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(( x +1/ x )) P (1,3) ∪(3, ∞) B g ( x )=√ ln (( x 2+3 x -2/ x +1)) Q (-∞, 2) C h ( x )=(1/ ln (( x -1)2)) R (-∞,-(1/2)] D φ( x )= ln (√ x 2+12-2 x ) S [-3,-1) ∪[1, ∞)

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(( x +1/ x )) P (1,3) ∪(3, ∞) B g ( x )=√ ln (( x 2+3 x -2/ x +1)) Q (-∞, 2) C h ( x )=(1/ ln (( x -1)2)) R (-∞,-(1/2)] D φ( x )= ln (√ x 2+12-2 x ) S [-3,-1) ∪[1, ∞) (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

Tardigrade Admin · Accepted Answer

(A)f(x)= sin -1((x+1/x)) ; -1 ≤ (x+1/x) ≤ 1 ∴ (x+1/x)-1 ≤ 0 ; (1/x) ≤ 0 ⇒ x<0 text and 0 ≤ 1+(x+1/x) ; 0 ≤ (2 x+1/x) ⇒ (2 x+1/x) ≥ 0

Hence domain of f(x) is (-∞,-(1/2)] (B) ln (x2+3 x-2/x+1) ≥ 0 ; (x2+3 x-2/x+1) ≥ 1 ; (x2+3 x-2-x-1/x+1) ≥ 0 ; (x2+2 x-3/x+1) ≥ 0 (( x +3)( x -1)/ x +1) ≥ 0

hence ⇒ (S) (C) ln (( x -1/2)) ≠ 0 ⇒ ( x -1/2) ≠ 1 ⇒ x ≠ 3 also (x-1/2)>0 ⇒ x>1 (1, ∞)- 3 ⇒ (D) for x ≤ 0, √ x 2+12>2 x always true text for x>0 x2+12>4 x2 0>3 x2-12 ⇒ x2-4<0 ∴ x ∈(0,2) ⇒ operatornamedomain(-∞, 2) ⇒ text (Q) ]

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