Question

If $lo{g}_{1/2}\frac{x^2+6x+9}{2\left(x+1\right)}<-lo{g}_2\left(x+1\right),$ then x lies in the interval

• A. $\left(-1,-1+2\sqrt{2}\right)$
• B. $\left(1,-2\sqrt{2},2\right)$
• C. $\left(-1,\infty \right)$
• D. None of these

Answer 1

Answer: (A)

The log functions are defined if $\frac{x^2+6x+9}{2\left(x+1\right)}>0$ and
$x+1>0$
$\Rightarrow \frac{{\left(x+3\right)}^2}{2\left(x+1\right)}>0$ and $x+1>0$
$\Rightarrow x>-1$
Now the inequality is $lo{g}_{2^{-1}}\frac{x^2+6x+9}{2\left(x+1\right)}<-lo{g}_2\left(x+1\right)$
$\Rightarrow -lo{g}_2\frac{x^2+6x+9}{2\left(x+1\right)}<-lo{g}_2\left(x+1\right)$
$\Rightarrow lo{g}_2\frac{x^2+6x+9}{2\left(x+1\right)}>lo{g}_2\left(x+1\right)$
$\Rightarrow \frac{x^2+6x+9}{2\left(x+1\right)}>\left(x+1\right)$
$\Rightarrow \frac{-x^2+2x+7}{2\left(x+1\right)}>0$
$\Rightarrow \left(x+1\right)\left(x^2-2x-7\right)<0$
$\Rightarrow x^2-2x-7<0\left[\because x+1>0\right]$
$\Rightarrow -1-2\sqrt{2}<x<-1+2\sqrt{2}$,
but $x>-1\Rightarrow -1<x<-1+2\sqrt{2}$