# Indian Statistical Institute, ISI BStat & BMath 2018 Solutions & Hints

## Problem 1.

Find all pairs with real, satisfying the equations:

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 1 UGB Solution & Discussion Video:

#### Hint 1

Zeros of the sine function are of the form

#### Hint 2

Consider the cases and .

#### Hint 1

In the case apply the triangle inequality.

#### Full Solution

Zeros of the sine function are of the form So,

Consider the case Then and

We obtain two solutions: and

Let then From the triangle inequality we obtain

But and there are no solution with

So, there are two such pairs: and

## Problem 2.

Suppose that and are two chords of a circle intersecting at a point It is given that cm and cm. Moreover, the

area of the triangle is cm Find the area of the triangle

## Diagram

#### Hint 1

Can you figure out that triangles triangles and are similar?

#### Hint 2

Find out what happens to the ratio of the area of two similar triangle. Figure out the coefficient of similarity.

#### Hint 3

Did you really needed this hint?
If you have figured out the ratio as in hint 2 then what are you waiting for?
You know the area of triangle .

#### Full Solution

Consider triangles and Consider angles and They are subtended by the same chord and are located on the same side of the chord (since and intersect). So, Angles and are vertical, so Triangles and are similar by the equality of two angles. Coefficiant of similarity is found from the ratio of sides: the side is opposite to the angle the side is opposite to the angle So, the coefficient of similarity is The area of the triangle is times bigger than the area of the triangle area of

## Problem 3.

Let be a continuous function such that for all and

for all

Show that is a constant function.

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 3 UGB Solution & Discussion Video:

#### Hint 1

Let For any natural number set

#### Full Solution

Let For any natural number set Then

Taking the limit as and using continuity of we get

For all and is a constant function.

## Problem 4.

Let be a continuous function such that for all

Show that the function defined by the equation

is a constant function.

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 4 UGB Solution & Discussion Video:

#### Hint 1

Try to see if you have all the conditions satisfied to apply Leibniz integral rule.

If you don’t know what Leibniz integral rule is then read: Leibntz Integral rule.

#### Hint 2

The function is continuous on

#### Hints 3

If you are convinced that you can apply Leibntz integral rule then calculate the derivative of .

#### Full Solution

The function is continuous on So, is continuously differentiable on We calculate the derivative of

So, is a constant function.

## Problem 5.

Let be a differentiable function such that its derivative is a continuous function. Moreover, assume that for all ,

Define a sequence of real numbers by:

Prove that there exists a positive real number such that for all

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 5 UGB Solution & Discussion Video:

#### Hint 1

Try to apply Mean Value Theorem.

Let By the mean value theorem, there is such that

So,

For any we have

#### Hint 3

By induction try to verify that for all

#### Full Solution

Let By the mean value theorem, there is such that

So,

For any we have

By induction we verify that for all

For inequality turns into equality. Assume that

Then

Hence, for all we have

Now, by the triangle inequality

So, for all we have

The needed statement holds with

## Problem 6.

Let be real numbers such that for all there

exist triangles of side lengths Prove that the triangles are

isosceles.

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 6 UGB Solution & Discussion Video:

#### Hint 1

Assume Then also Since are the sides of a triangle, the triangle inequality holds:

#### Hint 2

It follows that for all

#### Hint 3

However, by assumption and

Now try to apply limits on and

#### Full Solution

Assume Then also Since are the sides of a triangle, the triangle inequality holds:

It follows that for all

However, by assumption and So,

Passing to the limit in (1) we obtain which is impossible. So, and all triangles with sides are isosceles.

## Problem 7.

Let be such that

Prove that

(i) is odd,

(ii) is divisible by 4,

(iii) is divisible by

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 7 UGB Solution & Discussion Video:

#### Hint 1

Substitute in the equation

#### Hint 2

Since is odd, write we get

#### Hint 3

Substitute . We must prove that is divisible by Expand

is odd, so

The second summand is divisible by so it is enough to check that is divisible by

#### Full Solution

(i) Substitute in the equation

It follows that is odd, hence is odd.

(ii) Since is odd, write we get

So,

One of the numbers is even, so the product is divisible by and is divisible by 4.

(iii) Substitute . We must prove that is divisible by Expand

is odd, so

The second summand is divisible by so it is enough to check that is divisible by From part (ii) we have so From the initial equation

Then

expand

is even, so

and the expression is also divisible by

## Problem 8.

Let . Let be an matrix such that

for all Suppose that

Show that is a multiple of

### Indian Statistical Institute, ISI Subjective BStat & BMath 2018 Problem 8 UGB Solution & Discussion Video:

#### Hint 1

Applying the definition of the matrix with we get that for all

#### Hint 2

Since we can choose such that Then

Consider following quantities:

is the number of indices such that

is the number of indices such that

is the number of indices such that

is the number of indices such that

Clearly,

#### Hint 3

Further,

is the number of indices such that

is the number of indices such that

is the number of indices such that

is the number of indices such that

Using these quantities we calculate sums in (2):

#### Full Solution

Applying the definition of the matrix with we get that for all

Since we can choose such that Then

Consider following quantities:

is the number of indices such that

is the number of indices such that

is the number of indices such that

is the number of indices such that

Clearly,

Further,

is the number of indices such that

is the number of indices such that

is the number of indices such that

is the number of indices such that

Using these quantities we calculate sums in (2):

We get following relations:

Adding first and second we get

Adding first and third we get

Adding second and third we get

It follows that is a multiple of

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