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Question : 56 of 120
Marks:
+1,
-0
Solution:
=− =x++(1−x) =y+x+(1+x−y) then
.(×) =|| ⇒.(×)=[1(1+x−y)−x(1−x)] −0−1(x2−y) ⇒.(×) =1+x−y−x+x2−x2+y ⇒.(×)=1 Hence,
.(×) depends on neither on
x nor on
y.
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