Question

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

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

Answer 1

Answer: (A)

$S=\left\{1\right\}$

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

$\Rightarrow 2+{\log }_2\sqrt{x+1}>1+{\log }_2\sqrt{4-x^2}[\because {\log }_{1/a}b=-{\log }_{a}b]$

Above equation is valid when $x+1>0$ $\Rightarrow x>-1$

and $4-x^2>0$
Therefore, $x\in (-1,2)$ ...(1)
Since, $2+{\log }_2\sqrt{x+1}>1+{\log }_2\sqrt{4-x^2}$

${\log }_2\sqrt{\frac{4-x^2}{x+1}}<1[\because \log a-\log b=\log \frac{a}{b}]$

$\Rightarrow \frac{4-x^2}{x+1}<2$
$\Rightarrow {(x+1)}^2>3$
$\Rightarrow x>\sqrt{3}-1$ or $x>-1-\sqrt{3}$
Therefore, from (1) $x\in (-1,2)$
Since, $S=\{x\in N\}$
$\Rightarrow S=\left\{1\right\}$
Ans: A