Match the following in their domains

	Column-I		Column-II
A)	$f\left(x\right)={\sin }^{2}2x-2{\sin }^{2}x$	P)	Range contains no natural number
B)	$f\left(x\right)=[\left|\sin x\right|+\left|\cos x\right|]$ where [.] denotes GIF	Q)	Range contains at least one integer
C)	$f\left(x\right)=\sqrt{\ln \left(\cos \left(\sin x\right)\right)}$	R)	Many one but not even function
D)	$f\left(x\right)={\left(\sin x\right)}^0+{\left(\cos x\right)}^0$	S)	Both many one and even function
		T)	Periodic but not odd function

$$\begin{gathered} A) f\left(x\right)={\sin }^{2}2x-2{\sin }^{2}x=1-{\cos }^{2}2x-(1-\cos 2x) \\ =-{\cos }^{2}2x+\cos 2x \\ \therefore f(-x)=f\left(x\right)\to even function. \end{gathered}$$

Even function is symmetric about y-axis, so if any horizontal line cuts it, it cuts at more than one point.

Hence $f\left(x\right)$ is many one

$f(x+\pi )=f\left(x\right)\to$periodic with period $\pi$

$f\left(x\right)=-[{\cos }^{2}2x-\cos 2x]=-[{(\cos 2x-\frac{1}{2})}^2-\frac{1}{4}]=\frac{1}{4}-{(\cos 2x-\frac{1}{2})}^2$

$f_{\min }=\frac{1}{4}-{(-1-\frac{1}{2})}^2=-2,f_{\max }=\frac{1}{4}-0=\frac{1}{4}$

Range $=[-2,\frac{1}{4}]$

Range contains no natural number, but contains atleast one integer. 

$\left(B\right) f\left(x\right)=\frac{4}{\pi }{\sin }^{-1}\left(\sin \pi x\right)$

Range is $[\frac{4}{\pi }\times \frac{-\pi }{2},\frac{4}{\pi }\times \frac{\pi }{2}]=[-2,2]\to contains atleast one integer$

$f(-x)=\frac{4}{\pi }{\sin }^{-1}(\sin 1-\pi x)=-f\left(x\right)\to odd function$

Also, $f\left(x\right)$ is periodic, Hence any horizontal line can cut it at more than are point, so it is many one

Ans : B-Q,R

$\left(C\right) f\left(x\right)=\sqrt{\ln \left(\cos \left(\sin x\right)\right)}$

$f(-x)=\sqrt{\ln \left(\mathrm{cossin}(-x)\right)}=\sqrt{\mathrm{lncos}\left(\sin x\right)}=f\left(x\right)\to$even function Graph of Even function is symmetric about y-axis, so it is many one. For $f\left(x\right)$ to be defined $\ln \left(\cos \left(\sin x\right)\right)\ge 0\Rightarrow \cos \left(\sin x\right)\ge 1$

But $\cos \theta$ cannot exceed 1

$\therefore \cos \left(\sin x\right)=1\Rightarrow \sin x=0\Rightarrow x=0$

Ans : C-P,Q,S

(D) conceptual