# Two Problems from IIT-JEE

IIT-JEE (Joint Entrance Examination for admission to Indian Institutes of Technology) is the most prestigious examination at 10+2 level in India. The questions are supposed to be tough and require ingenuity on the part of the student. There is a booming ecosystem of books, coaching classes and correspondence classes around IIT-JEE. But I find this ecosystem to be quite detrimental in improving the abilities of the student.

Today I will discuss two problems from the mathematics paper which I think are worth discussing here. The reason behind choosing these two problems is the fact that both of them are quite old (one from IIT-JEE 1981 and another from IIT-JEE 2001), but inspite of them being solved I have not yet found a proper solution to them in any of the books especially being published for IIT-JEE. This also shows the flaw in the ecosystem developed around IIT-JEE which focuses only on problem solving without providing any conceptual framework.

To begin with, let me start with an easy (seemingly) problem from Maths paper of IIT JEE 1981.
Problem 1: Let $f(x)$ be a function defined for all $x \in \mathbb{R}$ such that $f(x + y) = f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$. Let $f'(0) = 3$ and $f(5) = 2$. Find the value of $f^{\prime}(5)$.

It is easy to observe that $f^{\prime}(x)$ exists for all $x \in \mathbb{R}$. For, we have \begin{align} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\ &= \lim_{h \to 0}\frac{f(x)f(h) - f(x)f(0)}{h}\notag\\ &= \lim_{h \to 0}f(x) \cdot \frac{f(h) - f(0)}{h} = f(x)f'(0) = 3f(x)\notag \end{align} and therefore $f^{\prime}(5) = 3f(5) = 3 \cdot 2 = 6$. This is exactly the solution provided in all the books which I have seen in the Indian market.

If we just dig slightly deeper we will find something really strange. A simple look at the functional equation $$f(x + y) = f(x)f(y)$$ gives us a hint that $f(x)$ is an exponential function. To make the argument rigorous we observe that $$\frac{d}{dx}\left(\frac{f(x)}{e^{3x}}\right) = \frac{e^{3x}f'(x) - 3e^{3x}f(x)}{e^{6x}} = 0$$ as $f'(x) = 3f(x)$. It follows that $f(x) = ke^{3x}$ where $k$ is some constant. Now $f(0) = k$ and since $f(0) = f(0 + 0) = f(0)f(0)$ it follows that $k = k \cdot k = k^{2}$. Thus either $k = 0$ or $k = 1$. If $k = 0$ then $f(x) = 0$ for all $x \in \mathbb{R}$ and therefore $f'(x) = 0$ for all $x \in \mathbb{R}$, which is not the case here. We thus have $k = 1$ and $f(x) = e^{3x}$.

Thus $f(5) = e^{15} > e > 2$ and there is no way that $f(5) = 2$. The function mentioned in the problem does not exist. But I don't think even the examiner had this thought in his mind. I had encountered this problem way back in 1997 from some book for IIT-JEE and had told the above solution to many people but no one was convinced that the books or the JEE question were wrong. There are scores of books on calculus for IIT-JEE which I have seen, but none of them have expressed the point which I have made about this question.

This does not sound good as the current students at 10+2 level are being programmed to act like mechanical robots and crack JEE. Especially in the field of calculus most of the students are conceptually bankrupt and for no fault of theirs. The books available for competitions simply try to teach them algebraical manipulations along with magical formulas of calculus and expect the student to be a master in the art of solving problems. Next they are provided with lots and lots of problems (mostly solved and few unsolved something like 70/30 ratio) so that they can tackle any problem which is similar to the ones provided. An average calculus book is therefore spanning around 1000 to 1200 pages. Coaching classes are even worse as they provide even larger set of problems and tests and the student acts more like a forced laborer trying to keep all the problems and their solutions in his head.

The second problem I have chosen is from IIT-JEE 2001 and I happened to come across it a few days ago when I was teaching calculus to my cousin. (After clearing JEE in 1998, I didn't have the chance to look at JEE problems with that much curiosity)
Problem 2: Let $f(x)$ be a non-negative continuous function defined for $x \geq 0$ and $F(x) = \int_{0}^{x}f(t)dt$. Also let there be a constant $c > 0$ such that $f(x) \leq cF(x)$ for all $x \geq 0$. Prove that $f(x) = 0$ for all $x \geq 0$.

All the solutions which I have seen till now proceed in very crazy fashion. First they assume that $f(x) = 0$ for all $x \geq 0$ and then there is nothing to prove. If that's not the case they assume (by magic and not logic) that $f(x) > 0$ for all $x \geq 0$. This is real crazy stuff. If its not the case that $f(x) = 0$ for all $x \geq 0$ then it means that there is some number say $x_{0} > 0$ such that $f(x_{0}) > 0$. I doubt the authors of these books really understand what they are writing.

Moreover I think that trying to prove that $F(x) = 0$ for all $x \geq 0$ is easier than directly focusing on $f(x)$ as the anti-derivative $F(x)$ is a smoother function with many nice properties like:
1. $F(x)$ is differentiable for $x \geq 0$ (and therefore continuous too).
2. $F(x)$ is increasing (not in the strict sense) function of $x$ for $x \geq 0$. (As $F^{\prime}(x) = f(x) \geq 0$.)

Clearly $F(0) = 0$. Since $F(x)$ is increasing therefore $F(x) \geq 0$ for all $x \geq 0$. Let us assume that there is a number $b > 0$ such that $F(b) > 0$. Let us now examine the behavior of $F(x)$ in the interval $[0, b]$. Since $F(0) = 0$ there must be a last value of $x \in [0, b]$ such that $F(x) = 0$ (this is a very subtle point and conveys much more than the intermediate value theorem for continuous functions, see this post). Let this value of $x$ be denoted by $a$. Clearly $0 \leq a < b$ and $F(a) = 0$ and $F(x) \neq 0$ for any value of $x \in (a, b]$. Since $F(x) \geq 0$ it follows that $F(x) > 0$ for all $x \in (a, b]$.

Let $x \in (a, b]$. Then we have \begin{align} &f(x) \leq cF(x)\notag\\ &\Rightarrow F'(x) \leq cF(x)\notag\\ &\Rightarrow \frac{F'(x)}{F(x)} \leq c\notag\\ &\Rightarrow \int_{x}^{b}\frac{F'(t)}{F(t)}\,dt \leq \int_{x}^{b}c\,dt\notag\\ &\Rightarrow \log F(b) - \log F(x) \leq c(b - x)\notag \end{align} If we now let $x \to a+$ we see that the LHS of above inequality tends to $\infty$ and the RHS tends to a positive number $c(b - a)$. This contradiction proves that that there is no number $b > 0$ such that $F(b) > 0$. Thus $F(x) = 0$ for all $x \geq 0$ and consequently its derivative $f(x) = 0$ for all $x \geq 0$.

The subtle point mentioned in bold is not presented in common textbooks of calculus. What is sometimes given without proof is the following:
If $f(x)$ is continuous in closed interval $[a, b]$ and $f(a)f(b) < 0$ there is a number $c \in (a, b)$ such that $f(c) = 0$.

Without using the fact that continuous functions take their values for the first and last time in a closed interval it is difficult to solve the above problem. At least I am not aware of any solution which avoids this rarely mentioned property of continuous function. Readers are welcome to provide any other solution to this problem.

P.S.: I received another solution to the second problem in the comments which I found so beautiful as to include in the post itself. Thanks to Milind Hegde for sharing this solution which avoids some deep theorems which I have used in my solution.

As before the idea is to prove that the antiderivative $F(x) = 0$. But here we use a trick of multiplying by $e^{-cx} > 0$. Thus we have the following derivation: \begin{align} &f(x) \leq cF(x)\text{ for all }x \geq 0\notag\\ &\Rightarrow F'(x) \leq cF(x)\notag\\ &\Rightarrow e^{-cx}F'(x) - ce^{-cx}F(x) \leq 0\notag\\ &\Rightarrow \frac{d}{dx}\{e^{-cx}F(x)\} \leq 0\notag\\ &\Rightarrow \int_{0}^{x}\frac{d}{dt}\{e^{-ct}F(t)\}\,dt \leq \int_{0}^{x}0\,dt = 0\notag\\ &\Rightarrow e^{-cx}F(x) - e^{-c\cdot 0}F(0) \leq 0\notag\\ &\Rightarrow e^{-cx}F(x) \leq 0\notag\\ &\Rightarrow F(x) \leq 0\notag \end{align} and since we already know that $F(x) \geq 0$ it follows that $F(x) = 0$ and therefore $f(x) = F'(x) = 0$ for all $x \geq 0$. A generalization of this problem has been discussed on MSE and a beautiful answer is also presented on MSE.

#### 21 comments :: Two Problems from IIT-JEE

1. The function may not exist, but if that option is not given, you have to choose the best answer, right? Possibly everyone was looking for that answer.

Well first of all it was not an objective question where we had to choose one or more answers. A proper solution should examine all the possibilities. For example consider the current trend of objective questions with one or more correct answers and we are given the following options as answers:

a) 6
b) 12
c) f’(5) does not exist
d) f(x) does not exist

In such a scenario, the correct answers should be c) and d), but a casual student (or authors of JEE books) would simply choose a) and get no marks for it.

What surprises me about this question is the fact that even famous authors have overlooked this simple solution which I have provided.

3. To add to my previous comment, one of the JEE books (which provides wrong solution to problem 1) gives the following question in exercise:

If f(x + y) = f(x)f(y) and f’(x) exists for some value of x, then prove that either f(x) = 0 for all x, or f(x) = e^(kx) for some constant k.

Also in this exercise we don’t need to assume the existence of derivative. Continuity at a single point suffices to prove that f(x) = 0 for all x, or f(x) = e^(kx) for some constant k.

4. I think for young mind appearing for IIT-JEE detailed analytic proof might appear diificult. But candidate who has conceptual understanding of definite integral (i.e area under a curve) will simplify the given problem statement in this fashion

1, there is a contunious function f(x) in the first quadrant (x axis included)
2. F(x) represents the area under the curve defined by f(x) from 0 to x
3. it is given that nuemerical value of f(x) at any point p is 0 at any point.

now clearly f(0) cannot be greater than zero, because area enclosed by the curve at x=0 is zero, so if f(0)>0 then condition mentioned in point 3 is violated.

now let us analyse the point in the immediate positive neigbourhood of point 0, say h where limit h->0

so the area bounded by f(x) in range [0,h] can be approximated as area of a triangle formed
A = (1/2)*(h)*f(h) (assuming f(h) to be some finite +ve number)

(as f(0)=0 so the figure looks like a triangle rather than a trapezium, this is the concept that student learns while he learns to see definite integral as limit of a summation of a series)

=> F(h) =(1/2)*(h)*f(h) < f(h) (as (h/2) = f(h)…………(ii)

as derived in (i) F(h)=1

=>c*(h/2)*f(h)>=f(h) ……………[replacing F(h) = (h/2)*f(h)]
=>c*h>=2 or f(h)=0

now if f(h) is +ve, then c->infinity as h->0 hence f(h) cannot be +ve, this implies that under the given condition (i.e point no. 3) only possible value of f(h)=0.

now that we have proven that f(h)=0 we can apply similar logic for a point h’ in the immediate right neighbourhood of h to derive that f(h’)=0. By same logic we can prove that f(x)=0 at any point.

Hence f(x)=0

This might appear a bit more crude than the solution provided by you but for a JEE candidate visualisation of this problem (i.e point 3) is the crux of the problem ..after that he can build on to it.

I hope you might find some value in my solution as this can be more appealing to the young minds where they find the practical meaning of this problem when they would realise that this complicated and dry looking problem statement, when simplified means that numerical value of f(x) has to be greater than numerical value of the area bounded by it at some point or the other.

anytomdickandhary

(you can mail me at anytomdickandhary@yahoo.com)

old friend

5. Please delete my previous comment as point no.3 did not get published correctly hence posting it again

I think for young mind appearing for IIT-JEE detailed analytic proof might appear diificult. But candidate who has conceptual understanding of definite integral (i.e area under a curve) will simplify the given problem statement in this fashion

1, there is a contunious function f(x) in the first quadrant (x axis included)
2. F(x) represents the area under the curve defined by f(x) from 0 to x
3. it is given that nuemerical value of f(x) at any point p is 0 at any point.

now clearly f(0) cannot be greater than zero, because area enclosed by the curve at x=0 is zero, so if f(0)>0 then condition mentioned in point 3 is violated.

now let us analyse the point in the immediate positive neigbourhood of point 0, say h where limit h->0

so the area bounded by f(x) in range [0,h] can be approximated as area of a triangle formed
A = (1/2)*(h)*f(h) (assuming f(h) to be some finite +ve number)

(as f(0)=0 so the figure looks like a triangle rather than a trapezium, this is the concept that student learns while he learns to see definite integral as limit of a summation of a series)

=> F(h) =(1/2)*(h)*f(h) < f(h) (as (h/2) = f(h)…………(ii)

as derived in (i) F(h)=1

=>c*(h/2)*f(h)>=f(h) ……………[replacing F(h) = (h/2)*f(h)]
=>c*h>=2 or f(h)=0

now if f(h) is +ve, then c->infinity as h->0 hence f(h) cannot be +ve, this implies that under the given condition (i.e point no. 3) only possible value of f(h)=0.

now that we have proven that f(h)=0 we can apply similar logic for a point h’ in the immediate right neighbourhood of h to derive that f(h’)=0. By same logic we can prove that f(x)=0 at any point.

Hence f(x)=0

This might appear a bit more crude than the solution provided by you but for a JEE candidate visualisation of this problem (i.e point 3) is the crux of the problem ..after that he can build on to it.

Also, one additional point that i would like to make here is that you proof rests on an very fundamental assumption that f(x) is a non-stochastic function. A non-stochatic function can be continuous as well, so based on problem definition someone as through as you (please take it as a sign of respect that I have for you) cannot assume it to be a non-stochastic function while providing the solution. But surprisingly the problem statement would hold for stochastic function as well, and in such a scenario the solution that I have provided would be a bit more robust (not the most robust!) than yours

I hope you might find some value in my solution as this can be more appealing to the young minds where they find the practical meaning of this problem when they would realise that this complicated looking problem statement when simplified means that numerical value of f(x) has to be greater than numerical value of the area bounded by it at some point or the other.

anytomdickandhary

(you can mail me at anytomdickandhary@yahoo.com)

old friend

6. for some reason point no.3 is not getting published properly when I publish the entire post…. hence re-posting just the point no. 3. would be really thankful if you can correct/delete the previous incorrect post

3. it is given that nuemerical value of f(x) at any point p is < = c*(area enclosed by the f(x) between 0 and p)

anytomdickandhary

(you can mail me at anytomdickandhary@yahoo.com)

old friend

7. for some reasons (absolutely unknow to me) the post is not getting published correctly … there are so many unwanted errors that come after publishing even after correcting it….. can i have your email ID to share my solution with you.

old friend

8. @old friend

It seems that you need to type the mathematical symbols using $\LaTeX$ and if you are not comfortable with it, then you can try to put the mathematical symbols in double quotes. Also I would like to know who you are. Only then I can provide you email.

9. Old friend, I understand and admire your intuitive solution, but it is precisely this kind of intuitive reasoning which needs to be made rigorous. What is really lacking in mathematics education in secondary and higher secondary classes is the rigor which is absolutely essential for development of mathematical skills. Moreover rigor is something completely different from formalism (which I hate absolutely).

Another point is about the assumption related to non-stochastic function. First of all I have no idea about non-stochastic. I am not well versed in any higher level mathematics courses. However I am still convinced that I am not assuming anything more than given in the statement of the problem.

10. Thanks Paramanand, I was sure that you would be able to get the intuitive logic mentioned in my post even when some unwanted errors creeped in. And I agree completely with you that intuitive logic needs to be more rigorous.

Based on my past teaching of maths(elementary level) to students what I have realized is that people apply their intuitive capabilities only when they are confronted with a problem which is challenging just enough to make them believe that they understand the problem statement and they would be able to solve the problem if they stretch themselves a bit. If the problem statement/definition appears too complicated they leave it and the journey of developing mathematical capability takes a halt.

Once the intuitive reasoning is developed well, it normally happens that person is able to visualize the problem better and after that when we present the rigor s/he is able to appreciate the rigor. What I mean to say here is that intuitive reasoning is the bridge that helps a person reach the island of mathematical rigor. For few gifted ones, like you, the bridge of intuitive reason is not required to cover the gap from understanding the problem statement to the rigor of the analysis.

regarding stochastic function. let me put a very simple problem (assuming that you are not aware of this and want to know what it is), say a fly is moving along x direction but it continuously and randomly keeps buzzing up and down along y axis (essentially the path of the fly would appear like a graph of a stock price movement over a day). Now, how do you find the area under the curve defined by the path that fly took?

In this case you would realize that the concept definite integral as limit of sum would not work fine, because the curve is not smooth enough to allow you the approximation as infinitesimal rectangles/trapezium. This due to ‘jagged’ nature of the function plot that you cannot find the slope this function at any point as well. This kind of problem has led to the development of stochastic calculus.

Now come back to the problem, if f(x) is defined as the path of the fly, then f(x) would be continuous and F(x) would be finite for all values of x, but you cannot take the derivative of F(x) now, because a plot of F(x) also would be jagged and rough as is the case with f(x). In such a scenario your solution would break down. I think I have given you a food for thought and you would enjoy this new recepie.

And its okay if you dont share your email ID. For now let me remain an Old friend. I shall disclose my identity when right time comes. As of now all I will say is that I have been a great admirer of yours for a long time and I continue to be one.

Old Friend

old friend

11. @old friend

There is some bit wrong about the the stochastic part you are talking about. If there is a function $f(x)$ which is continuous (you must understand that continuous functions can be weird enough like Weierstrass function which is continuous everywhere but differentiable nowhere) on some interval $[a, b]$ then clearly the integral $F(x) = \int_{a}^{x}f(t)\,dt$ is differentiable in $[a, b]$. Moreover the idea is that if $f(x)$ is integrable on $[a, b]$ (integrable is equivalent to continuity almost everywhere i.e continuous at all points except points belonging to set of measure zero) and if $x = c$ is a point of continuity of $f(x)$ then $F^{\prime}(c) = f(c)$. Hence in the example you quote $F(x)$ will always have derivatives.

12. I think I haven’t been able to put my thought using right example, please allow me sometime to find a better example and better way of expressing my thought.

Old Friend

old friend

13. say function is defined as

f(x) = x for “0<=x<=1" and 2x-1 for "1<x<=2"

f(x) is continuous in [0,2]

F(x) exists at all the points in [0,2] and is continuous as well, but does the derivative of F(x) at x=1 exist?

am i missing something?

Old friend.

old friend

14. yes it does exist….I made an calculation error. :)

old friend

15. @old friend

dont worry creating examples, use theory to understand that whenever $f(x)$ is continuous then $F(x)$ is differentiable and its derivative at such a point $c$ is $f(c)$. This is the fundamental theorem of calculus.

16. @paramanand

its nice to be getting refreshed with the fundamental theorems of mathematics once again. May be your blog will help me refresh long forgotten things. Thanks once again.

Btw could you help me learn how to use LATEX, any good site/resource that helps.

Old Friend.

old friend

17. Old friend, thanks for the complements. I have deleted your comments about the problem of floors because this is not really looking like a comment related to this post. Please understand that blog comments are not a medium for general discussion like a forum. Also I think we can now close this comment thread. About learning $\LaTeX$ you need to google for “LaTex tutorials”.

18. There is an alternate solution to the second problem, which I found in KD Joshi's Educative JEE Mathematics (An excellent book, nothing like the other JEE books commonly found).

$$F(x) = \int_0^x f(t)\,\mathrm dt \implies F'(x) = f(x)$$
As $f$ is non-negative, we get that $F(x)\geq0$. Also $F(0) = 0$.

Then from the given inequality we get,
$$f(x) = F'(x) \leq cF(x)$$

Multiplying both sides by $e^{-cx}>0$,
$$e^{-cx}F'(x) - ce^{-cx}F(x) \leq 0\\ \implies\mathrm d(e^{-cx}F(x))\leq0$$
Now integrating the two sides from $0$ to $x$, we get that for all $x\geq0$
$$e^{-cx}F(x) - e^{0}F(0)\leq0\\ \implies e^{-cx}F(x) \leq0\\ \implies F(x)\leq0$$

From earlier we have that $F(x) \geq 0$ and so we get that $F(x) = 0$. From this we get by differentiating that $f(x) = 0.$

19. @ Milind Hegde

Your solution is better than the one I presented because it avoid the difficult theorems which I have used in my solution. I have not read the book by KD Joshi so I can not comment on the whole book, but this particular solution is very good. Thanks for sharing it.

20. Dear Paramanand,
I find your objection to Problem 1 perfect and this one is one of my favorites to be pointed out as one of the very few mistakes the JEE has had in the past. There are other problems in JEE in the past which are inconsistent like the one here (Take for eg Prob 3 of JEE 2003, Q.66 JEE 2011 paper 1, Q.60 JEE 2012 Paper 2, Q.55 JEE 2015 Paper I just to mention a few which are not highlighted in standard books), but it only takes an intriguing mind like yours to point out one. Even the renowned teachers in the coaching field fail to acknowledge the same. I'll not comment on the standard JEE books in the market (which are utterly careless) as they're not my type either, but I should point it out that even the AIEEE has been extremely careless in this regard. They did give this exact same question in an objective exam without any change or realizing any mistake. Every year, such subtle errors are part of the exam system partly because the paper setters (non IIT) themselves are not sound in their own subject. Take for example, this year in JEE Mains Offline/online, there were total 6 questions I can figure out which are subtly wrong and absolutely no one pointed this out (except for one question in offline paper) This is a pity in a country like ours where more focus is on solution focus approach rather than problem focused. No wonder therefore we produce a huge chink of mediocre engineers.

21. excellent sir,true ,many times we assume things which may not be obvious.only after seeing your results i realize even i didn't notice that,and i can clearly understand that this mistake is neither restricted to mathematics nor to iit jee.