Question

Number of integral values of $x$, that satisfy ${\log }_{0.3}(x^2-x+1)>0$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

Consider ${\log }_{0.3}(x^2-x+1)$
$x^2-x+1>0\forall x\in R\text{as}D<0$
$\Rightarrow x\in R$

Now, ${\log }_{0.3}(x^2-x+1)>0$
$\Rightarrow x^2-x+1<1[\because a=0.3<1]$
$\Rightarrow x^2-x<0$
$\Rightarrow x(x-1)<0$
$\Rightarrow 0<x<1$
So, there is no integral value of $x$ satisfying the inequality.