[" Hence (d) holds."],[" Example "3" .Let "A=[x:x in R,|x|<1];],[" B "=[x:x in R,|x-1|>=1]" and "],[" A "uu B=R-D" ,then the set "D" is "],[[" A "U=1

Let A={x:x in R,-1 =2} and A uu B=R-D then the set D is

Let A={x : x in R,-1 lt x lt 1} , B={x : x in R, x le0 or x ge 2} and AuuB=R-D , then find set D

Let A={x : x in R,-1 lt x lt 1} , B={x : x in R, x le0 or x ge 2} and AuuB=R-D , then find set D

Let P = { x : x in R , |x| Q = {x:x in R , |x -1 |ge 2} P uu Q = R - S then set S is : (where R is the set of real number)

Let P = { x : x in R , |x| Q = {x:x in R , |x -1 |ge 2} P uu Q = R - S then set S is : (where R is the set of real number)

Let f : D -> R , where D is the domain off. Find the inverse of f(x) = 1 - 2^-x

Let f(x)=x-[x],x in R,backslash then backslash f'((1)/(2)) is (3)/(2) b.1 c.0 d.-1

If set A and B are defined as A = {(x,y)|y = 1/x, 0 ne x in R}, B = {(x,y)|y = -x , x in R,} . Then (a) A ∩ B = A (b)A ∩ B = B (c)A ∩ B = ϕ (d)A ∪ B = A