Question

Column I contains functions and column II contains their natural domains. Exactly one entry of column II matches with exactly one entry of column I.

Column I		Column II
A	$ f ( x )=\sin ^{-1}\left(\frac{ x +1}{ x }\right)$	P	$(1,3) \cup(3, \infty)$
B	$ g ( x )=\sqrt{\ln \left(\frac{ x ^2+3 x -2}{ x +1}\right)}$	Q	$(-\infty, 2)$
C	$h ( x )=\frac{1}{\ln \left(\frac{ x -1}{2}\right)}$	R	$\left(-\infty,-\frac{1}{2}\right]$
D	$ \phi( x )=\ln \left(\sqrt{ x ^2+12}-2 x \right)$	S	$[-3,-1) \cup[1, \infty)$

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

Answer 1

Answer: (A)

(A)$f(x)=\sin ^{-1}\left(\frac{x+1}{x}\right) ; -1 \leq \frac{x+1}{x} \leq 1 $
$\therefore \frac{x+1}{x}-1 \leq 0 ; \frac{1}{x} \leq 0 \Rightarrow x<0 $
$\text { and } 0 \leq 1+\frac{x+1}{x} ; 0 \leq \frac{2 x+1}{x} \Rightarrow \frac{2 x+1}{x} \geq 0$

Hence domain of $f(x)$ is $\left(-\infty,-\frac{1}{2}\right]$
(B)$\ln \frac{x^2+3 x-2}{x+1} \geq 0 ; \frac{x^2+3 x-2}{x+1} \geq 1 ; \frac{x^2+3 x-2-x-1}{x+1} \geq 0 ; \frac{x^2+2 x-3}{x+1} \geq 0$
$\frac{( x +3)( x -1)}{ x +1} \geq 0$

hence $\Rightarrow$ (S)
(C)$\ln \left(\frac{ x -1}{2}\right) \neq 0 \Rightarrow \frac{ x -1}{2} \neq 1 \Rightarrow x \neq 3$
also $\frac{x-1}{2}>0 \Rightarrow x>1$
$(1, \infty)-\{3\} \Rightarrow$
(D) for $x \leq 0, \sqrt{ x ^2+12}>2 x$ always true
$\text { for } x>0 x^2+12>4 x^2 $
$ 0>3 x^2-12 \Rightarrow x^2-4<0$
$\therefore x \in(0,2) \Rightarrow \operatorname{domain}(-\infty, 2) \Rightarrow \text { (Q) ] }$