Question

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

• A. $S=\left\{1\right\}$
• B. $S=Z$
• C. $S=N$
• D. $S=\{1,2,3\}$

Answer 1

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

$S=\{x\in N:2+{\log }_2\sqrt{x+1}>1-{\log }_{1/2}\sqrt{4-x^2}\}$
We have,
$2+{\log }_2\sqrt{x+1}>1-{\log }_{1/2}\sqrt{4-x^2}$
$\Rightarrow x+1>0,4-x^2>0$
$\Rightarrow x>-1,-2<x<2$
$\Rightarrow -1<x<2...\left(1\right)$

Now,
$1+{\log }_2\sqrt{x+1}>{\log }_2\sqrt{4-x^2}$
$\Rightarrow {\log }_{2}2\sqrt{x+1}>{\log }_2\sqrt{4-x^2}$
$\Rightarrow 2\sqrt{x+1}>\sqrt{4-x^2}$
Squaring both the sides
$\Rightarrow 4(x+1)>4-x^2$
$\Rightarrow x^2+4x>0$
$\Rightarrow x>0$ or $x<-4...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 0<x<2$