Q67 / 160 · TS EAMCET 2011 · Mathematics

$\begin{aligned} & \frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\ & \Rightarrow A+C=\end{aligned}$

  • A
    4
  • B
    5
  • C
    6
  • Correct answer
    D
    8

Solution

Given, $\begin{array}{r}\frac{x^2+x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2} \\ +\frac{C}{x-3}\end{array}$ $\begin{aligned} \Rightarrow x^2+x & +1=A(x-2)(x-3) \\ & +B(x-1)(x-3)+C(x-1)(x-2)\end{aligned}$ Put $x=1,2$ and 3 respectively we get $\begin{array}{lc} & 1+1+1=A(1-2)(1-3) \\ \Rightarrow & 3=2 A \Rightarrow A=\frac{3}{2} \\ & 2^2+2+1=B(2-1)(2-3) \\ \Rightarrow & 7=-B \Rightarrow B=-7 \\ \text { and } & 3^2+3+1=C(3-1)(3-2) \end{array}$ $\begin{array}{ll}\Rightarrow & 13=2 C \Rightarrow C=\frac{13}{2} \\ \therefore & A+C=\frac{3}{2}+\frac{13}{2}=8\end{array}$

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

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