# ISI MMath Solutions and Discussion 2019 : PMB

## ISI MMath PMB 2019 Subjective Questions along with hints and solutions and discussions

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**Problem 1.**

Let be a continuous function satisfying

Show that is a bounded function on and attains a maximum or a minimum. Give an example to show that it attains a maximum but not a minimum.

### Topic: Real Analysis

Difficulty level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 1 Hints along with Solution & Discussion**

#### Hint 1

Assume that is non-negative, i.e. for all If for all then obviously is bounded and attains its maximal (and minimal) value at each point. Let for some By assumption, there exists such that for all and all Let be the restriction of to is a continuous function on a compact segment.

#### Hint 2

Let be a continuous function satisfying

The function is a non-negative continuous function satisfying

#### Hint 3

For the Example part, consider the function

#### Full Solution

### Step 1:

Assume that is non-negative, i.e. for all If for all then obviously is bounded and attains its maximal (and minimal) value at each point. Let for some By assumption, there exists such that for all and all Let be the restriction of to is a continuous function on a compact segment. Hence, is bounded and attains its maximal value on i.e. there exists such that We check that the function attains its maximal value at Let Then

Let or Then and

So, for each

is bounded and attains its maximal value.

### Step 2:

Let be a continuous function satisfying

The function is a non-negative continuous function satisfying

By step 1, is bounded and attains its maximal value. It means that there exists such that for all

So, for all

The function is bounded. If then and attains its maximal value. If then and attains its minimal value.

### Step 3:

Consider the function Then is a continuous non-negative function,

In particular, Further,

Function attains its maximal value at However, for all and the function does not attain its minimal value.

**Problem 2.**

Let be a continuous function such that Show

that

### Topic: Real Analysis

Difficulty Level: Easy

**Indian Statistical Institute, ISI MMath 2019 Problem 2 Hints along with Solution & Discussion**

#### Hint 1

Function is bounded, as it is a continuous function on a compact Let Fix By assumption, there exists such that for all Let be such that

#### Hint 2

Let and If then

#### Hint 3

If then

#### Full Solution

Function is bounded, as it is a continuous function on a compact Let Fix By assumption, there exists such that for all Let be such that

Let and If then

If then

In any case,

and

Since is arbitrary, this proves the needed convergence.

**Problem 3.**

Let be a twice continuously differentiable function and

suppose If for all then prove that

for all

### Topic: Real Analysis

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 3 Hints along with Solution & Discussion**

#### Hint 1

For every we have by the Newton-Leibniz formula

#### Hint 2

Applying the Newton-Leibniz rule again we get

#### Hint 3

In particular, for each we have

#### Full Solution

For every we have by the Newton-Leibniz formula

Applying the Newton-Leibniz rule again we get

In particular, for each we have

## Problem 4.

Suppose is a function defined by

(a). Find all such that has a nonzero directional derivative at with respect to the direction

(b). Is continuous at Justify your answer.

### Topic: Multivariable Calculus

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 4 Hints along with Solution & Discussion**

#### Hint 1

Part a.

Let Then the directional derivative at with respect to the direction is

#### Hint 2

Let Then the directional derivative at with respect to the direction is

#### Hint 3

Part b.

If is continuous at , then for any sequence we have Let

#### Full Solution

Part a.

Let Then the directional derivative at with respect to the direction is

Let Then the directional derivative at with respect to the direction is

has a nonzero directional derivative at with respect to the direction if and only if and

Part b.

If is continuous at , then for any sequence we have Let Then

So, is not continuous at

**Problem 5.**

Let be a subset of a compact metric space Assume that for

every continuous function , the restriction of to attains

a maximum on Prove that is compact.

### Topic: Metric Space

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 5 Hints along with Solution & Discussion**

#### Hint 1

A subset of a compact metric space is compact if and only if it is a closed subset. So, it is enough to check that is a closed subset of

#### Hint 2

Let Consider function is a continuous function on

#### Hint 3

Indeed, by the triangle inequality,

So, attains a maximum on . Since it follows that there exists a point such that

#### Full Solution

A subset of a compact metric space is compact if and only if it is a closed subset. So, it is enough to check that is a closed subset of Let Consider function is a continuous function on Indeed, by the triangle inequality,

So, attains a maximum on . Since it follows that there exists a point such that Then and So, and is closed.

**Problem 6.**

Let be a non-abelian group of order , where are primes.

(a). How many elements of have order

(b). How many elements of have order

### Topic: Group Theory

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 6 Hints along with Solution & Discussion**

#### Hint 1

The order of an element divides the order of a group. So each element of is of order or Since is non-abelian, it has no elements of order By Cauchy’s theorem, there exists an element of order Then each of the elements

is of order .

#### Hint 2

Assume that is of order and Consider subgroups

and

#### Hint 3

Then is a subgroup of strictly smaller than (as ). The order of is prime, hence Consider cosets

The size of each coset is and there are cosets. hence two of different cosets intersect:

#### Full Solution

The order of an element divides the order of a group. So each element of is of order or Since is non-abelian, it has no elements of order By Cauchy’s theorem, there exists an element of order Then each of the elements

is of order .

Assume that is of order and Consider subgroups

and

Then is a subgroup of strictly smaller than (as ). The order of is prime, hence Consider cosets

The size of each coset is and there are cosets. hence two of different cosets intersect:

We deduce that there exists such that But then which contradicts the fact that is of order

Conclusion: all elements of order are among

(a). There are elements of order

(b). There are elements of order

## Problem 7.

Prove or disprove the following statement: The ring is isomorphic to a product of fields.

### Topic: Field Theory

Difficulty Level: Hard

**Indian Statistical Institute, ISI MMath 2019 Problem 7 Hints along with Solution & Discussion**

#### Hint 1

We prove that is isomorphic to a product of three fields. Consider the factorization

#### Hint 2

Consider ideals

#### Hint 3

Relations

imply that ideals are pairwise coprime.

Now try to apply the Chinese remainder theorem.

#### Full Solution

We prove that is isomorphic to a product of three fields. Consider the factorization

Consider ideals Relations

imply that ideals are pairwise coprime. By the Chinese remainder theorem,

Each of the multiples in the right-hand side is a field, as it is a quotient of by an irreducible polynomial. In fact,

The ring is isomorphic to a product of three fields

## Problem 8.

Let be a symmetric matrix with real entries such that for some Show that

### Topic: Linear Algebra

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 8 Hints along with Solution & Discussion**

#### Hint 1

A symmetric real matrix is diagonalizable

#### Hint 2

there exists an invertible matrix and a diagonal matrix such that

#### Hint 3

Then

#### Full Solution

A symmetric real matrix is diagonalizable: there exists an invertible matrix and a diagonal matrix such that

Then

So,

and and

## Problem 9.

Suppose and are two matrices with real entries such that

the sum of their ranks is strictly less than Show that there exists a

nonzero column vector such that

### Topic: Linear Algebra

Difficulty Level: Easy

**Indian Statistical Institute, ISI MMath 2019 Problem 9 Hints along with Solution & Discussion**

#### Hint 1

Consider the matrix

The rank of is the maximal number of linearly independent rows of

#### Hint 2

Then

#### Hint 3

Consequently, columns of are linearly dependent, and there exists a nonzero vector such that

It means that

#### Full Solution

Consider the matrix

The rank of is the maximal number of linearly independent rows of Then

Consequently, columns of are linearly dependent, and there exists a nonzero vector such that

It means that

## Problem 10.

Suppose there are persons in a party. Every pair of persons meet each other with probability independently of the other pairs.

Let be the number of people the person meets in the party. For all with and for all

show that

### Topic: Probability

Difficulty Level: Medium

**Indian Statistical Institute, ISI MMath 2019 Problem 10 Hints along with Solution & Discussion**

#### Hint 1

Introduce two events:

Assume the event happens. Then meets and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

#### Hint 2

Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

#### Hint 3

Finally,

#### Full Solution

Introduce two events:

Assume the event happens. Then meets and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

Finally,

Assume the event happens. Then does not meet and meets other persons (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

Further, meets persons except (from possible persons). There are choices of persons except that meets. The probability that meets persons except is

Finally,

We deduce that

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Sir,

I am from howrah, West Bengal. There is a kind request to you from my side that if u please teach me as a private tutor for preparation for ISI and CMI, I would be highly obliged. I also like your problems and try to figure it out. But I think it would be more clear if you kind see to my above request.

Regards,

Soumyajit Ganguly

Please check your email. I have sent you all the details regarding our classes.

Sir

I am from howrah, West Bengal. There is a kind request to you from my side that if u please teach me as a private tutor for preparation for ISI and CMI, I would be highly obliged. I also like your problems and try to figure it out. But I think it would be more clear if you kind see to my above request.

Please check your email. I have sent you all the details regarding our classes.

Hi!

Can you please conduct few mock tests for ISI M.MATH entrance?

We will try our best to make the mock test course. But currently, we are busy with making some other courses. We will keep you updated.

Sir, it would be really very helpful if you can create a mock test course for this year’s entrance.

In 5th question I got it h(x)<h(x○)=1 but I didn't got how u proved then that C is closed

Showing that C(Bar)=C, that is C contains all it’s limit points. The last line there would be C(bar) but the latex is not taking the symbol bar.

For problem 3, can we use the Taylor’s theorem instead? like for f(x)= f(0) + xf'(0) + (x^2/2)f”(c), for c belongs to (0,x) where for x<0 the case is same because in either way f'(0) = 0 = f(0)

so when we apply |f(x)| =|x^2f"(c)|/2 < 1/2

For problem 3, can we use the Taylor’s theorem instead? like for , for and for , the case is same because in either way

so when we apply

For problem 3, can we use the Taylor’s theorem instead? like for , for and for , the case is same because in either way

so when we apply

#1) similar sol. for question 3rd:

i.e. Maclaurin’s theorem can we used.

#2) and i think for question no. 10th: probability section i.e. in case of P[B] , there is a calculation mistake …

if not then i have doubt how { (1-p)^ 2n – k – l – 3 } comes?

because i concluded it is not -3 , it would be -4.

please correct me.

#3) for question no. 4th:

can we check limit along the curve y = x^2/3 which can also shows that f is not continuous?

#4) for |G| = pq

if we take a example of o(G) = 10 = 1*2*5

let p=2 < q=5 and both are prime.

and there is an element of order 1 (identity element) and it is non abelian group i.e. no element exist of order 10.

now phi(q) == phi(5) == (5^1-5^0) == 4

that is we can say only four elements are there of order q. And remaining elements could be of order p i.e. there are 4 elements of

order q.

Also q=5 implies that no. of elements of order q = q -1

And no. of elements of order p == o(G) – identity element – (q-1) == (p-1)q

can it be possible ?

if not give the query please.

sorry for #2)

it was my mistake.

For the 6th problem, It easily follows from the Sylow’s theorem.

No. of q-sylow subgroup is, for some and i.e there exist a unique subgroup of order . so there are elements of order

and we know that every non-identity element in has order either or . So the remaining elements must have order

Therefore the no. of elements having order q is

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You will have to join the MMath group here. I teach for MMath entrance exam. No there is no mock tests for ISI MMath.

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You can text me on WhatsApp @ 9700803692.

For problem 9, Here’s an alternative solution:

if then we are done. Otherwise assume that , also note that then . So we have as dim(Ker (A) \cap ker(B)) =0 but from the given condition we have we have i.e but from the above inequality we get a contradiction. Hence . Therefore the result follows

For problem 9, Here’s an alternative solution:

if then we are done. Otherwise assume that , also note that then . So we have as but from the given condition we have we have i.e but from the above inequality we get a contradiction. Hence . Therefore the result follows