Q61 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics

If $\frac{x^4}{\left(x^2+1\right)(x-2)}=f(x)+\frac{A x+B}{x^2+1}+\frac{C}{x-2}$, then $f(14)+2 A-B=$

  • Correct answer
    A
    $5 C$
  • B
    $4 C$.
  • C
    $6 C$
  • D
    $7 C$

Solution

Given that $\frac{x^4}{\left(x^2+9\right)(x-2)}=\frac{x^4-1+1}{\left(x^2+1\right)(x-2)}$ $\begin{aligned} & =\frac{x^2-1}{x-2}+\frac{1}{\left(x^2+1\right)(x-2)} \\ & =\frac{x^2-4+3}{x-2}+\frac{1}{5}\left\{\frac{1}{x-2}-\frac{x+2}{x^2+1}\right\} \\ & =x+2+\frac{3}{x-2}+\frac{1}{5(x-2)}-\frac{1}{5} \frac{(x+2)}{x^2+1} \\ & =x+2+\frac{16}{5(x-2)}-\frac{1}{5} \frac{(x+2)}{\left(x^2+1\right)} \\ & f(x)=x+2 \\ & C=\frac{16}{5}, A=\frac{-1}{5}, B=\frac{-2}{5}, f(14)=16 \\ & f(14)+2 A-B=16-\frac{2}{5}+\frac{2}{5}=16=5 C .\end{aligned}$

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

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