solve in R i: ⌊ x ⌊ x⌋⌋= 3x ii : ⌊x ⌋^( 2) −3 ⌊x ⌋ +2 ≤ 0 −−−−−−


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Question Number 166892 by mnjuly1970 last updated on 01/Mar/22

$s o l v e i n R$ $i : ⌊ x ⌊ x ⌋ ⌋ = 3 x$ $i i : ⌊ x ⌋^{2} - 3 ⌊ x ⌋ + 2 ⩽ 0$ $- - - - - -$
	Answered by mathsmine last updated on 01/Mar/22
	$[x[x]]=3x [x]=n x=n+f,f∈[0,1[ ⇒[(n+f)n]=3n+3f ⇒n^2 +[nf]=3n+3f n^2 −3n=3f−[nf]≤3+n ⇒n^2 −4n−3≤0⇒(n−2)^2 −7≤0 ⇒n∈[2−(√7),2+(√7)]n∈[0,4] n=0⇒f=0⇒x=0 n=1⇒1+[f]=3f+3⇒2=[f]−3f impossibl n=2⇒4+[2f]=6+3f⇒2=[2f]−3f impossible n=3⇒0=3f−[3f]⇒[3f]=3f∈{0,1,2} f=0,f=(1/3),f=(2/3),x=3,x=3+(1/3),x=3+(2/3) n=4⇒16+[4f]=12+3f ⇒4=3f−[4f]?impossible x∈{0,3,((10)/3),((11)/3)}$
	$[x [x]] = 3 x$ $[x] = n$ $x = n + f, f \in [0, 1 [$ $\Rightarrow [(n + f) n] = 3 n + 3 f$ $\Rightarrow n^{2} + [n f] = 3 n + 3 f$ $n^{2} - 3 n = 3 f - [n f] ⩽ 3 + n$ $\Rightarrow n^{2} - 4 n - 3 ⩽ 0 \Rightarrow {(n - 2)}^{2} - 7 ⩽ 0$ $\Rightarrow n \in [2 - \sqrt{7}, 2 + \sqrt{7}] n \in [0, 4]$ $n = 0 \Rightarrow f = 0 \Rightarrow x = 0$ $n = 1 \Rightarrow 1 + [f] = 3 f + 3 \Rightarrow 2 = [f] - 3 f i m p o s s i b l$ $n = 2 \Rightarrow 4 + [2 f] = 6 + 3 f \Rightarrow 2 = [2 f] - 3 f i m p o s s i b l e$ $n = 3 \Rightarrow 0 = 3 f - [3 f] \Rightarrow [3 f] = 3 f \in {0, 1, 2}$ $f = 0, f = \frac{1}{3}, f = \frac{2}{3}, x = 3, x = 3 + \frac{1}{3}, x = 3 + \frac{2}{3}$ $n = 4 \Rightarrow 16 + [4 f] = 12 + 3 f$ $\Rightarrow 4 = 3 f - [4 f] ? i m p o s s i b l e$ $x \in {0, 3, \frac{10}{3}, \frac{11}{3}}$
	Answered by mathsmine last updated on 01/Mar/22

	$i i$ $T^{2} - 3 T + 2 ⩽ 0$ $(T - 1) (T - 2) ⩽ 0, T \in [1, 2]$ $[x] \in [1, 2] \Rightarrow x \in [1, 3 [$
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