solve (1/([x]))+(1/([2x]))={x}+(1/3)


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 204329 by mr W last updated on 13/Feb/24

$s o l v e \frac{1}{[x]} + \frac{1}{[2 x]} = {x} + \frac{1}{3}$
	Answered by deleteduser1 last updated on 13/Feb/24

	$x c a n n o t b e n e g a t i v e, o t h e r w i s e, L . H . S a n d R . H . S$ $w i l l h a v e o p p o s i t e s i g n s . x c a n n o t a l s o g r o w$ $a r b i t r a r i l y l a r g e, o t h e r w i s e \frac{1}{[x]} + \frac{1}{[2 x]} ≪ \frac{1}{3}$ $L e t x ⩾ 5 \Rightarrow [x] ⩾ 5 \Rightarrow [2 x] ⩾ 10 \Rightarrow \frac{1}{[x]} + \frac{1}{[2 x]} ⩽ \frac{3}{10} < \frac{1}{3}$ $A l s o [x] \neq 0 . S o c o n s i d e r x \in [1, 5)$ $w h e n [x] = 4, t h e n C a s e I ({x} < 0.5) : \frac{1}{4} + \frac{1}{8} = {x} + \frac{1}{3}$ $\Rightarrow {x} = \frac{1}{24} \Rightarrow x = 4 \frac{1}{24}$ $C a s e I I ({x} > 0.5) : \frac{1}{4} + \frac{1}{9} - \frac{1}{3} = {x} = \frac{1}{36} < 0.5 \to\leftarrow$ $w h e n [x] = 3, C a s e I : \frac{1}{3} + \frac{1}{6} = {x} + \frac{1}{3} \Rightarrow {x} = \frac{1}{6}$ $\Rightarrow x = 3 \frac{1}{6} . . S i m i l a r l y, w e g e t {x} < 0.5 f o r C a s e I I$ $w h e n [x] = 2, C a s e I : \frac{1}{2} + \frac{1}{4} - \frac{1}{3} = {x} = \frac{5}{12} \Rightarrow x = 2 \frac{5}{12}$ $[x] = 1 \Rightarrow {x} ⩾ 1 \to\leftarrow$ $\Rightarrow x = 4 \frac{1}{24}, 3 \frac{1}{6}, 2 \frac{5}{12}$
	Commented by mr W last updated on 13/Feb/24

	$g r e a t s o l u t i o n!$
	Answered by mr W last updated on 13/Feb/24
	$(1/3)≤{x}+(1/3)<1+(1/3)=(4/3) say [x]=n, f={x}, then x=n+f 2n≤[2x]=2n+[2f]<2n+2 (1/([x]))+(1/([2x]))≤(1/n)+(1/(2n))=(3/(2n)) (3/(2n))≥(1/3) ⇒n≤(9/2) ⇒n≤4 (1/([x]))+(1/([2x]))>(1/n)+(1/(2(n+1)))=((3n+2)/(2n(n+1))) ((3n+2)/(2n(n+1)))<(4/3) ⇒8n^2 −n−6>0 ⇒n>((1+(√(1^2 +4×8×6)))/(2×8))≈0.93 ⇒n≥1 ⇒ or n<((1−(√(1^2 +4×8×6)))/(2×8))≈−0.806 ⇒n≤−1 ⇒1≤n≤4 (1/n)+(1/(2n+[2f]))=f+(1/3) case 0≤f<(1/2): f=(1/n)+(1/(2n))−(1/3) n=1: f=(7/6)>1 rejected n=2: f=(5/(12)) ⇒x=2(5/(12)) ✓ n=3: f=(1/6) ⇒x=3(1/6) ✓ 4=4: f=(1/(24)) ⇒x=4(1/(24)) ✓ case (1/2)≤f<1: f=(1/n)+(1/(2n+1))−(1/3) n=1: f=1+(1/3)−(1/3)=1 rejected n=2: f=(1/2)+(1/5)−(1/3)=((11)/(30))<(1/2) rejected n=3: f=(1/3)+(1/7)−(1/3)=(1/7)<(1/2) rejected n=4: f=(1/4)+(1/9)−(1/3)=(1/(36))<(1/2) rejected summary: x=2(5/(12)), 3(1/6), 4(1/(24))$
	$\frac{1}{3} ⩽ {x} + \frac{1}{3} < 1 + \frac{1}{3} = \frac{4}{3}$ $s a y [x] = n, f = {x}, t h e n x = n + f$ $2 n ⩽ [2 x] = 2 n + [2 f] < 2 n + 2$ $\frac{1}{[x]} + \frac{1}{[2 x]} ⩽ \frac{1}{n} + \frac{1}{2 n} = \frac{3}{2 n}$ $\frac{3}{2 n} ⩾ \frac{1}{3} \Rightarrow n ⩽ \frac{9}{2} \Rightarrow n ⩽ 4$ $\frac{1}{[x]} + \frac{1}{[2 x]} > \frac{1}{n} + \frac{1}{2 (n + 1)} = \frac{3 n + 2}{2 n (n + 1)}$ $\frac{3 n + 2}{2 n (n + 1)} < \frac{4}{3} \Rightarrow 8 n^{2} - n - 6 > 0$ $\Rightarrow n > \frac{1 + \sqrt{1^{2} + 4 \times 8 \times 6}}{2 \times 8} \approx 0.93 \Rightarrow n ⩾ 1$ $\Rightarrow o r n < \frac{1 - \sqrt{1^{2} + 4 \times 8 \times 6}}{2 \times 8} \approx - 0.806 \Rightarrow n ⩽ - 1$ $\Rightarrow 1 ⩽ n ⩽ 4$ $\frac{1}{n} + \frac{1}{2 n + [2 f]} = f + \frac{1}{3}$ $c a s e 0 ⩽ f < \frac{1}{2} :$ $f = \frac{1}{n} + \frac{1}{2 n} - \frac{1}{3}$ $n = 1 : f = \frac{7}{6} > 1 r e j e c t e d$ $n = 2 : f = \frac{5}{12} \Rightarrow x = 2 \frac{5}{12} ✓$ $n = 3 : f = \frac{1}{6} \Rightarrow x = 3 \frac{1}{6} ✓$ $4 = 4 : f = \frac{1}{24} \Rightarrow x = 4 \frac{1}{24} ✓$ $c a s e \frac{1}{2} ⩽ f < 1 :$ $f = \frac{1}{n} + \frac{1}{2 n + 1} - \frac{1}{3}$ $n = 1 : f = 1 + \frac{1}{3} - \frac{1}{3} = 1 r e j e c t e d$ $n = 2 : f = \frac{1}{2} + \frac{1}{5} - \frac{1}{3} = \frac{11}{30} < \frac{1}{2} r e j e c t e d$ $n = 3 : f = \frac{1}{3} + \frac{1}{7} - \frac{1}{3} = \frac{1}{7} < \frac{1}{2} r e j e c t e d$ $n = 4 : f = \frac{1}{4} + \frac{1}{9} - \frac{1}{3} = \frac{1}{36} < \frac{1}{2} r e j e c t e d$ $s u m m a r y : x = 2 \frac{5}{12}, 3 \frac{1}{6}, 4 \frac{1}{24}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum