Question Number 204329 by mr W last updated on 13/Feb/24
![solve (1/([x]))+(1/([2x]))={x}+(1/3)](https://www.tinkutara.com/question/Q204329.png)
$${solve}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}=\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by AST last updated on 13/Feb/24
![x cannot be negative,otherwise,L.H.S and R.H.S will have opposite signs. x cannot also grow arbitrarily large,otherwise (1/([x]))+(1/([2x]))≪(1/3) Let x≥5⇒[x]≥5⇒[2x]≥10⇒(1/([x]))+(1/([2x]))≤(3/(10))<(1/3) Also [x]≠0. So consider x∈[1,5) when [x]=4,then Case I({x}<0.5): (1/4)+(1/8)={x}+(1/3) ⇒{x}=(1/(24))⇒x=4(1/(24)) Case II({x}>0.5): (1/4)+(1/9)−(1/3)={x}=(1/(36))<0.5→← when [x]=3,Case I: (1/3)+(1/6)={x}+(1/3)⇒{x}=(1/6) ⇒x=3(1/6).. Similarly,we get {x}<0.5 for Case II when [x]=2, Case I:(1/2)+(1/4)−(1/3)={x}=(5/(12))⇒x=2(5/(12)) [x]=1⇒{x}≥1→← ⇒x=4(1/(24)),3(1/6),2(5/(12))](https://www.tinkutara.com/question/Q204332.png)
$${x}\:{cannot}\:{be}\:{negative},{otherwise},{L}.{H}.{S}\:{and}\:{R}.{H}.{S} \\ $$$${will}\:{have}\:{opposite}\:{signs}.\:{x}\:{cannot}\:{also}\:{grow} \\ $$$${arbitrarily}\:{large},{otherwise}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\ll\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Let}\:{x}\geqslant\mathrm{5}\Rightarrow\left[{x}\right]\geqslant\mathrm{5}\Rightarrow\left[\mathrm{2}{x}\right]\geqslant\mathrm{10}\Rightarrow\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\leqslant\frac{\mathrm{3}}{\mathrm{10}}<\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Also}\:\left[{x}\right]\neq\mathrm{0}.\:{So}\:{consider}\:{x}\in\left[\mathrm{1},\mathrm{5}\right)\: \\ $$$${when}\:\left[{x}\right]=\mathrm{4},{then}\:\:{Case}\:{I}\left(\left\{{x}\right\}<\mathrm{0}.\mathrm{5}\right):\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}=\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\left\{{x}\right\}=\frac{\mathrm{1}}{\mathrm{24}}\Rightarrow{x}=\mathrm{4}\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${Case}\:{II}\left(\left\{{x}\right\}>\mathrm{0}.\mathrm{5}\right):\:\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{3}}=\left\{{x}\right\}=\frac{\mathrm{1}}{\mathrm{36}}<\mathrm{0}.\mathrm{5}\rightarrow\leftarrow \\ $$$${when}\:\left[{x}\right]=\mathrm{3},{Case}\:{I}:\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}=\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\left\{{x}\right\}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\mathrm{3}\frac{\mathrm{1}}{\mathrm{6}}..\:{Similarly},{we}\:{get}\:\left\{{x}\right\}<\mathrm{0}.\mathrm{5}\:{for}\:{Case}\:{II} \\ $$$${when}\:\left[{x}\right]=\mathrm{2},\:{Case}\:{I}:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}=\left\{{x}\right\}=\frac{\mathrm{5}}{\mathrm{12}}\Rightarrow{x}=\mathrm{2}\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\left[{x}\right]=\mathrm{1}\Rightarrow\left\{{x}\right\}\geqslant\mathrm{1}\rightarrow\leftarrow \\ $$$$\Rightarrow{x}=\mathrm{4}\frac{\mathrm{1}}{\mathrm{24}},\mathrm{3}\frac{\mathrm{1}}{\mathrm{6}},\mathrm{2}\frac{\mathrm{5}}{\mathrm{12}} \\ $$
Commented by mr W last updated on 13/Feb/24
$${great}\:{solution}! \\ $$
Answered by mr W last updated on 13/Feb/24
![(1/3)≤{x}+(1/3)<1+(1/3)=(4/3) say [x]=n, f={x}, then x=n+f 2n≤[2x]=2n+[2f]<2n+2 (1/([x]))+(1/([2x]))≤(1/n)+(1/(2n))=(3/(2n)) (3/(2n))≥(1/3) ⇒n≤(9/2) ⇒n≤4 (1/([x]))+(1/([2x]))>(1/n)+(1/(2(n+1)))=((3n+2)/(2n(n+1))) ((3n+2)/(2n(n+1)))<(4/3) ⇒8n^2 −n−6>0 ⇒n>((1+(√(1^2 +4×8×6)))/(2×8))≈0.93 ⇒n≥1 ⇒ or n<((1−(√(1^2 +4×8×6)))/(2×8))≈−0.806 ⇒n≤−1 ⇒1≤n≤4 (1/n)+(1/(2n+[2f]))=f+(1/3) case 0≤f<(1/2): f=(1/n)+(1/(2n))−(1/3) n=1: f=(7/6)>1 rejected n=2: f=(5/(12)) ⇒x=2(5/(12)) ✓ n=3: f=(1/6) ⇒x=3(1/6) ✓ 4=4: f=(1/(24)) ⇒x=4(1/(24)) ✓ case (1/2)≤f<1: f=(1/n)+(1/(2n+1))−(1/3) n=1: f=1+(1/3)−(1/3)=1 rejected n=2: f=(1/2)+(1/5)−(1/3)=((11)/(30))<(1/2) rejected n=3: f=(1/3)+(1/7)−(1/3)=(1/7)<(1/2) rejected n=4: f=(1/4)+(1/9)−(1/3)=(1/(36))<(1/2) rejected summary: x=2(5/(12)), 3(1/6), 4(1/(24))](https://www.tinkutara.com/question/Q204341.png)
$$\frac{\mathrm{1}}{\mathrm{3}}\leqslant\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}}<\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${say}\:\left[{x}\right]={n},\:{f}=\left\{{x}\right\},\:{then}\:{x}={n}+{f} \\ $$$$\mathrm{2}{n}\leqslant\left[\mathrm{2}{x}\right]=\mathrm{2}{n}+\left[\mathrm{2}{f}\right]<\mathrm{2}{n}+\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\leqslant\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}}=\frac{\mathrm{3}}{\mathrm{2}{n}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}{n}}\geqslant\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{n}\leqslant\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow{n}\leqslant\mathrm{4} \\ $$$$\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}>\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}\left({n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}\left({n}+\mathrm{1}\right)}<\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\mathrm{8}{n}^{\mathrm{2}} −{n}−\mathrm{6}>\mathrm{0}\: \\ $$$$\Rightarrow{n}>\frac{\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}×\mathrm{8}×\mathrm{6}}}{\mathrm{2}×\mathrm{8}}\approx\mathrm{0}.\mathrm{93}\:\Rightarrow{n}\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{or}\:{n}<\frac{\mathrm{1}−\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{4}×\mathrm{8}×\mathrm{6}}}{\mathrm{2}×\mathrm{8}}\approx−\mathrm{0}.\mathrm{806}\:\Rightarrow{n}\leqslant−\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}\leqslant{n}\leqslant\mathrm{4} \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\left[\mathrm{2}{f}\right]}={f}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${case}\:\mathrm{0}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${f}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${n}=\mathrm{1}:\:{f}=\frac{\mathrm{7}}{\mathrm{6}}>\mathrm{1}\:{rejected} \\ $$$${n}=\mathrm{2}:\:{f}=\frac{\mathrm{5}}{\mathrm{12}}\:\Rightarrow{x}=\mathrm{2}\frac{\mathrm{5}}{\mathrm{12}}\:\checkmark \\ $$$${n}=\mathrm{3}:\:{f}=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{x}=\mathrm{3}\frac{\mathrm{1}}{\mathrm{6}}\:\checkmark \\ $$$$\mathrm{4}=\mathrm{4}:\:{f}=\frac{\mathrm{1}}{\mathrm{24}}\:\Rightarrow{x}=\mathrm{4}\frac{\mathrm{1}}{\mathrm{24}}\:\checkmark \\ $$$${case}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{f}<\mathrm{1}: \\ $$$${f}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${n}=\mathrm{1}:\:{f}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{1}\:{rejected} \\ $$$${n}=\mathrm{2}:\:{f}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{11}}{\mathrm{30}}<\frac{\mathrm{1}}{\mathrm{2}}\:{rejected} \\ $$$${n}=\mathrm{3}:\:{f}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{7}}<\frac{\mathrm{1}}{\mathrm{2}}\:{rejected} \\ $$$${n}=\mathrm{4}:\:{f}=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{36}}<\frac{\mathrm{1}}{\mathrm{2}}\:{rejected} \\ $$$$ \\ $$$${summary}:\:{x}=\mathrm{2}\frac{\mathrm{5}}{\mathrm{12}},\:\mathrm{3}\frac{\mathrm{1}}{\mathrm{6}},\:\mathrm{4}\frac{\mathrm{1}}{\mathrm{24}} \\ $$