Question

Column I		Column II
A	$f(x)={\sin }^{-1}\left(\frac{2}{|\sin x-1|+|\sin x+1|}\right)$	P	$f(x)$ is many one
B	$f(x)={\cos }^{-1}(|x-1|-|x-2|)$	Q	Domain of $f(x)$ is $R$
C	$f(x)={\sin }^{-1}\left(\frac{\pi }{|{\sin }^{-1}x-(\pi /2)|+|{\sin }^{-1}x+(\pi /2)|}\right)$	R	Range contain only irrational number
D	$f(x)=\cos ({\cos }^{-1}\left|x[)+{\sin }^{-1}(\sin x)-{\mathrm{cosec}}^{-1}(\mathrm{cosec}x)+{\mathrm{cosec}}^{-1}\right|x\mid$	S	$f(x)$ is even.

• A. (A) P, Q, R, P; (B) P, Q; (C) P, R, S; (D) P, R, S
• B. (A) P, Q, R, Q; (B) P, Q; (C) P, R, S; (D) P, R, S
• C. (A) P, Q, R, R; (B) P, Q; (C) P, R, S; (D) P, R, S
• D. (A) P, Q, R, S; (B) P, Q; (C) P, R, S; (D) P, R, S

Answer 1

Answer: (D)

(A) Let $|\sin x-1|+|\sin x+1|=y$
if $\sin x=t\Rightarrow |t-1|+|t+1|=y$
if $t\in [-1,1]\Rightarrow y=2$
if $t\in (-\infty ,-1]\cup [1,\infty )\Rightarrow y>2$
$t\in [-1,1]\Rightarrow \sin x=t\Rightarrow x\in R\Rightarrow y=2$
$t\in (-\infty ,-1)\cup (1,\infty )\Rightarrow$ not possible because $\sin x\in [-1,1]$
$\therefore f(x)={\sin }^{-1}(2/2)={\sin }^{-1}(1)=\pi /2=$ constant (periodic)
$\Rightarrow P,Q,R,S$
(B) $\text{ Let }|x-1|-|x-2|=y$
$\text{ if }x\in (-\infty ,1]\Rightarrow y=-1$
$\text{ if }x\in [1,2]\Rightarrow y=[-1,1]$
$\text{ if }x\in [2,\infty )\Rightarrow y=1$
$\therefore f(x)={\cos }^{-1}(|x-1|-|x-2|)$
$\text{ domain is }R\text{, range is }[0,\pi ];\therefore P,Q\text{ is correct }$
$P,Q$ is correct
(C) Let $\left|{\sin }^{-1}x-(\pi /2)\right|+\left|{\sin }^{-1}x+(\pi /2)\right|=y$
$$\begin{gathered} \text{ let }{\sin }^{-1}x=t\Rightarrow t\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right] \\ <br/>\left[t-\frac{\pi }{2}\right]+\left[t+\frac{\pi }{2}\right]=y \end{gathered}$$
$t\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\Rightarrow y=\pi$
$\therefore f(x)={\sin }^{-1}\left(\frac{\pi }{\pi }\right)=\frac{\pi }{2}$
$\therefore f(x)\text{ is periodic and constant function also range contain only irrational number. }$
$\Rightarrow P,R,S$
$\therefore f(x)$ is periodic and constant function also range contain only irrational number.
$\Rightarrow$ P, R, S
(D) Domain is $\{\pm 1\}$
$f(x)=\frac{\pi }{2}+1$ for $x=\pm 1$
range contain only irrational value and also constant function.
$\Rightarrow P,R,S]$