Question

The solution set of the inequality $x{\log }_{\frac{1}{10}}(x^2+x+1)>0$ is

• A. $(\infty ,0)$
• B. $(0,1)$
• C. $(-\infty ,-1)$
• D. $(-2,-1)$

Answer 1

Answer: (C), (D)

The correct options are
C $(-\infty ,-1)$
D $(-2,-1)$

$xl{n}_{\frac{1}{10}}(x^2+x+1)>0$
$x\frac{ln(x^2+x+1)}{-ln\left(10\right)}>0$
Or
$xln(x^2+x+1)\frac{1}{2.303}<0$
Now, in
$x^2+x+1$
$D<0$
Hence
$x^2+x+1>0$ for all $xϵR$.
$x^2+x+1=1$
Implies $x=0$ and $x=-1$.
Thus for $xln(x^2+x+1)\frac{1}{2.303}<0$ then
$x<-1$ and $x<0$
Hence
$xϵ(-\infty ,-1)$