# ISI MMath Solutions and Discussion 2020 : PMB

## ISI MMath PMB 2020 Subjective Questions, solutions and discussions

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### Question Paper ISI M.Math 2020 PMA Objective Questions Solutions and Discussions: Click Here

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**Problem 1.**

(a) Let be a sequence of continuous real-valued functions on converging uniformly on to a function . Suppose for all there exists such that . Show that there exists such that .

(b) Give an example of a sequence of continuous real-valued functions on converging uniformly on to a function , such that for each there exists satisfying , but satisfies for all .

### Topic: Real Analysis, Difficulty level: Medium

#### Full Solution

(a)

The segment is compact. Hence, the sequence has a convergent subsequence:

The function is continuous, since it is a uniform limit of continuous functions.

Let There exists such that for all

There exists such that for all Then we estimate

(b)

Let We have

Also, However, for all

It follows that Since is arbitrary, we deduce that

**Problem 2.**

Let be continuous function. Show that

### Topic: Real Analysis, Difficulty Level: Hard

**Problem 3.**

(a) Let be a twice continuously differentiable function. Show that

for all .

(b) Show that if further satisfies

for all , then there exist such that for all .

### Topic: Real Analysis, Difficulty Level: Medium

## Problem 4.

Let be twice continuously differentiable function. Show that if is bounded and for all then must be constant.

### Topic: Real Analysis, Difficulty Level: Medium

**Problem 5.**

Let be a real matrix such that , where is the identity matrix

(a) Show that if and , then the vectors are linearly independent

(b) Show that there exists an invertible real matrix such that

### Topic: Multivariable Calculus, Difficulty Level: Medium

#### Full Solution

(a)

Assume there are scalars such that

Apply to this equality:

From these two identities we get

(b)

Let be the matrix with the first column given by the vector and the second column given by the vector

is invertible, since is a basis of Let Then

So,

Since we get and Linear independence of and are proved.

**Problem 6.**

Suppose is a -dimensional real vector space and is a linear map such that and .

(a) Show that there exists a vector such that the set is a basis of .

(b) Suppose is another linear map such that and . Show that there exists an invertible linear map such that .

### Topic: Linear Algebra, Difficulty Level: Medium

#### Full Solution

(a)

Let be such that Assume the system is linearly dependent. Then there are scalars such that

Apply to this identity and use the fact

Apply to this identity and use the fact

(b)

Let be such that is a basis of Let

be the matrix whose columns are given by vectors The matrix is invertible. We have

It follows that

for some invertible matrix .

If is another linear map of such that and then for some invertible matrix we have

So,

and

for

But So, and consequently Linear independence of is proved.

## Problem 7.

Let be a field, and let be the ring . Let be the ideal generated by . Find all maximal ideals of the ring .

### Topic: Ring Theory, Difficulty Level: Hard

#### Full Solution

Consider the mapping given by

is a ring homomorphism. The kernel of coincides with Indeed,

Given consider Then

So, maps onto By the isomorphism theorem, is isomorphic to . Consider a maximal ideal of Consider arbitrary nonzero Then

If and then and In this case is not a maximal ideal. So, either

or

Conclusion. There are two maximal ideals in One is generated by another is generated by

## Problem 8.

Let be a finite group, and let be a normal subgroup of , Let be a Sylow -subgroup of

(a) Show that for all , there exists such that .

(b) Let . Let be the set . Show that .

Here is a less elegant solution to Problem 4 which does not require integration:

We first show that is identically zero. Suppose for some real number . Since is continuous, there is an open interval containing such that for all . Thus, is strictly increasing on . Let with . Then .

By Taylor's Theorem, there is such that

Noting that , \frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(x_1)+\frac{(x_2-x_1)}{2}f''(d)f'(d).f”(x)=0xf(x)=Ax+BA,BfA=0f$ is a constant function.

I am a little confused about Problem 3 part (a). It can be solved using only L-Hopital’s Rule without using continuity of second derivative. Is there any reason that hypothesis was included?

You are using which can be guranteed by continuity of