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16 Jul 2018, 20:54
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
82% (01:23) correct
18%
(01:18)
wrong
based on 1143
sessions
History
Date
Time
Result
Not Attempted Yet
16 Jul 2018, 22:57
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Bunuel
two cases
1) 2x+3 >0
so \(2x+3<6..............2x<3.............x<\frac{3}{2}\)
2) 2x+3<0
so \(2x+3>-6..............2x>-9.............x>\frac{-9}{2}\)
integers value-
-4, -3, -2, -1, 0, 1 ------ 6 values
C
General Discussion
16 Jul 2018, 23:16
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Bunuel
Given, \(|2x + 3 | < 6\)
Or, \(|x-(\frac{-3}{2})|< 3\) (Dividing both sides of the inequality by 2)
Or, \((\frac{-3}{2}-3) < x < (\frac{-3}{2}+3)\)
Or, -\(4.5<x<1.5\)
So, Possible integer values of x: -4, -3, -2, -1, 0, 1
So, no of integers are 6.
Ans. (C)
