X€[-1;0] and -2y€[-2;2] then #(x-2y)^2#€...?

Question

831 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-21T04:54:23

# (x-2y)^2 in [0,9]#.

#x in [-1,0] rArr -1lexle0.............<<1>>#.

#-2y in [-2,2] rArr -2le-2yle2...................<<2>>#.

#:. <<1>>+<<2>> rArr -1-2lex-2yle0+2, ie., #

# -3lex-2yle2#.

#rArr (x-2y) in [-3,2]=[-3,0]uu[0,2]#.

#rArr (x-2y) in [-3,0], or, (x-2y) in [0,2]#.

#"If, "(x-2y) in [-3,0], -3le(x-2y)le0#

#rArr0le(x-2y)^2le9, or,#

#(x-2y)^2 in [0,9].............................<<3>>#.

#"Similarly, "(x-2y) in [0,2]rArr (x-2y)^2 in [0,4]...<<4>>#.

Combining #<<3,4>>#, we find,

#(x-2y)^2 in [0,9]uu[0,4]#.

#rArr (x-2y)^2 in [0,9]#.