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 is?

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Please find below the solution to the asked query :

x+y=a   ...1ax-y=1    ...2Adding 1 and 2x+ax=a+1x1+a=a+1x=a+11+a>0     Since lines intersects in first qauadrantMultiplying 1 with a and subtracting it from 2So ,-y-ay=1-aay -1-a=1-aay=1-aa-1+a=aa-11+a>0    Since lines intersects in first qauadrantAs a+11+a>0  , aa-11+a>0 and a+1>01+a>0   and   aa-1>0a>-1   and   aa>1As aa>1 and a0So , a>0Then aa=a2>1a>1a1,   ANS...
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