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TIFR 2014 Problem 13 Solution - Supremum of Function

TIFR 2014 Problem 13 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Let (S) be the set of all tuples ((x,y)) with (x,y) non-negative real numbers satisfying (x+y=2n) ,for a fixed (n\in\mathbb{N}). Then the supremum value of (x^2y^2(x^2+y^2)) on the set (S) is:

A. (3n^6)

B. (2n^6)

C. (4n^6)

D. (n^6)


Write the expression in terms of (x) only by substituting (y=2n-x).

Let (f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)). Here, (x\in[0,2n]).

Note that for (x=0) or (x=2n) the function (f(x)=0). Also, (f) is positive everywhere else on the interval.

So we want to find (sup{f(x)|x\in (0,2n)}). Note that it exists because the interval is compact and (f) is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval ((0,2n)).


Now take derivative.




Now, (f'(x)=0) if and only if (4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0).

Note that ([\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0) for all (x\in (0,2n)).

Therefore, (f'(x)=0) if and only if (x=n). If now, (x) is slightly bigger than (n) then (\frac{f'(x)}{f(x)}<0) and since (f(x)>0) we have (f'(x)<0) in that case. And if (x) is slightly smaller than (n) then (f'(x)>0).

This proves that indeed the point (x=n) is a point of maxima.

Therefore, the supremum value is (f(n)=2n^6). So the correct answer is option B.


  • What is this topic: Real Analysis
  • What are some of the associated concept: Supremum Property, Maxima-Minima
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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