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Home / Doubts and Solutions/JEE/Mathematics

the lines x+y=|a| and ax-y=1  intersect each other in the first quadrant .The set of all possible values of a is in the interval.
a.(1,-1)
b.(0,infinity)
c.(1,infinity)
d.(-1,infinity)

Asked by saraswatmajumder 27th October 2017, 10:29 PM
Answered by Expert
Answer:
begin mathsize 16px style straight x plus straight y equals open vertical bar straight a close vertical bar ax minus straight y equals 1 Solving space these space equations space simultaneously space you space get space straight x equals fraction numerator open vertical bar straight a close vertical bar plus 1 over denominator 1 plus straight a end fraction comma space straight y equals fraction numerator straight a open vertical bar straight a close vertical bar minus 1 over denominator 1 plus straight a end fraction As space both space are space in space the space first space quadrant space both space are space greater space than space zero. 1 plus straight a greater than 0 space and space straight a open vertical bar straight a close vertical bar minus 1 greater than 0 After space solving space this space we space get space straight a greater than 1 straight a element of open parentheses 1 comma infinity close parentheses end style
Answered by Expert 13th December 2017, 6:10 PM
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