E-StoreHome » Ask & Discuss » Mathematics. » Differential Calculus « Back to Discussion
Differential Calculus
Comments (3)
In many of these problems, a useful strategy is try to obtain an equation like {x} = "something" and exploit the fact that 0<={x}<1.
Here LHS = {x} + 2x = {x} + 2([x]+{x}] = 2[x] +3{x}
So the equation can be written as 3 {x} = 2[x] - 2. or ![\{x\} = \frac{2[x]-2}{3}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/8/5/085f8fad1afb0617b183ee4e61495811fb73a39a.gif)
So {x} is a rational number of the form n/3 where n is a non-negative even integer. So n = 0 or 2.
n = 0 gives x =1.
n = 2 gives 8/3
as we know...{x}=x-[x] (x+1)-[x+1]+2x=4[x+1]-6 3x+1=5[x+1]-6 3x+1=5([x]+1)-6 3([x]+{x})=5[x]-2 3{x}=2[x]-2................(1) as we know..0<={x}<1 0<=3{x}<3........................(2) on substitutin...(1) we get.. 1<=[x]<5/2...so we conclude that...[x]=1 or 2. hence from (1)..3{x}=2*1-2=0 or {x}=0 or 2/3. so x={x}+[x]...1+0 or 2+2/3 i.e 1,8/3 are solutions..... nudge me if i m wrong....
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
TopicsMathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
1675 Students Currently Online |
Connect with Us  |
 |
see, x=[x] +{x}.
now, use this:
(x+1) -[x+1] + 2x = 4[x+1] - 6
3x +7=5[x+1].
3x +7=5[x] + 5.
3x +2 = 5[x].
Now, draw the graphs of y=3x+2 and y=5[x].
the intersection points will give you the desired solutions. You can try hit n trial also.
like 8/3 and 1 are two solutions. I haven't checked for more.
but for clarity try to draw graphs.
If still any doubt exists, do tell us.