Question

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

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

Answer 1

Answer: (A), (C)

$ \log _{\frac{1}{2}}^2(\sin \theta)+1=\log _{\frac{1}{2}} x $ .....(i)
$\log _{\frac{1}{2}} x \leq 2 \log _{\frac{1}{2}} \sin \theta \ldots \ldots \text { (ii) }$
$1+\log _{\frac{1}{2}}^2(\sin \theta) \leq 2 \log _{\frac{1}{2}}(\sin \theta)$
$\log _{\frac{1}{2}}^2(\sin \theta)-2 \log _{\frac{1}{2}}(\sin \theta)+1 \leq 0 $
$\left(\log _{\frac{1}{2}} \sin \theta-1\right)^2 \leq 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.