Question

If $|{\log }_{2}x+1|+|1-({\log }_{2}x{)}^2|=|{\log }_{2}x+({\log }_{2}x{)}^2|,$ then the true set of values of $x$ is $\left\{\lambda \right\}\cup [\mu ,\infty ).$ Then

• A. $2\lambda -3=0$
• B. $\mu -2=0$
• C. $2\lambda -\mu =0$
• D. $4\lambda -\mu =0$

Answer 1

Answer: (B), (D)

$4\lambda -\mu =0$

$|{\log }_{2}x+1|+|1-({\log }_{2}x{)}^2|=|{\log }_{2}x+({\log }_{2}x{)}^2|$
$\Rightarrow |{\log }_{2}x+1|+\left|\right({\log }_{2}x{)}^2-1|=|{\log }_{2}x+({\log }_{2}x{)}^2|$
$\therefore ({\log }_{2}x+1)\left(\right({\log }_{2}x{)}^2-1)\ge 0$
$\Rightarrow ({\log }_{2}x+1{)}^2({\log }_{2}x-1)\ge 0$
$\Rightarrow {\log }_{2}x=-1$ or ${\log }_{2}x\ge 1$
$\Rightarrow x\in \left\{\frac{1}{2}\right\}\cup [2,\infty )$
$\Rightarrow \lambda =\frac{1}{2},\mu =2.$