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

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

ISI BMath & BSTAT 2020 Subjective Questions UGB: 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, then only look at the solution of 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 like my videos and subscribe me on youtube to get more such videos in the future. 

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

Leave a Reply