213.Let [x1-x]=a and [x]=b
we get a+1a=1b
→(a+1)(1-b)=1
Solutions are (a,b)=(0,0)=(-2,2) since both a and b are integers.
x=0 U[2,3]
\hspace{-16}$(1)\;\; Find all Triplet $x,y,z$ such that \\\\ $\lfloor x \rfloor -y=2\lfloor y \rfloor -z=3\lfloor z \rfloor -x=\frac{2004}{2005}$\\\\\\ (2)\;\; Find all Real no. $x$ such that\\\\ $\frac{x}{x+4}=\frac{5\lfloor x \rfloor -7}{7\lfloor x \rfloor -5}$\\\\\\ (3)\;\; Find all real no. $x$ such that\\\\ $\lfloor \frac{x}{1-x}\rfloor=\frac{\lfloor x \rfloor}{1-\lfloor x \rfloor}$
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5 Answers
21
1057question 2:
\frac{x}{x+4}=\frac{5\left[x \right]-7}{7\left[x \right]-5} \: this\: evaluates\: to\:
2x([x]+1)=-4(5[x]+7)
since rhs is a integer x has to be a integer...
replacing [x] with x we get
x2+11x+14=0
therefore x=-11±√652
2621)[x]-y = 2004/2005
or ([x]-[y] ) - {y} = 2004/2005 .
hence {y} = 1/2005
similarly from the other two eqns, {x}={y}={z}=1/2005
now putting [x]=x-1/2005 and likewise for y and z ,
we get
(x,y,z) = (4011/2005 , 2006/2005 , 2006/2005)
