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Explore the Back-StoryThis is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section

**Answer Key**

Problem 1 -> A | Problem 7 -> C | 13. | Problem 19 -> B | Problem 25 -> C |

Problem 2 -> D | 8. | Problem 14 -> C | Problem 20 -> D | Problem 26 -> D |

Problem 3 -> | 9. | Problem 15 -> B | Problem 21 -> B | Problem 27 -> A |

Problem 4 -> | 10. | Problem 16 -> A | Problem 22 -> C | Problem 28 -> |

Problem 5 -> C | 11. | Problem 17 -> C | Problem 23 -> B | Problem 29 -> B |

Problem 6 -> D | 12 | Problem 18 -> A | Problem 24 -> C | Problem 30 -> |

Any positive real number can be expanded as

for some , where each . In the above-described expansion of , the smallest positive integer such that is:

**(A) 3** **< —Answer**

(B) 2

(C) 1

(D) 4

**Solution**

Binary represation of is (very standard Method to Convert Dercimal to Binary)

So,

So, we got (A)

Suppose, for some . Then equals

(A)

(B)

(C) **(D)** **<— Answer**

**Solution**

So,

So we got, (D)

The locus of points in the complex plane satisfying is

(A) a straight line

(B) a pair of straight lines

(C) a circle

(D) a parabola

**Solution**

Amongst all polynomials with real coefficients satisfying for all , what is the maximum possible value of ?

(A)

(B)

(C)

(D) 1

**Solution**

Let denote the set of integers. Let be such that for all . If , then equals

(A) 840

(B) 844**(C) 843 <— Answer**

(D) 842

**Solution**

Now,

Again,

Again,

(C)

Let and be two matrices such that . Which of the following must be true?

(A) and are zero matrices.

(B) is the zero matrix.

(C) **(D)** **<— Answer**

**Solution**

So,

(D)

Let be a permutation of the numbers . The number of arrangements with

and equals:

(A)

(B)

(C)

(D)

**Solution**

is the smallest so,

So, if we just choose first 5 it is enough.

So, as is fixed

(C)

The sides of a regular hexagon is extended by doubling them to form a bigger hexagon as in the figure below.

Then the ratio of the areas of the bigger to the smaller hexagon is:

(A)

(B) 3

(C)

(D) 4

In how many ways can we choose from the set such that are in arithmetic progression?

(A) 135

(B) 145

(C) 155

(D) 165

Suppose the numbers 71,104 and 159 leave the same remainder when divided by a certain number . Then, the value of must equal:

(A) 53

(B) 48

(C) 37

(D) 23

If are positive real numbers such that , then the maximum possible value of is:

(A) 12240

(D) 13824

(C) 10656

(D) 8640

What is the minimum value of the function for real ?

(A) 3

(B) 5

(C) 6

(D) 8

Consider a differentiable function . Assume the function satisfies

for all and all .

(A) attains its maximum but not its minimum on the set .

(B) attains its minimum but not maximum on the set .

(C) If attains either its maximum or its minimum on the set , then it must be constant.

(D) attains both its maximum and its minimum on the set .

A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance below the ladders?

(A) 3

(B) **(C)** **<— Answer**

(D)

**Solution**

, in

, in

So, from the previous relations we get, .

So,

feet (C)

Let be the two tangents to the ellipse . What is the value of

(A) **(B)** **<— Answer**

(C)

(D) 1

**Solution**

We want

So differentiating with respect of we get

for to be 1 we get

Substituting

Let ,

Now, (B)

In the figure below, is a square and is a triangle with given sides inscribed as in the figure. Find the length .

**(A)** **<— Answer**

(B)

(C)

(D)

**Solution**

(A)

Let and be two non-zero polynomials such that the degree of is less than or equal to the degree of , and for . Which of the following must be true?

(A) degree of

(B) degree of **(C)** degree of **<— Answer**

(D) degree of

**Solution**

, for

Hence the polynonial, has at least 11 roots as so

Now, if , then which is a contradiction.

Degree of

For , let be defined

Then

**(A) equals 0 <— Answer**

(B) equals

(C) equals

(D) does not exist

**Solution**

, as

A magic square is a rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square

(A) 90**(B) 96 <— Answer**

(C) 94

(D) 99

**Solution**

Asrume the sum is and as shown. Fill the rest of the square setting the diagonals equal to we get.

So we get that

The number of positive integers less than or equal to 22 such that 7 divides is

(A) 7

(B) 8

(C) 9**(D) 10 <— Answer**

**Solution**

checking

Now we see that only is permitted

Thus,

In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:

(A) 33**(B) 36 <— Answer**

(C) 39

(D) 41

**Solution**

Set of right handers

= Set of left handers

Thenefore,

Moreover,

Hence,

Let be the cube roots of unity. Then the product

(B) **(C)** <— **Answer**

(D)

**Solution**

mod 3

mod 3

Product

The function in the interval

*(A) exactly one point of local maximum and no points of local minimum. ***(B)** exactly one point of local minimum and no points of local maximum. **<— Answer***(C) points of local maximum as well as local minimum. **(D) neither a point of local maximum nor a point of local minimum.*

**Solution**

The number of triples of positive integers satisfying the equation

and such that , equals:

(A) 3

(B) 2**(C) 1 <— Answer**

(D) 0

**Solution**

,

So,

So, (C)

An urn contains 30 balls out of which one is special. If 6 of these balls are taken out at random, what is the probability that the special ball is chosen?

(A)

(B) **(C)** **<— Answer**

(D)

**Solution**

A triangle has sides of lengths units. Then, the radius of its inscribed circle is :

(A)

(B)

(C) **(D)** **<— Answer**

**Solution**

Two ships are approaching a port along straight routes at constant velocities. Initially, the two ships and the port formed an equilateral triangle. After the second ship travelled , the triangle became right-angled.

When the first ship reaches the port, the second ship was still from the port. Find the initial distance of the ships from the port.**(A)** ** <— Answer**

(B)

(C)

(D)

**Solution**

Let

Ship 1

Ship 2

When travelled km, travelled \& when travelled km, travelled . Since they we moving at constant velocites.

So,

So, (A)

If are real numbers, what is the least possible value of

(A)

(B)

(C)

(D)

**Solution**

By AM-GM Inequality we get,

Equality is achieved when are in AP eg (C)

The range of values that the function

takes as varies over all real numbers in the domain of is:

(A)

(B)

(C)

(D)

**Solution**

In the following diagram, four triangles and their sides are given. Areas of three of them are also given. Find the area of the remaining triangle.

(B) 13

(C) 14

orms a squole of

(A) 12

(D) 15

Note. In this question-paper, denotes the set of real numbers.

Consider a board having 2 rows and columns. Thus there are cells in the board. Each cell is to be filled in by 0 or 1 .

(a) In how many ways can this be done such that each row sum and each column sum is even?

(b) In how many ways can this be done such that each row sum and each column sum is odd?

Solution

Consider the function

where is an integer. Show that has a unique minimum and find the point where the minimum is attained.

Solution

Consider the parabola and the straight line : . Let be a variable point on . Draw the two tangents from to and let and denote the two points of contact on . Let be the mid-point of the line segment joining and . Find the locus of as moves along .

Solution

Let be an odd degree polynomial in with real coefficients. Show that the equation has at least as many distinct real roots as the equation .

Solution

Since and both are odd degree polynomial it must have at least one real root.

Suppose, the distinct real roots of be

Roots of happens for all such that

Now, are different odd degree polynomials therefore there shoul we at least distinct real roots of . They should be distinct because are themselves distinct.

For any positive integer , and , let denote the number of divisors of of the form (including 1 and ). Define, for any positive integer .

Find the values of and .

Solution

(i) Factors of :

(ii)

factors without 3 as a factor are only considered:

Consider a sequence of points in the plane such that are non-collinear and for every is the midpoint of the line segment joining and . Let denote the line segment joining and . Prove the following:

(a) The area of the triangle formed by the points converges to zero as goes to infinity.

(b) The point lies on .

**Solution**

**(a)** Let us denole the complex no. connespida to by .

Then

satisfies the recurrence relation

It suffices to find three independent solutions to to find the genenal solution. Let's try to find solutions of the form .

Let

So the roots are

Thenefone, the general sol is

The (complex) constants can be adjusted to make

As

Thus

Thenefore, eventually all points are contained in a ball of radius about , no matter how small might be. Thus, the anea of must go to zero.

**(b)**

7. Let

Calculate for all .

Find the minimum value of

Find the smallest positive real number such that the following inequality holds

This is a work in progress. Please come back for the solutions. You can also suggest your solutions in the comment section

**Answer Key**

Problem 1 -> A | Problem 7 -> C | 13. | Problem 19 -> B | Problem 25 -> C |

Problem 2 -> D | 8. | Problem 14 -> C | Problem 20 -> D | Problem 26 -> D |

Problem 3 -> | 9. | Problem 15 -> B | Problem 21 -> B | Problem 27 -> A |

Problem 4 -> | 10. | Problem 16 -> A | Problem 22 -> C | Problem 28 -> |

Problem 5 -> C | 11. | Problem 17 -> C | Problem 23 -> B | Problem 29 -> B |

Problem 6 -> D | 12 | Problem 18 -> A | Problem 24 -> C | Problem 30 -> |

Any positive real number can be expanded as

for some , where each . In the above-described expansion of , the smallest positive integer such that is:

**(A) 3** **< —Answer**

(B) 2

(C) 1

(D) 4

**Solution**

Binary represation of is (very standard Method to Convert Dercimal to Binary)

So,

So, we got (A)

Suppose, for some . Then equals

(A)

(B)

(C) **(D)** **<— Answer**

**Solution**

So,

So we got, (D)

The locus of points in the complex plane satisfying is

(A) a straight line

(B) a pair of straight lines

(C) a circle

(D) a parabola

**Solution**

Amongst all polynomials with real coefficients satisfying for all , what is the maximum possible value of ?

(A)

(B)

(C)

(D) 1

**Solution**

Let denote the set of integers. Let be such that for all . If , then equals

(A) 840

(B) 844**(C) 843 <— Answer**

(D) 842

**Solution**

Now,

Again,

Again,

(C)

Let and be two matrices such that . Which of the following must be true?

(A) and are zero matrices.

(B) is the zero matrix.

(C) **(D)** **<— Answer**

**Solution**

So,

(D)

Let be a permutation of the numbers . The number of arrangements with

and equals:

(A)

(B)

(C)

(D)

**Solution**

is the smallest so,

So, if we just choose first 5 it is enough.

So, as is fixed

(C)

The sides of a regular hexagon is extended by doubling them to form a bigger hexagon as in the figure below.

Then the ratio of the areas of the bigger to the smaller hexagon is:

(A)

(B) 3

(C)

(D) 4

In how many ways can we choose from the set such that are in arithmetic progression?

(A) 135

(B) 145

(C) 155

(D) 165

Suppose the numbers 71,104 and 159 leave the same remainder when divided by a certain number . Then, the value of must equal:

(A) 53

(B) 48

(C) 37

(D) 23

If are positive real numbers such that , then the maximum possible value of is:

(A) 12240

(D) 13824

(C) 10656

(D) 8640

What is the minimum value of the function for real ?

(A) 3

(B) 5

(C) 6

(D) 8

Consider a differentiable function . Assume the function satisfies

for all and all .

(A) attains its maximum but not its minimum on the set .

(B) attains its minimum but not maximum on the set .

(C) If attains either its maximum or its minimum on the set , then it must be constant.

(D) attains both its maximum and its minimum on the set .

A straight road has walls on both sides of height 8 feet and 4 feet respectively. Two ladders are placed from the top of one wall to the foot of the other as in the figure below. What is the height (in feet) of the maximum clearance below the ladders?

(A) 3

(B) **(C)** **<— Answer**

(D)

**Solution**

, in

, in

So, from the previous relations we get, .

So,

feet (C)

Let be the two tangents to the ellipse . What is the value of

(A) **(B)** **<— Answer**

(C)

(D) 1

**Solution**

We want

So differentiating with respect of we get

for to be 1 we get

Substituting

Let ,

Now, (B)

In the figure below, is a square and is a triangle with given sides inscribed as in the figure. Find the length .

**(A)** **<— Answer**

(B)

(C)

(D)

**Solution**

(A)

Let and be two non-zero polynomials such that the degree of is less than or equal to the degree of , and for . Which of the following must be true?

(A) degree of

(B) degree of **(C)** degree of **<— Answer**

(D) degree of

**Solution**

, for

Hence the polynonial, has at least 11 roots as so

Now, if , then which is a contradiction.

Degree of

For , let be defined

Then

**(A) equals 0 <— Answer**

(B) equals

(C) equals

(D) does not exist

**Solution**

, as

A magic square is a rectangular array of positive integers such that the sum of the three numbers in any row, any column or any of the two major diagonals, is the same. For the following incomplete magic square

(A) 90**(B) 96 <— Answer**

(C) 94

(D) 99

**Solution**

Asrume the sum is and as shown. Fill the rest of the square setting the diagonals equal to we get.

So we get that

The number of positive integers less than or equal to 22 such that 7 divides is

(A) 7

(B) 8

(C) 9**(D) 10 <— Answer**

**Solution**

checking

Now we see that only is permitted

Thus,

In a class of 45 students, three students can write well using either hand. The number of students who can write well only with the right hand is 24 more than the number of those who write well only with the left hand. Then, the number of students who can write well with the right hand is:

(A) 33**(B) 36 <— Answer**

(C) 39

(D) 41

**Solution**

Set of right handers

= Set of left handers

Thenefore,

Moreover,

Hence,

Let be the cube roots of unity. Then the product

(B) **(C)** <— **Answer**

(D)

**Solution**

mod 3

mod 3