Q59 / 160 · AP EAMCET 2024 (20 May Shift 1) · Mathematics

If $\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)\left(x^2+b^2\right)}$ then $\mathrm{C}=$

  • A
    $\frac{-1}{a^2+b^2}$
  • B
    $\frac{1}{a^2+b^2}$
  • Correct answer
    C
    $\frac{-a}{a^2+b^2}$
  • D
    $\frac{a}{a^2+b^2}$

Solution

$\frac{A}{(x-a)}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)\left(x^2+b^2\right)}$ $\Rightarrow A\left(x^2+b^2\right)+(B x+C)(x-a)=1$ $\Rightarrow(A+B) x^2+(C-a B) x+A b^2-a C=1$ On comparing we get, $A+B=0$ ....(i) $C-a B=0$ ....(ii) $A b^2-a C=1$ ....(iii) From (i) and (ii), $C+a A=0 \Rightarrow A=\frac{-C}{a}$; Put $A=\frac{-C}{a}$ in equation (iii), $\frac{-C}{a} b^2-a C=1 \Rightarrow-C\left(\frac{b^2+a^2}{a}\right)=1 \Rightarrow C=\frac{-a}{a^2+b^2}$

Hence, the correct answer is option: 'C'.

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