Question

If ${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$, then $x\in$

• A. $(1,2]$
• B. $[3,4)$
• C. $(1,3]$
• D. $[1,4)$

Answer 1

Answer: (A), (B)

$[3,4)$

${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$
For $\log$ to be defined,
$4-x>0$ and $x-1>0$
$\Rightarrow x\in (1,4)\dots \left(1\right)$

Now, ${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$
$\Rightarrow -{\log }_2(4-x)\ge -{\log }_{2}2+{\log }_2(x-1)\Rightarrow {\log }_2(x-1)+{\log }_2(4-x)\le 1\Rightarrow {\log }_2\left[\right(x-1\left)\right(4-x\left)\right]\le 1\Rightarrow -x^2+5x-4\le 2\Rightarrow x^2-5x+6\ge 0\Rightarrow x\in (-\infty ,2]\cup [3,\infty )\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$x\in (1,2]\cup [3,4)$