Question

The domain and range of the function $f(x)=\operatorname{cosec}^{-1} \sqrt{\log _{\frac{3-4 \sec x}{1-2 \sec x}} 2}$ are respectively

• A. $R ;\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
• B. $R ^{+} ;\left(0, \frac{\pi}{2}\right)$
• C. $\left(2 n \pi-\frac{\pi}{2}, 2 n \pi+\frac{\pi}{2}\right)-\{2 n \pi\} ;\left(0, \frac{\pi}{2}\right)$
• D. $\left(2 n \pi-\frac{\pi}{2}, 2 n \pi+\frac{\pi}{2}\right)-\{2 n \pi\} ;\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\{0\}$

Answer 1

Answer: (C)

$\text { note that, } \log _{\frac{4 \sec x-3}{2 \sec x-1}} 2 \geq 1 \text { and } 1<\frac{4 \sec x-3}{2 \sec x-1} \leq 2 $
$\text { But } \frac{4 \sec x-3}{2 \sec x-1} \neq 2 \text { as in this case }-3=-2 \text { (not possible) }$
$\text { now } \frac{4 \sec x-3}{2 \sec x-1}-2<0$
$\frac{-1}{2 \sec x-1}<0 \Rightarrow \frac{1}{2 \sec x-1}>0 \text { or } 2 \sec x-1>0 $
$\Rightarrow \sec x>1 / 2 \text { which is alway True }$
$\text { (if } \sec x=1 \text { then base is } 1 \Rightarrow x \neq 0 \text { ) } $
$\text { and } \frac{4 \sec x-3}{2 \sec x-1}-1>0 $
$\frac{2 \sec x-2}{2 \sec x-1}>0 \Rightarrow \frac{\sec x-1}{2 \sec x-1}>0$

$\Rightarrow \sec x>1 \text { which is always true ....(2) } $
$\text { from (1) and (2) } x \text { must lies in } 1^{\text {st }} \text { or } 4^{\text {th }} \text { quad. except zero } $
$-\frac{\pi}{2}< x <\frac{\pi}{2}-\{0\} \text { or } 2 n \pi-\frac{\pi}{2}< x <2 n \pi+\frac{\pi}{2}-\{2 n \pi\} $
$\text { range is obv. }\left(0, \frac{\pi}{2}\right) $