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 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

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.

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

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

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.

Full Solution

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.

 

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

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

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

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

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 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 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

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.

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 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 2

Full Solution

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

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

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 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

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

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\}.

 

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

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

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

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

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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24 Comments

  • 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.

  • 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

  • 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.

  • Hi!
    Can you please conduct few mock tests for ISI M.MATH 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

  • 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 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}

  • 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}

  • #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, 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)

  • 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.

  • 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

  • Great work…
    Can you provide your what’s app group?

  • 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

  • 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

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