Question

The domain of the function, $y=\sqrt{\frac{1}{2}{\log }_{4}16-{\log }_8\left(x^2-4x+3\right)}$ is

• A. $\left[-1,1\right]\cup \left(3,5\right]$
• B. $\left[-1,1\right]\cup \left[3,5\right)$
• C. $\left[-1,1\right)\cup \left(3,5\right]$
• D. None of the above

Answer 1

Answer: (C)

$\left[-1,1\right)\cup \left(3,5\right]$

To define logarithmic function,

${\log }_8(x^2-4x+3)$

$\Rightarrow x^2-4x+3>0$

$\Rightarrow (x-3)(x-1)>0$

$\Rightarrow x\in (-\infty ,1)\cup (3,\infty )$

and to define square root,

$\frac{1}{2}{\log }_{4}16-{\log }_8(x^2-4x+3)\ge 0$

$\Rightarrow \frac{2}{2}{\log }_{4}4-{\log }_8(x^2-4x+3)\ge 0$

$\Rightarrow 1-{\log }_8(x^2-4x+3)\ge 0$

$\Rightarrow (x^2-4x+3)\le 8$

$\Rightarrow x^2-4x-5\le 0$

$\Rightarrow (x-5)(x+1)\le 0$

$\Rightarrow x\in (-1,5)$

Taking intersection of both sets,we get,

$\Rightarrow x\in [-1,1)\cup (3,5]$