If $\sqrt{9x^2+6x+1} < (2-x)$,then:

  • A
    $x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$
  • B
    $x \in \left(-\frac{3}{2}, \frac{1}{4}\right]$
  • C
    $x \in \left[-\frac{3}{2}, \frac{1}{4}\right)$
  • D
    $x < \frac{1}{4}$

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