The fraction $\frac{x^2}{(x-a)(x-b)}$ is

  • A
    always a proper partial fraction
  • B
    always an improper partial fraction
  • C
    a proper partial fraction for certain values of $a, b$ only
  • D
    an improper partial fraction for certain values of $a, b$ only

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Similar Questions

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

$\frac{x^2 + 13x + 15}{(2x + 3)(x + 3)^2} = $

If $\frac{x^3 - 6x^2 + 10x - 2}{x^2 - 5x + 6} = f(x) + \frac{A}{x - 2} + \frac{B}{x - 3}$,then $f(x) = $

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2A-3B+C=$

Let $\frac{1}{(x^2-3)^2} = \frac{A_1}{x-\sqrt{3}} + \frac{A_2}{(x-\sqrt{3})^2} + \frac{A_3}{x+\sqrt{3}} + \frac{A_4}{(x+\sqrt{3})^2}$. Then,consider the following statements:
$(i)$ All the $A_i$'s are not distinct
(ii) There exists a pair,$A_p$ and $A_q$ such that $A_p^2 = A_q^2$ $(p \neq q)$
(iii) $\sum_{i=1}^4 A_i = \frac{1}{6}$
(iv) $\sum_{i=1}^4 A_i = 1$
Which one of the following is true?

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