Question

If ${\log }_{\frac{1}{2}}^2(\sin \theta )+1={\log }_{\frac{1}{2}}x$ and ${\log }_{2}x\ge 2{\log }_2(\sin \theta )$, then

• A. ${\cot }^2\theta +\frac{1}{x}=7$
• B. $8x\sin \theta =2$
• C. $\sum _{r=1}^{\infty }{\left(\frac{x}{\sin \theta }\right)}^r=1$
• D. $\sum _{r=1}^{\infty }{\left(\frac{x}{\sin \theta }\right)}^r=2$

Answer 1

Answer: (A), (C)

${\log }_{\frac{1}{2}}^2(\sin \theta )+1={\log }_{\frac{1}{2}}x$ .....(i)
${\log }_{\frac{1}{2}}x\le 2{\log }_{\frac{1}{2}}\sin \theta \dots \dots \text{ (ii) }$
$1+{\log }_{\frac{1}{2}}^2(\sin \theta )\le 2{\log }_{\frac{1}{2}}(\sin \theta )$
${\log }_{\frac{1}{2}}^2(\sin \theta )-2{\log }_{\frac{1}{2}}(\sin \theta )+1\le 0$
${\left({\log }_{\frac{1}{2}}\sin \theta -1\right)}^2\le 0\Rightarrow {\log }_{\frac{1}{2}}\sin \theta =1$
$\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6}$
$\text{ and }{\log }_{\frac{1}{2}}x=2\Rightarrow x=\frac{1}{4}.$
Now, verify the options.