Q51 / 160 · TS EAMCET 2008 · Mathematics

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

  • A
    $4 C$
  • B
    $4 C+1$
  • C
    $3 C$
  • Correct answer
    D
    $2 C$

Solution

$$ \frac{x^2+x+1}{x^2+2 x+1}=1-\frac{x}{x^2+2 x+1} $$ 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.

Hence, the correct answer is option: 'D'.

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