Indian Statistical Institute, ISI MMATH 2023 PMB Solutions & Discussions

ISI MMath PMB 2023 Subjective Questions, solutions and discussions

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Problem 1.

Let S=\left\{(x, y, z) \in \mathbb{R}^3 \mid x, y, z>0, x y+y z+z x=1\right\}. Prove that there exists (a, b, c) \in S such that abc \geq x y z for all (x, y, z) \in S.

Topic: Set Theory, Inequalitites

Difficulty level: Easy

Solution:

Let (x,y,z)\in S. By the AM-GM inequality

    \[(xyz)^{2/3}=\sqrt[3]{(xyz)^2}=\sqrt[3]{xyyzzx}\leq \frac{xy+yz+zx}{3}=\frac{1}{3}.\]

Hence, for all (x,y,z)\in S

    \[xyz\leq \frac{1}{3^{3/2}}.\]

Equality in the AM-GM inequality \sqrt[n]{a_1\ldots a_n}\leq \frac{a_1+\ldots+a_n}{n} is achieved if and only if all numbers a_1,\ldots,a_n are equal. Hence, to get the maximal possible product xyz among all (x,y,z)\in S we need to look for (x,y,z)\in S with identical coordinates. If x=y=z>0 and xy+yz+zx=1, then 3x^2=1 and x=\frac{1}{3^{1/2}}. Let a=b=c=\frac{1}{3^{1/2}}. Since

    \[ab+bc+ca=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1,\]

we have (a,b,c)\in S. Further,

    \[abc=\frac{1}{3^{3/2}}\geq xyz\]

for all (x,y,z)\in S.

Problem 2.

Prove that x^n is a solution of the differential equation

    \[y^{\prime \prime}+P(x) y^{\prime}+Q(x) y=0\]

on [-1,1] for some nonzero real polynomials P(x), Q(x) if and only if n=1.

Topic: Differential Equations

Difficulty Level: Medium

Solution:

In this solution n is assumed to be an integer.} Since y=x^n must be twice differentiable on [-1,1], we need to have n\geq 0. If n=0, then

    \[y''+P(x)y'+Q(x)y=Q(x).\]

If Q(x)=0 for x\in[-1,1] then Q(x)=0 for all x\in \mathbb{R}, which is not the case.

Let n=1. If we take y=x, P(x)=x and Q(x)=-1, then

    \[y''+P(x)y'+Q(x)y=x-x=0.\]

So, y=x is a solution to the equation of interest.

Assume that y=x^n, n\geq 2 is a solution of

    \[y''+P(x)y'+Q(x)y=0, \ x\in [-1,1].\]

Then

    \[n(n-1)x^{n-2}+nx^{n-1}P(x)+x^nQ(x)=0, \ x\in [-1,1].\]

The function R(x)=n(n-1)x^{n-2}+nx^{n-1}P(x)+x^nQ(x) is a polynomial that is equal to zero on the interval [-1,1]. It follows that all coefficients of R are equal to zero. However, the coefficient of R(x) near x^{n-2} equals n(n-1)\ne 0. The obtained contradiction shows that the only function y=x^n which is a solution of the equation

    \[y''+P(x)y'+Q(x)y=0, \ x\in [-1,1],\]

is y=x (i.e. when n=1).

Problem 3.

Let A=\left\{(x, y, z) \in \mathbb{R}^3 \mid x^2+y^2=1\right\} and B=\mathbb{R}^2 \backslash\{(0,0)\}. Consider A and B as metric spaces via their inclusions in \mathbb{R}^3 and \mathbb{R}^2 respectively. Construct a bijection f: A \rightarrow B such that both, f and f^{-1} are contimuous.

Topic: Metric Space

Difficulty Level: Medium

Solution:

Consider the mapping f:A\to B defined by

    \[f(x,y,z)=(e^zx,e^zy).\]

This mapping is well defined, as

    \[|f(x,y,z)|=e^z>0, \ f(x,y,z)\ne (0,0).\]

The mapping f is continuous, as functions (x,y,z)\mapsto e^zx and (x,y,z)\mapsto e^zy are continuous. The mapping g:B\to A defined by

    \[g(u,v)=\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}},\ln \sqrt{u^2+v^2}\right)\]

is also continuous (as u^2+v^2>0 for (u,v)\in B). In fact, g=f^{-1}. Indeed,

    \[g(f(x,y,z))=g\left(e^zx,e^zy\right)=\]

since e^{2z}x^2+e^{2z}y^2=e^{2z}

    \[=\left(\frac{e^zx}{\sqrt{e^{2z}}},\frac{e^zy}{\sqrt{e^{2z}}},\ln\sqrt{e^{2z}}\right)=(x,y,z),\]

and

    \[f(g(u,v))=f\left(\frac{u}{\sqrt{u^2+v^2}},\frac{v}{\sqrt{u^2+v^2}},\ln \sqrt{u^2+v^2}\right)=\]

    \[=\left(e^{\ln \sqrt{u^2+v^2}}\frac{u}{\sqrt{u^2+v^2}}, e^{\ln \sqrt{u^2+v^2}}\frac{v}{\sqrt{u^2+v^2}}\right)=(u,v).\]

So, the mapping f is the needed mapping.

Problem 4.

Let n \geq k \geq 2. A system consists of n components, each of which functions independently with probability 1 / 2. The system is said to function if exactly k of the n components function. Compute the conditional probability that at least one of the first two components function given that the system functions.

Topic: Probability

Difficulty Level: Easy

Solution:

The system functions if and only if k of the n components function, and other n-k components do not function. There are {n\choose k} ways to choose k components that function, hence the probability that the system functions equals {n\choose k}2^{-n}. The system functions while first two components do not function if and only if there are exactly k components among components 3,4,\ldots,n that function. So, the probability that the system functions and first two components do not function equals
{n-2\choose k}2^{-n}. It follows that the probability that first two components do not function conditionally on the event that the system functions equals

    \[\frac{{n-2\choose k}2^{-n}}{{n\choose k}2^{-n}}=\frac{{n-2\choose k}}{{n\choose k}}.\]

Correspondigly, the probability that at least one of the first two components function given that the system functions equals

    \[1-\frac{{n-2\choose k}}{{n\choose k}}.\]

Problem 5.

Let T: \mathbb{R}^d \rightarrow \mathbb{R}^d be a linear transfarmation where d is a positive integer. Prove that T^m\left(\mathbb{R}^d\right)=T^d\left(\mathbb{R}^d\right) for all m \geq d.

Topic: Linear Algebra

Difficulty Level: Medium

Solution:

Denote by T^0 the identical operator. Since T^{m+1}(\mathbb{R}^d)=T^m(T(\mathbb{R}^d))\subset T^m(\mathbb{R}^d), we have that

    \[T^0(\mathbb{R}^d)\supset T^1(\mathbb{R}^d)\supset \ldots \supset T^d(\mathbb{R}^d)\supset T^{d+1}(\mathbb{R}^d).\]

Let D_n denote the dimension of the subspace T^n(\mathbb{R}^d). Then

    \[d=D_0\geq D_1\geq \ldots \geq D_d\geq D_{d+1}\geq 0.\]

If all inequalities are strict, then

    \[D_1\leq d-1, D_2\leq d-2, \ldots, D_{d+1}\leq d-d-1\leq -1,\]

which is impossible. So, there exists n\in \{0,\ldots,d\} such that D_n=D_{n+1}, hence T^n(\mathbb{R}^d)=T^{n+1}(\mathbb{R}^d). We will verify by induction that for all k\geq 0 T^{n+k}(\mathbb{R}^d)=T^n(\mathbb{R}^d). For k=0 this is obvious. If T^{n+k}(\mathbb{R}^d)=T^n(\mathbb{R}^d), then

    \[T^{n+k+1}(\mathbb{R}^d)=T(T^{n+k}(\mathbb{R}^d))=T(T^{n}(\mathbb{R}^d))=T^{n+1}(\mathbb{R}^d)=T^{n}(\mathbb{R}^d),\]

where the latter equality follows from the choice of n. So, for all m\geq d(\geq n),

    \[T^{m}(\mathbb{R}^d)=T^{n+m-n}(\mathbb{R}^d)=T^{n}(\mathbb{R}^d)=T^{n+d-n}(\mathbb{R}^d)=T^{d}(\mathbb{R}^d).\]

Problem 6.

Let B be a square matrix with values in \{-1,+1,0\} such that every column has at most one +1 and at most one -1 . Show that \operatorname{det}(B) \in\{-1,+1,0\}.

Topic: Linear Algebra

Difficulty Level: Medium

Solution:

Denote by n the size of B, i.e. B is n\times n matrix. We will prove the result by induction on n. If n=1, then

    \[B=\begin{bmatrix} b_{11}\end{bmatrix}\]

and \det B=b_{11}\in \{-1,0,+1\}. Assume the needed statement is true for n-1 and let

    \[B=\begin{bmatrix} b_{11} & b_{12} & \ldots & b_{1n} \\ b_{21} & b_{22} & \ldots & b_{2n} \\ \ldots & \ldots & \ldots & \ldots \\ b_{n1} & b_{n2} & \ldots & b_{nn} \end{bmatrix}.\]

By assumption, each column has at most one +1 and at most one -1. Consider following cases.

Case 1. There exists a column with n-1 zero elements. In other words, either all elements of the column are equal to zero, or there exists one element which is equal to +1 or to -1. Then there are indices k,l\in \{1,\ldots,n\} such that

    \[b_{kl}\in \{-1,0,+1\}, \ b_{il}=0, \ i\ne k.\]

Expanding the determinant with respect to l-th column we find that

    \[\det B=\pm b_{kl} \det C,\]

where the matrix C is obtained by deleting k-th row and j-th column from the matrix B. Every column of the matrix C is contained in a column of the matrix B. Hence, every column of the matrix C contains at most one +1 and at most one -1. By inductive assumption, \det C\in \{-1,0,+1\} and \det B\in \{-1,0,+1\}.

Case 2. Every column of the matrix B contains exactly one +1 and exactly one -1. Consider the sum of all rows of the matrix B. For each j\in \{1,\ldots,n\}

    \[\sum^n_{i=1}b_{ij}=0.\]

Indeed, in the latter sum there is one +1, one -1, and n-2 zeros. Hence, rows of the matrix B are linearly dependent and \det B=0.

In any case, \det B\in \{-1,0,+1\}.

Problem 7.

Let G be a finite group having an odd number of elements. Suppose a is an element of G of order 3 such that the cyclic subgroup \langle a\rangle generated by a is normal in G. Prove that a commutes with every element of G.

Topic: Group Theory

Difficulty Level: Easy

Solution:

Let g\in G. Since \langle a\rangle is a normal subgroup of G,

    \[g\langle a \rangle=\langle a \rangle g.\]

In particular, ga\in \{g,ag,a^2g\}.

If ga=g, then a=e, which is not the case.

Assume that ga=a^2g. The order of g is an odd number. Denote it by 2n+1. Then

    \[a=g^{2n}a^2g.\]

We will prove by induction that for k=0,\ldots,n, a=g^{2(n-k)}a^2g^{1+2k}. For k=0 this was already shown. Assume the result is proved for k<n. Then

    \[\begin{aligned} a&=g^{2(n-k)}a^2g^{1+2k}=g^{2(n-k)-1}(ga)ag^{1+2k}=g^{2(n-k)-1}a^2(ga)g^{1+2k} \\ &=g^{2(n-k)-1}a^4g^{2+2k}=g^{2(n-k)-1}ag^{2+2k}=g^{2(n-(k+1))}(ga)g^{2+2k}\\ &=g^{2(n-(k+1))}a^2g^{3+2k}=g^{2(n-(k+1))}a^2g^{1+2(k+1)} \end{aligned}\]

Hence, the result is true for all k=0,\ldots,n. Taking k=n we get

    \[a=a^2g^{1+2n}=a^2, \ a=e,\]

which is impossible. So, ga\ne a^2 g.

Hence, ga=ag and a commutes with all elements of G.

Problem 8.

Let p, q be odd primes such that p-1 divides q-1. If n is a positive integer coprime to p q, prove that p q divides n^{q-1}-1.

Topic: Number Theory

Difficulty Level: Easy

Solution:

Remark. We need to assume that p<q. If p=q, then the statement is false. For example, we can take p=q=3, n=2. Then n^{q-1}-1=2^2-1=3 is not divisible by pq=9.}

By Fermat’s little theorem q divides n^{q-1}-1. So, it is enough to prove that p divides n^{q-1}-1. Since p-1 divides q-1, it follows that q-1=k(p-1) for some integer k\geq 2. Hence

    \[n^{q-1}-1=n^{k(p-1)}-1=(n^{p-1}-1)\left(n^{(k-1)(p-1)}+n^{(k-2)(p-1)}+\ldots +1\right).\]

It follows that n^{p-1}-1 divides n^{q-1}-1. By Fermat’s little theorem p divides n^{p-1}-1. Hence, p divides n^{q-1}-1 and pq divides n^{q-1}-1.

Indian Statistical Institute, ISI, M.Math 2021 PMA Objective Questions Solutions and Discussions: Click Here

To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

To view ISI MMATH 2018 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