Column-I Column-IIA)Domain of sin−1(x2−3x+2x2+2x+7)p)R−{1,2}B)Domain of f(x)=1|x|−xq)(−∞,−5]∪[5,∞)C)Domain of f(x)=x2−25r)[−1,∞)D)Domain of f(x)=x2−1x2−3x+2s)(−∞,0)

A)x2−3x+2x2+2x+7≥−1 and x2−3x+2x2+2x+7≤1Hence domain x∈[−1,∞) B)|x|−x>0 ⇒|x|>x x<0 x∈(−∞,0) C)x2−25≥0 x∈(−∞,−5]∪[5,∞) D) x2−3x+2≠0 x≠1,2

Column-I Column-IIA)Domain of sin−1(x2−3x+2x2+2x+7)p)R−{1,2}B)Domain of f(x)=1|x|−xq)(−∞,−5]∪[5,∞)C)Domain of f(x)=x2−25r)[−1,∞)D)Domain of f(x)=x2−1x2−3x+2s)(−∞,0)

A)x2−3x+2x2+2x+7≥−1 and x2−3x+2x2+2x+7≤1Hence domain x∈[−1,∞) B)|x|−x>0 ⇒|x|>x x<0 x∈(−∞,0) C)x2−25≥0 x∈(−∞,−5]∪[5,∞) D) x2−3x+2≠0 x≠1,2

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Column-I

Column-II
A)
Domain of $\sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$
p)
$R - {1, 2}$
B)
Domain of $f (x) = \frac{1}{\sqrt{| x | - x}}$
q)
$(- \infty, - 5] \cup [5, \infty)$
C)
Domain of $f (x) = \sqrt{x^{2} - 25}$
r)
$[- 1, \infty)$
D)
Domain of $f (x) = \frac{x^{2} - 1}{x^{2} - 3 x + 2}$
s)
$(- \infty, 0)$

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$(A \to r); (B \to s); (C \to q); (D \to p)$

$(A \to r); (B \to s); (C \to p); (D \to q)$

$(A \to s); (B \to r); (C \to q); (D \to p)$

$(A \to s); (B \to r); (C \to p); (D \to q)$

answer is A.

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Detailed Solution

$A) \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \geq - 1 and \frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7} \leq 1$
Hence domain $x \in [- 1, \infty)$
$B) | x | - x > 0$
$\Rightarrow | x | > x$
$x < 0$
$x \in (- \infty, 0)$
$C) x^{2} - 25 \geq 0$
$x \in (- \infty, - 5] \cup [5, \infty)$
$D) x^{2} - 3 x + 2 \neq 0$
$x \neq 1, 2$

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	Column-I		Column-II
A)	Domain of $\sin^{- 1} (\frac{x^{2} - 3 x + 2}{x^{2} + 2 x + 7})$	p)	$R - {1, 2}$
B)	Domain of $f (x) = \frac{1}{\sqrt{\| x \| - x}}$	q)	$(- \infty, - 5] \cup [5, \infty)$
C)	Domain of $f (x) = \sqrt{x^{2} - 25}$	r)	$[- 1, \infty)$
D)	Domain of $f (x) = \frac{x^{2} - 1}{x^{2} - 3 x + 2}$	s)	$(- \infty, 0)$

Column-I Column-IIA)Domain of sin−1(x2−3x+2x2+2x+7)p)R−{1,2}B)Domain of f(x)=1|x|−xq)(−∞,−5]∪[5,∞)C)Domain of f(x)=x2−25r)[−1,∞)D)Domain of f(x)=x2−1x2−3x+2s)(−∞,0)

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