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Question: Answered & Verified by Expert
If $\frac{5 x^2+2}{x^3+x}=\frac{A_1}{x}+\frac{A_2 x+A_3}{x^2+1}$, then $\left(A_1, A_2, A_3\right)=$
MathematicsBasic of MathematicsAP EAMCETAP EAMCET 2017 (26 Apr Shift 1)
Options:
  • A (0, 2, 3)
  • B (3, 0, 2)
  • C (2, 3, 0)
  • D (2, 0, 3)
Solution:
1521 Upvotes Verified Answer
The correct answer is: (2, 3, 0)
$\begin{aligned}
& \text {Given that, } \frac{5 x^2+2}{x^3+x}=\frac{A_1}{x}+\frac{A_2 x+A_3}{x^2+1} \\
& \Rightarrow \frac{5 x^2+2}{x\left(x^2+1\right)}=\frac{A_1\left(x^2+1\right)+\left(A_2 x+A_3\right) x}{x\left(x^2+1\right)} \\
& \Rightarrow 5 x^2+2=A_1\left(x^2+1\right)+A_2 x^2+A_3 x \\
& \Rightarrow 5 x^2+2=\left(A_1+A_2\right) x^2+A_3 x+A_1
\end{aligned}$
On comparing the coefficients of $x^2, x$ and constant term, we get
$\therefore \quad A_1=2 ; A_2=3 \text { and } A_3=0$$\begin{aligned}
& \text {Given that, } \frac{5 x^2+2}{x^3+x}=\frac{A_1}{x}+\frac{A_2 x+A_3}{x^2+1} \\
& \Rightarrow \frac{5 x^2+2}{x\left(x^2+1\right)}=\frac{A_1\left(x^2+1\right)+\left(A_2 x+A_3\right) x}{x\left(x^2+1\right)} \\
& \Rightarrow 5 x^2+2=A_1\left(x^2+1\right)+A_2 x^2+A_3 x \\
& \Rightarrow 5 x^2+2=\left(A_1+A_2\right) x^2+A_3 x+A_1
\end{aligned}$
On comparing the coefficients of $x^2, x$ and constant term, we get
$\therefore A_1=2 ; A_2=3 \text { and } A_3=0$

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