(x^2+1)(x^2+3)}\) is proper. So the "1 +" term in option (d) should not be there. This means there is likely an error in option (d) being marked correct for the given expression, or the expression in the question is a typo. Let's check the form in option (c) which is \(\frac{Ax}{x^2+1} + \frac{Bx}{x^2+3}\). This means B and D in the general form are zero. If \(f(x)\) is an odd function, then \(B=0\) and \(D=0\). Let \(g(x) = x^3\). \(g(-x) = (-x)^3 = -x^3 = -g(x)\). So numerator is odd. Let \(h(x) = (x^2+1)(x^2+3)\). \(h(-x) = ((-x)^2+1)((-x)^2+3) = (x^2+1)(x^2+3) = h(x)\). So denominator is even. Then \(f(x) = \text{odd}/\text{even} = \text{odd}\). If \(f(x)\) is odd, then its partial fraction decomposition should also be odd. A term like \(\frac{Ax+B}{x^2+1}\) is odd if and only if \(B=0\). (Because \(\frac{Ax}{x^2+1}\) is odd, \(\frac{B}{x^2+1}\) is even). So, if \(f(x)\) is odd, then B=0 and D=0. The form would be \(\frac{Ax}{x^2+1} + \frac{Cx}{x^2+3}\). This matches the structure of option (c). Let's actually find A and C for form (c): \(\frac{x^3}{(x^2+1)(x^2+3)} = \frac{Ax}{x^2+1} + \frac{Cx}{x^2+3}\) Divide by x (assuming \(x \neq 0\)): \(\frac{x^2}{(x^2+1)(x^2+3)} = \frac{A}{x^2+1} + \frac{C}{x^2+3}\) Let \(y = x^2\). Then \(\frac{y}{(y+1)(y+3)} = \frac{A}{y+1} + \frac{C}{y+3}\). Multiply by \((y+1)(y+3)\): \(y = A(y+3) + C(y+1)\). Set \(y=-1\): \(-1 = A(-1+3) + C(0) \Rightarrow -1 = 2A \Rightarrow A = -1/2\). Set \(y=-3\): \(-3 = A(0) + C(-3+1) \Rightarrow -3 = -2C \Rightarrow C = 3/2\). So, \(A=-1/2\) and \(C=3/2\). These are non-zero real numbers. Thus, the form \(\frac{Ax}{x^2+1} + \frac{Cx}{x^2+3}\) is correct. This is option (c). The checkmark in the image is on (d). If (d) \(1 + \dots\) is correct, the original fraction must have been improper, e.g., \(\frac{x^4+x^3+\dots}{(x^2+1)(x^2+3)}\). Given the question as written, \(f(x)\) is proper and odd, so its partial fraction expansion should be of the form given in (c). There's a discrepancy with the marked answer. I will use the derived correct form. \[ \boxed{\frac{Ax}{x^2+1} + \frac{Cx}{x^2+3} \text{ for some } A,B \in R \setminus \{0\} \text{ (Matches form of option c)}} \] If forced to pick an option that includes a "1+", it implies the problem statement had a typo and was an improper fraction.