# Indian Statistical Institute, ISI BStat & BMath 2023 UGA Solutions & Discussions

## Problem 1

For a real number

if and only if

(a)

(b)

(c) or

(d) or

## Solution:

We note that is a root of Hence, we can factor the expression as follows:

This expression is positive for and for

Answer: (D)

## Problem 2

Diefne a polynomial by

for all , where the right hand side above is a determinant. Then the roots of are of the form

(a) where and is a square root of

(b) where are disdinct.

(c) where

(d) for some

## Solution:

Multiplying the first column by and subtracting it from the second and the third columns we rewrite the determinant as

So, the roots of are

Answer: (B)

## Problem 3

Let be the set of those real numbers for which the identity

is valid, and the quantities on both sides are finite. Then

(a) is the empty set

(b)

(c)

(d)

## Solution:

If then and the series diverges. If then and the series diverges. Assuming that for integer values of
we will prove that the identity holds. The series is a geometric series with the start value and common ratio

Answer: (B)

## Problem 4

The number of consecutive zeroes adjacent to the dignt in the unit’s place of is

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

By the binomial formula,

Consider first three terms separately:

Note that is divisible by It follows that ends in

Answer: (A)

## Problem 5

Consider a right angled triangle whose hypotenuse is of length 1. The bisector of intersects at . If is of length , then what is the length of CD?

(a)

(b)

(c)

(d)

## Solution:

Denote the length of by

If is the angle then the angle is From triangle we find that

From triangle we find that

Hence,

Answer: (A)

## Problem 6

Consider a triangle with vertices and , Let be the area of the triangle and be the area of the circumcircle of the triangle. Then equals

(a)

(b)

(c)

(d)

## Solution:

The triangle is right angled. Indeed, vectors and are orthogonal:

The length of the hypotenuse equals Hence the radius of the circumcircle is and the area of the circumcircle is

Length of catheti are

Hence the area of the triangle is

We can find the ratio:

Answer: (B)

## Problem 7

Let be continuous functions from to itself

If is the derivative of , then for .

(a) .
(b) .
(c) .
(d) .

## Solution:

The derivative of is computed as the derivative of a composite function (using the rule of differentiation of an integral):

Further, derivative of is also computed as the derivative of a composite function:

We get

Answer: (D)

## Problem 8

How many numbers formed by rearranging the digits of 234578 are divisible by

(a) ;

(b) ;

(c) ;

(d) .

## Solution:

Consider one number obtained by rearraging digits This number is divisible by if and only if sums of digits and differ by a multiple of The smallest possible sum of three digits among is The largest possible sum of three digits among is No sum of three digits among can be equal to the sum of other three digits. Indeed, there are three even and three odd digits. If the sum of three digits among is even, then the sum of other three digits must be odd, and vice versa. The difference between two such sums is at most Hence, we will have divisibility by if and only if either are chosen from and are chosen from or are chosen from and are chosen from To get divisibility by we must have Hence, we will have divisibility by if and only if are chosen from and are chosen from Thus, there are

numbers formed by rearraging the digits that are divisible by

Answer: (B)

## Problem 9

Let

and

How many elements does have?

(a)

(b)

(c)

(d)

## Solution:

Assume that Then for some real

and

It follows that

The latter equality is true if and only if

for some integer Equivalently,

Since we get

The only such integer is Hence,

There is only one element in namely

Answer: (B)

The limit

equals

(a)

(b)

(c)

(d)

## Solution:

We take the logarithm and perform the following transformations:

The first summand converges to zero, while the second summand is the integral sum for the integral

Hence,

integrate by parts with

Taking exponent we obtain the needed limit:

Answer: (D)

## Problem 11

Suppose and are positive integers. If and are divided by 7 , then the respective remainders are 2 and 5 . If is divided by 7 , then the remainder equals

(a)

(b)

(c)

(d)

Modulo we have

Hence,

Answer: (B)

The value of

equals

(a) .
(b)
(c) .
(d) .

## Solution:

The expression under consideration is the real part of

by the binomial identity

where the latter equality follows from the relation Hence, the value of the sum is

Answer: (C)

## Problem 13

For real numbers , consider the system of equations

If denotes the set of all real solutions of the above system of equations, then the number of elements in can never be

(a)
(b)
(c)
(d)

## Solution:

The number of solutions to the system of equations

can never be In the following we assume that solutions exist.

Indeed, the first equation is either always holds (if ), or represents a line (if and are not simultaneously zeroes), or represents a circle. Consider the latter case when Then the first equation takes the form

where (note that as we assume that there are solutions to this equation).

Similarly, the second equation is either always holds (if ), or represents a line (if and are not simultaneously zeroes), or represents a circle (if ).

If one of the equations always holds, then the other has infinitely many solutions (as it is either always true, or represents a line, or represents a circle). If both are equations of lines, then either these lines coincide and we get infinitely many solutions, or these lines are different and we get at most one solution. If both are equations of circles, then either these circles coincide and we get infinitely many solutions, or these circles are different and we get at most two solutions. If one equation represents a line, and the other represents a circle, we get at most two solutions.

Answer (D)

## Problem 14

The limit

(a) equals 0 .
(b) equals .
(c) equals 1.
(d) does not exist.

## Solution:

Rewrite the expression as

It can be bounded as follows

Hence, the limit is zero, as

Answer: (A)

## Problem 15

Let be a positive integer having 27 divisors including 1 and , which are denoted by . Then the product of equals

(a) .
(b) .
(c) .
(d) .

## Solution:

For a positive integer denote by the number of positive divisors of (including and ). is a positive divisor of if and only if is a positive divisor of Hence,

It follows that the product of all positive divisors of equals In our case we get

Answer: (C)

## Problem 16

Suppose is a continuous function which has exactly one local maximum. Then which of the following is true?

(a) cannot have a local minimum.
(b) must have exactly one local minimum.
(c) must have at least two local minima.
(d) must have either a global maximum or a local minimum.

## Solution:

Let be a single point of local maximum of Necessarily, is a point of strict local maximum. Assume that has no global maximum. Then there exists a point such that If then, since for some the function attains a local minimum in If then, since for some the function attains a local minimum in So, necessarily has a point of local minimum.

Answer: (D)

## Problem 17

Suppose is such that the imaginary part of is non-zero and . Then

equals

(a) .
(b) .
(c)
(d) .

## Solution:

Summing the geometric series we get

we note that

Answer: (D)

## Problem 18

Let be a twice differentiable one-to-one function. If and

then

equals

(a) -25 .
(b) 25 .
(c) -31
(d) 31.

## Solution:

We change variables Then If then If then Hence

integrate by parts with

Answer: (B)

## Problem 19

If is a continuous function such that

then for all ,

(a) .
(b) .
(c) .
(d) .

## Solution:

At first we note that The function is differentiable, hence is differentiable as well. For we find from the given equation that

i.e.

Answer: (D)

## Problem 20

If denotes the largest integer less than or equal to , then

equals

(a) .
(b) .
(c) .
(d) .

## Solution:

At first we note that The function is differentiable, hence is differentiable as well. For we find from the given equation that

i.e.

{\bf Answer:} (D)

\item The number is not an integer. Indeed,

The latter expression is not integer, as is irrational.

Let Then is an integer such that We note that Hence, The number is integer. Indeed,

Further,

or

The integer part of equals

Answer: (C)

The Limit

equals

(a)

(b)

(c)

(d)

## Solution:

Denote Then

The logarithm of this expression equals

Since it follows that

Hence,

Answer: (B)

## Problem 22

In the following figure, is a quarter-circle. The unshaded region is a circle to which and are tangents.

If is of length 10 and is parallel to , then the area of the shaded region in the above figure equals

(a)

(b)

(c)

(d)

## Solution:

Let be the radius of the quarter-circle, and be the radius of the inscribed circle. Consider the right triangle Its hypotenuse equals and its catheti equal and Hence, The area of the quarter-circle equals

The area of the inscribed circle equals Hence the area of the shaded region equals

Answer: (A)

## Problem 23

Three left brackets and three right brackets have to be arranged in such a way that if the brackets are serially counted from the left, then the number of right brackets counted is always less than or equal to the number of left brackets counted. In how many ways can this be done?

(a)

(b)

(c)

(d)

## Solution:

Denote the left bracket by L and the right bracket by R. Let and be the number, respectively, of left and right brackets among first counted. The first bracket from the left must be L, otherwise we would have If the last bracket is then Hence, the last bracket must be R. The case corresponds to the satisfactory configuration LLLRRR. If then which is not satisfactory. Let Then the first three brackets are either LLR or LRL. Since the last bracket is R, the fourth and fifth bracekts are either LR or RL. In any case we get a satisfactory configuration. So, there are configurations.

Answer: (C)

## Problem 24

The polynomial is divisible by

(a) .
(b) .
(c) .
(d) .

We note that and

In particular,

Further,

and

Finally,

Answer: (A)

## Problem 25

Suppose and

If for all , and , then

(a)
(b) .
(c)
(d) .

## Solution:

The function is a non-zero quadratic function. It is non-linear (as ) and has a unique root (as ). So, is a square:

Since we deduce that

Answer: (A)

## Problem 26

As in the following figure, the straight line lies in the second quadrant of the -plane and makes an angle with the negative half of the -axis, where .

The line segment of length 1 slides on the -plane in such a way that is always on and on the positive side of the -axis. The locus of the mid-point of is

(a) .
(b)
(c) .
(d) .

## Solution:

Denote the length of by and the length of by Consider the right triangle

The length of equals The length of equals Hence,

Coordinates of are Coordinates of are If are coordinates of the mid-point of then

Hence,

This equality can be transformed as follows

Answer: (A)

## Problem 27

Suppose that where are real numbers with . The equation has exactly two distinct real solutions. If is the derivative of , then which of the following is a possible graph of ?

(a)

(b)

(c)

(d)

## Solution:

The function is a cubic polynomial with real coefficients. Since has exactly two distinct real roots, all roots are real and one is of order Hence, a possible behavior of is that increases to one of the roots, then decreases, and then increases intersecting the axis again. Correspondingly, the derivative is positive, then negative, then positive.

Answer: (B)

## Problem 28

Consider the function defined by

where is a square root of -1 . Then

(a) is one-to-one and onto.
(b) is one-to-one but not onto.
(c) is onto but not one-to-one.
(d) is neither one-to-one nor onto.

## Solution:

We note that In particular, and is not onto.Since and are -periodic, is not one-to-one:

Answer: (D)

## Problem 29

Suppose is a non-decreasing function. Consider the following two cases:

In which of the above cases it is necessarily true that there exists an with ?

(a) In both cases.
(b) In neither case.
(c) In Case 1. but not necessarily in Case 2.
(d) In Case 2. but not necessarily in Case 1 .

## Solution:

Consider a non-decreasing function such that Assume that for all We will prove by induction that for all Indeed, Since we deduce that Assume that Then Since we deduce that So, for all But this contradicts the property So, in the Case 1 there must exist such that

Consider a function defined by

Then is non-decreasing, but for all

Answer: (C)

## Problem 30

How many functions , which satisfy

are there?

(a)
(b)
(c)
(d)

## Solution:

Denote Then we need to find the number of integers solutions of the equation

such that Equivalently, if we define then we need to find the number of integers solutions of the equation

where all The number of solutions is

Answer: (C)