Consider the statement, "Either -2 ≤ x ≤ -1 or 1 ≤ x ≤ 2" The negation of this statement is (A) x<-2 or 2<x or -1<x<1 (B) x<-2 or 2<x (C) -1<x<1 (D) x ≤ -2 or 2 ≤ x or -1<x<1

Consider the statement,

"Either -2 ≤ x ≤ -1 or 1 ≤ x ≤ 2"

The negation of this statement is

(A) x<-2 or 2<x or -1<x<1 (B) x<-2 or 2<x

(C) -1<x<1 (D) x ≤ -2 or 2 ≤ x or -1<x<1 
by

1 Answer

Answer :

(A) x<-2 or 2<x or -1<x<1 
by

Related questions

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The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w} (C) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

Description : The grammar with production rules S → aSb |SS|λ generates language L given by: (A) L = {w∈{a, b}* | na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} (B) L = {w∈{a, b}* | na(w) = nb(w) and na(v ... (D) L = {w∈{a, b}* | na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w}

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