Indian Statistical Institute, ISI BStat & BMath 2021 UGB Solutions & Discussions

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ISI BMath & BSTAT 2021 Subjective Questions UGB: solutions and discussions

Indian Statistical Institute,ISI BMATH/BSTAT 2023 All UGA Objective Solutions: Click Here

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.

There are three cities each of which has exactly the same number of citizens, say $n$ . Every. citizen in each city has exactly a total of $n+1$ friends in the other two cities. Show that there exist three people, one from each city, such that they are friends. We assume that friendship is mutual (that is, a symmetric relation).

Topic: The Extremal Principle
Difficulty level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 1 UGB Solution

Solution:

From the $3n$ citizens, take one who has a maximum number $k$ of acquaintances from one of the two other cities. Suppose it is citizen $A$ from the first city, who knows $k$ citizens from the second city. Then $A$ knows $(n+1-k)$ citizens from the third city, $n+1-k \geq 1$ since $k \leq n$ . Consider citizen $B$ from the third city, who knows $A$ . If $B$ knows at least one citizen $C$ from the $k$ acquaintances of $A$ in the second city, then $\{A, B, C\}$ is a triple of mutual acquaintances. But if $B$ knows none of the $k$ acquaintances of $A$ in the second city, then, in this city he does not know more than $(n-k)$ citizens, and hence, in the first city, he does not know less than $n+1-(n-k)=k+1$ citizens, which contradicts the choice of $k$ .

Video Solution:

Problem 2.

Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be a function satisfying $f(0) \neq 0=f(1)$ . Assume also that $f$ satisfies equations (A) and (B) below.

$\begin{aligned} f(x y) &=f(x)+f(y)-f(x) f(y) \\ f(x-y) f(x) f(y) &=f(0) f(x) f(y) \end{aligned}$

for all integers $x, y$ .
(i) Determine explicitly the set $\{f(a): a \in \mathbb{Z}\}$ .
(ii) Assuming that there is a non-zero integer $a$ such that $f(a) \neq 0$ , prove that the set $\{b: f(b) \neq 0\}$ is infinite.

Topic: Functional Equations
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 2 UGB Solution.

Solution:

Let $x=y=0$ in (A):

$f(0)=f(0)+f(0)-f(0)^2, f(0)^2=f(0).$

Since $f(0)\ne 0$ we deduce $f(0)=1.$

Let $y=0$ in (B):

$f(x)^2f(0)=f(x)f(0)^2, f(x)^2=f(x).$

Either $f(x)=0$ or $f(x)=1.$

The function $f$ takes only two values $0$ and $1$ (both values are really in the range of $f$ , because $f(0)=1,$ $f(1)=0$ ).

$\{f(a):a\in\mathbb{Z}\}=\{0,1\}.$

Assume that there exist a non-zero $a\in\mathbb{Z}$ such that $f(a)\ne 0.$ Then $a\ne 1.$ Take $x=y=-1$ in (A):

$0=f(1)=2f(-1)-f(-1)^2.$

It follows that $f(-1)\ne 1,$ hence $f(-1)=0$ and $a\ne -1.$ So, $|a|>1.$ Further, from (A) we find that if $f(a)=f(b)=1,$ then

$f(ab)=f(a)+f(b)-f(a)f(b)=1.$

By induction, $f(a^k)=1$ for all $k\geq 1$ and the set
$\{b:f(b)\ne 0\}$ is infinite.

Video Solution:

Problem 3.

Prove that every positive rational number can be expressed uniquely as a finite sum of the form

$a_{1}+\frac{a_{2}}{2 !}+\frac{a_{3}}{3 !}+\cdots+\frac{a_{n}}{n !},$

where $a_{n}$ are integers such that $0 \leq a_{n} \leq n-1$ for all $n>1$ .

Topic: Number Theory
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 3 UGB Solution.

Solution:

Let $r$ be a positive rational number. Define $a_1=[r],$ so that

$0\leq r-a_1<1.$

Assume that $a_1,\ldots,a_m$ are defined in such a way that

$0\leq r-\sum^m_{k=1}\frac{a_k}{k!}<\frac{1}{m!}.$

Define

$a_{m+1}=\left[(m+1)!\left(r-\sum^m_{k=1}\frac{a_k}{k!}\right)\right].$

Since

$0\leq (m+1)! \left(r-\sum^m_{k=1}\frac{a_k}{k!}\right)<m+1,$

we necessarily have $a_{m+1}\in \{0,\ldots,m\}.$ So, there exists a sequence of integers $(a_n:n\geq 1)$ such that

$a_1\geq 0, 0\leq a_n\leq n-1, \ n>1$

and for all $n\geq 1$

$0\leq r-\sum^n_{k=1}\frac{a_k}{k!}<\frac{1}{n!}.$

For large enough $n$ we have $n!r\in \mathbb{Z},$ so

$n! \left(r-\sum^n_{k=1}\frac{a_k}{k!}\right)\in [0,1)\cap \mathbb{Z}.$

The latter is only possible when

$r=\sum^n_{k=1}\frac{a_k}{k!}.$

Now assume that there are two representations

$\sum^n_{k=1}\frac{a_k}{k!}=\sum^n_{k=1}\frac{b_k}{k!}$

with integer coefficients $a_k,b_k$ such that $a_1,b_1\geq 0,$ $a_k,b_k\in \{0,\ldots,k-1\}$ for $k>1.$ Let $j$ be the first index such that $a_j\ne b_j.$ By symmetry we can assume that $a_j<b_j.$ Then

$\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{a_k}{k!}=\frac{b_j}{j!}+\sum^n_{k=j+1}\frac{b_k}{k!}.$

Consider inequalities

$\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{a_k}{k!}\leq \frac{a_j}{j!}+\sum^n_{k=j+1}\frac{k-1}{k!}=$

$=\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{1}{(k-1)!}-\sum^n_{k=j+1}\frac{1}{k!}=$

$=\frac{a_j}{j!}+\frac{1}{j!}-\frac{1}{n!}<\frac{a_j+1}{j!}\leq \frac{b_j}{j!}+\sum^n_{k=j+1}\frac{b_k}{k!}.$

Hence, the equality $\sum^n_{k=1}\frac{a_k}{k!}=\sum^n_{k=1}\frac{b_k}{k!}$ is impossible.

Video Solution:

Problem 4.

Let $g:(0, \infty) \rightarrow(0, \infty)$ be a differentiable function whose derivative is continuous, and such that $g(g(x))=x$ for all $x>0 .$ If $g$ is not the identity function, prove that $g$ must be strictly decreasing.

Topic: Calculus
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 4 UGB Solution.

Solution:

We differentiate the relation $g(g(x))=x:$

$g'(g(x))g'(x)=1.$

$g'(x)\ne 0$ for all $x>0,$ hence $g'$ preserves the sign on $(0,+\infty).$ Assume that $g'(x)>0$ for all $x>0.$ Then $g(x)$ is strictly increasing. For some $x$ we have $g(x)\ne x$ (as $g$ is not the identity function). If $g(x)>x,$ then

$g(g(x))>g(x)>x,$

but this contradicts $g(g(x))=x.$ If $g(x)<x,$ then

$g(g(x))<g(x)<x,$

but this also contradicts $g(g(x))=x.$ So, the assumption $g'(x)>0$ leads to a contradiction. Then $g'(x)<0$ for all $x>0$ and $g$ is strictly decreasing.

Video Solution:

Problem 5.

Let $a_{0}, a_{1}, \cdots, a_{19} \in \mathbb{R}$ and

$P(x)=x^{20}+\sum_{i=0}^{19} a_{i} x^{i}, \quad x \in \mathbb{R} .$

If $P(x)=P(-x)$ for all $x \in \mathbb{R}$ , and

$P(k)=k^{2}, \text { for } k=0,1,2 \cdots, 9$

then find

$\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}$

Topic: Calculus
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 5 UGB Solution.

Solution:

The relation $P(x)=P(-x)$ implies that

$x^{20}+\sum^{19}_{i=0}a_i x^i=x^{20}+\sum^{19}_{i=0}(-1)^ia_i x^i,$

$\sum^{19}_{i=0} (1-(-1)^i)a_i x^i=0.$

All coefficients $a_i$ that correspond to odd indices $i$ are equal to zero. So,

$P(x)=x^{20}+\sum^{9}_{i=0}a_{2i}x^{2i}.$

Consider polynomial $Q(x)=x^{10}+\sum^{9}_{i=0}a_{2i}x^{i}-x.$ By assumption,

$Q(k^2)=k^{20}+\sum^9_{i=0}a_{2i} k^{2i}-k^2=P(k)-k^2=0, 0\leq k\leq 9.$

Observe that $a_2-1$ is the coefficient near $x$ in $Q(x).$ $10$ roots of the polynomial $Q$ are known:

$Q(x)=x (x-1)(x-2^2)\ldots (x-9^2).$

In particular, $a_0=0$ and

$a_2-1=-\prod^9_{k=1} k^2, \ a_2=1-(9!)^2=-362879.$

Then $P(x)=-362879x^2+\sum^{19}_{i=3}a_i x^i+x^{20}$ and

$\lim_{x\to 0}\frac{P(x)}{\sin^2 x}=\lim_{x\to 0}\frac{x^2}{\sin^2 x}\lim_{x\to 0}\frac{P(x)}{x^2}=-362879.$

Video Solution:

Problem 6.

If a given equilateral triangle $\Delta$ of side length $a$ lies in the union of five equilateral triangles of side length $b$ , show that there exist four equilateral triangles of side length $b$ whose union contains $\Delta$ .

Topic: Geometry
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 6 UGB Solution.

Solution:

We show that necessarily $b\geq \frac{a}{2}.$ Assume that $b<\frac{a}{2}.$ Consider an equilateral triangle $\Delta_b$ of side length $b.$ The distance between any two points in $\Delta_b$ is $\leq b.$ Now consider an equilateral triangle $\Delta_a$ of side length $a.$ Consider the following configuration:

There are six points in the triangle with pairwise distances $\geq \frac{a}{2}>b.$ Hence, if $b<\frac{a}{2}$ then one needs at least six equilateral triangles of side length $b$ to cover an equilateral triangle of side length $a.$ So, $b\geq \frac{a}{2}$ and the configuration above shows that $4$ equilateral triangles of side length $b$ are enough to cover an equilateral triangle of side length $a.$

Video Solution:

Problem 7.

Let $a, b, c$ be three real numbers which are roots of a cubic polynomial, and satisfy $a+b+c=6$ and $a b+b c+a c=9 .$ Suppose $a<b<c .$ Show that

$0<a<1<b<3<c<4$

Topic: Polynomials
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 7 UGB Solution.

Solution 1:

We have

$P(x)=(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(ab+ac+bc)-abc=x^3-6x^2+9x-abc.$

The derivative of $P(x):$

$P'(x)=3x^2-12x+9=3(x-1)(x-3).$

The polynomial $P$ is increasing on $(-\infty,1],$ decreasing on $[1,3],$ increasing on $[3,\infty).$ Then necessarily

$a<1<b<3<c.$

Further,

$P(3)=27-54+27-abc=-abc<0,$

i.e. $abc>0$ and $a>0;$ and

$P(1)=4-abc>0, abc<4.$

Finally,

$P(4)=64-24+36-abc=76-abc>72>0$

and $c<4.$

Video Solution:

Problem 8.

A pond has been dug at the Indian Statistical Institute as an inverted truncated pyramid with a square base (see figure below). The depth of the pond is $6 \mathrm{~m}$ . The square at the bottom has side length $2 \mathrm{~m}$ and the top square has side length $8 \mathrm{~m} .$ Water is filled in at a rate of $\frac{19}{3}$ cubic meters per hour. At what rate is the water level rising exactly 1 hour after the water started to fill the pond?

Topic: Calculus
Difficulty Level:

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 8 UGB Solution.

Solution :

Denote by $V(t)$ the volume of the water after $t$ hours the water started to fill the pond. Then $V(t)=\frac{19}{3}t$ (cubic meters). Let $h(t)$ be the level of the water and $a(t)$ be the side of the square at the top of the water after $t$ hours the water started to fill the pond.

Extend the truncated pyramid to get the non-truncated pyramid and let $x$ be the height of the added part. Then we have the following picture

Similarity of triangles implies relations

$\frac{x}{x+h(t)}=\frac{2}{a(t)}$

$\frac{x+h(t)}{x+6}=\frac{a(t)}{8}.$

Multiplying them we find

$\frac{x}{x+6}=\frac{1}{4}, \ x=2.$

Hence,

$a(t)=h(t)+2.$

Now,

$V(t)=\frac{h(t)}{3}(4+a(t)^2+2a(t))=\frac{h(t)}{3}(4+h(t)^2+4h(t)+4+2h(t)+4)=$

$=\frac{h(t)}{3}(h(t)^2+6h(t)+12)=\frac{1}{3}((h(t)+2)^3-8).$

Differentiate this:

$V'(t)=(h(t)+2)^2h'(t).$

Then

$h'(1)=\frac{V'(1)}{(h(1)+2)^2}=\frac{V'(1)}{(3V(1)+8)^{2/3}}=\frac{\frac{19}{3}}{9}=\frac{19}{27} \mbox{ (meters per hour)}$

Video Solution:

9 Comments

Abhinav

July 20, 2021 at 8:29 pm Reply

In two questions I messed up in the last line where we finally write the answer .
Like instead of 19/27 or 19/3(3)^2 in the last question , is wrote 19/3(3) forgot the square , other wise all was fine until the last line …..
And I’m polynomial question , I wrote 1-(9!)^2 instead of plus sign in last again , how much marks would that cost , can I get a 9/10 probably ,so email me
- Shubhrojyoti Dhara
  
  July 21, 2021 at 9:40 am Reply
  
  I feel sir made a mistake out there because the constant term of Q(x) is indeed -(9!)^2 as putting 0 in (x-1)(x-4)(x-9)……(x-81)….would give us (-1)(-4)(-9)….(-81) = -(9!)^2
  - FractionsHub
    
    July 22, 2021 at 8:55 pm Reply
    
    Yes, there were some calculation errors. Thanks for notifying me. Corrected.
    - But sir the polynomial should be x^2(x^2 - 1^2)(x^2-2^2)...(x^2-9^2)+x^2. So the answer should be 1-(9!)^2
      
      July 23, 2021 at 9:42 am Reply
      
      But sir the polynomial should be x^2(x^2 – 1^2 )(x^2 – 2^2)… (x^2 – 9^2)+x^2.so the answer should be 1-(9!)^2.
      ( I have not given the exam this year, I read in class 12and will give isi entrance on 2022 though I have solved this in home and got this expression of the polynomial as result. Please inform me if it is wrong
Ripunjay Dwivedi

July 22, 2021 at 12:10 pm Reply

sir, in the 4th question, since, g(g(x)) = x, g(x) is a function which is its own inverse. there are three such operations which have these property, which is, they cancel their own effect when acted an even number of times, which are as follows:
(i) the identity function : doing nothing to the input.
(ii) g(x) = -x : multiplication by -1 is cancelled by again multiplying it by -1.
and (iii) taking the reciprocal, i.e., g(x) = 1/x or a/x.
since, g(x) is defined from (o, infinity) to (o, infinity) and is not the identity function, it is the reciprocal i.e., g(x) = a/ x , for some positive, a.
so its derivative is -a/x^2 which is always negative as x^2 and a are always positive. so, the function g(x) is strictly decreasing.

is this correct, sir?
But sir the polynomial should be x^2(x^2 - 1^2)(x^2-2^2)...(x^2-9^2)+x^2. So the answer should be 1-(9!)^2

July 23, 2021 at 9:44 am Reply

But sir the polynomial should be x^2(x^2 – 1^2 )(x^2 – 2^2)… (x^2 – 9^2)+x^2.so the answer should be 1-(9!)^2.
( I have not given the exam this year, I read in class 12and will give isi entrance on 2022 though I have solved this in home and got this expression of the polynomial as result. Please inform me if it is wrong
IR08 Milan Paul

July 28, 2021 at 7:14 pm Reply

Sir for problem 5, if we do not write the final answer(-362879) how much mark will be deduced?
khushal

December 1, 2021 at 11:55 am Reply

In question number 5,what concept did you use and it from what book,i have read enough but this was out of the blue.
thank you in advance
- FractionsHub
  
  December 28, 2021 at 12:05 pm Reply
  
  You can’t get everything in a book. You have to come up with some ideas too. Anyways this is basic polynomial stuff.

Indian Statistical Institute, ISI BStat & BMath 2021 UGB Solutions & Discussions

ISI BMath & BSTAT 2021 Subjective Questions UGB: solutions and discussions

Indian Statistical Institute,ISI BMATH/BSTAT 2023 All UGA Objective Solutions: Click Here

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.

Topic: The Extremal Principle Difficulty level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 1 UGB Solution

Solution:

Video Solution:

Problem 2.

Topic: Functional Equations Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 2 UGB Solution.

Solution:

Video Solution:

Problem 3.

Topic: Number Theory Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 3 UGB Solution.

Solution:

Video Solution:

Problem 4.

Topic: Calculus Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 4 UGB Solution.

Solution:

Video Solution:

Problem 5.

Topic: Calculus Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 5 UGB Solution.

Solution:

Video Solution:

Problem 6.

Topic: Geometry Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 6 UGB Solution.

Solution:

Video Solution:

Problem 7.

Topic: Polynomials Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 7 UGB Solution.

Solution 1:

Video Solution:

Problem 8.

Topic: Calculus Difficulty Level:

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 8 UGB Solution.

Solution :

Video Solution:

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Topic: The Extremal Principle
Difficulty level: Medium

Topic: Functional Equations
Difficulty Level: Medium

Topic: Number Theory
Difficulty Level: Medium

Topic: Calculus
Difficulty Level: Easy

Topic: Calculus
Difficulty Level: Medium

Topic: Geometry
Difficulty Level: Medium

Topic: Polynomials
Difficulty Level: Easy

Topic: Calculus
Difficulty Level: