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)

$[3,4)$

${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$
$\Rightarrow 4-x>0$ and $x-1>0$
$\Rightarrow x\in (1,4)\cdots \left(1\right)$

Now,
${\log }_{1/2}(4-x)(x-1)\ge {\log }_{1/2}2$
As the base of the log function is less than $1$, so the inequality will reverse,
$\Rightarrow (4-x)(x-1)\le 2$
$\Rightarrow x^2-5x+6\ge 0$
$\Rightarrow (x-3)(x-2)\ge 0$
$\Rightarrow x\ge 3$ or $x\le 2\cdots \left(2\right)$

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