Question

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

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

Answer 1

The correct option is C $(2n\pi -\frac{\pi }{2},2n\pi +\frac{\pi }{2})-\left\{2n\pi \right\},n\in Z;(0,\frac{\pi }{2})$

Since in ${\text{cosec}}^{-1}t$
$t\in (-\infty ,-1]\cup [1,\infty )$
So, $\sqrt{t}\in [1,\infty )$
$\sqrt{{\log }_{\left(\frac{3-4\sec x}{1-2\sec x}\right)}2}\ge 1$
Squaring both sides,
${\log }_{\left(\frac{3-4\sec x}{1-2\sec x}\right)}2\ge 1$
and base of $\log$
$1<\frac{3-4\sec x}{1-2\sec x}<2$

Now, $\frac{3-4\sec x}{1-2\sec x}<2$
$\Rightarrow \frac{3-4\sec x}{1-2\sec x}-2<0$
$\Rightarrow \frac{1}{1-2\sec x}<0$
$\Rightarrow 1-2\sec x<0$
$\Rightarrow \sec x>\frac{1}{2}\cdots \left(1\right)$
(If $\sec x=1$, then base is $1$ $\Rightarrow x\ne 2n\pi$)

And
$\frac{3-4\sec x}{1-2\sec x}>1$
$\Rightarrow \frac{3-4\sec x}{1-2\sec x}-1>0$
$\Rightarrow \frac{1-\sec x}{1-2\sec x}>0$
$\Rightarrow \frac{\sec x-1}{2\sec x-1}>0$
$\Rightarrow \sec x-1>0$
$\Rightarrow \sec x>1\cdots \left(2\right)$

Clearly, from equation $\left(1\right)$ and $\left(2\right),$
$x$ must lie in $1^{\text{st}}$ or $4^{\text{th}}$ quadrant.
So, domain $=(2n\pi -\frac{\pi }{2},2n\pi +\frac{\pi }{2})-\left\{2n\pi \right\}$
and the range is $(0,\frac{\pi }{2})$ since $t$ does not take negative values.