Question

Let $\alpha$ be a solution of the equation $2[x+32]=3[x-64]$ where $[x]$ is the greatest integer less than or equal to $x$ and let $\beta=\prod\limits_{r=1}^9 \sin \left(\frac{2 r-1}{18}\right) \pi$, then

• A. $[\alpha]=[\beta]$
• B. $\alpha=\frac{2051}{8}$
• C. $[\alpha]\left[\frac{1}{\beta}\right]=1$
• D. $\left[\frac{1}{\alpha}\right]+\left[\frac{1}{\beta}\right]=2^8$

Answer 1

Answer: (B), (D)

We have
$2[ x ]+64=3[ x ]-192 \Rightarrow[ x ]=256 \Rightarrow x \in[256,257)$
Also,
$\beta=\sin \frac{\pi}{18} \cdot \sin \frac{3 \pi}{18} \ldots \ldots \sin \frac{9 \pi}{18} \ldots \ldots \sin \frac{17 \pi}{18} $
$=\sin ^2 \frac{\pi}{18} \cdot \sin ^2 \frac{3 \pi}{18} \cdot \sin ^2 \frac{5 \pi}{18} \cdot \sin ^2 \frac{7 \pi}{18}=\sin ^2 10^{\circ} \cdot \sin ^2 30^{\circ} \cdot \sin ^2 50^{\circ} \cdot \sin ^2 70^{\circ}$
$\frac{1}{4}\left(\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70\right)^2=\frac{1}{4}\left(\frac{\sin 30^{\circ}}{4}\right)^2$
$\therefore \quad \beta=\frac{1}{4}\left(\frac{1}{8}\right)^2=\frac{1}{256}$