• Support PF! Buy your school textbooks, materials and every day products Here!

Arithmetic Progression question

  • Thread starter Saitama
  • Start date
  • #1
3,812
92

Homework Statement


If x ε R, the numbers 51+x+51-x, a/2, 25x
+25-x form an AP, then 'a' must lie in the interval:-
a)[1,5)
b)[2,5]
c)[5,12]
d)[12,∞)

Homework Equations


Not required


The Attempt at a Solution


I substituted y=5x.
The terms are in AP, so the common difference is constant. That is:-
[tex]\frac{a}{2}-5y-\frac{5}{y}=y^2+\frac{1}{y^2}-\frac{a}{2}[/tex]
[tex]\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}[/tex]

Now i am stuck here. :(
Please somebody help.

Thanks! :smile:
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94
Think about the domain of each variable. [itex]x\in R[/itex] thus [itex]y=5^x[/itex] is in which set? And what about a?
You may want to graph the function [tex]a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}[/tex] to get an idea of what the answer might be.
 
  • #3
3,812
92
Think about the domain of each variable. [itex]x\in R[/itex] thus [itex]y=5^x[/itex] is in which set? And what about a?
Thanks for the reply Mentallic. :smile:

y=5x in (0,∞) set. Right?

Then a should also lie in the same set.

Can you tell me an other way except graphing? Because this question is from a test paper and i don't think graphing the function would do me any good because i suck at graphing. :D
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,257
974

Homework Statement


If x ε R, the numbers 51+x+51-x, a/2, 25x
+25-x form an AP, then 'a' must lie in the interval:-
a)[1,5)
b)[2,5]
c)[5,12]
d)[12,∞)

Homework Equations


Not required


The Attempt at a Solution


I substituted y=5x.
The terms are in AP, so the common difference is constant. That is:-
[tex]\frac{a}{2}-5y-\frac{5}{y}=y^2+\frac{1}{y^2}-\frac{a}{2}[/tex]
[tex]\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}[/tex]

Now i am stuck here. :(
Please somebody help.

Thanks! :smile:
Now substitute 5x back in for y.
[itex]a=5^{2 x}+(5)5^{x}+(5)5^{-x}+5^{-2 x}[/itex]​

See if you can find a minimum and/or a maximum for this.

(It's an even function.)
 
  • #5
3,812
92
Now substitute 5x back in for y.
[itex]a=5^{2 x}+(5)5^{x}+(5)5^{-x}+5^{-2 x}[/itex]​

See if you can find a minimum and/or a maximum for this.

(It's an even function.)
Hi SammyS! :)

Even function? What do you mean by that?

And i am not sure about how to find maximum and minimum of a function. I just read it once.
Do we find minimum or maximum like this: First we take the derivative of the function and then substitute it to zero.?
 
Last edited:
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,257
974
Hi SammyS! :)

Even function? What do you mean by that?

And i am not sure about how to find maximum and minimum of a function. I just read it once.
Do we minimum or maximum like this: First we take the derivative of the function and then substitute it to zero.?
If you know how to take the derivative, then do that and set it equal to zero. (This is the precalculus section.)

An even function is one whose graph is symmetric with respect to the y-axis.

You can also represent this as the sum of two hyperbolic cosine functions.

5 = eln(5), so

[itex]\displaystyle a=5^{2 x}+5^{-2 x}+(5)5^{x}+(5)5^{-x}[/itex]
[itex]\displaystyle =e^{2 x\,\ln5}+e^{-2 x\,\ln5}+(5)e^{x\,\ln5}+(5)e^{-x\,\ln5}[/itex]

[itex]\displaystyle =2\cosh(2 x\,\ln5)+2(5)\cosh(x\,\ln5)[/itex]​

Now you can graph this.
 
  • #7
3,812
92
If you know how to take the derivative, then do that and set it equal to zero. (This is the precalculus section.)

An even function is one whose graph is symmetric with respect to the y-axis.

You can also represent this as the sum of two hyperbolic cosine functions.

5 = eln(5), so

[itex]\displaystyle a=5^{2 x}+5^{-2 x}+(5)5^{x}+(5)5^{-x}[/itex]
[itex]\displaystyle =e^{2 x\,\ln5}+e^{-2 x\,\ln5}+(5)e^{x\,\ln5}+(5)e^{-x\,\ln5}[/itex]

[itex]\displaystyle =2\cosh(2 x\,\ln5)+2(5)\cosh(x\,\ln5)[/itex]​

Now you can graph this.
Lol, i don't know about hyperbolic functions. :rofl:

Sorry, but i did not know that we will have to use Calculus in this question, so i posted it in precalculus section. :)

Ok i go like this:-
I find the derivative of the given function which comes out to be:-
[tex]25^xlog_e25-25^{-x}log_e25+5(5^xlog_e5)-5(5^{-x}log_e5)[/tex]

When i substitute it to zero, i get stuck.. Here are the steps:-
[tex]log_e25(25^x-25^{-x})=-5log_e5(5^x-5^{-x})[/tex]
[tex]\frac{25^x-25^{-x}}{5^x-5^{-x}}=\frac{-5}{2}[/tex]

What should i do next?
 
  • #8
Mentallic
Homework Helper
3,798
94
Thanks for the reply Mentallic. :smile:

y=5x in (0,∞) set. Right?

Then a should also lie in the same set.
Yes, that's right :smile:

Can you tell me an other way except graphing? Because this question is from a test paper and i don't think graphing the function would do me any good because i suck at graphing. :D
Oh, yep ofcourse. Graphing calculators are just the cheap and brainless way of figuring out the answer anyway :wink:

Looking at the function [tex]a=5^{2x}+5^{-2x}+5^{1+x}+5^{1-x}[/tex]

finding a turning point for this function we set the derivative equal to zero. In short, by using the rational root theorem, you should find [itex]5^x=\pm 1[/itex] but 5x>0 so 5x=1 is the only valid solution, thus x=0 is where the turning point occurs. You can go further and show this is a minimum turning point, so at x=0, [itex]a=5^0+5^0+5^{1+0}+5^{1+0}=12[/itex] and clearly a is always increasing since all the powers are greater than zero, and 5x approaches infinite as x approaches infinite, so we have the answer.
 
  • #9
Mentallic
Homework Helper
3,798
94
I find the derivative of the given function which comes out to be:-
[tex]25^xlog_e25-25^{-x}log_e25+5(5^xlog_e5)-5(5^{-x}log_e5)[/tex]
Yes good work, that's correct.

When i substitute it to zero, i get stuck..
Because you shouldn't be plugging in x=0, that will give you the gradient of the original function at x=0, what you instead want to do is solve the derivative equal to 0. Multiply through by 25x and to make things simpler, let y=5x. Then use the rational root theorem to find the roots.
 
  • #10
3,812
92
Yes, that's right :smile:

Oh, yep ofcourse. Graphing calculators are just the cheap and brainless way of figuring out the answer anyway :wink:
We aren't allowed calculators during an exam. :rofl:


Yes good work, that's correct.


Because you shouldn't be plugging in x=0, that will give you the gradient of the original function at x=0, what you instead want to do is solve the derivative equal to 0. Multiply through by 25x and to make things simpler, let y=5x. Then use the rational root theorem to find the roots.
I did not plugged x=0, i solved the derivative equal to 0.
Sorry for my confusing English. :redface:

Here i go again:-
let y=5x
then continuing my above equation,
[tex]\frac{y^2-y^{-2}}{y-y^{-1}}=\frac{-5}{2}[/tex]
Solving i get:-
[tex]\frac{y^2+1}{y}=\frac{-5}{2}[/tex]
[tex]\Rightarrow 2y^2+5y+2=0[/tex]

Solving i get, y=-2 and -1/2.
that is, 5^x=-2 and -1/2.
This is funny, we need to have y>0. :D

I don't understand where i am wrong.
 
  • #11
3,812
92
Bump :)
 
  • #12
Mentallic
Homework Helper
3,798
94
I did not plugged x=0, i solved the derivative equal to 0.
Sorry for my confusing English. :redface:
Ahh ok, then your problem lies elsewhere :tongue:

Here i go again:-
let y=5x
then continuing my above equation,
[tex]\frac{y^2-y^{-2}}{y-y^{-1}}=\frac{-5}{2}[/tex]
Solving i get:-
[tex]\frac{y^2+1}{y}=\frac{-5}{2}[/tex]
How did you get from the first equation to the next? I suggest you get rid of the fractions, so that involves multiplying through by y2. You'll have a polynomial of degree 4 to solve, which is why I mentioned the rational root theorem.
 
  • #13
ehild
Homework Helper
15,427
1,827

[tex]\Rightarrow a=y^2+\frac{1}{y^2}+5y+\frac{5}{y}[/tex]


You can see at once that a can go to infinity.

To find the minimum of a, find the derivative with respect to y. You did it with respect to x, but it differs only by a factor of 5xln(5)

You get the equation

[tex]2y-\frac{2}{y^3}+5-\frac{5}{y^2}=0[/tex]

factor out 5 and 2y:

[tex]5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^4})=0[/tex]

Notice that [tex]1-\frac{1}{y^4}=(1-\frac{1}{y^2})(1+\frac{1}{y^2})[/tex]

[tex]5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^2})(1+\frac{1}{y^2})=0[/tex]

[tex](1-\frac{1}{y^2})(5+2y(1+\frac{1}{y^2}))=0[/tex]
One solution is y2-1=, y=1. The other solution of
[tex]5+2y(1+\frac{1}{y^2})=0[/tex]
leads to negative roots, which are inappropriate. So the only extremum is at y=1, and it is minimum.

ehild
 
  • #14
3,812
92
Ahh ok, then your problem lies elsewhere :tongue:


How did you get from the first equation to the next? I suggest you get rid of the fractions, so that involves multiplying through by y2. You'll have a polynomial of degree 4 to solve, which is why I mentioned the rational root theorem.
Lol, i have never used that theorem. :D
I think i will have to refer wiki.

You can see at once that a can go to infinity.

To find the minimum of a, find the derivative with respect to y. You did it with respect to x, but it differs only by a factor of 5xln(5)

You get the equation

[tex]2y-\frac{2}{y^3}+5-\frac{5}{y^2}=0[/tex]

factor out 5 and 2y:

[tex]5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^4})=0[/tex]

Notice that [tex]1-\frac{1}{y^4}=(1-\frac{1}{y^2})(1+\frac{1}{y^2})[/tex]

[tex]5(1-\frac{1}{y^2})+2y(1-\frac{1}{y^2})(1+\frac{1}{y^2})=0[/tex]

[tex](1-\frac{1}{y^2})(5+2y(1+\frac{1}{y^2}))=0[/tex]
One solution is y2-1=, y=1. The other solution of
[tex]5+2y(1+\frac{1}{y^2})=0[/tex]
leads to negative roots, which are inappropriate. So the only extremum is at y=1, and it is minimum.

ehild
Thank you ehild. I get it now.
I was getting only negative roots, now my problem is solved, thank you. :smile:
 
  • #15
ehild
Homework Helper
15,427
1,827
You are welcome.
Advice: never ever simplify with something which can be zero.

ehild
 
  • #16
3,812
92
You are welcome.
Advice: never ever simplify with something which can be zero.

ehild
Thanks for the advice.
I will remember that. :)
 

Related Threads on Arithmetic Progression question

  • Last Post
Replies
4
Views
1K
Replies
3
Views
788
Replies
5
Views
1K
  • Last Post
Replies
3
Views
602
  • Last Post
Replies
4
Views
634
  • Last Post
Replies
2
Views
2K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
4
Views
695
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
611
Top