Question

If ${\log }_{\frac{1}{2}}\left(x^2-5x+7\right)>0$, then an exhaustive range of values of $x$ is

• A. $(-\infty ,2)\cup (3,\infty )$
• B. $(2,3)$
• C. $(-\infty ,1)\cup (1,2)\cup (2,\infty )$
• D. None of these

Answer 1

The correct option is B

$(2,3)$

Explanation for the correct answer:

Given: ${\log }_{\frac{1}{2}}\left(x^2-5x+7\right)>0$

$$\begin{gathered} \Rightarrow x^2-5x+7<{\left(\frac{1}{2}\right)}^0\left[\because {\log }_a\left(b\right)=c\Rightarrow b=a^c\right] \\ \Rightarrow x^2-5x+7<1 \\ \Rightarrow x^2-5x+6<0 \\ \Rightarrow x^2-2x-3x+6<0 \\ \Rightarrow x(x-2)-3(x-2)<0 \\ \Rightarrow (x-2)(x-3)<0 \end{gathered}$$

From the options we can say that the above condition is satisfied by option(B) i.e. $(2,3)$

Hence, option(B) i.e.$(2,3)$ is correct.