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Question Number 110980 by mohammad17 last updated on 01/Sep/20

Commented by mohammad17 last updated on 01/Sep/20

$h e l p m e s i r$
Commented by Her_Majesty last updated on 01/Sep/20

$Q 5 i s w r o n g . {w e}^{'} r e f r e e t o c h o o s e a, b, c w i t h$ $a \neq 0 . i . e . a = 7, b = - 7, c = 5 \Rightarrow c i s n o t t h e$ $m i d p o i n t o f [7, - 7]$
	Answered by Her_Majesty last updated on 01/Sep/20
	$Q3 ∣x∣>0 ⇒ −∣x∣<0 ∀x≠0 ⇒ domain=range=0 Q4 i lim_(x→+∞) ((∣x^2 −4∣)/(2x^2 −3))=lim_(x→+∞) ((x^2 −4)/(2x^2 −3))=(1/2) ii lim_(x→2) ((∣x^2 −4∣)/(x−2)) { ((−2<x<2; −((x^2 −4)/(x−2))=−(((x−2)(x+2))/(x−2))=−(x+2))),((x>2; ((x^2 −4)/(x−2))=(((x−2)(x+2))/(x−2))=x+2)) :} ⇒ lim_(x→2^− ) =−4 ≠ lim_(x→2^+ ) =4 ⇒ lim doesn′t exist$
	$Q 3$ $∣ x ∣> 0 \Rightarrow - ∣ x ∣< 0 \forall x \neq 0$ $\Rightarrow d o m a i n = r a n g e = 0$ $Q 4$ $i$ ${l i m}_{x \to + \infty} \frac{∣ x^{2} - 4 ∣}{2 x^{2} - 3} = {l i m}_{x \to + \infty} \frac{x^{2} - 4}{2 x^{2} - 3} = \frac{1}{2}$ $i i$ ${l i m}_{x \to 2} \frac{∣ x^{2} - 4 ∣}{x - 2} {\begin{cases} - 2 < x < 2; - \frac{x^{2} - 4}{x - 2} = - \frac{(x - 2) (x + 2)}{x - 2} = - (x + 2) \\ x > 2; \frac{x^{2} - 4}{x - 2} = \frac{(x - 2) (x + 2)}{x - 2} = x + 2 \end{cases}$ $\Rightarrow {l i m}_{x \to 2^{-}} = - 4 \neq {l i m}_{x \to 2^{+}} = 4$ $\Rightarrow l i m {d o e s n}^{'} t e x i s t$
	Answered by Her_Majesty last updated on 01/Sep/20

	$Q 6$ $(x < 0 \land y < 0) \lor (x > 0 \land y > 0) \Rightarrow x / y > 0$ $(x < 0 \land y > 0) \lor (x > 0 \land y < 0) \Rightarrow x / y < 0$ $e a s y t o c o m p l e t e$ $Q 7$ $x > 0 \land y > 0 \Rightarrow x + y > 0 \Rightarrow∣ x ∣= x \land ∣ y ∣= y \land ∣ x + y ∣= x + y$
	Commented by mohammad17 last updated on 01/Sep/20

	$s i r c a n y o u c o m p l e t e Q 6 p l e a s e$
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