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02 Dec 2020, 10:08
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I understand how you arrive at (x - 3/2)(x-3) < 0 and that one is positive and one is negative. But how do you determine that 3/2<x<3 as opposed to x<3/2 and x>3 ?
29 Jan 2021, 01:10
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what is the value of integer x = ?
s1: 2(x)^2 + 9 < 9x
2(x)^2 - 9x + 9 < 0
2(x)^2 - 6x - 3x + 9 < 0
2x (x - 3) - 3 (x - 3) < 0
(2x - 3) (x - 3) < 0
the Factors must have opposite signs to be less than 0, so it must be 1 of 2 cases:
Case 1: x - 3 > 0 ------ and -------- 2x - 3 < 0
which means: x > 3 ----- and ------ x < 3/2
since X can not be larger than 3 and less than 3/2, it must be Case 2
Case 2: x - 3 < 0 ----- and ------ 2x - 3 > 0
x < 3 ----- and ----- x > 3/2
3/2 < X < 3
1.5 < X < 3
the only integer Value in this Range is 2.
thus, X must = 2.
S1 Sufficient
S2: [x + 10] = 2x + 8
since the Output of the Absolute Value Expression is always NON-negative, it must be True that:
2x + 8 >/= 0
x >/= (-)4
Since X must be greater than or equal to (-)4, it means that the Quantity inside the Modulus must be NON-Negative. we can thus remove the Absolute Value Brackets:
x + 10 = 2x + 8
x = 2
S2 Sufficient alone also
D - Each STatement Sufficient alone
s1: 2(x)^2 + 9 < 9x
2(x)^2 - 9x + 9 < 0
2(x)^2 - 6x - 3x + 9 < 0
2x (x - 3) - 3 (x - 3) < 0
(2x - 3) (x - 3) < 0
the Factors must have opposite signs to be less than 0, so it must be 1 of 2 cases:
Case 1: x - 3 > 0 ------ and -------- 2x - 3 < 0
which means: x > 3 ----- and ------ x < 3/2
since X can not be larger than 3 and less than 3/2, it must be Case 2
Case 2: x - 3 < 0 ----- and ------ 2x - 3 > 0
x < 3 ----- and ----- x > 3/2
3/2 < X < 3
1.5 < X < 3
the only integer Value in this Range is 2.
thus, X must = 2.
S1 Sufficient
S2: [x + 10] = 2x + 8
since the Output of the Absolute Value Expression is always NON-negative, it must be True that:
2x + 8 >/= 0
x >/= (-)4
Since X must be greater than or equal to (-)4, it means that the Quantity inside the Modulus must be NON-Negative. we can thus remove the Absolute Value Brackets:
x + 10 = 2x + 8
x = 2
S2 Sufficient alone also
D - Each STatement Sufficient alone
Poojita
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06 Apr 2021, 09:30
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what a quality question!!!
