1. Tardigrade
  2. Question
  3. Mathematics
  4. Column I Column II A f(x)= sin -1((2/| sin x-1|+| sin x+1|)) 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((π/| sin -1 x-(π / 2)|+| sin -1 x+(π / 2)|)) R Range contain only irrational number D f(x)= cos ( cos -1|x[)+ sin -1( sin x)- operatornamecosec-1( operatornamecosec x)+ operatornamecosec-1| x mid. S f(x) is even.

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Q.
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 $f(x)=\sin ^{-1}\left(\frac{\pi}{\left|\sin ^{-1} x-(\pi / 2)\right|+\left|\sin ^{-1} x+(\pi / 2)\right|}\right)$ R Range contain only irrational number
D $f(x)=\cos \left(\cos ^{-1}\left|x[)+\sin ^{-1}(\sin x)-\operatorname{cosec}^{-1}(\operatorname{cosec} x)+\operatorname{cosec}^{-1}\right| x \mid\right.$ S $f(x)$ is even.

Inverse Trigonometric Functions

Solution:

(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 $\left|\sin ^{-1} x -(\pi / 2)\right|+\left|\sin ^{-1} x +(\pi / 2)\right|= y$
$\text { let } \sin ^{-1} x=t \Rightarrow t \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\
{\left[t-\frac{\pi}{2}\right]+\left[t+\frac{\pi}{2}\right]=y} $
$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.
$\Rightarrow P , R , S ]$