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Question Number 166892 by mnjuly1970 last updated on 01/Mar/22
solve in R i: ⌊ x ⌊ x⌋⌋= 3x ii : ⌊x ⌋^( 2) −3 ⌊x ⌋ +2 ≤ 0 −−−−−−
solveinRi:⌊x⌊x⌋⌋=3xii:⌊x⌋2−3⌊x⌋+2⩽0−−−−−−
Answered by mathsmine last updated on 01/Mar/22
[x[x]]=3x [x]=n x=n+f,f∈[0,1[ ⇒[(n+f)n]=3n+3f ⇒n^2 +[nf]=3n+3f n^2 −3n=3f−[nf]≤3+n ⇒n^2 −4n−3≤0⇒(n−2)^2 −7≤0 ⇒n∈[2−(√7),2+(√7)]n∈[0,4] n=0⇒f=0⇒x=0 n=1⇒1+[f]=3f+3⇒2=[f]−3f impossibl n=2⇒4+[2f]=6+3f⇒2=[2f]−3f impossible n=3⇒0=3f−[3f]⇒[3f]=3f∈{0,1,2} f=0,f=(1/3),f=(2/3),x=3,x=3+(1/3),x=3+(2/3) n=4⇒16+[4f]=12+3f ⇒4=3f−[4f]?impossible x∈{0,3,((10)/3),((11)/3)}
[x[x]]=3x[x]=nx=n+f,f∈[0,1[⇒[(n+f)n]=3n+3f⇒n2+[nf]=3n+3fn2−3n=3f−[nf]⩽3+n⇒n2−4n−3⩽0⇒(n−2)2−7⩽0⇒n∈[2−7,2+7]n∈[0,4]n=0⇒f=0⇒x=0n=1⇒1+[f]=3f+3⇒2=[f]−3fimpossibln=2⇒4+[2f]=6+3f⇒2=[2f]−3fimpossiblen=3⇒0=3f−[3f]⇒[3f]=3f∈{0,1,2}f=0,f=13,f=23,x=3,x=3+13,x=3+23n=4⇒16+[4f]=12+3f⇒4=3f−[4f]?impossiblex∈{0,3,103,113}
Answered by mathsmine last updated on 01/Mar/22
ii T^2 −3T+2≤0 (T−1)(T−2)≤0,T∈[1,2] [x]∈[1,2]⇒x∈[1,3[
iiT2−3T+2⩽0(T−1)(T−2)⩽0,T∈[1,2][x]∈[1,2]⇒x∈[1,3[

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