Question

The solution set of $1+{\log }_5(x^2+1)\ge {\log }_5(x^2+4x+1)$ contains

• A. $(-\infty ,-2-\sqrt{3})$
• B. $(-2+\sqrt{3},\infty )$
• C. $(-\infty ,-2+\sqrt{3})$
• D. $(-2-\sqrt{3},-2+\sqrt{3})$

Answer 1

Answer: (A), (B)

$(-2+\sqrt{3},\infty )$

${\log }_5(x^2+4x+1)$ is defined for $x^2+4x+1>0$
$\Rightarrow (x+2{)}^2>3$
$\text{i.e.,}x\in (-\infty ,-2-\sqrt{3})\cup (-2+\sqrt{3},\infty )$

We have,
$1+{\log }_5(x^2+1)\ge {\log }_5(x^2+4x+1)$
$\Rightarrow {\log }_{5}5+{\log }_5(x^2+1)\ge {\log }_5(x^2+4x+1)$
$\Rightarrow 5(x^2+1)\ge x^2+4x+1$
$\Rightarrow 4x^2-4x+4\ge 0$
$\Rightarrow x^2-x+1\ge 0,\Rightarrow {(x-\frac{1}{2})}^2+\frac{3}{4}\ge 0$
which is true for all $x\in R$

Hence, the solution set of the given inequality is
$(-\infty ,-2-\sqrt{3})\cup (-2+\sqrt{3},\infty )$