Question

If ${\log }_2(1+x^{-1})-{\log }_{\frac{1}{2}}(1+\frac{x}{4})\ge 1$, then $x$ lies in the interval

• A. $(5,\infty )$
• B. $(\sqrt{5},5)$
• C. $(4,\infty )$
• D. none of these

Answer 1

The correct option is D none of these

Now domain for the above expression will be
$1+\frac{1}{x}>0$
$1>\frac{-1}{x}$
$x\in (-1,\infty )-\left\{0\right\}$ ...(i)

And
$1+\frac{x}{4}>0$
$\frac{x}{4}>-1$
$x>-4$...(ii)
From i and ii, we get the domain as $(-1,\infty )-\left\{0\right\}$.
${\log }_2(1+\frac{1}{x})-{\log }_2(1+\frac{x}{4})\ge 1[\because {\log }_{1/a}b=-{\log }_{a}b]$
${\log }_2\left(\frac{1+\frac{1}{x}}{1+\frac{x}{4}}\right)\ge 1$

$[\because \log a-\log b=\log \frac{a}{b}]$
$\frac{1+\frac{1}{x}}{1+\frac{x}{4}}\ge 2$
$\frac{x+1}{x}\ge 2+\frac{x}{2}$
$x+1\ge \frac{x^2}{2}+2x$
$2x+2\ge x^2+4x$
$x^2+2x-2\le 0$
$(x+1{)}^2-3\le 0$
$(x+1{)}^2\le 3$
$-\sqrt{3}\le (x+1)\le \sqrt{3}$
$-(1+\sqrt{3})\le x\le \sqrt{3}-1$
Hence
$x\in [-(1+\sqrt{3}),\sqrt{3}-1]$.
However domain is $(-1,\infty )-\left\{0\right\}$.
Hence the inequality is true for
$x\in (-1,\sqrt{3}-1]-\left\{0\right\}$.