Question Number 22463 by Tinkutara last updated on 18/Oct/17
![Solve for real x: (1/([x])) + (1/([2x])) = (x) + (1/3), where [x] is the greatest integer less than or equal to x and (x) = x − [x], [e.g. [3.4] = 3 and (3.4) = 0.4].](Q22463.png)
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:{x}: \\ $$$$\frac{\mathrm{1}}{\left[{x}\right]}\:+\:\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\:=\:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}, \\ $$$$\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x}\:\mathrm{and}\:\left({x}\right)\:=\:{x}\:−\:\left[{x}\right], \\ $$$$\left[\mathrm{e}.\mathrm{g}.\:\left[\mathrm{3}.\mathrm{4}\right]\:=\:\mathrm{3}\:\mathrm{and}\:\left(\mathrm{3}.\mathrm{4}\right)\:=\:\mathrm{0}.\mathrm{4}\right]. \\ $$
Answered by ajfour last updated on 18/Oct/17
![let x=n+f ⇒ (1/n)+(1/(2n+[2f]))=f+(1/3) Case I: f< 1/2 ⇒ (3/(2n))=f+(1/3) ⇒ 0≤ (3/(2n))−(1/3) <(1/2) or ⇒ (1/3) ≤ (3/(2n)) <(5/6) ⇒ (9/5) < n≤(9/2) ⇒ n=2, 3, 4 corresponding f=(3/(2n))−(1/3) f= (5/(12)), (1/6), (1/(24)) So x_1 =2+(5/(12))= ((29)/(12)) x_2 =3+(1/6)=((19)/6) x_3 =4+(1/(24))=((97)/(24)) Case II: if (1/2)≤ f <1 (1/n)+(1/(2n+1))=f+(1/3) As f ≥1/2 (1/2)≤ ((3n+1)/(n(2n+1)))−(1/3) < 1 ⇒ (5/6) ≤ ((3n+1)/(n(2n+1))) <(4/3) condition (a): ⇒ 10n^2 +5n ≤ 18n+6 or 10n^2 −13n−6 ≤ 0 10(n−((13)/(20)))^2 ≤ 6+((169)/(40)) ⇒ ((13)/(20))−(√((409)/(400))) ≤ n ≤ ((13)/(20))+(√((409)/(400))) ⇒ n=0, 1 condition (b): 8n^2 +4n > 9n+3 ⇒ 8n^2 −5n−3 > 0 ⇒ 8(n−(5/(16)))^2 > 3+((25)/(32)) n < (5/(16))−((11)/(16)) and n > (5/(16))+((11)/(16)) ⇒ n< −(6/(16)) and n > 1 Intersection of the two conditions gives no solution for n for this case. So finally x=((29)/(12)), ((19)/6), ((97)/(24)) .](Q22464.png)
$${let}\:{x}={n}+{f} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\left[\mathrm{2}{f}\right]}={f}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${Case}\:{I}:\:\:\:{f}<\:\mathrm{1}/\mathrm{2} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{3}}{\mathrm{2}{n}}={f}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{0}\leqslant\:\:\frac{\mathrm{3}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{3}}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\:\:\:\:\Rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{2}{n}}\:<\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{9}}{\mathrm{5}}\:<\:{n}\leqslant\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow\:{n}=\mathrm{2},\:\mathrm{3},\:\mathrm{4} \\ $$$${corresponding}\:{f}=\frac{\mathrm{3}}{\mathrm{2}{n}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:{f}=\:\frac{\mathrm{5}}{\mathrm{12}},\:\frac{\mathrm{1}}{\mathrm{6}},\:\frac{\mathrm{1}}{\mathrm{24}} \\ $$$${So}\:{x}_{\mathrm{1}} =\mathrm{2}+\frac{\mathrm{5}}{\mathrm{12}}=\:\frac{\mathrm{29}}{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:{x}_{\mathrm{2}} =\mathrm{3}+\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{19}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:{x}_{\mathrm{3}} =\mathrm{4}+\frac{\mathrm{1}}{\mathrm{24}}=\frac{\mathrm{97}}{\mathrm{24}} \\ $$$${Case}\:{II}:\:{if}\:\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\:{f}\:<\mathrm{1}\: \\ $$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}={f}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${As}\:\:\:\:{f}\:\geqslant\mathrm{1}/\mathrm{2} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant\:\frac{\mathrm{3}{n}+\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{3}}\:<\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{5}}{\mathrm{6}}\:\leqslant\:\frac{\mathrm{3}{n}+\mathrm{1}}{{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:<\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${condition}\:\left({a}\right): \\ $$$$\Rightarrow\:\:\:\mathrm{10}{n}^{\mathrm{2}} +\mathrm{5}{n}\:\leqslant\:\mathrm{18}{n}+\mathrm{6} \\ $$$${or}\:\:\:\mathrm{10}{n}^{\mathrm{2}} −\mathrm{13}{n}−\mathrm{6}\:\leqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{10}\left({n}−\frac{\mathrm{13}}{\mathrm{20}}\right)^{\mathrm{2}} \leqslant\:\mathrm{6}+\frac{\mathrm{169}}{\mathrm{40}} \\ $$$$\:\Rightarrow\:\frac{\mathrm{13}}{\mathrm{20}}−\sqrt{\frac{\mathrm{409}}{\mathrm{400}}}\:\leqslant\:{n}\:\leqslant\:\frac{\mathrm{13}}{\mathrm{20}}+\sqrt{\frac{\mathrm{409}}{\mathrm{400}}}\: \\ $$$$\Rightarrow\:\:\:\:\:\:{n}=\mathrm{0},\:\mathrm{1} \\ $$$${condition}\:\left({b}\right): \\ $$$$\:\:\mathrm{8}{n}^{\mathrm{2}} +\mathrm{4}{n}\:>\:\mathrm{9}{n}+\mathrm{3} \\ $$$$\Rightarrow\:\:\:\mathrm{8}{n}^{\mathrm{2}} −\mathrm{5}{n}−\mathrm{3}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{8}\left({n}−\frac{\mathrm{5}}{\mathrm{16}}\right)^{\mathrm{2}} >\:\mathrm{3}+\frac{\mathrm{25}}{\mathrm{32}} \\ $$$${n}\:<\:\frac{\mathrm{5}}{\mathrm{16}}−\frac{\mathrm{11}}{\mathrm{16}}\:{and}\:\:{n}\:>\:\frac{\mathrm{5}}{\mathrm{16}}+\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\Rightarrow\:\:\:{n}<\:−\frac{\mathrm{6}}{\mathrm{16}}\:{and}\:\:{n}\:>\:\mathrm{1} \\ $$$${Intersection}\:{of}\:{the}\:{two}\:{conditions} \\ $$$${gives}\:{no}\:{solution}\:{for}\:{n}\:{for}\:{this} \\ $$$${case}. \\ $$$$\boldsymbol{{So}}\:{finally}\:\boldsymbol{{x}}=\frac{\mathrm{29}}{\mathrm{12}},\:\frac{\mathrm{19}}{\mathrm{6}},\:\frac{\mathrm{97}}{\mathrm{24}}\:. \\ $$
Commented by Tinkutara last updated on 18/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$