Q42 / 160 · TS EAMCET 2009 · Mathematics

$\left\{x \in R: \frac{2 x-1}{x^3+4 x^2+3 x} \in R\right\}$ equals

  • A
    $R-\{0\}$
  • B
    $R-\{0,1,3\}$
  • Correct answer
    C
    $R-\{0,-1,-3\}$
  • D
    $R-\left\{0,-1,-3,+\frac{1}{2}\right\}$

Solution

$\begin{aligned} & \text { Let } A=\left\{x \in R: \frac{2 x-1}{x^3+4 x^2+3 x}\right\} \\ & \text { Now, } x^3+4 x^2+3 x=x\left(x^2+4 x+3\right) \\ & =x(x+3)(x+1) \\ & \therefore \quad A=R-\{0,-1,-3\} \\ & \end{aligned}$

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

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