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Question

The lines x+y=|a| and axy=1 intersect each other in the first quadrant. Then the set of all possible values of a in the interval are
  1. [1,)
  2. (1,)
  3. (1,)
  4. (1,1]

A
(1,)
B
(1,)
C
(1,1]
D
[1,)
Solution
Verified by Toppr

Given x+y=|a| and axy=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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