Q54 / 160 · AP EAMCET 2006 · Mathematics

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

  • Correct answer
    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
    $x < \frac{1}{4}$

Solution

$\sqrt{9 x^2+6 x+1} < (2-x)$ $\Rightarrow \quad \sqrt{(3 x+1)^2} < (2-x)$ $\Rightarrow \quad \pm(3 x+1) < 2-x$ Taking + ve sign $3 x+1 < 2-x$ $\Rightarrow \quad x < \frac{1}{4}$ Taking -ve sign $-3 x-1 < 2-x$ $\Rightarrow \quad-2 x < 3$ $\Rightarrow \quad x>-\frac{3}{2}$ $\therefore \quad x \in\left(-\frac{3}{2}, \frac{1}{4}\right)$

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

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