ISI MMath Solutions and Discussion 2019 : PMB

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

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Problem 1.

Let $f : \mathbb{R}\to \mathbb{R}$ be a continuous function satisfying

$\lim_{x\to \infty}f(x)=\lim_{x\to -\infty}f(x)=0.$

Show that $f$ is a bounded function on $\mathbb{R}$ 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

Hint 2

Let $f : \mathbb{R}\to \mathbb{R}$ be a continuous function satisfying

$\lim_{x\to \infty}f(x)=\lim_{x\to -\infty}f(x)=0.$

The function $h(x)=|f(x)|$ is a non-negative continuous function satisfying

$\lim_{x\to \infty}h(x)=\lim_{x\to -\infty}h(x)=0.$

Hint 3

For the Example part, consider the function $f(x)=\frac{1}{1+x^2}.$

Full Solution

Step 1:

Assume that $f$ is non-negative, i.e. $f(x)\geq 0$ for all $x\geq 0.$ If $f(x)=0$ for all $x\in \mathbb{R},$ then obviously $f$ is bounded and attains its maximal (and minimal) value at each point. Let $f(x_0)>0$ for some $x_0\in\mathbb{R}.$ By assumption, there exists $C>|x_0|$ such that for all $x>C$ and all $x<-C,$ $0\leq f(x)<f(x_0).$ Let $g(x)$ be the restriction of $f$ to $[-C,C].$ $g:[-C,C]\to \mathbb{R}$ is a continuous function on a compact segment. Hence, $g$ is bounded and attains its maximal value on $[-C,C],$ i.e. there exists $x_1\in [-C,C],$ such that $g(x_1)=\max_{x\in [-C,C]}g(x).$ We check that the function $f$ attains its maximal value at $x_1.$ Let $x\in [-C,C].$ Then

$f(x)=g(x)\leq g(x_1)=f(x_1).$

Let $x>C$ or $x<-C.$ Then $x_0\in [-C,C]$ and

$f(x)<f(x_0)=g(x_0)\leq g(x_1)=f(x_1).$

So, for each $x\in\mathbb{R}$

$0\leq f(x)\leq f(x_1).$

$f$ is bounded and attains its maximal value.

Step 2:

Let $f : \mathbb{R}\to \mathbb{R}$ be a continuous function satisfying

$\lim_{x\to \infty}f(x)=\lim_{x\to -\infty}f(x)=0.$

The function $h(x)=|f(x)|$ is a non-negative continuous function satisfying

$\lim_{x\to \infty}h(x)=\lim_{x\to -\infty}h(x)=0.$

By step 1, $h$ is bounded and attains its maximal value. It means that there exists $x_1\in \mathbb{R}$ such that for all $x\in \mathbb{R}$

$0\leq h(x)=|f(x)|\leq h(x_1)=|f(x_1)|.$

So, for all $x\in \mathbb{R},$

$-|f(x_1)|\leq f(x)\leq |f(x_1)|.$

The function $f$ is bounded. If $f(x_1)\geq 0,$ then $f(x_1)=|f(x_1)|$ and $f$ attains its maximal value. If $f(x_1)\leq 0,$ then $f(x_1)=-|f(x_1)|$ and $f$ attains its minimal value.

Step 3:

Consider the function $f(x)=\frac{1}{1+x^2}.$ Then $f$ is a continuous non-negative function,

$\lim_{x\to \infty}f(x)=\lim_{x\to -\infty}f(x)=0.$

In particular, $\inf_{x\in\mathbb{R}}f(x)=0.$ Further,

$0\leq f(x)\leq 1=f(0).$

Function $f$ attains its maximal value at $x=0.$ However, for all $x>0,$ $f(x)>0$ and the function $f$ does not attain its minimal value.

Problem 2.

Let $g: [0, 1] \to \mathbb{R}$ be a continuous function such that $g(1) = 0.$ Show

that

$\sup_{x\in[0,1]}|x^ng(x)|\to 0 \mbox{ as } n\to\infty.$

Topic: Real Analysis
Difficulty Level: Easy

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

Hint 1

Function $g$ is bounded, as it is a continuous function on a compact $[0,1].$ Let $C=\sup_{x\in[0,1]}|g(x)|<\infty.$ Fix $\epsilon>0.$ By assumption, there exists $\delta\in (0,1)$ such that for all $x\in (1-\delta,1],$ $|g(x)|\leq \epsilon.$ Let $N\geq 1$ be such that $C(1-\delta)^N\leq \epsilon.$

Hint 2

Let $n\geq N$ and $x\in [0,1].$ If $x>1-\delta,$ then

$|x^ng(x)|\leq |g(x)|\leq \epsilon.$

Hint 3

If $0\leq x\leq 1-\delta,$ then

$|x^ng(x)|\leq (1-\delta)^n|g(x)|\leq (1-\delta)^NC\leq \epsilon.$

Full Solution

Let $n\geq N$ and $x\in [0,1].$ If $x>1-\delta,$ then

$|x^ng(x)|\leq |g(x)|\leq \epsilon.$

If $0\leq x\leq 1-\delta,$ then

$|x^ng(x)|\leq (1-\delta)^n|g(x)|\leq (1-\delta)^NC\leq \epsilon.$

In any case,

$|x^ng(x)|\leq \epsilon$ and

$\sup_{x\in[0,1]}|x^ng(x)|\leq \epsilon, \ n\geq N.$

Since $\epsilon>0$ is arbitrary, this proves the needed convergence.

Problem 3.

Let $f :\mathbb{R}\to\mathbb{R}$ be a twice continuously differentiable function and

suppose $f(0)=f'(0)=0.$ If $|f''(x)|\leq 1$ for all $x\in \mathbb{R},$ then prove that

$|f(x)|\leq 1/2$ for all $x\in[-,1,1].$

Topic: Real Analysis
Difficulty Level: Medium

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

Hint 1

For every $x\in \mathbb{R},$ we have by the Newton-Leibniz formula

$|f'(x)|=|f'(x)-f'(0)|=\left|\int^x_0 f''(y)dy\right|\leq \int^{|x|}_0 |f''(y)|dy\leq |x|.$

Hint 2

Applying the Newton-Leibniz rule again we get

$|f(x)|=|f(x)-f(0)|=\left|\int^x_0 f'(y)dy\right|\leq \int^{|x|}_0 ydy=\frac{x^2}{2}.$

Hint 3

In particular, for each $x\in [-1,1]$ we have

$|f(x)|\leq \frac{x^2}{2}\leq \frac{1}{2}.$

Full Solution

For every $x\in \mathbb{R},$ we have by the Newton-Leibniz formula

$|f'(x)|=|f'(x)-f'(0)|=\left|\int^x_0 f''(y)dy\right|\leq \int^{|x|}_0 |f''(y)|dy\leq |x|.$

Applying the Newton-Leibniz rule again we get

$|f(x)|=|f(x)-f(0)|=\left|\int^x_0 f'(y)dy\right|\leq \int^{|x|}_0 ydy=\frac{x^2}{2}.$

In particular, for each $x\in [-1,1]$ we have

$|f(x)|\leq \frac{x^2}{2}\leq \frac{1}{2}.$

Problem 4.

Suppose $f : \mathbb{R}^2\to \mathbb{R}$ is a function defined by

$f(x,y)=\begin{cases} \frac{x^2y^3}{x^4+y^6}, \mbox{ if } x\ne 0, y\in\mathbb{R}, \\ 0, \mbox{ if } x=0, y\in\mathbb{R}. \end{cases}$

(a). Find all $(a, b) \in \mathbb{R}^2\setminus \{(0,0)\}$ such that $f$ has a nonzero directional derivative at $(0, 0)$ with respect to the direction $(a, b).$

(b). Is $f$ continuous at $(0, 0)?$ 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 $a=0,$ $b\ne 0.$ Then the directional derivative at $(0,0)$ with respect to the direction $(a,b)$ is

$\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)-f(0,0)}{\epsilon}=\lim_{\epsilon\to 0}\frac{f(0,\epsilon b)}{\epsilon}=\lim_{\epsilon\to 0}0=0.$

Hint 2

Let $a\ne 0.$ Then the directional derivative at $(0,0)$ with respect to the direction $(a,b)$ is

$\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)-f(0,0)}{\epsilon}=\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)}{\epsilon}=$

$=\lim_{\epsilon\to 0}\frac{\epsilon^5 a^2 b^3}{\epsilon(\epsilon^4 a^4 +\epsilon^6 b^6)}=\lim_{\epsilon\to 0}\frac{a^2 b^3}{ a^4 +\epsilon b^6}=\frac{b^3}{a^2}.$

Hint 3

Part b.

If $f$ is continuous at $(0,0)$ , then for any sequence $(x_n,y_n)\to 0$ we have $f(x_n,y_n)\to 0.$ Let $x_n=\frac{1}{n^3},$ $y_n=\frac{1}{n^2}.$

Full Solution

Part a.
Let $a=0,$ $b\ne 0.$ Then the directional derivative at $(0,0)$ with respect to the direction $(a,b)$ is

$\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)-f(0,0)}{\epsilon}=\lim_{\epsilon\to 0}\frac{f(0,\epsilon b)}{\epsilon}=\lim_{\epsilon\to 0}0=0.$

Let $a\ne 0.$ Then the directional derivative at $(0,0)$ with respect to the direction $(a,b)$ is

$\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)-f(0,0)}{\epsilon}=\lim_{\epsilon\to 0}\frac{f(\epsilon a,\epsilon b)}{\epsilon}=$

$=\lim_{\epsilon\to 0}\frac{\epsilon^5 a^2 b^3}{\epsilon(\epsilon^4 a^4 +\epsilon^6 b^6)}=\lim_{\epsilon\to 0}\frac{a^2 b^3}{ a^4 +\epsilon b^6}=\frac{b^3}{a^2}.$

$f$ has a nonzero directional derivative at $(0, 0)$ with respect to the direction $(a, b)$ if and only if $a\ne 0$ and $b\ne 0.$

Part b.

If $f$ is continuous at $(0,0)$ , then for any sequence $(x_n,y_n)\to 0$ we have $f(x_n,y_n)\to 0.$ Let $x_n=\frac{1}{n^3},$ $y_n=\frac{1}{n^2}.$ Then

$f(x_n,y_n)=\frac{n^{-6}n^{-6}}{n^{-12}+n^{-12}}=\frac{1}{2}\not\to 0.$

So, $f$ is not continuous at $(0,0).$

Problem 5.

Let $C$ be a subset of a compact metric space $(X, d).$ Assume that for

every continuous function $h : X \to\mathbb{R}$ , the restriction of $h$ to $C$ attains

a maximum on $C.$ Prove that $C$ 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 $C$ is a closed subset of $X.$

Hint 2

Let $x_0\in \bar{C}.$ Consider function $h(x)=1-d(x,x_0).$ $h$ is a continuous function on $X.$

Hint 3

Indeed, by the triangle inequality,

$|h(x)-h(y)|=|d(x,x_0)-d(y,x_0)|\leq d(x,y).$

So, $h$ attains a maximum on $C$ . Since $h(x)\leq 1=h(x_0),$ it follows that there exists a point $x_1\in C$ such that $h(x_1)=1.$

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 $C$ is a closed subset of $X.$ Let $x_0\in \bar{C}.$ Consider function $h(x)=1-d(x,x_0).$ $h$ is a continuous function on $X.$ Indeed, by the triangle inequality,

$|h(x)-h(y)|=|d(x,x_0)-d(y,x_0)|\leq d(x,y).$

So, $h$ attains a maximum on $C$ . Since $h(x)\leq 1=h(x_0),$ it follows that there exists a point $x_1\in C$ such that $h(x_1)=1.$ Then $d(x_1,x_0)=0$ and $x_0=x_1\in C.$ So, $\bar{C}\subset C$ and $C=\bar{C}$ is closed.

Problem 6.

Let $G$ be a non-abelian group of order $pq$ , where $p < q$ are primes.

(a). How many elements of $G$ have order $q?$

(b). How many elements of $G$ have order $p?$

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 $G$ is of order $1,p,q$ or $pq.$ Since $G$ is non-abelian, it has no elements of order $pq.$ By Cauchy’s theorem, there exists an element $x$ of order $q.$ Then each of the elements

$x,x^2,x^3,\ldots,x^{q-1}$

is of order $q$ .

Hint 2

Assume that $y$ is of order $q,$ and $y\ne x^j,$ $0\leq j\leq q-1.$ Consider subgroups

$H_1=\{e,x,x^2,x^3,\ldots,x^{q-1}\}$

and

$H_2=\{e,y,y^2,y^3,\ldots,y^{q-1}\}.$

Hint 3

Then $H_1\cap H_2$ is a subgroup of $H_2,$ strictly smaller than $H_2$ (as $y\not\in H_1$ ). The order of $H_2$ is prime, hence $H_1\cap H_2=\{e\}.$ Consider cosets

$H_1,yH_1,y^2H_1,\ldots,y^{q-1}H_1.$

The size of each coset is $q,$ and there are $q$ cosets. $|G|=pq<q^2$ hence two of different cosets intersect:

$y^iH_1\cap y^jH_1\ne \emptyset, \ o\leq i<j\leq q-1.$

Full Solution

$x,x^2,x^3,\ldots,x^{q-1}$

is of order $q$ .

Assume that $y$ is of order $q,$ and $y\ne x^j,$ $0\leq j\leq q-1.$ Consider subgroups

$H_1=\{e,x,x^2,x^3,\ldots,x^{q-1}\}$

and

$H_2=\{e,y,y^2,y^3,\ldots,y^{q-1}\}.$

Then $H_1\cap H_2$ is a subgroup of $H_2,$ strictly smaller than $H_2$ (as $y\not\in H_1$ ). The order of $H_2$ is prime, hence $H_1\cap H_2=\{e\}.$ Consider cosets

$H_1,yH_1,y^2H_1,\ldots,y^{q-1}H_1.$

The size of each coset is $q,$ and there are $q$ cosets. $|G|=pq<q^2$ hence two of different cosets intersect:

$y^iH_1\cap y^jH_1\ne \emptyset, \ o\leq i<j\leq q-1.$

We deduce that there exists $l\in \{1,\ldots,q-1\}$ such that $y^j\in H_1.$ But then $y^j\in H_1\cap H_2=\{e\},$ $y^j=e$ which contradicts the fact that $y$ is of order $q.$

Conclusion: all elements of order $q$ are among $x,x^2,x^3,\ldots,x^{q-1}.$

(a). There are $q-1$ elements of order $q.$

(b). There are $pq-1-(q-1)=(p-1)q$ elements of order $p.$

Problem 7.

Prove or disprove the following statement: The ring $\mathbb{Q}[X]/(X^4-1)$ 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 $\mathbb{Q}[X]/(X^4-1)$ is isomorphic to a product of three fields. Consider the factorization

$X^4-1=(X-1)(X+1)(X^2+1).$

Hint 2

Consider ideals $I_1=(X-1),$ $I_2=(X+1),$ $I_3=(X^2+1).$

Hint 3

Relations

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

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

imply that ideals $I_1,I_2,I_3$ are pairwise coprime.

Now try to apply the Chinese remainder theorem.

Full Solution

We prove that $\mathbb{Q}[X]/(X^4-1)$ is isomorphic to a product of three fields. Consider the factorization

$X^4-1=(X-1)(X+1)(X^2+1).$

Consider ideals $I_1=(X-1),$ $I_2=(X+1),$ $I_3=(X^2+1).$ Relations

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

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

imply that ideals $I_1,I_2,I_3$ are pairwise coprime. By the Chinese remainder theorem,

$\mathbb{Q}[X]/(X^4-1)\cong \mathbb{Q}[X]/(X-1)\times \mathbb{Q}[X]/(X+1)\times \mathbb{Q}[X]/(X^2+1).$

Each of the multiples in the right-hand side is a field, as it is a quotient of $\mathbb{Q}[X]$ by an irreducible polynomial. In fact,

$\mathbb{Q}[X]/(X-1)\cong \mathbb{Q}, \mathbb{Q}[X]/(X+1)\cong \mathbb{Q}, \mathbb{Q}[X]/(X^2+1)\cong \mathbb{Q}(i).$

The ring $\mathbb{Q}[X]/(X^4-1)$ is isomorphic to a product of three fields $\mathbb{Q}\times \mathbb{Q}\times \mathbb{Q}(i).$

Problem 8.

Let $M$ be a symmetric matrix with real entries such that $M^k = 0$ for some $k\in \mathbb{N}.$ Show that $M = 0.$

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 $C$ and a diagonal matrix $D$ such that

$M=CDC^{-1}, D=\begin{pmatrix} \lambda_1 & 0 & 0 & \ldots & 0 \\ 0& \lambda_2 & 0 & \ldots & 0 \\ 0 & 0 & \lambda_3 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \lambda_n \end{pmatrix}$

Hint 3

Then

$M^k=CD^kC^{-1}=0.$

Full Solution

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

Then

$M^k=CD^kC^{-1}=0.$

So,

$D^k=\begin{pmatrix} \lambda^k_1 & 0 & 0 & \ldots & 0 \\ 0& \lambda^k_2 & 0 & \ldots & 0 \\ 0 & 0 & \lambda^k_3 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \lambda^k_n \\ \end{pmatrix}=0$

and $\lambda_1=\lambda_2=\ldots=\lambda_n=0,$ $D=0$ and $M=0.$

Problem 9.

Suppose $A$ and $B$ are two $n\times n$ matrices with real entries such that

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

nonzero column vector ${\bf x}\in \mathbb{R}^n$ such that $A{\bf x} = B{\bf x} = {\bf 0}.$

Topic: Linear Algebra
Difficulty Level: Easy

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

Hint 1

Consider the matrix

$C=\begin{pmatrix} A \\ B\end{pmatrix}.$

The rank of $C$ is the maximal number of linearly independent rows of $C.$

Hint 2

Then

$\mbox{rank} C\leq \mbox{rank} A+\mbox{rank} B<n.$

Hint 3

Consequently, columns of $C$ are linearly dependent, and there exists a nonzero vector ${\bf x}\in\mathbb{R}^n,$ such that

$C{\bf x}=\begin{pmatrix} A{\bf x} \\ B{\bf x}\end{pmatrix}={\bf 0}.$

It means that $A{\bf x} = B{\bf x}={\bf 0}.$

Full Solution

Consider the matrix

$C=\begin{pmatrix} A \\ B\end{pmatrix}.$

The rank of $C$ is the maximal number of linearly independent rows of $C.$ Then

$\mbox{rank} C\leq \mbox{rank} A+\mbox{rank} B<n.$

Consequently, columns of $C$ are linearly dependent, and there exists a nonzero vector ${\bf x}\in\mathbb{R}^n,$ such that

$C{\bf x}=\begin{pmatrix} A{\bf x} \\ B{\bf x}\end{pmatrix}={\bf 0}.$

It means that $A{\bf x} = B{\bf x}={\bf 0}.$

Problem 10.

Suppose there are $n$ persons in a party. Every pair of persons meet each other with probability $p\in (0,1)$ independently of the other pairs.

Let $N(i)$ be the number of people the $i^{th}$ person meets in the party. For all $i, j \in \{1, 2, . . . , n\}$ with $i\ne j$ and for all $k, l \in \{1, 2, . . . , n − 2\},$

show that

$P [N(i) = k, N(j) = l] = {n-2\choose k-1}{n-2\choose l-1}p^{k+l-1}(1-p)^{2n-k-l-2}$

$+{n-2\choose k}{n-2\choose l}p^{k+l}(1-p)^{2n-k-l-3}.$

Topic: Probability
Difficulty Level: Medium

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

Hint 1

Introduce two events:

$A=\{N(i)=k,N(j)=l, i \mbox{ meets } j\},$

$B=\{N(i)=k,N(j)=l, i \mbox{ does not meet } j\}.$

Assume the event $A$ happens. Then $i$ meets $j$ and $i$ meets $k-1$ other persons (from $n-2$ possible persons). There are ${n-2\choose k-1}$ choices of persons except $j$ that $i$ meets. The probability that $i$ meets $k-1$ persons except $j$ is

${n-2\choose k-1}p^{k-1}(1-p)^{n-k-1}.$

Hint 2

Further, $j$ meets $l-1$ persons except $i$ (from $n-2$ possible persons). There are ${n-2\choose l-1}$ choices of persons except $i$ that $j$ meets. The probability that $j$ meets $l-1$ persons except $i$ is

${n-2\choose l-1}p^{l-1}(1-p)^{n-l-1}.$

Hint 3

Finally,

$P[A]=p\cdot {n-2\choose k-1}p^{k-1}(1-p)^{n-k-1}\cdot {n-2\choose l-1}p^{l-1}(1-p)^{n-l-1}=$

$={n-2\choose k-1}\cdot {n-2\choose l-1}p^{k+l-1}(1-p)^{2n-k-l-2}$

Full Solution

Introduce two events:

$A=\{N(i)=k,N(j)=l, i \mbox{ meets } j\},$

$B=\{N(i)=k,N(j)=l, i \mbox{ does not meet } j\}.$

${n-2\choose k-1}p^{k-1}(1-p)^{n-k-1}.$

${n-2\choose l-1}p^{l-1}(1-p)^{n-l-1}.$

Finally,

$P[A]=p\cdot {n-2\choose k-1}p^{k-1}(1-p)^{n-k-1}\cdot {n-2\choose l-1}p^{l-1}(1-p)^{n-l-1}=$

$={n-2\choose k-1}\cdot {n-2\choose l-1}p^{k+l-1}(1-p)^{2n-k-l-2}$

Assume the event $B$ happens. Then $i$ does not meet $j$ and $i$ meets $k$ other persons (from $n-2$ possible persons). There are ${n-2\choose k}$ choices of persons except $j$ that $i$ meets. The probability that $i$ meets $k$ persons except $j$ is

${n-2\choose k}p^{k}(1-p)^{n-k-2}.$

Further, $j$ meets $l$ persons except $i$ (from $n-2$ possible persons). There are ${n-2\choose l}$ choices of persons except $i$ that $j$ meets. The probability that $j$ meets $l$ persons except $i$ is

${n-2\choose l}p^{l}(1-p)^{n-l-2}.$

Finally,

$P[B]=(1-p)\cdot{n-2\choose k}p^{k}(1-p)^{n-k-2}\cdot {n-2\choose l}p^{l}(1-p)^{n-l-2}=$

$={n-2\choose k} {n-2\choose l}p^{k+l}(1-p)^{2n-k-l-3}.$

We deduce that

$P [N(i) = k, N(j) = l] =P[A]+P[B]={n-2\choose k-1}\cdot {n-2\choose l-1}p^{k+l-1}(1-p)^{2n-k-l-2}+$

${n-2\choose k} {n-2\choose l}p^{k+l}(1-p)^{2n-k-l-3}.$

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26 Comments

FractionsHub

May 6, 2020 at 12:41 am Reply

I have tried my best to explain the solutions. If you have some other solutions, please feel free to add them in the comment section. My website is latex enabled so that you can type all the mathematical equations. If you face any problem understanding the solutions, then comment here, and I will get back to you as soon as I can. Also if you find any mathematical error or any typing error then please do comment here.
Soumyajit ganguly

May 24, 2020 at 2:36 am Reply

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.
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Soumyajit Ganguly
- FractionsHub
  
  May 24, 2020 at 3:06 am Reply
  
  Please check your email. I have sent you all the details regarding our classes.
Soumyajit ganguly

May 24, 2020 at 2:37 am Reply

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.
- FractionsHub
  
  May 24, 2020 at 3:06 am Reply
  
  Please check your email. I have sent you all the details regarding our classes.
Anish Ray

June 26, 2020 at 1:02 pm Reply

Hi!
Can you please conduct few mock tests for ISI M.MATH entrance?
- FractionsHub
  
  June 26, 2020 at 5:54 pm Reply
  
  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.
  - Anish Ray
    
    June 26, 2020 at 6:57 pm Reply
    
    Sir, it would be really very helpful if you can create a mock test course for this year’s entrance.
Aviral

September 1, 2020 at 12:16 am Reply

In 5th question I got it h(x)<h(x○)=1 but I didn't got how u proved then that C is closed
- FractionsHub
  
  September 1, 2020 at 2:23 am Reply
  
  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.
Lucas

September 6, 2020 at 7:50 pm Reply

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
Lucas

September 6, 2020 at 7:52 pm Reply

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 $c\in (0,x)$ and for $x<0$ , the case is same because in either way $f'(0) = 0 = f(0)$
so when we apply $|f(x)| = \dfrac{|x^2 f"(c)|}{2} \leq \dfrac{1}{2}$
Lucas

September 6, 2020 at 7:53 pm Reply

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 $c\in (0,x)$ and for $x<0$ , the case is same because in either way $f'(0) = 0 = f(0)$
so when we apply $|f(x)| = \dfrac{|x^2 f"(c)|}{2} \leq \dfrac{1}{2}$
M d sharma

September 12, 2020 at 2:49 pm Reply

#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.
M D sharama

September 12, 2020 at 2:53 pm Reply

sorry for #2)
it was my mistake.
Lucas

September 13, 2020 at 10:06 pm Reply

For the 6th problem, It easily follows from the Sylow’s theorem.
No. of q-sylow subgroup is, $n_q= qk+1$ for some $k\in \mathbb{Z}$ and $n_q|p \implies k=0 \implies n_q=1$ i.e there exist a unique subgroup of order $q$ . so there are $q-1$ elements of order $q$
and we know that every non-identity element in $g\in G$ has order either $p$ or $q$ . So the remaining elements must have order $p$
Therefore the no. of elements having order q is $pq-1 - (q-1) = q(p-1)$
Subrata Mahata

April 2, 2021 at 9:24 pm Reply

Sir,I am from Midnapore college ,W.B and I have decided to appear in ISI.math entrance.But I have no guide.So,I am quite confused about what type of preparation and study is needed.Please suggest me something in this regard.
- FractionsHub
  
  April 3, 2021 at 1:45 am Reply
  
  I have already written everything you need to know on how to prepare for the ISI M.Math entrance exam here: https://fractionshub.com/courses/i-s-i-m-math-i-i-t-jam-exam-preparation/
Devansh Gupta

June 7, 2021 at 10:03 pm Reply

Hello SIr, I am preparing for ISI MMath Entrance. Can you please provide my some training for problem solving. Is there any mock test series for ISI MMath
- FractionsHub
  
  June 7, 2021 at 10:27 pm Reply
  
  You will have to join the MMath group here. I teach for MMath entrance exam. No there is no mock tests for ISI MMath.
Vikas Vaibhav

June 8, 2021 at 11:46 am Reply

Great work…
Can you provide your what’s app group?
- FractionsHub
  
  June 8, 2021 at 3:15 pm Reply
  
  You can text me on WhatsApp @ 9700803692.
Lucas Haobam

July 14, 2021 at 12:53 am Reply

For problem 9, Here’s an alternative solution:
if $Ker (A) \cap Ker(B) \neq {0}$ then we are done. Otherwise assume that $Ker (A) \cap Ker(B) = {0}$ , also note that $Ker(A+B) \subset \mathbb{R}^{n}$ then $dim(Ker(A+B)) \leq n$ . So we have $\eta(A) + \eta (B) \leq n$ as dim(Ker (A) \cap ker(B)) =0 but from the given condition we have $\rho(A) + \rho(B) < n$ we have $n - \eta(A) + n-\eta(B)<n$ i.e $n< \eta(A) + \eta(B)$ but from the above inequality we get a contradiction. Hence $ker(A) \cap Ker (B) \neq {0}$ . Therefore the result follows
Lucas Haobam

July 14, 2021 at 1:17 am Reply

For problem 9, Here’s an alternative solution:
if $Ker (A) \cap Ker(B) \neq \{0\}$ then we are done. Otherwise assume that $Ker (A) \cap Ker(B) = \{0\}$ , also note that $Ker(A+B) \subset \mathbb{R}^{n}$ then $dim(Ker(A+B)) \leq n$ . So we have $\eta(A) + \eta (B) \leq n$ as $\dim(Ker (A) \cap ker(B)) =0$ but from the given condition we have $\rho(A) + \rho(B) < n$ we have $n - \eta(A) + n-\eta(B)<n$ i.e $n< \eta(A) + \eta(B)$ but from the above inequality we get a contradiction. Hence $ker(A) \cap Ker (B) \neq \{0\}$ . Therefore the result follows
Divya Ruhela

January 12, 2023 at 1:09 am Reply

Sir ,your solution is really very helpful..
But if you provide us crash course for preparing M math ..
It will be very helpful…
T

May 12, 2023 at 3:15 pm Reply

For problem 2 can we use sandwich theorem instead,
Since g : [0,1] –> R, it is a continuous function from a compact set, thus its image will be compact.
|g(x)| 0
Let k be the supremum
for x belongs to [0,1)
-M(x^n) <= (x^n) g(x) <= k <= M (x^n)
applying limit and sandwich theorem, we get k tends to 0, and for x=1 since g(1)=0, thus (1^n)(g(1))=0, thus the limit always tends to zero.

ISI MMath Solutions and Discussion 2019 : PMB

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

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

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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 Analysis Difficulty level: Medium

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

Hint 1

Hint 2

Hint 3

Full Solution

Step 1:

Step 2:

Step 3:

Problem 2.

Topic: Real Analysis Difficulty Level: Easy

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

Hint 1

Hint 2

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Problem 3.

Topic: Real Analysis Difficulty Level: Medium

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

Hint 1

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Hint 3

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Problem 4.

Topic: Multivariable Calculus Difficulty Level: Medium

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

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Hint 3

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Problem 5.

Topic: Metric Space Difficulty Level: Medium

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

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Hint 3

Full Solution

Problem 6.

Topic: Group Theory Difficulty Level: Medium

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

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Hint 2

Hint 3

Full Solution

Problem 7.

Topic: Field Theory Difficulty Level: Hard

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

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Hint 3

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Problem 8.

Topic: Linear Algebra Difficulty Level: Medium

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

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Hint 3

Full Solution

Problem 9.

Topic: Linear Algebra Difficulty Level: Easy

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

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Hint 3

Full Solution

Problem 10.

Topic: Probability Difficulty Level: Medium

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

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Hint 3

Full Solution

Topic: Real Analysis
Difficulty level: Medium

Topic: Real Analysis
Difficulty Level: Easy

Topic: Real Analysis
Difficulty Level: Medium

Topic: Multivariable Calculus
Difficulty Level: Medium

Topic: Metric Space
Difficulty Level: Medium

Topic: Group Theory
Difficulty Level: Medium

Topic: Field Theory
Difficulty Level: Hard

Topic: Linear Algebra
Difficulty Level: Medium

Topic: Linear Algebra
Difficulty Level: Easy

Topic: Probability
Difficulty Level: Medium