Question

If ${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$, then $x$ belongs to which interval?

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

Answer 1

Answer: (A), (B)

$(1,2]$
$[3,4)$

${\log }_{1/2}(4-x)\ge {\log }_{1/2}2-{\log }_{1/2}(x-1)$
Where, $4-x>0$ & $x-1>0\Rightarrow x\in (1,4)\Rightarrow \left(1\right)$
$\Rightarrow {\log }_2(4-x)\le {\log }_{2}2-{\log }_2(x-1)[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$
$\Rightarrow {\log }_2(4-x)\le {\log }_2\frac{2}{x-1}[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow (4-x)(x-1)\le 2$
$\Rightarrow x^2-5x+6\ge 0$
$\Rightarrow x\ge 3$ or $x\le 2\Rightarrow \left(2\right)$
Therefore, using (1) and (2)
$x\in (1,2]\cup [3,4)$
Ans: A,B