Q59 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
$\begin{aligned}
& \text { If } \frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)}=\frac{A x+B}{x^2+3}+\frac{C x+D}{x^2+2}+ \\
& \frac{E x^3+F x^2+G x+H}{x^4+x^2} \text { then (E + F) (C + D) (A) }=
\end{aligned}$
- A$-\frac{1}{4}$
- B$-\frac{3}{4}$
- C$\frac{3}{4}$
- Correct answerD$\frac{1}{4}$
Solution
$\begin{aligned} & \text { } \frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)} \\ & =\frac{A^{\prime}}{x}+\frac{B^{\prime}}{x^2}+\frac{C^{\prime} x+D^{\prime}}{x^2+1}+\frac{E^{\prime} x+F^{\prime}}{x^2+2}+\frac{G^{\prime} x+H^{\prime}}{x^2+3} \\ & \Rightarrow x+2=x\left(x^2+1\right)\left(x^2+2\right)\left(x^2+3\right) A^{\prime} \\ & +\left(x^2+1\right)\left(x^2+2\right)\left(x^2+3\right) B^{\prime}+\left(C^{\prime} x+D\right) x^2\left(x^2+2\right) \\ & \left(x^2+3\right) +x^2\left(x^2+1\right)\left(x^2+3\right)\left(E^{\prime} x+F^{\prime}\right)+x^2\left(x^2+1\right)\left(x^2+\right. \\ & 2)\left(G^{\prime} x+H^{\prime}\right)\end{aligned}$
$\begin{aligned} & \Rightarrow x+2=x^7\left(A^{\prime}+C^{\prime}+E^{\prime}+G^{\prime}\right)+x^6\left(B^{\prime}+D^{\prime}+F^{\prime}+G^{\prime}\right) \\ & +x^5\left(6 A^{\prime}+5 C^{\prime}+4 E^{\prime}+3 G^{\prime}\right)+x^4\left(6 B^{\prime}+5 D^{\prime}+4 F^{\prime}+3 H^{\prime}\right)+x^3 \\ & \left(11 A^{\prime}+6 C^{\prime}+3 E^{\prime}+2 G^{\prime}\right)+x^2(11 B+6 D+3 F+2 H)+6 x A+6 B\end{aligned}$
On comparing coefficients $\begin{aligned} & A^{\prime}+C^{\prime}+E^{\prime}+G^{\prime}=0, B^{\prime}+D^{\prime}+F^{\prime}+H^{\prime}=0 \\ & 6 A^{\prime}+5 C^{\prime}+4 F^{\prime}+3 G^{\prime}=0,6 B^{\prime}+5 D^{\prime}+4 F^{\prime}+3 H^{\prime}=0 \\ & 11 A^{\prime}+6 C^{\prime}+3 E^{\prime}+2 G^{\prime}=0,11 B^{\prime}+6 D^{\prime}+3 F^{\prime}+2 H^{\prime}=0 \\ & 6 A^{\prime}=1,6 B^{\prime}=2 \end{aligned}$
$\therefore$ On solving we get $\begin{aligned} & A^{\prime}=\frac{1}{6}, B^{\prime}=\frac{1}{3}, C^{\prime}=\frac{-1}{2}, D^{\prime}=-1 \\ & E^{\prime}=\frac{1}{2}, F^{\prime}=1, G^{\prime}=\frac{-1}{6}, H^{\prime}=\frac{-1}{3} \\ & \therefore \frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)} \end{aligned}$
$\begin{aligned} & =\frac{1}{6 x}+\frac{1}{3 x^2}-\frac{x+2}{2\left(x^2+1\right)}+\frac{x+2}{2\left(x^2+2\right)}-\frac{x+2}{6\left(x^2+3\right)} \\ & =\frac{-\frac{x}{6}-\frac{1}{3}}{\left(x^2+3\right)}+\frac{\frac{x}{2}+1}{x^2+2}+\frac{\frac{-x^3}{3}-\frac{2 x^2}{3}+\frac{x}{6}+\frac{1}{3}}{\left(x^4+x^2\right)} \end{aligned}$
On comparing with $\frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)}$
$=\frac{A x+B}{x^2+3}+\frac{C x+D}{x^2+2}+\frac{E x^3+F x^2+G x+H}{x^4+x^2}$
we get, $A=\frac{-1}{6}, B=\frac{-1}{3}, C=\frac{1}{2}, D=1, E=\frac{-1}{3}$, $F=\frac{-2}{3}, G=\frac{1}{6}$ and $H=\frac{1}{3}$ So, $(E+F)(C+D)(A)=\left(\frac{-1}{3}-\frac{2}{3}\right)\left(\frac{1}{2}+1\right)\left(\frac{-1}{6}\right)=\frac{1}{4}$.
$\begin{aligned} & \Rightarrow x+2=x^7\left(A^{\prime}+C^{\prime}+E^{\prime}+G^{\prime}\right)+x^6\left(B^{\prime}+D^{\prime}+F^{\prime}+G^{\prime}\right) \\ & +x^5\left(6 A^{\prime}+5 C^{\prime}+4 E^{\prime}+3 G^{\prime}\right)+x^4\left(6 B^{\prime}+5 D^{\prime}+4 F^{\prime}+3 H^{\prime}\right)+x^3 \\ & \left(11 A^{\prime}+6 C^{\prime}+3 E^{\prime}+2 G^{\prime}\right)+x^2(11 B+6 D+3 F+2 H)+6 x A+6 B\end{aligned}$
On comparing coefficients $\begin{aligned} & A^{\prime}+C^{\prime}+E^{\prime}+G^{\prime}=0, B^{\prime}+D^{\prime}+F^{\prime}+H^{\prime}=0 \\ & 6 A^{\prime}+5 C^{\prime}+4 F^{\prime}+3 G^{\prime}=0,6 B^{\prime}+5 D^{\prime}+4 F^{\prime}+3 H^{\prime}=0 \\ & 11 A^{\prime}+6 C^{\prime}+3 E^{\prime}+2 G^{\prime}=0,11 B^{\prime}+6 D^{\prime}+3 F^{\prime}+2 H^{\prime}=0 \\ & 6 A^{\prime}=1,6 B^{\prime}=2 \end{aligned}$
$\therefore$ On solving we get $\begin{aligned} & A^{\prime}=\frac{1}{6}, B^{\prime}=\frac{1}{3}, C^{\prime}=\frac{-1}{2}, D^{\prime}=-1 \\ & E^{\prime}=\frac{1}{2}, F^{\prime}=1, G^{\prime}=\frac{-1}{6}, H^{\prime}=\frac{-1}{3} \\ & \therefore \frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)} \end{aligned}$
$\begin{aligned} & =\frac{1}{6 x}+\frac{1}{3 x^2}-\frac{x+2}{2\left(x^2+1\right)}+\frac{x+2}{2\left(x^2+2\right)}-\frac{x+2}{6\left(x^2+3\right)} \\ & =\frac{-\frac{x}{6}-\frac{1}{3}}{\left(x^2+3\right)}+\frac{\frac{x}{2}+1}{x^2+2}+\frac{\frac{-x^3}{3}-\frac{2 x^2}{3}+\frac{x}{6}+\frac{1}{3}}{\left(x^4+x^2\right)} \end{aligned}$
On comparing with $\frac{x+2}{\left(x^2+3\right)\left(x^4+x^2\right)\left(x^2+2\right)}$
$=\frac{A x+B}{x^2+3}+\frac{C x+D}{x^2+2}+\frac{E x^3+F x^2+G x+H}{x^4+x^2}$
we get, $A=\frac{-1}{6}, B=\frac{-1}{3}, C=\frac{1}{2}, D=1, E=\frac{-1}{3}$, $F=\frac{-2}{3}, G=\frac{1}{6}$ and $H=\frac{1}{3}$ So, $(E+F)(C+D)(A)=\left(\frac{-1}{3}-\frac{2}{3}\right)\left(\frac{1}{2}+1\right)\left(\frac{-1}{6}\right)=\frac{1}{4}$.
Hence, the correct answer is option: 'D'.