Q59 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
If $\frac{2 x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1}$, then $A+B+C+D=$
- A1
- B2
- C$\frac{3}{4}$
- Correct answerD$\frac{1}{2}$
Solution
If $\frac{2 x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1}$
$\Rightarrow \quad 2 x+1=A(x-1)\left(x^2+1\right)+B\left(x^2+1\right)$ $+(C x+D)(x-1)^2$
$\Rightarrow \quad 2 x+1=x^3(A+C)+x^2(-A+B-2 C+D)$ $+x(A+C-2 D)+(-A+B+D)$
On comparing coefficients of $x^3, x^2, x$ and constant terms both sides we get.
$A+C=0$ $\ldots(\mathrm{i})$
$-A+B-2 C+D=0$ $\ldots(\mathrm{ii})$
$A+C-2 D=2$ $\ldots(\mathrm{iii})$
and $\quad-A+B+D=1$ $\ldots(\mathrm{iv})$
On solving Eqs. (i), (ii), (iii) and (iv), we get
$A=\frac{-1}{2}, B=\frac{3}{2}, C=\frac{1}{2} \text { and } D=-1$
$\therefore \quad A+B+C+D=\frac{-1}{2}+\frac{3}{2}+\frac{1}{2}-1=\frac{1}{2}$
Hence, the correct answer is option: 'D'.