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11 Nov 2023, 08:39
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
89% (00:56) correct
11%
(01:14)
wrong
based on 321
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Not Attempted Yet
On the number line shown above, the shaded interval is the graph of which of the following inequalities?
(A) |x| ≤ 4
(B) |x| ≤ 6
(C) |x − 2| ≤ 4
(D) |x − 1| ≤ 5
(E) |x + 2| ≤ 4
Attachment:
Number line.png [ 5 KiB | Viewed 4070 times ]
12 Nov 2023, 07:31
Kudos
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Bunuel
OA is E
Posted from my mobile device
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Screenshot_20231112_200037_Samsung Notes.jpg [ 193.25 KiB | Viewed 3602 times ]
09 Jun 2025, 19:27
Kudos
Bookmarks
Region given in the line is -6 ≤ x ≤ 2
(A) |x| ≤ 4
This is of the form |x| ≤ A and can be opened as -A ≤ x ≤ A
=> -4 ≤ x ≤ 4
This is partially in the range and partially not, so FALSE
(B) |x| ≤ 6
=> -6 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(C) |x − 2| ≤ 4
=> -4 ≤ x - 2 ≤ 4
Adding 2 on all the sides we get
=> -4 + 2 ≤ x - 2 + 2 ≤ 4 + 2
=> -2 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(D) |x − 1| ≤ 5
=> -5 ≤ x - 1 ≤ 5
Adding 1 on all the sides we get
=> -5 + 1 ≤ x - 1 + 1 ≤ 5 + 1
=> -4 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(E) |x + 2| ≤ 4
=> -4 ≤ x + 2 ≤ 4
Subtracting -2 from all the sides we get
=> -4 - 2 ≤ x + 2 - 2 ≤ 4 - 2
=> -6 ≤ x ≤ 2
=> TRUE
So, Answer will be E
Hope it helps!
Playlist on Solve problems on Absolute Value here
Watch the following video to MASTER Absolute Values
(A) |x| ≤ 4
This is of the form |x| ≤ A and can be opened as -A ≤ x ≤ A
=> -4 ≤ x ≤ 4
This is partially in the range and partially not, so FALSE
(B) |x| ≤ 6
=> -6 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(C) |x − 2| ≤ 4
=> -4 ≤ x - 2 ≤ 4
Adding 2 on all the sides we get
=> -4 + 2 ≤ x - 2 + 2 ≤ 4 + 2
=> -2 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(D) |x − 1| ≤ 5
=> -5 ≤ x - 1 ≤ 5
Adding 1 on all the sides we get
=> -5 + 1 ≤ x - 1 + 1 ≤ 5 + 1
=> -4 ≤ x ≤ 6
This is partially in the range and partially not, so FALSE
(E) |x + 2| ≤ 4
=> -4 ≤ x + 2 ≤ 4
Subtracting -2 from all the sides we get
=> -4 - 2 ≤ x + 2 - 2 ≤ 4 - 2
=> -6 ≤ x ≤ 2
=> TRUE
So, Answer will be E
Hope it helps!
Playlist on Solve problems on Absolute Value here
Watch the following video to MASTER Absolute Values
