Q61 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $\frac{3 x^4-2 x^2+1}{(x-2)^4}=A+\frac{B}{x-2}+\frac{C}{(x-2)^2}+\frac{D}{(x-2)^3}+\frac{E}{(x-2)^4}$, then $2 A+3 B-C-D+E=$
- A0
- B1
- C-11
- Correct answerD-39
Solution
$\frac{3 x^4-2 x^2+1}{(x-2)^4}$
$\begin{aligned}
& =A+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}+\frac{D}{(x-2)^3}+\frac{E}{(x-2)^4} \\
& \text { L.H.S }=\frac{3 x^4-2 x^2+1}{(x-2)^4}=3+\frac{24 x^3-74 x^2+96 x-47}{(x-2)^4} \\
& \Rightarrow \mathrm{~A}=3 \\
& \text { Let } \frac{24 x^3-74 x^2+96 x-47}{(x-2)^4} \\
& =\frac{B}{(x-2)}+\frac{C}{(x-2)^2}+\frac{D}{(x-2)^3}+\frac{E}{(x-2)^4} \\
& \Rightarrow 24 x^3-74 x^2+96 x-47 \\
& =\mathrm{B}(x-3)^3+\mathrm{C}(x-2)^2+\mathrm{D}(\mathrm{x}-2)+\mathrm{E} \\
& \Rightarrow 24 x^3-74 x^2+96 x-47 \\
& \quad=\mathrm{B} x^3+(-6 \mathrm{~B}+\mathrm{C}) x^2+(12 \mathrm{~B}-4 \mathrm{C}+\mathrm{D}) x-8 \mathrm{~B}
\end{aligned}$
On comparing, we get $\mathrm{B}=24$ $\begin{aligned} & -6 B+C=-74 \Rightarrow C=70 \\ & 12 B-4 C+D=96 \Rightarrow D=88 \\ & -8 B+4 C-2 D+E=-47 \Rightarrow E=41 \end{aligned}$
Now, $2 A+3 B-C-D+E=-39$
On comparing, we get $\mathrm{B}=24$ $\begin{aligned} & -6 B+C=-74 \Rightarrow C=70 \\ & 12 B-4 C+D=96 \Rightarrow D=88 \\ & -8 B+4 C-2 D+E=-47 \Rightarrow E=41 \end{aligned}$
Now, $2 A+3 B-C-D+E=-39$
Hence, the correct answer is option: 'D'.