Question Number 172452 by mnjuly1970 last updated on 27/Jun/22
$$ \\ $$$$\:\:\:\:\:\mathrm{I}{f}\:\:,\:{f}\left({x}\right)=\:\mid\:{x}\:−\:{k}\lfloor{x}\rfloor\mid\:{be}\: \\ $$$$\:{one}\:−{to}−{one}\:{function}\:{on}\:\left(\mathrm{0},\mathrm{2}\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:{find}\:{the}\:{value}\:{of}\:\:\:\:\:''\:\:\:\:{k}\:\:\:''. \\ $$$$ \\ $$
Answered by mahdipoor last updated on 27/Jun/22
![D_f =(0,2) { ((D_1 =(0,1)⇒f_1 (x)=∣x∣ ⇒R_1 =(0,1) )),((D_2 =[1,2)⇒f_2 (x)=∣x−k∣ ⇒)) :} R_2 = { (([1−k,2−k) k≤1)),(((k−2,k−1] k≥2)),(((0,max(∣1−k∣,∣2−k∣) 1<k<2)) :} R_1 ∩R_2 =∅ ⇒k≤0 or k≥3](Q172466.png)
$${D}_{{f}} \:=\left(\mathrm{0},\mathrm{2}\right)\: \\ $$$$\begin{cases}{{D}_{\mathrm{1}} =\left(\mathrm{0},\mathrm{1}\right)\Rightarrow{f}_{\mathrm{1}} \left({x}\right)=\mid{x}\mid\:\Rightarrow{R}_{\mathrm{1}} =\left(\mathrm{0},\mathrm{1}\right)\:}\\{{D}_{\mathrm{2}} =\left[\mathrm{1},\mathrm{2}\right)\Rightarrow{f}_{\mathrm{2}} \left({x}\right)=\mid{x}−{k}\mid\:\Rightarrow}\end{cases}\: \\ $$$${R}_{\mathrm{2}} =\begin{cases}{\left[\mathrm{1}−{k},\mathrm{2}−{k}\right)\:\:\:\:\:{k}\leqslant\mathrm{1}}\\{\left({k}−\mathrm{2},{k}−\mathrm{1}\right]\:\:\:\:\:{k}\geqslant\mathrm{2}}\\{\left(\mathrm{0},{max}\left(\mid\mathrm{1}−{k}\mid,\mid\mathrm{2}−{k}\mid\right)\:\:\:\mathrm{1}<{k}<\mathrm{2}\right.}\end{cases} \\ $$$${R}_{\mathrm{1}} \cap{R}_{\mathrm{2}} =\varnothing\: \\ $$$$\Rightarrow{k}\leqslant\mathrm{0}\:{or}\:{k}\geqslant\mathrm{3}\: \\ $$