# ISI MStat 2016 (Sample) Problem 2 | Continuous function | PSB

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 2. This is based on application of continuity and integration.

## Problem- ISI MStat 2016 Problem 2

Let $$f:[-1,1] \rightarrow \mathbb{R}$$ be a continuous function. Suppose that $$f^{\prime}(x)$$ exists and $$f^{\prime}(x) \leq 1$$ for all $$x \in(-1,1)$$ . If $$f(1)=1$$ and $$f(-1)=-1,$$ prove that $$f(x)=x \text { for all } x \in[-1,1]$$.

## Solution

Given function is continuous and $$f^{\prime}(x)$$ exists .

Hence we can integrate $$f^{\prime}(x)$$ .

Another thing is given that $$f^{\prime}(x) \leq 1$$ for all $$x \in(-1,1)$$ ----(1) (say)

Now if we integrate both side of (1) from -1 to y , where $$y \in(-1,1)$$ then we get ,

$$\int^y_{-1} f^{\prime}(x) \,dx \leq \int^y_{-1} 1\,dx$$ , for all $$y \in(-1,1)$$

$$\Rightarrow f(y)-f(-1) \leq y+1 \Rightarrow f(y) \leq y$$ ,for all $$y \in(-1,1)$$ [ since $$f(-1)=-1$$ given ]---(2)

Again if we integrate both side of (1) from y to 1 ,where $$y \in(-1,1)$$ then we get,

$$\int^1_{y} f^{\prime}(x) \,dx \leq \int^1_{y} 1\,dx$$ , for all $$y \in(-1,1)$$

$$\Rightarrow f(1)-f(y) \leq 1-y \Rightarrow f(y) \geq y$$ ,for all $$y \in(-1,1)$$ [ since $$f(1)=1$$ given ] ---(3)

Hence from (2) & (3) we get $$f(y)=y$$ for all $$y \in(-1,1)$$ and $$f(-1)=-1 , f(1)=1$$ .

Therefore , $$f(x)=x$$ for all $$x \in[-1,1]$$ ( proved )

## Challenge Problem

$$f:(-1,1) \to (-1,1)$$ be continuous function such that $$f(x)=f(x^2)$$ for every x and $$f(0)=\frac{1}{2}$$

Then find f(x) .

# Telescopic Continuity | ISI MStat 2015 PSB Problem 1

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

## Problem- Telescopic Continuity

Let $$f: R \rightarrow R$$ be a function which is continuous at 0 and $$f(0)=1$$
Also assume that $$f$$ satisfies the following relation for all $$x$$ :
$$f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x$$ Find $$f(3)$$.

## Solution

$$f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3$$

$$f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}$$

$$f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}$$

$$\cdots$$

$$f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}$$

Add them all up. That's the telescopic elegance.

$$f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]$$

Observe that $$a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1$$ since, $$f(x)$$ is continuous at $$x=0$$.

Hence take limit $$n \to \infty$$ on $$[*]$$, and we get $$f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15$$.

## Food for Thought

• Find the general function from the given condition.
• $$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that $$g(0) =0$$.
• What if $$g(x)$$ is not continuous?

# Rolle's Theorem | IIT JAM 2017 | Problem 10

Try this problem from IIT JAM 2017 exam (Problem 10).This problem needs the concept of Rolle's Theorem.

## Rolle's Theorem | IIT JAM 2017 | Problem 10

$$f(x)=\left\{\begin{array}{ll}1+x & \text { if } x<0 \\ (1-x)(p x+q) & \text { if } x \geq 0\end{array}\right.$$

satisfies the assumptions of Rolle's theorem in the interval $[-1,1],$ then the ordered pair $(p, q)$ is

• $(2,-1)$
• $(-2,-1)$
• $(-2,1)$
• $(2,1)$

### Key Concepts

Real Analysis

Continuity / Differentiability

Mean-value theorem of differential calculus

Answer: $(2,1)$

IIT JAM 2017 , Problem 10

Real Analysis : Robert G. Bartle

## Try with Hints

Rolle's Theorem :

Let a function $f:[a, b] \rightarrow R$ be such that

1. $f$ is continuous on $[a, b]$
2. $f$ is differentiable at every point of $(a, b)$
3. $f(a)=f(b)$

Then there exists at least one point $c \in(a, b)$ such that $f^{\prime}(c)=0$

We can easily see that $3^{rd}$ assumption of Rolle's theorem is satisfied for $f(x)$ irrespective of the values of $p,q$.

Since $f(-1)=0=f(1)\quad \forall p,q$

Since $f(x)$ satisfies $1^{st}$ assumption, then

\begin{aligned}& \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)\\&\text { ie, } \lim _{x \rightarrow 0^{-}}(1+x)=\lim _{x \rightarrow 0^{+}}(1-x)(px+q)=q\\&\Rightarrow 1=q\end{aligned}

$L f^{\prime}(0)=R f^{\prime}(0) \cdots \cdots(*)$

\begin{aligned} \text{Now, } L f^{\prime}(0) &=\lim _{h \rightarrow 0^{-}} \frac{f(0+h)-f(0)}{h} \\&=\lim _{h \rightarrow 0^{-}} \frac{(1+h)-q}{h} \\ &=\lim _{h \rightarrow 0^{-}} \frac{1+h-1}{h}[\text{because } q=1] \\&=\lim_{h \to 0^{-}} \frac hh\\&=1\end{aligned}

\begin{aligned}\text{and, } R f^{\prime}(0)&=\displaystyle\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)\left(ph+q\right)-q}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{(1-h)(ph +1)-1}{h}\quad[\text{because } q=1]\\&=\lim _{h \rightarrow 0^{+}}\frac{ph+1- ph^{2}-h-1}{h}\\&=\lim _{h \rightarrow 0^{+}} \frac{h(p-ph-1)}{h}\\&=\lim_{h \ to 0^{+}} (p-ph-1)\\&=p-1\end{aligned}

Then by $(*) \text{we have}, \quad P-1=1 \Rightarrow P=2$

Then order pair $(p,q)\equiv (2,1)$ [ANS]

# Limit to Function - ISI UG 2019 Subj Problem 2

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $$f$$ : $$(0,1) \rightarrow \mathbb{R}$$ be defined by  $$f(x) = \lim_{n\to\infty} cos^n(\frac{1}{n^x})$$. (a) Show that $$f$$ has exactly one point of discontinuity. (b) Evaluate $$f$$ at its point of discontinuity.

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2019
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Calculus

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]6 out of 10

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Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Try to determine the Form of the Limit. Show that the limit is of the form $$1^\infty$$. Hence, try to find the Functional Form.

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

$$f(x) = \lim_{n\to\infty} (1 + (cos(\frac{1}{n^x}) - 1))^n = e^{(\lim_{n\to\infty}(cos(\frac{1}{n^x}) - 1).n}$$.

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

$${\lim_{n\to\infty}(cos(\frac{1}{n^x}) - 1).n = -\frac{1}{2}\lim_{n\to\infty}\frac{(sin^2(\frac{1}{2n^x}))}{(\frac{1}{2n^x})^2}.n^{(1-2x)} }$$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Prove that  $$f(x)$$ = $\left\{ \begin{array}{ll} 0 & 0 < x < \frac{1}{2} \\ \frac{1}{\sqrt{e}} & x = \frac{1}{2} \\ 1 & x > \frac{1}{2} \\ \end{array} \right.$

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problems

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# TIFR 2015 Problem 7 Solution -Increasing Function and Continuity

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TIFR 2015 Problem 7 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.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let $f$ and (g) be two functions from ([0,1]) to ([0,1]) with (f) strictly increasing. Which of the following statements is always correct?

A. If (g) is continuous, then (fog) is continuous

B. If  (f) is continuous, then (fog) is continuous

C. If (f) and (fog) is continuous, then (g) is continuous

D. If (g) and (fog) are continuous, then (f) is continuous

## Discussion:

A: Let (g(x)=x) for all (xin [0,1]).

(f(x)=x) for (xin [0,frac{1}{2}]) and (f(x)=5+x) for (xin (frac{1}{2},1]).

Then (fog=f) and (f) is not continuous.

So A is False.

B: Reverse (f) and (g) in A to show that B is False.

C: If (f) and (fog) are continuous then (f) is 1-1 (increasing), continuous map ([0,1]to [0,1]).

(A subset [0,1] ) be closed. Then (A) is compact. (Closed subsets of compact spaces are compact).

Therefore (f(A)) is compact. (continuous image of compact set is compact).

We have that (f(A)) is a compact subset of ([0,1]). Therefore (f(A)) is closed in ([0,1]). (compact subspace of Hausdorff space is closed).

Therefore, (f) is a closed map. So (f^{-1}) is continuous.

Hence (f^{-1}ofog=g) is continuous.

So, C is True.

D: Let (g(x)=frac{x}{4}) for all (xin [0,1]).

(f(x)=x) for (xin [0,frac{1}{2}]) and (f(x)=5+x) for (xin (frac{1}{2},1]).

Then (fog(x)=f(frac{x}{4})=frac{x}{4}) for all (xin [0,1]).

So (fog) is continuous but (f) is not continuous.

So, D is False.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity,Closed Set, Compact Set
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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# TIFR 2014 Problem 22 Solution -An application of Intermediate Value Theorem

TIFR 2015 Problem 7 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.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (f:\mathbb{R}^2 \to \mathbb{R} ) be a continuous map such that (f(x)=0 ) for only finitely many values of (x). Which of the following is true?

A. either (f(x) \le 0 )for all (x) or (f(x) \ge 0 ) for all (x).

B. the map (f) is onto

C. the map (f) is one-to-one

D. none of the above

## Discussion:

Let (f) be the map (f(x_1,x_2)=x_1^2 +x_2^2 ). Then (f) is zero at only (0,0). (f) is continuous because (x=(x_1,x_2) \to x_1 \to x_1^2 ) is continuous from (\mathbb{R}^2 \to \mathbb{R} \to \mathbb{R} ). i.e, each arrow is continuous. The first arrow is the projection map, and such maps are always continuous, and the second arrow is just squaring, which is continuous. And composition of continuous functions are continuous, so (x \to x_1^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}). Where here and henceforth (x=(x_1,x_2)\in \mathbb{R}^2 ).

Similar reasoning will show that (x \to x_2^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}).

Sum of continuous functions is continuous, so the map (x \to x_1^2+ x_2^2 ) is continuous function from (\mathbb{R}^2 \to \mathbb{R}).

This function (f) is not one-one since (f(1,0)=f(0,1)=1) and it is not onto since it only takes values in ([0,\infty ) ).

So we now are sure that B,C are false options.

We will prove A now.

Let (x=(x_1,x_2)) and (y=(y_1,y_2)) be two points in (\mathbb{R}^2) such that (f(x)>0) and (f(y)<0).

We will prove that this will imply infinitely many zeros in between (x) and (y). But wait a second... what does between mean in this context? For that we consider the paths between (x) and (y). Note that there are infinitely many paths between any two points in (\mathbb{R}^2). Further, we can in fact have infinitely many paths completely disjoint except for the initial and final points. We show that corresponding to each path (\alpha: [0,1] \to \mathbb{R}^2) which connects (x) and (y) we have a zero in the path. Since there are infinitely many disjoint paths, we get infinitely many distinct zeros for (f).

Now, (\alpha: [0,1] \to \mathbb{R}^2) is a continuous function , ( \alpha(0)=x ), (\alpha(1)=y).

Consider the composition (g=f o \alpha : [0,1] \to \mathbb{R}). (g) is continuous. g(0)=f(x)>0 and g(1)=f(y)<0.

Therefore by the intermediate value theorem, (g(c)=0) for some (c\in [0,1]).

That means, (f(\alpha(c))=0). And using the discussion above we get a contradiction.

This proved the option A.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity,Intermediate Value Theorem
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

# TIFR 2013 Problem 29 Solution -Continuity of function defined on rationals and irrationals

TIFR 2013 Problem 29 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 programe 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.

Also Visit: College Mathematics Program of Cheenta

## Problem:True/False?

Let $$f$$ be a function on the closed interval $$[0,1]$$ defined by

$$f(x)=x$$ if $$x$$ is rational and $$f(x)=x^2$$ if $$x$$ is irrational.

Then $$f$$ is continuous at 0 and 1.

## Hint:

Use the sequence criterion for continuity.

## Discussion:

Let $$x_n$$ be a sequence converging to $$0$$. Then $$x_n^2$$ also converges to $$0$$. Since the value of $$f$$ at $$x_n$$ is either of the two, $$f(x_n)\to 0$$ as $$n\to \infty$$. If this seem confusing, think in terms of $$\epsilon$$. For $$\epsilon >0$$, there exists $$n_1$$ and $$n_2$$ such that $$|x_n|< \epsilon$$ and $$|x_n^2|< \epsilon$$ for $$n>n_1$$ and $$n>n_2$$ respectively. Taking the maximum of $$n_1$$ and $$n_2$$ we get the N for which the condition in "epsilon definition" is satisfied.

The same argument applies for the continuity at 1.

The function is not continuous at any other point. Because if a rational sequence and an irrational sequence converge to $$x_0$$ and the function is continuous at that point then by sequential criterion, $$f(x_0)=x_0$$ due to the rational sequence and also $$f(x_0)=x_0^2$$ due to the irrational sequence. Therefore, $$x_0^2=x_0$$ and hence the only possible points of continuity is $$0$$ or $$1$$.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous function, Closed Interval, Sequential criterion
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

# Understanding the Infinitesimal

Understanding the Infinitesimal
Cheenta Notes in Mathematics

Let's discuss a beautiful idea related to progress in mathematics and understanding the infinitesimal.
Adding infinitely many positive quantities, you may end up having something finite. Greeks did not understand this very well. Archimedes had some ideas. Kerala school of mathematics under the leadership of Madhavacharya made real progress in refining this notion. They created the necessary groundwork for the advent of calculus later in Germany and Britain.

Indian Statistical Institute's 2017 entrance for B.Stat and B.Math had a very simple problem based on infinitesimals. Adding little things, can you end up having more than 'little'?

Here is the problem:

Given f : ℝ → ℝ be a continuous function such that for any two real numbers x and y,
|f(x) - f(y)| ≤ 7|x-y|201

Then:

(A) f(101) = f(202) + 8;
(B) f(101) = f(201) +1;
(C) f(101) = f(200) + 2;
(D) None of the above;

Before we solve the actual problem, lets have a fun detour. What if f is differentiable (the problem does not say that)? Then take y = x + δ where δ > 0 Clearly, by the given condition:

| f(x) - f(x + δ) | ≤ 7 |x - (x + δ) |201 = 7δ201
⇒ |f(x) - f(x+δ) |/δ ≤ 7 δ200
⇒ limδ → 0 |f(x) - f(x+δ) |/δ ≤ limδ → 0  7δ200 =  0
⇒ |f'(x)| = 0

That means, if f is differentiable, then it's derivative is 0, or in other words, it is a constant function. In that case f(x) = f(y) = c (a constant for all x and y). Hence none of the first three options would hold. (Interestingly enough this is sufficient to choose (D) as the correct option as differentiable functions are an important subclass of continuous functions).

However, we cannot assume differentiability as it is not mentioned in the problem. But now we have a hunch! We are already guessing that maybe f is not changing much.

Suppose we want to check:
| f(101) - f(202) |

Chop off the distance between 101 to 202 into intervals of 0.1 unit long. There are 1010 such intervals (this is the 'adding the little' part). Why did I choose 0.1 length? Well, that is because it is a fractional length and raising a fraction to large powers will make it even smaller.

Now note that:

|f(101) - f(202)|
= |f(101) - f(101.1) + f(101.1) - f(101.2) + ... + f(201.9) - f(202)|
≤  |f(101) - f(101.1)| + |f(101.1) - f(101.2)| + ... + |f(201.9) - f(202)|
≤ 7|101-101.1|201 + ... + 7|201.9-202|201
= 7 ( 0.1201 + ... + 0.1201
= 7 × 1010 × 0.1201
= 7070/10201

But that implies |f(101) - f(202)| is much smaller 1 let alone 8. Using this same technique you can make |f(101)  -  f(202| smaller than 1 and f(101) - f(200)| smaller than 2. Hence the answer is option (D).

More interestingly, can you make the difference between f(x) and f(y) arbitrarily small? If you can do that then f(x) will be a constant function! This is something for you to think about this week.

Here is a possible way to think about it:

Taking smaller intervals. For example, for integers x and y chop off |x-y| into intervals of length 1/n. There will be |x-y|/(1/n) = n|x-y| intervals. Then you can use the above algorithm to compute:

|f(x) - f(y)|
≤ 7 × n|x-y|/n201
= 7 × |x-y|/n200

No matter how large |x-y| is, (by the archimedean property) we can find a positive integer k such that

|x-y| < k × n

Try to finish it off from here. Let me know if you get anything.

# One-One function and differentiability

Let's understand one-one function and differentiability with the help of a problem. Try it yourself before reading the solution.

Let f be real valued, differentiable on (a, b) and $f'(x) \ne 0$ for all $x \in (a, b)$. Then f is 1-1.

True

Discussion:

Suppose f is not 1-1. Then there exists $x_1 , x_2 \in (a, b)$ such that $f(x_1 ) = f(x_2)$. Since f(x) is differentiable it must be continuous as well. Applying Rolles Theorem in the interval $(x_1 , x_2 )$ we conclude that there exists a number c in this interval such that f'(c) = 0. But this contradicts the given conditions. Hence f must be 1-1

Chinese Remainder Theorem

A Math Game in Symmetry - Video

# Fixed Point of continuous bounded function

Let's understand Fixed Point of continuous bounded function with the help of a problem. This problem is useful for College Mathematics.

f: $[0 , \infty ) to [0. \infty )$ is continuous and bounded then f has a fixed point.

True

Discussion: Consider the function g(x) = f(x) - x. Since f(x) and x are continuous then g(x) must be continuous. Since f(x) is bounded then there exists a M such that f(x) < M.

Now $f(0) \ge 0$ as the codomain is $[0, \infty )$ . Thus $g(0) = f(0) - 0 \ge 0$ . Also g(M) must be negative as f(M) < M. Since g(x) is continuous, by Intermediate Value Property of Continuous Functions g(x) must attain the value of 0 somewhere between x = 0 to x = M. Suppose that value is c.

Hence g(c) = f(c) - c = 0 or f(c) = c.