Question

[x]+[2x]+[3x]=4x. Sa se calculeze media aritmetica a solutiilor ecuatiei.​

Answer 1

$[x]+[2x]+[3x] = 4x \\ \\ x-1<[x] \leq x \\ 2x-1<[2x]\leq 2x \\ 3x-1< [3x]\leq 3x \\ \\ 6x-3 < [x]+[2x]+[3x]\leq 6x \\ 6x-3 < 4x\leq 6x \\ -3<-2x\leq 0\\ 3>2x\geq 0 \\\\ \dfrac{3}{2}> x\geq 0\\ \\ \Rightarrow x\in \Big[0,\dfrac{3}{2}\Big) \text{ dar }x < 1 \\ \\\Rightarrow x \in [0,1) \Rightarrow 4x \in [0,4)$

$4x\in \mathbb{Z} \Rightarrow 4x\in \{0,1,2,3\} \Rightarrow \\ \\ \Rightarrow x\in \Big\{0,\dfrac{1}{4},\dfrac{1}{2},\dfrac{3}{4}\Big\} \\ \\ \text{Observam ca }x = \dfrac{1}{4} \text{ nu verifica.}\\ \\ \Rightarrow S = \Big\{0,\dfrac{1}{2},\dfrac{3}{4}\Big\} \Rightarrow m.a. = \dfrac{0+\dfrac{1}{2}+\dfrac{3}{4}}{3} = \dfrac{5}{12}$