Question

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. $x<\frac{1}{4}$

Answer 1

Answer: (A)

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