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# TIFR 2014 Problem 21 Solution – Determining function

TIFR 2014 Problem 21 Solution has been written for TIFR entrance preparation series. A problem using the idea of Fundamental theorem of calculus.

TIFR 2014 Problem 21 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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## Problem:

Let (f:[0,1]\to [0,\infty ) ) be continuous. Suppose ( \int_{0}^{x} f(t)dt \ge f(x) ) for all (x\in [0,1]).

Then

A. no such function exists

B. there are infinitely many such functions

C. there is only one such function

D. there are exactly two such functions

Discussion:

Basically, the question is to find out how many such functions can exist.

Let (F(x)=\int_{0}^{x} f(t)dt ). That is, (F(x)) is the indefinite integral of (f).

We know from the fundamental theorem of calculus that:

(F'(x)=f(x)) .

So we have (F'(x) \ge F(x) ) for all (x\in [0,1] ).

We will manage this equation as we do in case of differential equations. Except that, we have an inequality here.

(F'(x)-F(x) \ge 0) for all (x\in [0,1] ).

Multiplying both sides by (e^{-x}) the inequality remain unchanged. This is because (e^{-x} >0 ).

(e^{-x}F'(x)-e^{-x}F(x) \ge 0) for all (x\in [0,1] ).

Now, ( (e^{-x}F(x))’ = e^{-x}F(x) -e^{-x}F'(x) ).

So we have (  (e^{-x}F(x))’ \le 0 ) for all (x\in [0,1] ).

Therefore, by taking integral from o to y and by using fundamental theorem of calculus:

(  e^{-y}F(y)-e^{0}F(0) \le 0 ) for all (y\in [0,1] ).

i.e, (  e^{-x}F(x) \le F(0) ) for all (x\in [0,1] ).

Also, note that by definition of (F), (F(0)=0). So we have

(  e^{-x}F(x) \le 0 ) for all (x\in [0,1] ).

But, since (e^{-x} >0 ) for all (x\in [0,1] ), we have (  F'(x) \le 0 )  for all (x\in [0,1] ).

So far, we have not used the fact that (f) is a non-negative function. Now we use it. Since (f(x) \ge 0) for all (x\in [0,1] ), therefore by monotonicity of the integral, (F) is an increasing function. This means (F'(x) \ge 0 ) for all (x\in [0,1] ).

By the two inequalities obtained above, we get (  F'(x) = 0 )  for all (x\in [0,1] ).

By the fundamental theorem (again!) we get (f(x)=F'(x)=0)  for all (x\in [0,1] ).

So there is only one such (f) namely the constant function (f=0).

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Fundamental theorem of calculus, Increasing Function
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

## One reply on “TIFR 2014 Problem 21 Solution – Determining function”

rajarshisays:

can we apply liebnitz rule at the first step ?

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