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If $\frac{x^2+x+1}{x^2+2 x+1}=A+\frac{B}{x+1}+\frac{C}{(x+1)^2}$, then $A-B$ is equal to
Options:
Solution:
2432 Upvotes
Verified Answer
The correct answer is:
$2 C$
$$
\frac{x^2+x+1}{x^2+2 x+1}=1-\frac{x}{x^2+2 x+1}
$$
Now, $\frac{x}{x^2+2 x+1}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}$
$$
\Rightarrow \quad x=A(x+1)+B
$$
On equating the coefficient of $x$ and constant, we get
$$
\begin{array}{llll}
\Rightarrow & A=1 & \text { and } & A+B=0 \\
\Rightarrow & A=1 & \text { and } & B=-1
\end{array}
$$
From Eq. (i),
$$
\begin{aligned}
& \frac{x^2+x+1}{x^2+2 x+1}=1-\frac{1}{(x+1)}+\frac{1}{(x+1)^2} \\
& \Rightarrow A+\frac{B}{x+1}+\frac{C}{(x+1)^2} \\
& =1-\frac{1}{x+1}+\frac{1}{(x+1)^2} \\
& \Rightarrow \quad A=1, B=-1 \text { and } C=1 \\
& \text { Now, } \quad A-B=1+1=2 \\
& =2 C \\
&
\end{aligned}
$$
(given)
Hence, option (d) is correct.
\frac{x^2+x+1}{x^2+2 x+1}=1-\frac{x}{x^2+2 x+1}
$$
Now, $\frac{x}{x^2+2 x+1}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}$
$$
\Rightarrow \quad x=A(x+1)+B
$$
On equating the coefficient of $x$ and constant, we get
$$
\begin{array}{llll}
\Rightarrow & A=1 & \text { and } & A+B=0 \\
\Rightarrow & A=1 & \text { and } & B=-1
\end{array}
$$
From Eq. (i),
$$
\begin{aligned}
& \frac{x^2+x+1}{x^2+2 x+1}=1-\frac{1}{(x+1)}+\frac{1}{(x+1)^2} \\
& \Rightarrow A+\frac{B}{x+1}+\frac{C}{(x+1)^2} \\
& =1-\frac{1}{x+1}+\frac{1}{(x+1)^2} \\
& \Rightarrow \quad A=1, B=-1 \text { and } C=1 \\
& \text { Now, } \quad A-B=1+1=2 \\
& =2 C \\
&
\end{aligned}
$$
(given)
Hence, option (d) is correct.
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