Indian Statistical Institute, ISI BStat & BMath 2020 Solutions & Discussions

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ISI BMath & BSTAT 2020 Subjective Questions UGB: solutions and discussions

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Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Problem 1.

Let i be a root of the equation x^2+1=0 and let \omega be a root of the equation x^2+x+1=0. Construct a polynomial

    \[f(x)=a_0+a_1x+\dots+a_nx^n\]

where a_0, a_1,\dots, a_n are all integers such that f(i+\omega)=0.

Topic: Polynomial
Difficulty level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 1 UGB Solution

Solution:

We compute powers of i+\omega:

    \[(i+\omega)^2=i^2+2i\omega+\omega^2=-1+2i\omega-\omega-1=-2-\omega+2i\omega\]

    \[(i+\omega)^3=(i+\omega)(-2-\omega+2i\omega)=-2i-i\omega+2i^2\omega-2\omega-\omega^2+2i\omega^2=\]

    \[=-2i-i\omega-2\omega-2\omega+\omega+1-2i(\omega+1)=1-4i-3\omega-3i\omega\]

    \[(i+\omega)^4=(i+\omega)(1-4i-3\omega-3i\omega)=i-4i^2-3i\omega-3i^2\omega+\omega-4i\omega-3\omega^2-3i\omega^2=\]

    \[=i+4-3i\omega+3\omega+\omega-4i\omega+3(\omega+1)+3i(\omega+1)=7+4i+7\omega-4i\omega\]

Then

    \[a_0+a_1(i+\omega)+a_2(i+\omega)^2+a_3(i+\omega)^3+a_4(i+\omega)^4=\]

    \[=(a_0-2a_2+a_3+7a_4)+(a_1-4a_3+4a_4)i+(a_1-a_2-3a_3+7a_4)\omega+(2a_2-3a_3-4a_4)i\omega\]

Equate all coefficients to zero:

    \[\begin{cases} a_0-2a_2+a_3+7a_4=0 \\ a_1-4a_3+4a_4=0 \\ a_1-a_2-3a_3+7a_4=0 \\ 2a_2-3a_3-4a_4=0 \end{cases}\]

Find an integer solution: a_0=1, a_1=4, a_2=5,a_3=2,a_4=1.

Answer: f(x)=1+4x+5x^2+2x^3+x^4.

Video Solution:

Problem 2.

Let a be a fixed real number. Consider the equation

    \[(x+2)^2(x+7)^2+a=0, \hspace{.1cm} x\in \mathbb{R}\]

where \mathbb{R} is the set of real numbers. For what values of a, will the equation have exactly one double root?

Topic: Polynomial
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 2 UGB Solution.

Solution:

Let g(x)=(x+2)^2(x+7)^2:

 

 

    \[g'(x)=2(x+2)(x+7)^2+2(x+2)^2(x+7)=2(x+2)(x+7)(2x+9)\]

g'(x)=0 if and only if x\in \{-7,-9/2,-2\}. Values x=-7 and x=-2 are roots of the equation g(x)=0. This equation has two double roots. Value x=-9/2 is the root of the equation

    \[g(x)=g(-9/2)=(5/2)^4\]

and this is the only double root of this equation.

Answer: a=-(5/2)^4

Video Solution:

Problem 3.

Let A and B be variable points on x-axis and y-axis respectively such that the line segment AB is in the first quadrant and of a fixed length 2d. Let C be the mid-point of AB and P be a point such that

(a) P and the origin are on the opposite sides of AB and,
(b) PC is a line segment of length d which is perpendicular to AB.
Find the locus of P.

Topic: Coordinate
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 3 UGB Solution.

Solution:

Let A has coordinates (x,0), and B has coordinates (0,y). Then x^2+y^2=4d^2. Coordinates of C are (x/2,y/2). The vector

    \[\vec{AB}=(-x,y).\]

The unit orthogonal vector is \vec{e}=(\frac{y}{\sqrt{x^2+y^2}},\frac{x}{\sqrt{x^2+y^2}})=\frac{1}{2d}(y,x). So, we find

    \[\vec{PC}=d\vec{e}=\frac{1}{2}(y,x)\]

and coordinates of P are

    \[\left(\frac{x}{2}+\frac{y}{2},\frac{y}{2}+\frac{x}{2}\right)=\left(\frac{x+y}{2},\frac{x+y}{2}\right).\]

The point P is situated on the line y=x. The maximal value of \frac{x+y}{2} subject to x^2+y^2=4d^2 is achieved when x=y=\sqrt{2}d, and is equal to \sqrt{2}d. The value of \frac{x+y}{2} subject to x^2+y^2=4d^2 is >d.

Answer: The locus of P is a segment \{(t,t):d< t\leq \sqrt{2}d\}.

Video Solution:

Problem 4.

Let a real-valued sequence \{x_n\}_{n\geq 1} be such that

    \[\lim_{n \to \infty} nx_n=0\]

Find all possible real values of t such that \lim_{n \to \infty}x_n(\log n)^t=0.

Topic: Calculus
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 4 UGB Solution.

Solution:

We check that for all a>0 and all x\geq 1:

    \[\frac{(\log x)^a}{x}\leq \frac{a^a}{e^a}.\]

Indeed,

    \[\frac{d}{dx}\frac{(\log x)^a}{x}=\frac{a(\log x)^{a-1}}{x^2}-\frac{(\log x)^a}{x^2}=\frac{(\log x)^{a-1}}{x^2}\left(a-\log x\right)>\]

    \[>0 \Leftrightarrow x<e^a.\]

So, the function x\to \frac{(\log x)^a}{x} is increasing on [1,e^a] and decreasing on [e^a,\infty):

    \[\frac{(\log x)^a}{x}\leq \frac{a^a}{e^a}.\]

Let t\in \mathbb{R} and choose a>\max(t,0). Then for all n>1

    \[0<\frac{(\log n)^t}{n}=\frac{(\log n)^a}{n}\frac{1}{(\log n)^{a-t}}\leq \frac{a^a}{e^a(\log n)^{a-t}}\to 0, \ n\to\infty.\]

It follows that

    \[x_n(\log n)^t=nx_n \frac{(\log n)^t}{n}\to 0\cdot 0 =0.\]

Answer: for all t\in\mathbb{R}, \lim_{n \to \infty}x_n(\log n)^t=0.

Video Solution:

Problem 5.

Prove that the largest pentagon (in terms of area) that can be inscribed in a circle of radius 1 is regular (i.e., has equal sides).

Topic: Geometry
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 5 UGB Solution.

Solution:

Separate the inscribed pentagon into 5 isosceles triangles.

 

 

 

 

 

Let \theta_j, 1\leq j\leq 5 be angles of triangles between equal sides. The length of these sides are equal to the radius of the circle R. The area of each triangle is \frac{1}{2}R^2\sin\theta_j, so, the area of the pentagon is

    \[A=\frac{R^2}{2}\sum^5_{j=1}\sin\theta_j.\]

We have \theta_j\in (0,\pi), 1\leq j\leq 5, and the function x\to \sin(x) is strictly concave of [0,\pi]. In particular

    \[\frac{\sum^5_{j=1}\sin\theta_j}{5}\leq \sin\frac{\sum^5_{j=1}\theta_j}{5}=\sin\frac{2\pi}{5},\]

since \sum^5_{j=1}\theta_j=2\pi. So, the area of the pentagon is

    \[A\leq \frac{5}{2}\sin\frac{2\pi}{5},\mbox{ since } R=1\]

the equality is achieved only when \theta_j=\frac{2\pi}{5}, i.e. when the pentagon is regular.

Video Solution:

Problem 6.

Prove that the family of curves

    \[\frac{x^2}{a^2+\lambda} +\frac{y^2}{b^2+\lambda}=1\]

satisfies

    \[\frac{dy}{dx}(a^2-b^2)=(x+y\frac{dy}{dx})(x\frac{dy}{dx}-y).\]

Topic: Calculus: Differential
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 6 UGB Solution.

Solution:

We differentiate the equality of the family of curves:

    \[\frac{2x}{a^2+\lambda}+\frac{2y\frac{dy}{dx}}{b^2+\lambda}=0,\]

    \[y\frac{dy}{dx}=-x\frac{b^2+\lambda}{a^2+\lambda}.\]

Then

    \[x+y\frac{dy}{dx}=x\left(1-\frac{b^2+\lambda}{a^2+\lambda}\right)=x\frac{a^2-b^2}{a^2+\lambda}\]

and

    \[x\frac{dy}{dx}-y=\frac{x}{y}\cdot y\frac{dy}{dx}-y=\]

    \[=-\frac{x^2}{y}\cdot \frac{b^2+\lambda}{a^2+\lambda}-y=-\frac{b^2+\lambda}{y}\left(\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}\right)=-\frac{b^2+\lambda}{y}\]

It follows that

    \[(x+y\frac{dy}{dx})(x\frac{dy}{dx}-y)=-x\frac{a^2-b^2}{a^2+\lambda}\cdot \frac{b^2+\lambda}{y}=\]

    \[=(a^2-b^2)\left(-\frac{x}{y}\cdot \frac{b^2+\lambda}{a^2+\lambda}\right)=\frac{dy}{dx}(a^2-b^2).\]

Video Solution:

Problem 7.

Consider a right-angled triangle with integer-valued sides a<b<c where a,b,c are pairwise co-prime. Let d=c-b. Suppose d divides a. \\
Then

(a) Prove that d\leq 2
(b) Find all such triangles (i.e., all possible triplets a,b,c) with perimeter less than 100.

Topic: Number Theory
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 7 UGB Solution.

Solution 1:

By the Pythagoras’ Theorem, c^2=a^2+b^2.

    \[a^2=c^2-b^2=(c-b)(c+b)=d(c+b)\]

d^2 divides a^2=d(c+b), hence d|c+b.

    \[c+b=kd, \ d+2b=kd, \ 2b=(k-1)d.\]

Let p\geq 3 be a prime factor of d. Then p divides d, p divides b, consequently p divides d+b=c. But b and c are co-prime, and we get a contradiction. So, the only possible prime factor of d is 2. The same reasoning show that the power of 2 in the prime factorization of d is at most 1. We have proved that either d=1 or d=2.

Let d=1. Then c=b+1.

    \[b^2+2b+1=c^2=a^2+b^2, \ a^2=2b+1.\]

It follows that a is odd, a=2k+1,

    \[4k^2+4k+1=2b+1, \ b=2k^2+2k.\]

We have

    \[a=2k+1, b=2k^2+2k, c=2k^2+2k+1.\]

The perimeter is

    \[2+6k+4k^2<100, \ 1\leq k\leq 4.\]

Corresponding triangles are

    \[(3,4,5),(5,12,13),(7,24,25),(9,40,41).\]

Let d=2. Thne c=b+2.

    \[b^2+4b+4=c^2=a^2+b^2, \ a^2=4b+4.\]

It follows that a is even, a=2k.

    \[4k^2=a^2=4b+4, b=k^2-1, c=k^2+1.\]

The perimeter is

    \[2k+2k^2<100, 2\leq k\leq 6.\]

For k=2 we have a>b. If k is odd, then b is even and a and b are not co-prime. So, in fact k=4,6.
Corresponding triangles are

    \[(8,15,17), (12,35,37).\]

Answer: there are six triangles that satisfy the condition:

    \[(3,4,5),(5,12,13),(7,24,25),(9,40,41), (8,15,17), (12, 35,37).\]

Video Solution:

Problem 8.

A finite sequence of numbers (a_1, \dots, a_n) is said to be alternating if

    \[a_1>a_2,\hspace{.2cm} a_2<a_3,\hspace{.2cm} a_3>a_4,\hspace{.2cm} a_4<a_5, \dots\]

    \[\mbox{or } a_1<a_2,\hspace{.2cm} a_2>a_3,\hspace{.2cm} a_3<a_4,\hspace{.2cm} a_4>a_5, \dots\]

How many alternating sequences of length 5, with distinct numbers a_1,a_2, \dots, a_5 can be formed such that a_i \in \{1,2,\dots,20\} for i=1, \dots, 5?

Topic: Combinatorics
Difficulty Level: Hard

Indian Statistical Institute, ISI Subjective BStat & BMath 2020 Problem 8 UGB Solution.

Solution :

We will say that a permutation a_1,\ldots,a_n of the set \{1,2,\ldots,n\} is an “up-down” permutation, if

    \[a_1<a_2,a_2>a_3,a_3<a_4,\ldots.\]

We will say that a permutation a_1,\ldots,a_n of the set \{1,2,\ldots,n\} is a “down-up” permutation, if

    \[a_1>a_2,a_2<a_3,a_3>a_4,\ldots.\]

Correspondence (a_1,\ldots,a_n)\to (n+1-a_1,\ldots,n+1-a_n) is a one-to-one correspondence between the set of “up-down” permutations and the set of “down-up” permutations. Let A_n denote the number of “up-down” permutation. Then 2A_n is the number of all alternating permutations of \{1,2,\ldots,n\}. Consider an alternating permutation (a_1,\ldots,a_n). Let a_k=n, so that a_{k-1}<a_k, a_k>a_{k+1}. If k is even, then (a_1,\ldots,a_{k-1}) is an “up-down” permutation of k-1 numbers (there are A_{k-1} such permutations), and (a_{k+1},\ldots,a_n) is an “up-down” permutation of n-k numbers (there are A_{n-k} such permutations). If k is odd, then (a_1,\ldots,a_{k-1}) is a “down-up” permutation of k-1 numbers (there are A_{k-1} such permutations), and (a_{k+1},\ldots,a_n) is a “up-down” permutation of n-k numbers (there are A_{n-k} such permutations). There are {n-1\choose k-1} ways to split numbers \{1,2,\ldots,n-1\} into groups by k-1 and n-k elements. Hence,

    \[2A_n=\sum^n_{k=1}{n-1 \choose k-1}A_{k-1}A_{n-k}.\]

Since A_0=A_1=1, we compute

    \[2A_2=2, A_2=1;\]

    \[2A_3=2A_2+2A^2_1=4, A_3=2;\]

    \[2A_4=2A_3+6A_1A_2=10, A_4=5;\]

    \[2A_5=2A_4+8A_1A_3+6A^2_2=32.\]

There are 32 permutations of the set of five numbers.

Additionally we have to choose 5 numbers from the set \{1,2,\ldots,20\}.

Answer: 32{20\choose 5}=496128.

Video Solution:

5 Comments

  • Souradeep Das
    September 26, 2020 at 5:46 pm Reply

    The sequence given is alternating so let us take ak=p such that k is even ,then for all ai will be less than p for i is even and for all aj will be greater than p for j is odd .
    Then comes the sequence of length L and there will be [L/2] no of ais(i=even) in the sequence and [L/2] or [L/2]+1]( even and odd respectively )
    Then the partition of the set {1,2…,n} in suchall ai way such ai will be chosen from {1,2..p} and aj from {p+1…n}
    And that p will range from [L/2] upto n – ([L/2]+1)
    The bound is there for ai and ajs
    By combinatorial arguemnt, since L is odd ,we will have [L/2]+ 1 no of ajs

    Summation i = [L/2] upto n-([L/2]+1) (i choose [L/2] ) (20-i choose [L/2]+1) ([L/2])!. ([L/2]1)!

    Regret for making the solution look messy ,but it would have looked better in copy format ,this is something i tried to derive ,might be wrong, thank in advanced for checking for my solution.

  • Supratim santra
    March 3, 2021 at 11:33 am Reply

    Hello Arkabrata da i am going to sit for M math examination this year . Would you kind enough to tell me about mock test series from RSM ? I want to buy them
    Please reply soon
    As early as possible

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