MATH GEEKS CHALLENGE!

[PenPoint]

Preface in a single sentence : This is NOT a homework.

 

 

 

There may be some wrong translations but it won't be a massive error

[PenPoint]

3 minutes ago, Kumaresh said:

I am getting a contradiction. Don't the conditions imply that alpha=beta when f(alpha)=f(beta)=g(alpha)=g(beta)=M ( local maxima )and applying (x-a)f(x)=g(x) since x>a for f(x) to exist..... Then beta-alpha=6√3 would not be possible.....

Um...

first of all alpha=/=beta.

second, f'(alpha)=g'(alpha)=M -> local maximum

[PenPoint]

1 minute ago, Kumaresh said:

f'(alpha)=g'(alpha)=M ? This question is worded weirdly.... f'(alpha)=0 and f''(alpha)<0 would imply that f(alpha) is a maxima M....... For an extremum of the function, derivative of the function must be zero...... If f'(alpha)=M ( maxima ), then f''(alpha)=0, we can't say anything about f(alpha). There is some confusion

Actually the question is originally not written in English. I translated into English so please understand that there may be some translation errors. :'(

 

Okay so... let's check it out. Oh and I'm not good at math

 

1. f'(alpha)=0? yes. 'cuz it's local maximum.

2. f''(alpha)<0? yes. Reason's same.

3. f(alpha)=M? yes. same.

4. At the extremum of the function(let's call x=k), yes, f'(k)=0.

5. f'(alpha)=M? Why? f'(alpha)=0, not M.

[kingkickolas]

8 hours ago, PenPoint said:

Actually the question is originally not written in English. I translated into English so please understand that there may be some translation errors. :'(

 

Okay so... let's check it out. Oh and I'm not good at math

 

1. f'(alpha)=0? yes. 'cuz it's local maximum.

2. f''(alpha)<0? yes. Reason's same.

3. f(alpha)=M? yes. same.

4. At the extremum of the function(let's call x=k), yes, f'(k)=0.

5. f'(alpha)=M? Why? f'(alpha)=0, not M.

@Kumaresh is right, f(A) = f(B) = g(A) = g(B) = M implies A = B, since g(x) = (x - a) f(x).

(Using A and B in place of alpha and beta.)

 

Maybe there's a translation error in the problem?

[PenPoint]

9 hours ago, kingkickolas said:

@Kumaresh is right, f(A) = f(B) = g(A) = g(B) = M implies A = B, since g(x) = (x - a) f(x).

(Using A and B in place of alpha and beta.)

 

Maybe there's a translation error in the problem?

I think there's a translation error. Sorry :'(

In the second condition, those conditions only fits in for function f(x), not g(x).

So f(A)=f(B)=M but g(A)=/=M, g(B)=/=M. Is everything alright now @Kumaresh?

[PenPoint]

Just now, Kumaresh said:

OK, I can solve the problem now. Will check the solution tonight.....

Phew... error solved

 

[kingkickolas]

I think there might still be a mistranslation somewhere in the question. (Sorry, I know translating can be difficult, especially for technical material!)

 

With the new information, I can rewrite g as a function of the free parameters alpha, beta, a, and M:

So, for any combination of alpha, beta, a, and M, there exists a function g (given above) such that f(x) = g(x)/(x-a) has local maxima at (alpha, M) and (beta, M). So therefore any value of M is permitted and there is no minimum for it.

[PenPoint]

3 hours ago, Kumaresh said:

-snip-

 

4 hours ago, kingkickolas said:

-snip-

 

I'm afraid to tell you both that there IS the minimum value of M.

And I'll give you a small tip: do you @Kumaresh really think that f(x) still have to be a cubic function?

And one more thing: this question has three main ways to solve.

[PenPoint]

Just now, Kumaresh said:

Then provide your solution please. I am sure that only if you multiply a cubic equation by a linear factor will you get a biquadratic polynomial. What else can you multiply to get it ?

We call f(x) as 'fractional function'.

[kingkickolas]

7 hours ago, PenPoint said:

 

 

I'm afraid to tell you both that there IS the minimum value of M.

And I'll give you a small tip: do you @Kumaresh really think that f(x) still have to be a cubic function?

And one more thing: this question has three main ways to solve.

f is only cubic if a is a root of g, which it isn't necessarily. 

 

But this is why I think there might still be a mistranslation error in the question. You can verify that the g in my post gives an f that satisfies all of the criteria, but permits any value of M (even negative M).

 

 

-- Edit --

Specifically, with the g from my previous post:

 

Edited March 12, 2017 by kingkickolas

[PenPoint]

5 hours ago, kingkickolas said:

-snip-

 

Hmm that's strange. I'll post the answer in a few hours.

[PenPoint]

@Kumaresh Nobody said that (x-a) should be a factor of g(x).

Actually when I saw this question for the first time, I thought the same thing.

 

And this question is not related to India at all XP

[kingkickolas]

@PenPoint, hey is there any update on this? Curious to see what you got 

[PenPoint]

1 hour ago, kingkickolas said:

@PenPoint, hey is there any update on this? Curious to see what you got 

Oh sorry. I forgot to post the answer ;-;

Answer is REALLY going to be posted in a few hours.

[kingkickolas]

Just now, PenPoint said:

Oh sorry. I forgot to post the answer ;-;

Answer is REALLY going to be posted in a few hours.

Haha no prob! :3

[PenPoint]

Dear @Kumaresh and @kingkickolas, here is the answer. Took me a lot... translation process is freakin' hard.

 

 

dif.pdf

 

Oh, and the source is CSAT of South Korea.

[kingkickolas]

9 hours ago, PenPoint said:

Dear @Kumaresh and @kingkickolas, here is the answer. Took me a lot... translation process is freakin' hard.

 

 

dif.pdf

 

Oh, and the source is CSAT of South Korea.

I have to agree, this is a very impressive bit of work! Well, until the last two lines...

 

Your conclusion is correct if A+B=0, but this is nowhere stated in the problem, and we cannot assume it! In fact, the conclusion is correct even if A+B is not 0, but we have not shown it yet!

 

What we have to do is find . That is, the value of g' at its local minimum (we know this occurs at the left-most point where g'' = 0, since g' is an overall-decreasing cubic). When you do this, you find that:

.

 

And when you plug this into g', you get, in general, that , which in the case for B = A + 6 \sqrt{3}, gives 216! So this is true even if A + B is not 0 (which we were not justified to assume anyway).

 

I wish I had stuck with this problem more, I just forgot to apply condition 3 in your OP to only x > a! What a fantastic problem though, I won't soon forget it!

[PenPoint]

3 hours ago, kingkickolas said:

-snip-

Actually you're right, there is no reason to assume α+β=0, but for the computational convenience! lol

Approximately top 1% of the whole students who took the test were only able to get this right.

 

[kingkickolas]

2 hours ago, PenPoint said:

Actually you're right, there is no reason to assume α+β=0, but for the computational convenience! lol

Approximately top 1% of the whole students who took the test were only able to get this right.

 

Yeah I don't envy them! That's a very difficult question, I have no idea how I would do this on a test!

[PenPoint]

Dear @Kumaresh and @kingkickolas, wanna try another question?

If any of you are interested, please leave a comment below

 

[kingkickolas]

1 hour ago, PenPoint said:

Dear @Kumaresh and @kingkickolas, wanna try another question?

If any of you are interested, please leave a comment below

 

Yeah definitely! You've got some good ones apparently!

[PenPoint]

@Kumaresh @kingkickolas Here you go :]

 

[kingkickolas]

6 minutes ago, PenPoint said:

@Kumaresh @kingkickolas Here you go :]

 

-snip-

Cool!

 

Clarification: Is the domain or range (or both) of h'' all of the reals? Also, f can have odd powers too, or just even?

[PenPoint]

21 minutes ago, kingkickolas said:

Cool!

 

Clarification: Is the domain or range (or both) of h'' all of the reals? Also, f can have odd powers too, or just even?

Yes, it is all of the reals. And f(x)=x^4+ax^3+bx^2+cx+d shape. Is everything clear?

[kingkickolas]

Just now, PenPoint said:

Yes, it is all of the reals. And f(x)=x^4+ax^3+bx^2+cx+d shape. Is everything clear?

Thanks! But is it the domain or range of h'' (or both)?