# Indian Statistical Institute, ISI BStat & BMath 2021 UGA Solutions & Discussions

## Problem 1

The number of ways one can express as a product of two numbers and , where , and , is
(A) .
(B) .
(C) .
(D) .

## Solution:

Denote If then each and where

Since we necessarily have or for each To construct the pair we have to choose either or for each Hence, there are choices. In the representation numbers and are distinct. Since and because of the condition only half of choices are satisfying. So we need to divide the number of choices by Finally, we must exclude the choice where all (because then ). The answer is

## Problem 2

The sum of all the solutions of in the interval is

(a) .

(b) .

(c) .

(d) .

## Solution:

Denote Then The equation becomes

Equivalently,

Its solutions are Corresponding values of are

## Problem 3

Let be a continuous function such that

and let for all integers . Then:

(a) exists and equals .

(b) does not exist.

(c) exists if and only if .

(d) exists and equals .

## Solution:

We verify by induction that Indeed, for this is obvious. If the result is true for then

Hence,

## Problem 4

Consider the curves and . The maximum number of disjoint regions into which these curves divide the -plane (excluding the curves themselves), is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

The equation of the first curve is

This is a circle with center and radius

The equation of the second curve is

This is an ellipse. The first curve is entirely inside this ellipse. Indeed, if

then hence and

The first two curves divide the plane into 3 regions.

The equation of the third curve is

This is a parabola with vertex (7,3) and which is rightward opened. Observe that the first curve and the third curve touch each other at the point which is the rightmost point of the first curve. Hence, the third curve does not divide the region of the circle inside the third curve, while it divides two other regions into two parts. Totally, we have 5 regions.

## Problem 5

A box has 13 distinct pairs of socks. Let denote the probability of having at least one matching pair among a bunch of socks drawn at random from the box. If is the maximum possible value of such that , then the value of is

(a) .

(b) .

(c) .

(d) .

## Solution:

We observe that is increasing with If we choose socks, then by the pigeonhole principle we neccesarily choose at least one matching pair. So, We compute There are choices of a bunch of 13 socks. There are no matches if we choose exactly one sock from each pair, i.e. there are choices when there are no matching pairs. Correspondingly,

## Problem 6

Let , be any real numbers. Then the maximum possible value of , over all points on the ellipse , must be

(a) .

(b)

(c) .

(d) .

## Solution:

By the Cauchy-Schwarz inequality

Moreover, if

then

i.e. the maximum possible value is

## Problem 7

Let , where is a fixed real number. The function is one-to-one if and only if

(a) or

(b) or .

(c) or .

(d) or

## Solution:

If the function is one-to-one, then it is either increasing or decreasing. Its derivative is

If then with in a countable set of points. is strcitly increasing.

If then with in a countable set of points. is strcitly decreasing.

The value of

is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

We use the formula

The quantity of interest is

## Problem 9

The volume of the region is

(a) ;

(b) ;

(c) ;

(d)

## Solution:

At first we note that the region is the rhombus with vertices Its area is When the cross-section of the region by the plane is the rhombus Its area is To get the volume of we integrate in

## Problem 10

Let be a twice differentiable function such that is positive for all , and suppose Which of the following is not a possible value of ?

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

The derivative is strictly increasing. By the mid-point theorem, there exists such that

So, for all we have and

is not a possible value of

## Problem 11

Let

and

Then,

(a) equals ;

(b) equals ;

(c) equals ;

(d) does not exist.

## Solution:

Consider the transformation

Hence,

## Problem 12

The number of different ways to colour the vertices of a square using one or more colours from the set \{Red, Blue, Green, Yellow , such that no two adjacent vertices have the same colour is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

There are 4 possibilities to choose a colour for the vertix Then there are 3 possibilities to choose a colour for the vertix If the vertex has the same colour as then there are 3 possibilities for the vertex In this case we have colourings. If the vertex has different colour than the vertex then there are 2 possibilities for the vertex and 2 possibilities for the vertex In this case we have colourings. Totally there are colourings.

## Problem 13

Define and , where is any prime number. Let . Then the set of possible values of is

(a) .

(b) .

(c) .

(d) .

## Solution:

Observe that

If and then If is odd, then 2= and are even, so If then
and So, we never have Possible values of are We check that all values appear.

If then

If then and

If then and

## Problem 14

Consider all matrices whose entries are distinct and taken from the set . The sum of determinants of all such matrices is

(a) ;

(b) ;

(c) ;

(d)

## Solution:

Let be the set of all matrices with all distinct values from the set For each such matrix denote by the matrix obtained by interchanging columns of the matrix . The transformation is a bijection of the set Hence

It follows that

## Problem 15

Let and be four non-negative real numbers where
1. The number of different ways one can choose these numbers such that is

(a) ;

(b) ;

(c) ;

(d) ;

## Solution:

Denote Then Further,

Hence,

All summands in the right-hand side are non-negative. The latter equality is possible if and only if all summands in the right-hand side are equal to zero. In every configuration some numbers are equal to zero (not all), and some are equal to Moreover, if numbers from are equal to zero , then There are possibilities to choose numbers from that are equal to zero.

## Problem 16

The polynomial has at least one real root if and only if

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

Consider equation (i.e. ). Consider the value of when the line touches the curve If is the common point of these curves, then derivatives of functions and are equal at

Hence, and When the curves and will intersect. In terms of the answer is

## Problem 17

The number of all integer solutions of the equation 2021 is

(a) ;

(b) ;

(c) ;

(d) ;

## Solution:

Denote Then

The right-hand side is even, so there are no integer solutions of the given equation.

## Problem 18

The number of different values of for which the equation 0 has two identical real roots is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

Let be the value of two identical real roots of the equation Let be the value of the third root. Then

Compare coefficients:

## Problem 19

Suppose is a twice differentiable function on such that

and

Then,
(a) is negative for all .

(b) is positive for all .

(c) for exactly one .

(d) for at least two .

## Solution:

Consider the function Then

The function is strictly convex. By assumption, so for all Correspondingly, for all

## Problem 20

Consider the following two subsets of :

Then

(a) is a circle, but is not a circle.

(b) is a circle, but is not a circle.

(c) and are both circles.

(d) Neither nor is a circle.

## Solution:

We check that Indeed, is a circle.

We check that is a circle. Consider points They belong to the circle Hence,

Denote so that If is a circle of center then and necessarialy, Further, hence and

Taking squares we find that

The radius of is then So, we will prove that Write The relation

is equivalent to i.e. The relation is equivalent to

and this is equivalent to

and to

and to

Finally, it means that

i.e. is a circle.

## Problem 21

For a positive integer , the equation

does not have a solution if and only if

(a) .

(b) is a prime number.

(c) is an odd number.

(d) is an even number not divisible by .

## Solution:

Rewrite the equation as

Assume that is odd, Then we can find the solution

Indeed,

Assume that is divisible by Then we can find the solution

Indeed,

Assume that where is odd. If there are solutions, then

i.e. either or is even. But,

so both numbers and are even and is divisible by This shows that in the case when is odd, there are no solutions.

## Problem 22

Let be any twice differentiable function such that its second derivative is continuous and

If
then

(a) for all .

(b) for all .

(c) for all .

(d) for all .

## Solution:

Observe that (otherwise does not exist in ). Since there exists such that

On the set we have So, the derivative preserves the sign on If then relations imply that is positive somewhere on But then it is positive everywhere on and for

On the set we have So, the derivative preserves the sign on If then relations imply that is negative somewhere on But then it is negative everywhere on and for

## Problem 23

Let us denote the fractional part of a real number by (note:
where is the integer part of ). Then,

(a) equals .

(b) equals .

(c) equals .

(d) does not exist.

## Solution:

We observe that

is an integer. Also, It follows that

## Problem 24

Let

How many roots does the equation have in the interval

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

For all we have

and So, if and only if The number of solutions of the equation coincides with the number of solutions of the equation so we need tofind the number of solutions of the equation

There are three solutions:

## Problem 25

For , the number of solutions of the equation

is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

and points are not solutions of the equation. Divide the equation by

Either or takes each value twice on There are 4 solutions of the initial equation.

## Problem 26

Let be a continuous function such that

for all . Suppose that is differentiable at and

Then, the value of is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

If for some then for all we have

which contradicts the assumption So, for all Denote Then is continuous and

It is well-known that So, By assumption,

So,

The expression

equals

(a)

(b) .

(c) .

(d) .

## Solution:

We use the relation for close to :

Take the and differentiate both sides of this equation:

Hence,

## Problem 28

If the maximum and minimum values of , as takes all real values, are and , respectively, then equals

(a) .

(b) .

(c)

(d) .

## Solution:

Denote Then Further,

Let The derivative of equals

So, Correspondingly,

## Problem 29

If two real numbers and satisfy , then the minimum possible value of is

(a) .

(b) .

(c) .

(d) .

## Solution:

Denote by the circle

If then

and the point belongs to the circle of radius and with center The point is in the interior of the circle . The smallest possible value of taken by points corresponds to the radius of a circle with center for which it touches the circle Consider the line the passes through centers of circles and

where The point belongs to the circle when

Hence, the minimal possible radius satisfies

Hence

## Problem 30

Define by

Then,

(a) is discontinuous.

(b) is continuous but not differentiable.

(c) is differentiable and its derivative is discontinuous.

(d) is differentiable and its derivative is continuous.

## Solution:

We compute the derivative of If then

If then

as and So,

The derivative is not continuous at as for we have

• MOHIT KUMAR JAIN

Sir, in solution 22, isn’t it that f(x)<f(0) for all x<0 since f' is negative in (-inf, 0).
Also as per me, answer should be option A- I arrived at the answer by checking for f(x)=(pi)*sin²x..hehe

• Think again if is negative then what will be the conclusion on the negative real axis. The function will be strictly decreasing. Now think graphically.

• Tejas Srivastava

Sir, any ideas about the cutoff for UGA and UGB B.Stat?

• Expect the cutoff to go around 70 to 80 for UGA and 4 to 5 problems for UGB.