Question

The number of ordered pairs $(x,y)$ satisfying $4({\log }_{2}x{)}^2+1=2{\log }_{2}y$ and ${\log }_2{x}^2\ge {\log }_{2}y$, is

• A. $1$
• B. $2$
• C. more than $2$ but finite
• D. infinite

Answer 1

The correct option is A $1$

We have,
$4({\log }_{2}x{)}^2+1=2{\log }_{2}y$ and ${\log }_2{x}^2\ge {\log }_{2}y$

$\Rightarrow 4({\log }_{2}x{)}^2+1\le 2{\log }_2{x}^2$
$\Rightarrow 4({\log }_{2}x{)}^2-4{\log }_{2}x+1\le 0$
$\Rightarrow (2{\log }_{2}x-1{)}^2\le 0\Rightarrow {\log }_{2}x=\frac{1}{2}\Rightarrow x=\sqrt{2}$

Since, here there is only one value of $x$ i.e., $x=\sqrt{2}$
$\Rightarrow 4({\log }_2\sqrt{2}{)}^2+1=2{\log }_{2}y$
$\Rightarrow 2=2{\log }_{2}y$
$\Rightarrow y=2$

Hence, there is only one ordered pair i.e., $(\sqrt{2},2).$