The lines x+y=|a| and ax−y=1 intersect each other in the first quadrant. Then the set of all possible values of a in the interval are
[1,∞)
(1,∞)
(−1,∞)
(−1,1]
A
(1,∞)
B
(−1,∞)
C
(−1,1]
D
[1,∞)
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Solution
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Given x+y=|a| and ax−y=1
solving these equations simultaneously,
we get,
x=|a|+11+a,y=a|a|−11+a
As both are in the 1st Quadrant both should be greater than 0,
1+a>0 and a|a|−1>0
a>−1 and a|a|>1
After solving this, we get
a>1, i.e aϵ(1,∞)
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