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	$g(x)=\sqrt{\ln \left(\frac{x^2+3x-2}{x+1}\right)}$	Q	$(-\infty ,2)$
C	$h(x)=\frac{1}{\ln \left(\frac{x-1}{2}\right)}$	R	$\left(-\infty ,-\frac{1}{2}\right]$
D	$\varphi (x)=\ln \left(\sqrt{x^2+12}-2x\right)$	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)$f(x)={\sin }^{-1}\left(\frac{x+1}{x}\right);-1\le \frac{x+1}{x}\le 1$
$\therefore \frac{x+1}{x}-1\le 0;\frac{1}{x}\le 0\Rightarrow x<0$
$\text{ and }0\le 1+\frac{x+1}{x};0\le \frac{2x+1}{x}\Rightarrow \frac{2x+1}{x}\ge 0$

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

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