ISI MMath Solutions and Discussion 2018 : PMB

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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.

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Problem 1.

Find the values of $a > 0$ for which the improper integral

$\int^\infty_0 \frac{\sin x}{x^a}dx$

converges.

Topic: Real Analysis
Difficulty level: Medium

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

Hint 1

The integral is defined as the limit of proper intergals

$\int^\infty_0 \frac{\sin x}{x^a}dx=\lim_{\delta\to 0, C\to\infty}\int^C_{\delta}\frac{\sin x}{x^a}dx=$

$=\lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx+\lim_{C\to \infty}\int^C_1\frac{\sin x}{x^a}dx$

Hint 2

Using the relation $\lim_{x\to 0}\frac{\sin x}{x}=1$ we find $\epsilon>0$ such that for all $x\in (0,\epsilon)$ we have $\frac{1}{2}<\frac{\sin x}{x}<2.$

Hint 3

Then for all $\delta<\delta'<\epsilon$ we have a two-sided estimate

$\left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\leq 2 \int^{\delta'}_\delta x^{1-a}dx$

and

$\left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\geq \frac{1}{2} \int^{\delta'}_\delta x^{1-a}dx$

Full Solution

The integral is defined as the limit of proper intergals

$\int^\infty_0 \frac{\sin x}{x^a}dx=\lim_{\delta\to 0, C\to\infty}\int^C_{\delta}\frac{\sin x}{x^a}dx=$

$=\lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx+\lim_{C\to \infty}\int^C_1\frac{\sin x}{x^a}dx$

Using the relation $\lim_{x\to 0}\frac{\sin x}{x}=1$ we find $\epsilon>0$ such that for all $x\in (0,\epsilon)$ we have $\frac{1}{2}<\frac{\sin x}{x}<2.$ Then for all $\delta<\delta'<\epsilon$ we have a two-sided estimate

$\left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\leq 2 \int^{\delta'}_\delta x^{1-a}dx$

and

$\left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\geq \frac{1}{2} \int^{\delta'}_\delta x^{1-a}dx$

So, existence of the limit $\lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx$ is equivalent to finiteness of the integral $\int^1_0 x^{1-a}dx.$ So,

$\exists \lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx\Leftrightarrow 1-a>-1\Leftrightarrow a<2.$

Let $1<C<C'.$ We have an estimate

$\left|\int^C_1 \frac{\sin x}{x^a}dx-\int^{C'}_1 \frac{\sin x}{x^a}dx\right|=\left| \int^{C'}_C\frac{\sin x}{x^a}dx\right|=$

integrate by parts with $u=x^{-a}, dv=\sin x dx,$ so that $v=-\cos x,$ $du=-ax^{-a-1}dx$

$=\left| \frac{\cos C}{C^a}-\frac{\cos C'}{(C')^a}-a\int^{C'}_C\frac{\cos x}{x^{a+1}}dx\right|\leq$

$\leq \frac{1}{C^a}+\frac{1}{(C')^a}+a\int^{\infty}_C\frac{dx}{x^{a+1}}=\frac{2}{C^a}+\frac{1}{(C')^a}\to 0, C,C'\to \infty.$

So, the limit $\lim_{C\to \infty}\int^C_1 \frac{\sin x}{x^a}dx$ exists for any $a>0.$

The improper integral $\int^\infty_0 \frac{\sin x}{x^a}dx$ converges if and only if $a<2.$

Problem 2.

Let $f : [0, 1] \to \mathbb{R}$ be a real-valued continuous function which is differentiable on $(0, 1)$ and satisfies $f(0) = 0.$ Suppose there exists a

constant $c\in (0,1)$ such that

$|f'(x)|\leq c|f(x)| \mbox{ for all } x\in (0,1).$

Show that $f(x) = 0$ for all $x\in [0,1]$ .

Topic: Real Analysis
Difficulty Level: Medium

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

Hint 1

Take $M=\max_{x\in (0,1)}|f(x)|$ and think of a way to bound $|f(x)|$ .

Hint 2

For any $x\in (0,1)$ we have

$|f(x)|=|f(x)-f(0)|=\left|\int^x_0 f'(y)dy\right|\leq c \int^x_0 |f(y)|dy\leq cx M\leq cM.$

Hint 3

Try to show that $M=0$ .

Full Solution

Denote $M=\max_{x\in (0,1)}|f(x)|.$ For any $x\in (0,1)$ we have

$|f(x)|=|f(x)-f(0)|=\left|\int^x_0 f'(y)dy\right|\leq c \int^x_0 |f(y)|dy\leq cx M\leq cM.$

Since the estimate holds for any $x\in (0,1)$ we take maximum in $x$ and get

$M\leq cM, \ (1-c)M\leq 0$

The assumption $0<c<1$ implies $0\leq M\leq 0,$ i.e. $M=0$ and $f(x) = 0$ for all $x\in [0,1]$ .

Problem 3.

Let $G$ be an abelian group of order $n.$

(a) If $f : G \to\mathbb{C}$ is a function, then prove that for all $h\in G$ ,

$\sum_{g\in G}f(g)=\sum_{g\in G}f(hg).$

(b) Let $\mathbb{C}^*$ be the multiplicative group of non-zero complex numbers and suppose $f : G\to \mathbb{C}^*$ is a homomorphism. Prove that

$\sum_{g\in G}f(g)=0 \mbox{ or } \sum_{g\in G}f(g)=n.$

$\sum_{g\in G}|f(g)|=n.$

Topic: Group Theory
Difficulty Level: Medium

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

Hint 1

(a) For any $h\in G$ the mapping $g\to hg$ is a bijection of $G$ .

Hint 2

(b) For any $h\in G$ we have

$f(h)\sum_{g\in G}f(g)=\sum_{g\in G}f(hg)=\sum_{g\in G}f(g)$

Hint 3

$|f(hg)|=|f(h)f(g)|=|f(h)||f(g)|$

implies that $g\to |f(g)|$ is also a homomorphism.

Full Solution

(a) For any $h\in G$ the mapping $g\to hg$ is a bijection of $G.$ When $g$ ranges through $G,$ the product $hg$ also ranges through $G:$

$\sum_{g\in G}f(hg)=\sum_{g\in G}f(g).$

(b) For any $h\in G$ we have

$f(h)\sum_{g\in G}f(g)=\sum_{g\in G}f(hg)=\sum_{g\in G}f(g)$

Taking the sum in $h\in G$ we get

$\sum_{h\in G}f(h)\sum_{g\in G}f(g)=n\sum_{g\in G}f(g).$

So,

$\left(\sum_{g\in G}f(g)\right)^2=n\sum_{g\in G}f(g),$

$\sum_{g\in G}f(g)\left(n-\sum_{g\in G}f(g)\right)=0.$

Either $\sum_{g\in G}f(g)=0$ or $\sum_{g\in G}f(g)=n.$

$|f(hg)|=|f(h)f(g)|=|f(h)||f(g)|$

implies that $g\to |f(g)|$ is also a homomorphism. By previous part, either

$\sum_{g\in G}|f(g)|=0$ or $\sum_{g\in G}|f(g)|=n.$ For all $g\in G$ we have $f(g)\neq 0,$ $|f(g)|>0$ and $\sum_{g\in G}|f(g)|>0.$ It follows that $\sum_{g\in G}|f(g)|=n.$

Problem 4.

(a) Is the ideal $I = (X + Y, X - Y )$ in the polynomial ring $\mathbb{C}[X, Y ]$ a prime ideal? Justify your answer.

(b) Is the ideal $I = (X + Y, X - Y )$ in the polynomial ring $\mathbb{Z}[X, Y ]$ 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

Hint 1

(a) Try to use the relation

$X=\frac{1}{2}(X+Y)+\frac{1}{2}(X-Y), Y=\frac{1}{2}(X+Y)-\frac{1}{2}(X-Y)$

Hint 2

Assume that $I$ is a prime ideal in $\mathbb{Z}[X,Y].$ Consider polynomial $P(X,Y)=2X=(X+Y)+(X-Y)\in I.$ Then either $2\in I$ or $X\in I.$

Hint 3

(b) Hint continued:

General element of $I$ is of the form

$(X+Y)Q(X,Y)+(X-Y)R(X,Y), \ Q(X,Y),R(X,Y)\in \mathbb{Z}[X,Y].$

Full Solution

(a) Relations

$X=\frac{1}{2}(X+Y)+\frac{1}{2}(X-Y), Y=\frac{1}{2}(X+Y)-\frac{1}{2}(X-Y)$

imply that $I=(X,Y).$ General element of $I$ is of the form $XP(X,Y)+YQ(X,Y),$ i.e.

$I=\{P\in \mathbb{C}[X,Y]: P(0,0)=0\}.$

If $PQ\in I,$ then $P(0,0)Q(0,0)=0$ which is possible only when $P(0,0)=0$ or $Q(0,0)=0.$ $I$ is a prime ideal.

(b) Assume that $I$ is a prime ideal in $\mathbb{Z}[X,Y].$ Consider polynomial $P(X,Y)=2X=(X+Y)+(X-Y)\in I.$ Then either $2\in I$ or $X\in I.$ General element of $I$ is of the form

$(X+Y)Q(X,Y)+(X-Y)R(X,Y), \ Q(X,Y),R(X,Y)\in \mathbb{Z}[X,Y].$

In particular, an element of $I$ vanished when $X=Y=0.$ So, $2\not\in I.$ It follows that $X\in I$ and

$X=(X+Y)Q(X,Y)+(X-Y)R(X,Y)=X(Q(X,Y)+R(X,Y))+Y(Q(X,Y)-R(X,Y)).$

Then, $Q(X,Y)=R(X,Y)$ and

$X=2Q(X,Y)X, 2Q(X,Y)=1,$

which is impossible in $\mathbb{Z}[X,Y].$ $I$ is not a prime ideal.

Problem 5.

Let $n \geq 2$ and $A$ be an $n\times n$ matrix with real entries. Let $\mbox{Adj} A$ denote the adjoint of $A,$ that is, the $(i, j)$ -th entry of $\mbox{Adj} A$ is the

$(j, i)$ -th cofactor of $A.$ Show that the rank of $\mbox{Adj} A$ is $0$ , $1$ or $n.$

Topic: Linear Algebra
Difficulty Level: Medium

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

Hint 1

If $\det A\ne 0,$ then the inverse of $A$ is given by $\mbox{Adj} A.$ In this case the rank of $\mbox{Adj}A$ is $n.$

Hint 2

Assume that the rank of $A$ is $\leq n-2.$ Then each $(n-1)\times (n-1)$ minor of $A$ is zero, and all elements of $\mbox{Adj} A$ are equal to zero. In this case the rank of $\mbox{Adj} A$ is equal to zero.

Hint 3

Assume that the rank of $A$ is equal to $n-1.$ Then there exists a nonzero $(n-1)\times (n-1)$ minor of $A,$ so the rank of $\mbox{Adj} A$ is $\geq 1.$ Denote by $a^{(ij)}$ the $(i,j)-$ th cofactor of $A$ . Then $(\mbox{Adj} A)_{ij}=a^{(ji)}.$ Consider the product

$A\cdot \mbox{Adj} A.$ Its $(i,j)-$ th entry is

$\sum^n_{k=1}a_{ik}( \mbox{Adj} A)_{kj}=\sum^n_{k=1}a_{ik}a^{(jk)}=0 \eqno(1)$

Full Solution

If $\det A\ne 0,$ then the inverse of $A$ is given by $\mbox{Adj} A.$ In this case the rank of $\mbox{Adj}A$ is $n.$

$A\cdot \mbox{Adj} A.$ Its $(i,j)-$ th entry is

$\sum^n_{k=1}a_{ik}( \mbox{Adj} A)_{kj}=\sum^n_{k=1}a_{ik}a^{(jk)}=0 \eqno(1)$

Indeed, if $i=j$ then the sum in (1) is the determinant of $A.$ If $i\ne j,$ then the sum in (1) is the determinant of the matrix $A$ after the $j-$ th row is replaced with the $i-$ th row. Such determinant is zero as the matrix has two identical rows. We see that $A\cdot \mbox{Adj} A=0.$ That is, each column of $\mbox{Adj} A$ is in the kernel of $A.$ But the rank of $A$ is $n-1$ , hence the dimension of the kernel of $A$ is $1$ and the dimension of the column range of $\mbox{Adj} A$ is $\leq 1.$ This proves that the rank of $\mbox{Adj} A$ 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 $X$ denote the random variable that takes the value $n$ if the first $n - 1$ draws yield red balls and the $n$ -th draw yields a blue ball.

(a) If $n\geq 1$ , find $P(X > n)$ .

(b) Show that the probability of a blue ball being chosen eventually is $1.$

Topic: Probability
Difficulty Level: Medium

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

Hint 1

(a) We use induction on $n.$ Let $n=1.$ The event $\{X>1\}$ means that the first draw yielded a red ball,

$P(X>1)=\frac{1}{2}.$

Hint 2

(b) If the blue ball is never chosen, then $\{X>n\}$ for all $n.$ Denote by $N$ the event when the blue ball is never chosen. Then

$P(N)\leq P(X>n)=\frac{1}{n+1}\to 0,\ n\to\infty.$

So, $P(N)=0.$

Hint 3

$P(X=n)=P(X>n-1)-P(X>n)=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}.$

Full Solution

(a) We use induction on $n.$ Let $n=1.$ The event $\{X>1\}$ means that the first draw yielded a red ball,

$P(X>1)=\frac{1}{2}.$

Assume that event $\{X>n\}$ happens. Then first $n$ draws yielded red balls, and before the $(n+1)-$ st draw there are $n+1$ red balls and $1$ blue ball in the urn. The probability to draw a red ball is $\frac{n+1}{n+2}.$ So,

$P(X>n+1)=P(X>n)P(X>n+1|X>n)=\frac{n+1}{n+2}P(X>n)$

If $P(X>n)=\frac{1}{n+1},$ then $P(X>n+1)=\frac{1}{n+2}.$ For $n=1$ the equality $P(X>n)=\frac{1}{n+1}$ is true. So, for all $n\geq 1$

$P(X>n)=\frac{1}{n+1}$

(b) If the blue ball is never chosen, then $\{X>n\}$ for all $n.$ Denote by $N$ the event when the blue ball is never chosen. Then

$P(N)\leq P(X>n)=\frac{1}{n+1}\to 0,\ n\to\infty.$

So, $P(N)=0.$

$P(X=n)=P(X>n-1)-P(X>n)=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}.$

Then

$E[X]=\sum^\infty_{n=1}nP(X=n)=\sum^\infty_{n=1}\frac{1}{n+1}=\infty.$

Problem 7.

A real number $x_0$ is said to be a limit point of a set $S\subseteq \mathbb{R}$ if every

neighbourhood of $x_0$ contains a point of $S$ other than $x_0.$ Consider the

set

$S=\{0\}\cup \left\{\frac{1}{m}+\frac{1}{n}:m,n\in\mathbb{N}\right\}$

(a) Show that $S$ contains infinitely many limit points of $S.$

(b) Show that $S$ is a compact subset of $\mathbb{R}.$

Topic: Real Analysis
Difficulty Level: Medium

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

Hint 1

(a) Observe that $\frac{1}{n}=\frac{1}{2n}+\frac{1}{2n}\in S.$ Let $\epsilon>0$ and $m>\frac{1}{\epsilon}.$ Then

$\frac{1}{n}<\frac{1}{n}+\frac{1}{m}<\frac{1}{n}+\epsilon.$

Hint 2

(b) The set $S$ is bounded: $S\subset [0,2].$ It remains to prove that $S$ is closed.

Hint 3

Full Solution

(a) Observe that $\frac{1}{n}=\frac{1}{2n}+\frac{1}{2n}\in S.$ Let $\epsilon>0$ and $m>\frac{1}{\epsilon}.$ Then

$\frac{1}{n}<\frac{1}{n}+\frac{1}{m}<\frac{1}{n}+\epsilon.$

Every neighborhood of $\frac{1}{n}$ contains a point from $S$ distinct from $\frac{1}{n}.$ Each point $\frac{1}{n}$ is a limit point of the set $S$ that belongs to the set $S.$

If $m>\frac{2}{\epsilon},$ then $\frac{1}{m}+\frac{1}{m}$ is a point of $S$ in the $\epsilon-$ neighborhood of $0$ . $0$ is a limit point of $S.$ We’ve proved that points

$\{0\}\cup \left\{\frac{1}{n}:n\geq 1\right\}$

are limit points of $S$ .
(b)

The set $S$ is bounded: $S\subset [0,2].$ It remains to prove that $S$ is closed. Suppose that $x\in \bar{S}\setminus S.$ Every neighborhood of $x$ contains a point $y \in S$ . But $x\not\in S$ and $y\ne x.$ Then $x$ is a limit point of $S.$ Since we assume $x\not \in S,$ it follows that $x>0.$ For every $k\geq 1$ find a point $\frac{1}{m_k}+\frac{1}{n_k}\in S,$ such that

$\left|x-\left(\frac{1}{m_k}+\frac{1}{n_k}\right)\right|<\frac{1}{k}.$

If both sequences $(m_k)_{k\geq 1}$ and $(n_k)_{k\geq 1}$ are bounded, then there exist values $m,n\geq 1$ such that $m_k=m$ and $n_k=n$ for infinitely many values of $k.$ But then $x=\frac{1}{m}+\frac{1}{n}\in S,$ which is not the case. Hence we can assume that $(n_k)_{k\geq 1}$ is unbounded. Let $(n_{k_j})_{j\geq 1}$ be a subsequence such that $\lim_{j\to\infty}n_{k_j}= \infty.$ If the sequence $(m_{k_j})_{j\geq 1}$ is bounded, there exists value $m\geq 1$ such that $m_{k_j}=m$ for infinitely many values of $j.$ But then

$x=\frac{1}{m}\in S.$ If the sequence $(m_{k_j})_{j\geq 1}$ is unbounded, we can pass to another subsequence $(m_{k'_j})_{j\geq 1}$ that converges to $\infty.$ Then we have $x=0.$ In any case $x\in S$ and $S$ is compact.

Problem 8.

(a) Let $\{f_n\}^\infty_{n=1}$ be a sequence of continuous functions on $[0, 1]$ such

that $\sum^\infty_{n=1} f_n$ converges uniformly on $(0, 1].$ Show that $\sum^\infty_{n=1}f_n(0)$

converges.

(b) Find the set $D$ of all points $x\in [0,1]$ such that the series

$\sum^\infty_{n=1}e^{-nx}\cos(nx)$

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

Topic: Real Analysis
Difficulty Level: Medium

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

Hint 1

(a) Denote $s_m(x)=\sum^m_{n=1}f_n(x).$ The sequence $(s_m)_{m\geq 1}$ converges uniformly on $(0,1]:$ there exists function $s$ on $(0,1]$ such that

$\lim_{m\to\infty}\sup_{0<x\leq 1}|s_m(x)-s(x)|=0.$

Hint 2

(a) Hint Continued: Verify that $\lim_{x\to 0}s(x)$ exists. Let $\epsilon>0.$ Find $m\geq 1$ such that

$\sup_{0<x\leq 1}|s_m(x)-s(x)|<\frac{\epsilon}{3}.$

Hint 3

(b) The series converges for all $x\in (0,1].$ Indeed,

$\sum^\infty_{n=1}|e^{-nx}\cos(nx)|\leq \sum^\infty_{n=1}e^{-nx}=\frac{e^{-x}}{1-e^{-x}}<\infty.$

Full Solution

(a) Denote $s_m(x)=\sum^m_{n=1}f_n(x).$ The sequence $(s_m)_{m\geq 1}$ converges uniformly on $(0,1]:$ there exists function $s$ on $(0,1]$ such that

$\lim_{m\to\infty}\sup_{0<x\leq 1}|s_m(x)-s(x)|=0.$

We verify that $\lim_{x\to 0}s(x)$ exists. Let $\epsilon>0.$ Find $m\geq 1$ such that

$\sup_{0<x\leq 1}|s_m(x)-s(x)|<\frac{\epsilon}{3}.$

Using continuity of $s_m$ find $\delta>0$ such that for all $x,y\in (0,\delta)$

$|s_m(x)-s_m(y)|<\frac{\epsilon}{3}.$

Then for all $x,y\in (0,\delta)$ we have

$|s(x)-s(y)|\leq |s(x)-s_m(x)|+|s_m(x)-s_m(y)|+|s_m(y)-s(y)|<\epsilon.$

Since $\epsilon>0$ is arbitrary, this shows that $\lim_{x\to 0}s(x)$ exists. We denote the limit by $s(0).$

Now, let $\epsilon>0.$ Find $m_0\geq 1$ such that for all $m\geq m_0$ and all $x\in(0,1]$ we have

$|s_m(x)-s(x)|\leq \epsilon.$

Passing to the limit as $x\to 0$ we get

$|s_m(0)-s(0)|\leq \epsilon.$

Since $\epsilon>0$ is arbitrary, this shows that

$s(0)=\lim_{m\to\infty}s_m(0)=\lim_{m\to\infty}\sum^m_{n=1}f_n(0),$

i.e. $\sum^\infty_{n=1}f_n(0)$

converges and has the sum $s(0)$ .

(b) The series converges for all $x\in (0,1].$ Indeed,

$\sum^\infty_{n=1}|e^{-nx}\cos(nx)|\leq \sum^\infty_{n=1}e^{-nx}=\frac{e^{-x}}{1-e^{-x}}<\infty.$

For $x=0$ the series diverges, as it is the series $1+1+1+\ldots.$ So, $D=(0,1].$ If the series converges uniformly, then it should also converge at $x=0.$ So, the series $\sum^\infty_{n=1}e^{-nx}\cos(nx)$ does not converge uniformly on $D.$

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To view previous year’s question papers: Click Here

4 Comments

Suchismita biswas

July 6, 2020 at 3:00 pm Reply

Sir,my daughter wants to prepare for m.math entrance examination..can you please help her with your classes and practice sets??
- FractionsHub
  
  July 6, 2020 at 4:32 pm Reply
  
  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.
Lucas Haobam

July 3, 2021 at 12:02 am Reply

For Problem 2, Can I solve this like this?

Let $x \in [0,1]$ then by using MVT there exists $t_1 \in (0,x)$ such that $f(x) - f(0) = f'(t_1)x$ i.e $|f(x)| = |f'(t_1)x| \leq c|f(t_1)|x$ . Similary there exists $t_2 \in (0,t_1)$ such that $f(t_1) = f'(t_2)t_1$ and applying modulus we get $|f(t_1)| = |f'(t_2)|t_1 \leq c|ft_2)|t_1$ So with the above inequalities we have $|f(x)| \leq c^2 |f(t_2)|t_1x \leq c^2|f(t_2)|$ . Proceeding this way we have $|f(x)| \leq c^n |f(x_n)|$ but since $f$ is continuous on $[0,1]$ so is $|f|$ so it attains the maximum say $M$ then the inequality becomes $|f(x)| \leq c^n|f(t_n)|\leq c^n M$ and we know that $c^n M \to 0$ . so for any $\epsilon>0$ there exists $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ we have $c^n M < \epsilon$ In particular $c^{n_0}M < \epsilon$ hence $|f(x)| \leq c^{n_0}M \leq \epsilon$ this implies $f(x)=0$ and since $x$ was arbitrary we have $f(x) = 0 \forall x \in [0,1]$
Lucas Haobam

July 3, 2021 at 12:03 am Reply

For Problem 2, Can I solve this like this?

Let $x \in [0,1]$ then by using MVT there exists $t_1 \in (0,x)$ such that $f(x) - f(0) = f'(t_1)x$ i.e $|f(x)| = |f'(t_1)x| \leq c|f(t_1)|x$ . Similary there exists $t_2 \in (0,t_1)$ such that $f(t_1) = f'(t_2)t_1$ and applying modulus we get $|f(t_1)| = |f'(t_2)|t_1 \leq c|ft_2)|t_1$ So with the above inequalities we have $|f(x)| \leq c^2 |f(t_2)|t_1x \leq c^2|f(t_2)|$ . Proceeding this way we have $|f(x)| \leq c^n |f(x_n)|$ but since $f$ is continuous on $[0,1]$ so is $|f|$ so it attains the maximum say $M$ then the inequality becomes $|f(x)| \leq c^n|f(t_n)|\leq c^n M$ and we know that $c^n M \to 0$ . so for any $\epsilon>0$ there exists $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$ we have $c^n M < \epsilon$ In particular $c^{n_0}M < \epsilon$ hence $|f(x)| \leq c^{n_0}M \leq \epsilon$ this implies $f(x)=0$ and since $x$ was arbitrary we have $f(x) = 0 \forall x \in [0,1]$

ISI MMath Solutions and Discussion 2018 : PMB

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

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.

Topic: Real AnalysisDifficulty level: Medium

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

Hint 1

Hint 2

Hint 3

Full Solution

Problem 2.

Topic: Real AnalysisDifficulty Level: Medium

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

Hint 1

Hint 2

Hint 3

Full Solution

Problem 3.

Topic: Group TheoryDifficulty Level: Medium

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

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Hint 3

Full Solution

Problem 4.

Topic: Ring TheoryDifficulty Level: Medium

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

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Hint 2

Hint 3

Full Solution

Problem 5.

Topic: Linear AlgebraDifficulty Level: Medium

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

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Hint 3

Full Solution

Problem 6.

Topic: ProbabilityDifficulty Level: Medium

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

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Hint 3

Full Solution

Problem 7.

Topic: Real AnalysisDifficulty Level: Medium

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

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Hint 3

Full Solution

Problem 8.

Topic: Real AnalysisDifficulty Level: Medium

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

Hint 1

Hint 2

Hint 3

Full Solution

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

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Topic: Real Analysis
Difficulty level: Medium

Topic: Real Analysis
Difficulty Level: Medium

Topic: Group Theory
Difficulty Level: Medium

Topic: Ring Theory
Difficulty Level: Medium

Topic: Linear Algebra
Difficulty Level: Medium

Topic: Probability
Difficulty Level: Medium

Topic: Real Analysis
Difficulty Level: Medium

Topic: Real Analysis
Difficulty Level: Medium