If $x$ is real, then the value of $\frac{x^2-3x+4}{x^2+3x+4}$ lies in the interval

  • A
    $[\frac{1}{3}, 3]$
  • B
    $[\frac{1}{5}, 5]$
  • C
    $[\frac{1}{6}, 6]$
  • D
    $[\frac{1}{7}, 7]$

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