Question

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

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

Answer 1

Answer: (B)

$\log _{\left(x+\frac{1}{x}\right)}(2 \sin \alpha-1) \leq 0$
$\because x+\frac{1}{x}>2$
so $ 2 \sin \alpha-1 \leq 1 \& 2 \sin \alpha-1>0$
$\Rightarrow 2 \sin \alpha \leq 2 \& \sin \alpha>\frac{1}{2}$
$\Rightarrow \sin \alpha \leq 1$
$\alpha \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$