# ISI MMath Solutions and Discussion 2018 : PMB

## ISI MMath PMB 2018 Subjective Questions along with hints and solutions and discussions

**Note: You must try the problems on your own before looking at the solutions. I have made the solutions so that you don’t have to join any paid program to get the solutions. After you have failed several times in your attempt to solve a question even with the hints provided, then only look at the solution to that problem. If you cannot solve a single question on your own then this exam is not for you. And if you like my work then please do share this among your friends and do comment below.Â Â **

### To view ISI MMATH 2020 solutions, hints, and discussions: Click Here

### To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

### To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

### To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

### To view previous year’s question papers: Click Here

**Problem 1.**

Find the values of for which the improper integral

converges.

### Topic: Real Analysis

Difficulty level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 1 Hints along with Solution & Discussion**

#### Full Solution

The integral is defined as the limit of proper intergals

Using the relation we find such that for all we have Then for all we have a two-sided estimate

and

So, existence of the limit is equivalent to finiteness of the integral So,

Let We have an estimate

integrate by parts with so that

So, the limit exists for any

The improper integral converges if and only if

**Problem 2.**

Let be a real-valued continuous function which is differentiable on and satisfies Suppose there exists a

constant such that

Show that for all .

### Topic: Real Analysis

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 2 Hints along with Solution & Discussion**

**Problem 3.**

Let be an abelian group of order

(a) If is a function, then prove that for all ,

(b) Let be the multiplicative group of non-zero complex numbers and suppose is a homomorphism. Prove that

(c) If is any homomorphism, then prove that

### Topic: Group Theory

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 3 Hints along with Solution & Discussion**

#### Full Solution

## Problem 4.

(a) Is the ideal in the polynomial ring a prime ideal? Justify your answer.

(b) Is the ideal in the polynomial ring a prime ideal? Justify your answer.

### Topic: Ring Theory

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 4 Hints along with Solution & Discussion**

#### Full Solution

(a) Relations

imply that General element of is of the form i.e.

If then which is possible only when or is a prime ideal.

(b) Assume that is a prime ideal in Consider polynomial Then either or General element of is of the form

In particular, an element of vanished when So, It follows that and

Then, and

which is impossible in is not a prime ideal.

**Problem 5.**

Let and be an matrix with real entries. Let denote the adjoint of that is, the -th entry of is the

-th cofactor of Show that the rank of is , or

### Topic: Linear Algebra

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 5 Hints along with Solution & Discussion**

#### Hint 2

#### Hint 3

#### Full Solution

If then the inverse of is given by In this case the rank of is

Assume that the rank of is Then each minor of is zero, and all elements of are equal to zero. In this case the rank of is equal to zero.

Assume that the rank of is equal to Then there exists a nonzero minor of so the rank of is Denote by the th cofactor of . Then Consider the product

Its th entry is

Indeed, if then the sum in (1) is the determinant of If then the sum in (1) is the determinant of the matrix after the th row is replaced with the th row. Such determinant is zero as the matrix has two identical rows. We see that That is, each column of is in the kernel of But the rank of is , hence the dimension of the kernel of is and the dimension of the column range of is This proves that the rank of is equal to 1.

**Problem 6.**

Suppose an urn contains a red ball and a blue ball. A ball is drawn at random and a ball of the same colour is added to the urn along with

the one that was drawn. This process is repeated indefinitely.

Let denote the random variable that takes the value if the first draws yield red balls and the -th draw yields a blue ball.

(a) If , find .

(b) Show that the probability of a blue ball being chosen eventually is

(c) Find

### Topic: Probability

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 6 Hints along with Solution & Discussion**

#### Hint 2

#### Full Solution

(a) We use induction on Let The event means that the first draw yielded a red ball,

Assume that event happens. Then first draws yielded red balls, and before the st draw there are red balls and blue ball in the urn. The probability to draw a red ball is So,

If then For the equality is true. So, for all

(b) If the blue ball is never chosen, then for all Denote by the event when the blue ball is never chosen. Then

So,

(c) For each we have

Then

## Problem 7.

A real number is said to be a limit point of a set if every

neighbourhood of contains a point of other than Consider the

set

(a) Show that contains infinitely many limit points of

(b) Show that is a compact subset of

(c) Find all limit points of

### Topic: Real Analysis

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 7 Hints along with Solution & Discussion**

#### Full Solution

(a) Observe that Let and Then

Every neighborhood of contains a point from distinct from Each point is a limit point of the set that belongs to the set

If then is a point of in the neighborhood of . is a limit point of We’ve proved that points

are limit points of .

(b)

The set is bounded: It remains to prove that is closed. Suppose that Every neighborhood of contains a point . But and Then is a limit point of Since we assume it follows that For every find a point such that

If both sequences and are bounded, then there exist values such that and for infinitely many values of But then which is not the case. Hence we can assume that is unbounded. Let be a subsequence such that If the sequence is bounded, there exists value such that for infinitely many values of But then

If the sequence is unbounded, we can pass to another subsequence that converges to Then we have In any case and is compact.

(c) Considerations of previous two parts shows that the limit points of are

## Problem 8.

(a) Let be a sequence of continuous functions on such

that converges uniformly on Show that

converges.

(b) Find the set of all points such that the series

converges. Does this series converge uniformly on ? Justify your answer.

### Topic: Real Analysis

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2018 Problem 8 Hints along with Solution & Discussion**

#### Full Solution

(a) Denote The sequence converges uniformly on there exists function on such that

We verify that exists. Let Find such that

Using continuity of find such that for all

Then for all we have

Since is arbitrary, this shows that exists. We denote the limit by

Now, let Find such that for all and all we have

Passing to the limit as we get

Since is arbitrary, this shows that

i.e.

converges and has the sum .

(b) The series converges for all Indeed,

For the series diverges, as it is the series So, If the series converges uniformly, then it should also converge at So, the series does not converge uniformly on

Sir,my daughter wants to prepare for m.math entrance examination..can you please help her with your classes and practice sets??

Hello, Yes I can help her out with my tutoring. There is a batch for M.Math where I teach every Sunday starting from 5 pm onwards. For further detail, you can contact me at +91-8247590347 or email me at admin@fractionshub.com.

For Problem 2, Can I solve this like this?

Let then by using MVT there exists such that i.e . Similary there exists such that and applying modulus we get So with the above inequalities we have . Proceeding this way we have but since is continuous on so is so it attains the maximum say then the inequality becomes and we know that . so for any there exists such that for all we have In particular hence this implies and since was arbitrary we have

For Problem 2, Can I solve this like this?

Let then by using MVT there exists such that i.e . Similary there exists such that and applying modulus we get So with the above inequalities we have . Proceeding this way we have but since is continuous on so is so it attains the maximum say then the inequality becomes and we know that . so for any there exists such that for all we have In particular hence this implies and since was arbitrary we have