Q78 / 180 · COMEDK 2018 · Mathematics
If $\frac{3 x^{2}-2 x+4}{(x-1)^{6}}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)^{2}}+\frac{A_{3}}{(x+1)^{3}}$ $+\frac{A_{4}}{(x+1)^{4}}+\frac{A_{5}}{(x+1)^{5}}+\frac{A_{6}}{(x+1)^{6}}$, then $\left(A_{1}+A_{2}+A_{2}-A_{4}-A_{6}\right)=$
- A$(0,0)$
- B$(-8,-12)$
- C$(8,-12)$
- Correct answerD$(-8,12)$
Solution
We have,
$$
\begin{aligned}
\frac{3 x^{2}-2 x+4}{(x-1)^{6}}=\frac{A_{1}}{x+1} &+\frac{A_{2}}{(x+1)^{2}}+\frac{A_{3}}{(x+1)^{3}} \\
&+\frac{A_{4}}{(x+1)^{4}}+\frac{A_{5}}{(x+1)^{5}}+\frac{A_{6}}{(x+1)^{6}} \ldots(\mathrm{i})
\end{aligned}
$$
Now, put $x=0$ in Eq. (i), we get
$$
A_{1}+A_{2}+A_{3}+A_{4}+A_{5}+A_{6}=4
$$
If $A_{1}+A_{3}+A_{5}=-8$
and $A_{2}+A_{1}+A_{6}=12$
satisfies the equation.
Hence, only option (i.e. $-8,12$ ) satisfies the solution.
Hence, the correct answer is option: 'D'.