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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 is the interval:
A. `[1,oo)`
B. `(-1,oo)`
C. `(-1,1)`
D. `(0,oo)`

1 Answer

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Best answer
Correct Answer - A
Given lines intersect at the point
`P((1+|a|)/(1+a),(a|a|-1)/(1+a))`
This point will lie in first quadrant iff
`1+a gt0 and a|a| -1 gt 0`
`implies a gt -1 and a|a| -1 ge 0`
`implies a^(2)-1 ge 0 " if " a ge 0 " or " , -a^(2)-1 gt 0 " if " 0 lt a lt -1`
`implies a in [1,oo)`

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