Question

If $ \sqrt{9{{x}^{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} $

Answer 1

Answer: (A)

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