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
(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
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
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
.
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
(b)
7. Let
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
(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