Question

Set of values of ${}^{\prime }{\alpha }^{\prime }$ in $[0,2\pi ]$ for which $m={\log }_{(x+\frac{1}{x})}(2\sin \alpha -1)\le 0$, is

• A. $\left[\frac{\pi }{6},\frac{5\pi }{6}\right]$
• B. $\left(\frac{\pi }{6},\frac{5\pi }{6}\right)$
• C. $(0,\frac{\pi }{6})\cup \left(\frac{5\pi }{6},\pi \right)$
• D. $\left(\frac{5\pi }{6},\frac{7\pi }{6}\right)$

Answer 1

The correct option is B $\left(\frac{\pi }{6},\frac{5\pi }{6}\right)$

As $x+\frac{1}{x}\le -2$ or $x+\frac{1}{x}\ge 2$

${\log }_{(x+\frac{1}{x})}(2\sin \alpha -1)\le 0$

$\Rightarrow 2\sin \alpha -1\le {(x+\frac{1}{x})}^0=1$

$\Rightarrow 2\sin \alpha -1\le 0$

$\Rightarrow \sin \alpha \le 0$ ...(1)

log is defined when

$2\sin \alpha -1<0\Rightarrow \sin \alpha <\frac{1}{2}$ ...(2)

From (1) and (2)

$\sin \alpha <\frac{1}{2}\Rightarrow \alpha \in (\frac{\pi }{6},\frac{5\pi }{6})$