Question

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

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

Answer 1

Answer: (C)

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

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