Question

If $S=\{x\in N:2+{\log }_2\sqrt{x+1}>1-{\log }_{\frac{1}{2}}\sqrt{4-x^2}\}$, then

• A. $S=\left\{1\right\}$
• B. $S=Z$
• C. $S=N$
• D. None of these

Answer 1

The correct option is A $S=\left\{1\right\}$

For $\log$ to be defined
$x+1>0\Rightarrow x>-1$ and
$x+1\ne 0\Rightarrow x\ne -1\cdots \left(1\right)$

Also,
$4-x^2>0\Rightarrow -2<x<2\cdots \left(2\right)\Rightarrow 2+{\log }_2\sqrt{x+1}>1-{\log }_{\frac{1}{2}}\sqrt{4-x^2}\Rightarrow 2+\frac{1}{2}{\log }_2\sqrt{x+1}>1+\frac{1}{2}{\log }_2\sqrt{4-x^2}\Rightarrow 1+\frac{1}{2}{\log }_2\sqrt{x+1}-\frac{1}{2}{\log }_2\sqrt{4-x^2}>0\Rightarrow {\log }_{2}2+\frac{1}{2}{\log }_2\sqrt{x+1}-\frac{1}{2}{\log }_2\sqrt{4-x^2}>0\Rightarrow {\log }_2(\frac{2\sqrt{x+1}}{\sqrt{4-x^2}})>0\Rightarrow (\frac{2\sqrt{x+1}}{\sqrt{4-x^2}})>1\Rightarrow 4(x+1)>4-x^2\Rightarrow x^2+4x>0\Rightarrow x(x+4)>0\Rightarrow x>0\text{or}x<-4\text{From (1), (2) and (3)}x\in (0,2)$