Solving Equations with Integer Solutions

[sankalpmittal]

Homework Statement

Find the domain of following functions : Note: [.] denotes greatest integer function. {.} denotes fractional part of a function.

1. f(x)=sin^-1([x]/{x})

2. √(2{x}²-3{x}+1)

3. Solve the equation :

y=[sinx+[sinx+[sinx]]]/3
[y+[y]]=2cosx

Homework Equations

The Attempt at a Solution

1. For this function to exist,

-1≤[x]/{x}≤1
As 0≤{x}<1

-{x}≤[x]≤{x}
Now [x]=x-{x}

So

0≤x≤2{x}

Now ??

2. For this to exist,

2{x}²-3{x}+1≥0

From here I found values of {x} but to no avail.

3. I do not know how to proceed. :(

Please help !

Thanks in advance...

 

[Saitama]

Homework Statement

Find the domain of following functions : Note: [.] denotes greatest integer function. {.} denotes fractional part of a function.

1. f(x)=sin^-1([x]/{x})

2. √(2{x}²-3{x}+1)

3. Solve the equation :

y=[sinx+[sinx+[sinx]]]/3
[y+[y]]=2cosx

Homework Equations

The Attempt at a Solution

1. For this function to exist,

-1≤[x]/{x}≤1
As 0≤{x}<1

-{x}≤[x]≤{x}
Now [x]=x-{x}

So

0≤x≤2{x}

Now ??

2. For this to exist,

2{x}²-3{x}+1≥0

From here I found values of {x} but to no avail.

3. I do not know how to proceed. :(

Please help !

Thanks in advance...

For the first question, it is obvious that it is not defined for integers.

I am unsure what would be the best approach. I would begin with considering different intervals for x. Check what happens to the ratio when 0<x<1, check similarly for 1<x<2 and others.

 

[sankalpmittal]

For the first question, it is obvious that it is not defined for integers.

I am unsure what would be the best approach. I would begin with considering different intervals for x. Check what happens to the ratio when 0<x<1, check similarly for 1<x<2 and others.

As per you the domain should be all fractions. But this is not true for negatives, as for the the greatest integer function "back-shifts".

So what is wrong in my approach ?

-1≤[x]/{x}≤1
As 0≤{x}<1

-{x}≤[x]≤{x}
Now [x]=x-{x}

So

0≤x≤2{x}

and

0≤2{x}<2

 

[Saitama]

As per you the domain should be all fractions. But this is not true for negatives, as for the the greatest integer function "back-shifts".

Nope, I never said that. I asked you to check for 0<x<1. I already know what happens in the case when -1<x<0.

So what is wrong in my approach ?

Not sure but it is usually a good idea to check specific intervals when you have to deal with these kind of functions.

 

[sankalpmittal]

Nope, I never said that. I asked you to check for 0<x<1. I already know what happens in the case when -1<x<0. Not sure but it is usually a good idea to check specific intervals when you have to deal with these kind of functions.

Case I: Does not work for integers.

Case II: Non integers.

Ok I considered 2 intervals.

at 0<x<1

at 1<x<2

And I got the answers. But I still can not get the actual mathematical procedure except plotting graph. I must wait till morning for more replies regarding first question.

Anyone else ?

 

[pasmith]

Homework Statement

Find the domain of following functions : Note: [.] denotes greatest integer function. {.} denotes fractional part of a function.

1. f(x)=sin^-1([x]/{x})

2. √(2{x}²-3{x}+1)

3. Solve the equation :

y=[sinx+[sinx+[sinx]]]/3
[y+[y]]=2cosx

Homework Equations

The Attempt at a Solution

1. For this function to exist,

-1≤[x]/{x}≤1
As 0≤{x}<1

-{x}≤[x]≤{x}

Therefore
<br /> -1 &lt; -\{x\} \leq [x] \leq \{x\} &lt; 1<br />

 

[sankalpmittal]

Therefore
<br /> -1 &lt; -\{x\} \leq [x] \leq \{x\} &lt; 1<br />

And hence from here, taking [x]=x-{x}

−1+{x}<0≤x≤2{x}<1+{x}

And so,

How can I get the values of x from here ? Thanks till now.

Edit: From this:

1+{x} is from [1,2). If x is always less than this, then x<1. Now x>{x}-1. {x}-1 is from [-1,0). If x be always greater than this, it has to be x>0. Hence combining the two, we get,

0<x<1

Which is the correct answer. Thanks to both the posters ! :)

Now for the second question ?

 

[sankalpmittal]

I give my attempt of second question a step forward :

2{x}²-3{x}+1≥0
2{x}²-2{x}-{x}+1≥0
(2{x}-1)({x}-1)≥0

So by method of intervals,
{x}≥1 OR {x}≤1/2
Also we know that,

0≤{x}<1

So using these two conditions, how to find for the values of x ? Please help !

 

[Saitama]

I give my attempt of second question a step forward :

2{x}²-3{x}+1≥0
2{x}²-2{x}-{x}+1≥0
(2{x}-1)({x}-1)≥0

So by method of intervals,
{x}≥1 OR {x}≤1/2
Also we know that,

0≤{x}<1

So using these two conditions, how to find for the values of x ? Please help !

So what is the range of {x}?

 

[sankalpmittal]

So what is the range of {x}?

Why, for this question it is, 0≤{x}≤1/2...

So what will be the values of x corresponding to this ?

 

[Saitama]

Why, for this question it is, 0≤{x}≤1/2...

Yes. :)

So what will be the values of x corresponding to this ?

{x} is periodic with period of 1, does that help?

Sketch a graph if that helps.

 

[sankalpmittal]

Yes. :)

Well good to know.. :)

{x} is periodic with period of 1, does that help?

Well I already know that. For all xεR, 0≤{x}<1
Now here we have 0≤{x}≤0.5. Making use of the knowledge that {x} has period 1, we can say that domain of x is,

xε[0,0.5]U[1,1.5]U[2,2.5],... it will keep moving on...U(0,-0.5]U[-1,-1.5]U...and so on...
But this is not the answer ! :(

Sketch a graph if that helps.

...Does not give answer... :(

 

[Saitama]

xε[0,0.5]U[1,1.5]U[2,2.5],... it will keep moving on...U(0,-0.5]U[-1,-1.5]U...and so on..

A much better way to write the above is
$$x \in \left[ m,m+\frac{1}{2} \right] \,\, \forall \,\, m \in \mathbb{Z}$$

EDIT: How do you get [-0.5,0]? What is the value of, say, {-0.1}?

But this is not the answer ! :(

What is the answer then?

 

[sankalpmittal]

A much better way to write the above is
$$x \in \left[ m,m+\frac{1}{2} \right] \,\, \forall \,\, m \in \mathbb{Z}$$

EDIT: How do you get [-0.5,0]? What is the value of, say, {-0.1}?


What is the answer then?

Sorry for my haste. The question explicitly stated that we have to restrict domain in [-1,1].
So we have,

x∈[m,m+1/2]∀m∈Z

putting, m=-1 and 0, m=1 lies outside [-1,1].

x∈[-1,-1/2]U[0,1/2]U{1}∀m∈Z , as 1 alone also satisfies the function.

Now can you give me a start for the third question ?

 

[Saitama]

Now can you give me a start for the third question ?

I am not entirely sure if there is a "proper" way to do it but currently, I cannot find a better method than checking the cases.

Look at the second equation, the left side is an integer so the right side must be also an integer.

The possible values the right side can have is -2,-1,0,1,2. When 2cosx=0, then sin(x)=1. Substitute this in the first equation and you get y=1. Check if this y satisfies the second equation. We find that when y=1, [y+[y]]=2 but right side is zero so there is no value of x satisfying this condition.

Similarly check for other cases.

Does the question restricts the domain? Can you re-check with the source?