EDUCATIVE COMMENTARY ON

ISI 2018 MATHEMATICS PAPERS

(Last Updated on 02/07/2018)

Contents

Introduction 1

Paper 1 3

Paper 2 41

Concluding Remarks 52

Introduction
The first draft of the Commentary Paper 1 was uploaded on 20/05/2018.
Commentary on Paper 2 is now added to it. Also the solution to Q.2 in Paper
1 is corrected, that to Q.19 revised and that to Q.29 in Paper 1 replaced by a
simpler solution. One reader, Saad Hassan also showed that in Q.5 in Paper
1, if we let g(n) be the number of permutations σ of the set {1, 2, . . . , n}
which satisfy the condition that |σ(i) − i| ≤ 1 for all i = 1, 2, . . . , n, then
g(n) satisfies a Finonacci relation. This has been added.
These commentaries started as an outgrowth of the author’s annual com-
mentaries on the Mathematics sections of the JEE Advanced Papers which
began in 2003. In the commentary on 2017 JEE, while commenting on the
declining quality of the JEE questions over the years, especially after the
JEE became totally Multiple Choice Questions format, a comparison was
drawn with the 2016 B.Sc. Entrance test of ISI (Indian Statistical Institute)
which had some imaginatively designed problems even within a limited scope
of the syllabus. Then to bring home the suggestion how the JEE could take
cue from ISI Entrance tests, commentaries on the 2016 and subsequently the
2014 ISI Papers were completed and displayed.
The quality of the questions in these two years had raised expectations.
But the first paper of 2015 turned out to be very disappointing. Also, a

1
look at the question papers prior to 2014 revealed certain common patterns
which were repeated. So the element of novelty was not so strong as it was
when the commentaries for 2016 and 2014 were written. To some extent
such repetitions are to be expected, given the limited syllabus and the large
number of questions to be asked. Instead of asking 30 questions in Paper 1
to be solved in two hours, it would be far better to ask only about 15 really
good questions.
As a result, the plan to write exhaustive commentaries on ISI Entrance
Tests was curtailed to only the recent ones, viz. from 2014 onwards. So far,
the commentaries from 2014 to 2017 ISI B.Sc. Entrance tests of ISI have
been displayed. The present one is a sequel for the year 2018.
As in the other commentaries, unless otherwise stated, all the references
are to the author’s Educative JEE Mathematics, published by Universities
Press, Hyderabad.
I am thankful to the persons who made the question papers available to
me as well as those who pointed out errors in the earlier draft. I am especially
thankful to Dr. Shailesh Shiral for the graph of y = sinx x in Q. 19 of Paper 1.
Readers are invited to send their comments and point out errors, if any, in
the commentary. They may be sent either by e-mail (kdjoshi314@gmail.com)
or by an SMS or a WhatsApp message on mobile (9819961036).

2
 ISI BStat-BMath-UGA-2018 Paper 1 with Comments

N.B. Each question has four options of which ONLY ONE is correct. There
are 4 marks for a correctly answered, 0 marks for an incorrectly answered
question and 1 mark for each unattempted question.

Q.1 Let 0 < x < 61 be a real number. When a certain biased dice is rolled,
a particular face F occurs with probability 61 − x and its opposite face
occurs with probability 61 + x. Recall that the opposite faces sum to 7
in any dice. Assume that the probability of obtaining the sum 7 when
two such dice are rolled is 13
96
. Then the value of x is
1 1 1 1
(A) 8
(B) 12
(C) 24
(D) 27
.

Answer and Comments: (A). Denote the face opposite to F by F ∗ .

Let A, A∗ and B, B ∗ be the other two pairs of opposite faces. When
two such dice are rolled, the possible outcomes are all 36 ordered pairs
of the form (X, Y ) where X, Y are (unrelated) faces of the dice. But
these 36 pairs are not equally likely as the dice is biased. We are given
13
P (E) = (1)
96
where E is the event that the two faces X, Y are opposite to each other.
This can happen in 6 possible ways: (A, A∗ ), (A∗ , A), (B, B ∗ ), (B ∗ , B),
1
(F, F ∗) and (F ∗ , F ). The probabilities of the first four cases are 36 each.
But those of the last two are ( 61 + x)( 61 − x) each. As these possibilities
are mutually exclusive, (1) gives an equation in x, viz.
1 1 13
+ 2( − x2 ) = (2)
9 36 96
Simplifying,
1 1 13 1 1 39 − 32 7
− x2 = ( − ) = × = (3)
36 2 96 9 2 288 576
Hence
1 7 16 − 7 9 1
x2 = − = = = (4)
36 576 576 576 64
which gives x = 18 .

3
 A good, simple problem on probability. The calculation involved is a
bit laborious and prone to error. But the numerical data is so adjusted
that x2 is the square of a rational. This will serve as a warning in case
some calculation goes wrong. The word ‘biased’ in the statement of
the problem is a general word which conveys the opposite of ‘fair’. It
applies to other devices such as coins too. But in the case of a dice,
when all the faces are not equally likely to appear, the dice is often
called a loaded dice, because one way to design such a dice is to make
the face opposite to the one you prefer to show slightly heavier than
the others.

Q.2 An office has 8 officers including two who are twins. Two teams, Red
and Blue, of 4 officers each are to be formed randomly. What is the
probability that the twins would be together in the Red team?
1 3 1 3
(A) 6
(B) 7
(C) 4
(D) 14

Answer and Comments: (D). Another probability problem,  but not

as exciting as the last one. The Red team can be formed in 84 i.e. in
8.7.6.5
= 70 ways. If both the twins are in it, then the remaining two
24  
officers can be chosen in 62 = 15 ways. So the desired probability is
15 3
70
= 14 .

Q.3 Suppose Roger has 4 identical green tennis balls and 5 identical red
tennis balls. In how many ways can Roger arrange these 9 balls in a
line so that no two green balls are next to each other and no three red
balls are together?

(A) 8 (B) 9 (C) 11 (D) 12

Answer and Comments: (D). A little narrative in the statement

of a problem often serves an important task of testing a candidate’s
ability to isolate the mathematical essence of a problem from a real life
situation. There is a distinct (and an alarmingly large) class of students
who simply dread what they call ‘word problems’. The reason is simple.
In a word problem, nobody tells you what to do. You have to figure it
out yourself! That is the very first step of applied mathematics.

4
 Even when the conversion of a word problem to a mathematical
one is fairly obvious, a little narrative can sometimes be refreshing.
Sometimes it might make an oblique reference to some popular figure
or sometimes the answer to the problem may come out to be consistent
with some convention or superstition associated with the narrative (as
when, in some problem, the ward number of a patient who did not
survive an operation comes out to be 13). And, occasionally, the nar-
rative may have a subtle humour which comes to surface when you find
the answer. In the classic Sanskrit treatise Leelavatee on arithmetic
by Bhaskaracharya, one of the problems asks to find how much money
a miser gave to a beggar as alms. There is nothing great about the
problem or the calculations per se. But the answer comes out to be
the smallest monetary unit at that time. Had it come out higher, the
miser would have to hide his face in shame!
But in the present problem, the narrative is serving no purpose. In
essence the problem asks for the number of all possible arrangements
of two kinds of objects, satisfying some restrictions. That these objects
are tennis balls has nothing to do with the essence of the problem. And
the introduction of the name Roger is absolutely high schoolish (actu-
ally, primary schoolish!). When the problem involves two (or more)
persons, one can understand giving them some sweet names instead of
the drab ones like A and B or X and Y . But in the present case, Roger
has no role. (The only conceivable link is that one of the paper setters
is a fan of the tennis great Roger Federer.) Thank God, the paper-
setters did not further specify that Roger got the red tennis balls from
his father and the green ones from his mother, in an attempt to make
the problem more appealing!
Getting down to the mathematical part of the problem, problems of
counting the number of linear arrangements of various types of objects
so that no two objects of a particular type are together are very popular
and there is a standard method to approach them. In the present
problem, suppose first that the restriction is merely that no two green
balls be adjacent. Then we first arrange the 5 red balls in a row leaving
gaps in between two adjacent ones and also two gaps at the two ends
where we can put the green balls as shown in the figure below.

1 2 3 4 5 6

5
 The four green balls are to be placed into these six gaps. If no two
green balls are to be next to each other, no gap can contain more than
one green ball. So the number of possible arrangements is the same as
the number of choosing 4 out of these 6 gaps and putting one green
ball
  in each one of them (leaving the other two vacant). This number
6
is 4 = 15.
But we have to exclude those arrangements in which three (or more)
red balls are together, with no intervening green ball. For example, we
must not leave gaps numbered 2 and 3 both vacant. Similarly, 3 and
4 cannot be both vacant and 4 and 5 cannot be both vacant. (It is all
right if 1 and 2 are both vacant or 5 and 6 are both vacant.) So, from
the 15 possible arrangements we got earlier, we have to exclude these
three. Hence the answer is 15 − 3 = 12.

A fairly standard problem about arrangements where no two objects

of a particular type are to appear together. If no two red balls are to
be together too, then there is only one arrangement, in which the gaps
numbered 2 to 5 are filled with one green ball each. More generally, if
there are m red and n green balls and no two balls of the same colour
are to be next to each other, then the number of such arrangements is
0 if |m − n| > 1, 1 if |m − n| = 1 and 2 if |m − n| = 0 (i.e. if m = n).
The problem gets very complicated if there are three or more types
of objects. Suppose for example, that we have m red, n green and p
blue balls and they are to arranged in a row with no two balls of the
same colour next to each other. Although special cases can be worked
out, there is no general formula as a function of m, n and p.

Q.4 The number of permutations σ of 1, 2, 3, 4 such that |σ(i) − i| < 2 for

every i = 1, 2, 3, 4 is

(A) 2 (B) 3 (C) 4 (D) 5.

Answer and Comments: (D). In all there are 4! = 24 permutations

of the set {1, 2, 3, 4}. We can take them one-by-one and see how many
of them meet the requirement. But this approach is bad for two reasons.
First, even though 4 is a small number, 4! is not. In fact, as n grows,
n! grows very fast, (faster than, say, 10n ), so even for a relatively small

6
value of n, a method based on an exhaustive checking all possible n!
permutations is exceedingly inefficient. Secondly, this approach would
also require a systematic ordering of all n! permutations lest we miss
out some of them. Although such an ordering is possible, that will
make the problem too complicated.
Another approach is to use that even though 4! is large, 4 is a small
number and 2 is even smaller. So, the restriction that |σ(i) − i| < 2
means, simply that σ(i) = i − 1, i or i + 1. So we can count the desired
permutations σ : {1, 2, 3, 4} −→ {1, 2, 3, 4} keeping track of these three
possibilities. For i = 1, there are only two possibilities, either σ(1) = 1
or σ(1) = 2. If σ(1) = 1, then 1 is a fixed point of σ and σ induces
a permutation of the set {2, 3, 4}. As this has 1 less element that the
original set, perhaps an inductive approach will work. On the other
hand, suppose that σ(1) = 2. Then σ(2) cannot be 2 and so has to
be either 1 or 3. If σ(2) = 1, then σ interchanges 1 and 2 and induces
a permutation of the set {3, 4} which is a smaller set. So again an
inductive approach seems feasible. But what if σ(2) = 3 (along with
σ(1) = 2)? We claim that this cannot happen. For, σ is a bijection.
So there is some i ∈ {1, 2, 3, 4} such that σ(i) = 1. But i 6= 1, 2 since
σ(1) = 2 and σ(2) = 3. So, i has to be either 3 or 4. But then
|σ(i) − i| = |1 − i| > 1, a contradiction.
Summing up, we have shown that only two possibilities can arise for
σ(1). Either σ(1) = 1 and σ permutes the remaining elements 2, 3, 4
among themselves or σ interchanges 1 and 2 and permutes the remain-
ing elements 3, 4 among themselves. In both the cases, the problem is
reduced to a similar problem for a set with fewer than 4 elements and
we can keep going.
We can now complete the solution by working out these tail ends
in both the cases. But what we have learnt so far can be used to
tackle a more general problem where 4 is replaced by any positive
integer n (but the requirement |σ(i) − i| ≤ 1 remains unchanged). So,
suppose σ : {1, 2, . . . , n} −→ {1, 2, . . . , n} is a bijection which satisfies
|σ(i) − i| ≤ 1 for all i = 1, 2, . . . , n. We consider the fixed points of
σ, i.e. points i for which σ(i) = i. σ may or may not have any fixed
points. For example the permutation σ = 2143 (which, in the full form
means σ(1) = 2, σ(2) = 1, σ(3) = 4 and σ(4) = 3) has no fixed points
but satisfies the condition in the problem as it merely interchanges

7
adjacent elements.
Now suppose j is any fixed point of σ, i.e. σ(j) = j. Let Aj =
{1, 2, . . . , j − 1} and Bj = {j + 1, j + 2, . . . , n}. For j = 1, the subset
Aj is empty, for j = n, the subset Bj is empty. But that makes no
difference. The point is that every element of Aj differs from every
element of Bj by at least 2. So σ cannot map an element of Aj to
an element of Bj or vice versa. Figuratively, the fixed point j is a
barrier which σ cannot cross. Hence σ permutes the elements of Aj
within themselves and similarly permutes the elements of Bj within
themselves.
Next consider the case where j and k are two fixed points of σ with
j < k and with no fixed point in between. Let C = {j + 1, j + 2, . . . , k −
1}. Again C may be empty if k = j + 1. But by the same reasoning
as above, σ cannot map any element of C to an element outside C or
vice versa. So, σ induces a permutation of the subset C. We further
show that this permutation must interchange adjacent pairs. Consider
j + 1 ∈ C. Then j + 2 is also in C as otherwise j + 2 = k, which would
mean that C = {j + 1} and would make j + 1 a fixed point of σ lying
between j and k, contradicting the assumption. Also σ(j + 1) = j + 2
since σ(j +1) can equal neither j nor j +1. (The latter possibility would
make j + 1 a fixed point of σ.) We now show that σ(j + 2) = j + 1. If
this is not so, then since j + 2 is not a fixed point, the only possibility
would be σ(j + 2) = j + 3. But then the (unique) element, say s in
C which maps to j + 1 under σ cannot be any of j, j + 1 and j + 2,
contradicting that |σ(s) −s| ≤ 1. Thus we get that σ interchanges j + 1
and j + 2, i.e. σ(j + 1) = j + 2 and σ(j + 2) = j + 1. We now start
with j + 3 (in case it is in C), show that σ(j + 3) = j + 4 and thereafter
σ(j + 4) = j + 3. In other words, σ interchanges j + 3 and j + 4. This
goes on till we exhaust all the k − j − 1 elements of C i.e. all elements
between the two adjacent fixed points j and k of σ. In particular this
also shows that |C| i.e. k − j is even (possibly 0).
The following diagram illustrates the case where j = 3 and k = 10.
The arrows join i to σ(i) for 3 ≤ i ≤ 10.
.
1 2
. .3 .4 .5 .
6 .7 8. .9 .
10 11
.
σ σ σ σ σ σ σ σ

. . . . . . . . . . .
1 2 3 4 5 6 7 8 9 10 11

8
 The general picture should be clear now. Assume σ is a permutation
of the set {1, 2, . . . , n} which satisfies the condition in the problem, viz.
|σ(i) − i| < 2 for all i. Let j1 , j2 , . . . , jr be the distinct fixed points of
σ where r ≥ 0 and 1 ≤ j1 < j2 < . . . jr ≤ n. Then σ fixes all these
elements. For notational uniformity, set j0 = 0 and jr+1 = n + 1. Then
the number of elements between any jp and jp+1 (where 0 ≤ p ≤ r)
is even (possibly 0) and σ interchanges these elements in pairs. In
particular, this implies that r has the same parity as n, i.e. either both
are even or both are odd. Moreover, jp has the same parity as p, for
p = 0, 1, . . . , r + 1. Hence the j’s are alternately even and odd.
It follows that every σ is uniquely determined by these integers
0 = j0 < j1 < j2 < . . . < jr < jr+1 = n + 1 which satisfy that jp has the
same parity as p for p = 0, 1, 2, . . . r, r + 1. Hence counting the number
of such permutations amounts to finding how many sequences of this
type there are. In the problem given, n = 4. So r can be either 4, 2
or 0. If r = 4, then σ is the identity permutation 1234 and we have
j1 = 1, j2 = 2, j3 = 3 and j4 = 4. The next case is r = 2. Because
of the requirement of alternating parity, we can have three possibilities
(i) j1 = 1, j2 = 2, in which case σ = 1243, or (ii) j1 = 1, j2 = 4 in
which case σ = 1324 and (iii) j1 = 3, j2 = 4, in which case σ = 2134.
Finally, we have to consider the case where r = 0, i.e. σ has no fixed
points. In this case j0 = 0, j1 = 5 and σ = 2143. Thus we see that
in all there are five permutations of {1, 2, 3, 4} of the desired type, viz.
1234, 1243, 1324, 2134 and 2143.

The answer in this special case, could, of course, have also been
obtained by trial. But once the essential idea strikes (viz. that σ must
permute the elements between two consecutive fixed points as adjacent
pairs), the general case is equally easy. This problem, therefore, is more
suited as a full length question (with a higher n than 4, say n = 8)
in Paper 2. A special consequence of the reasoning above, which is
interesting in its own right, is that if n is odd, then any permutation
σ of 1, 2, . . . , n which satisfies the condition |σ(i) − i)| < 2 for all i =
1, 2, . . . , n, must have at least one fixed point. Even this special case
would have been an interesting question for Paper 2.
Another interesting problem would be to let g(n) be the number of
all such permutations for a given n and ask to find a formula for g(n)

9
as a function of n. A direct answer here is not easy. But it is not hard
to show that g(n) satisfies a Fibonacci relation, viz.

g(n) = g(n − 1) + g(n − 2) (1)

for all n ≥ 2. This is easy to show. As we already noted, σ(1) is

either 1 or 2. In the first case the elements 2 to n are permuted among
themselves and this can be done in g(n − 1) ways. In the second case,
the elements 3 to n permute themselves in g(n − 2) ways. This proves
(1).
However, the answer is not quite the n-th Fibonacci number Fn .
These are defined recursively by Fn = Fn−1 + Fn−2 with the initial
conditions F0 = F1 = 1. In our problem the initial conditions are
g(1) = g(2) = 1. So the answer is not g(n) = Fn but g(n) = Fn+1 .
We skip the derivation of the formula for Fn . But using it the answer
comes out to be
 √ !n+1 √ !n+1 
1 1+ 5 1− 5
g(n) = √  −  (2)
5 2 2

Of course, one can verify (2) by induction on n, using the recurrence

relation (1). Indeed that is essentially the JEE 1981 problem in Com-
ment No. 11 of Chapter 4. In the problem that is asked here, we need
only g(4) which we calculated earlier as 5. This can also be derived
from (2) by noting that
√ √ √ √
(1 + 5)5 = 1 + 5 5 + 50 + 50 5 + 125 + 25 5 (3)
√ 4 √ √ √
and (1 − 5) = 1 − 5 5 + 50 − 50 5 + 125 − 25 5 (4)
√
Subtracting
√ (4) from (3) and dividing the difference by 25 5 i.e. by
32 5 gives g(4) = 5.
Finally, we mention one possible extension of the problem which
is really challenging. If σ satisfies |σ(i) − i| = 0 for every i, then it is
the identity permutation. In the present problem the condition on σ is
slightly relaxed. That is, |σ(i) − i| ≤ 1 for every i. What if we relax it
further to |σ(i) − i| ≤ 2 for every i? Then counting the number of all
permutations σ of 1, 2, . . . , n which satisfy it will not be so easy.

10
Q.5 Let f (x) be a degree 4 polynomial with real coefficients. Let z be the
number or real zeroes of f , and e be the number of local extrema (i.e.
local maxima or minima) of f . Which of the following is a possible
(z, e) pair?

(A) (4, 4) (B) (3, 3) (C) (2, 2) (D) (0, 0)

Answer and Comments: (B). In general f (x) has 4 roots. The

complex roots, if any, appear in conjugate pairs. So the number of real
roots is 4, 2 or 0. But in the first two cases, it is possible that two of
these roots coincide. As for the local extrema, they always occur at
the roots of f ′ (x), which is a polynomial of degree 3. Hence e can be
at most 3. This rules out (A). As for the remaining possibilities, we
assume, without loss of generality that the leading coefficient of f (x)
is positive. If not, we replace f by −f . That does not affect the zeros
of f . So z remains the same. Local maxima/mimima of −f are the
local minima/maxima of f respectively. So e also remains the same.
Now (D) is easily ruled out because f has an absolute minimum over
the the entire IR (as its leading coefficient is positive and degree even)
which is also a local minimum of f .
To see which of the remaining possibilities can hold, note that f ′ (x)
is a cubic polynomial. If e = 2, then f ′ (x) has only two zeros, say a and
b with a < b. One of these, say a, is a double root of f ′ (x). But then f ′
has the same sign on both the sides of a (near a) and hence a cannot
be a local extremum of f . That would mean e = 1, a contradiction.
Hence (C) cannot hold. As only one of the options is correct, it has to
be (B).

The argument above presupposes that only one of the four options
is correct. It is desirable to prove that (B) is indeed a possibility by
giving an actual example. If z = 3, then f must have one double root,
say b and two other simple roots, say a and c. The double root b is also
a simple root of f ′ and hence a local extremum of f . The other two
local extrema exist by Rolle’s theorem, which says that in between any
two zeros of f , there has to be a zero of f ′ . For an actual numerical
example, take the double root as 0 and the other two roots of f (x)
as ±1. Then f (x) = x2 (x − 1)(x + 1) = x4 − x2 . f ′ (x) = 4x3 − 2x
which vanishes at 0 and at ± √12 . So, f has three local extrema, viz. a

11
 local maximum at 0 and local minima at ± √12 . Hence (B) is an actual
possibility. The graph of y = f (x) is sketched below.
y

x
−1 O 1

Chances are that some students have already seen graphs of this
kind. They will immediately recognise that z = 3, e = 3 is a possibility.
No further work is needed because the question does not ask to rule
out the other possibilities logically, as we did above. So, the problem
is a give-away for such students.
Q.6 A number is called a palindrome if it reads the same backward or
forward. For example, 112211 is a palindrome. How many 6-digit
palindromes are divisible by 495 ?

(A) 10 (B) 11 (C) 30 (D) 45

Answer and Comments: (B). Let the palindrome be n = xyzzyx.

Since 495 = 11 × 9 × 5, we have to check divisibility by 11, 9 and 5. As
the sums of alternate digits are the same (viz. x + y + z), n is always
divisible by 11. 5 divides n if and only if x = 5. (The value x = 0
is not allowed as the leading digit of n cannot be 0.) Finally, n will
be divisible by 9 if and only if its digital sum 2(x + y + z) is divisible
by 9, which reduces to x + y + z being divisible by 9. So the problem
amounts to counting all triples (5, y, z) in which y, z can vary from 0
to 9 and 5 + y + z is a multiple of 9. Hence y + z is either 4 or 13.
The first case gives 5 possible solutions (with y = 0, 1, 2, 3, 4). In the
second case, the possibilities are y = 9, 8, 7, 6, 5, 4. Hence in all there
are 11 palindromes of the given type.

12
 A simple counting problem involving very elementary number theory
(specifically, the criteria for divisibility by 11, 9 and 5). Perhaps the
idea was merely to test a candidate’s ability to understand a newly
defined term. In that case, the given illustration is not sufficiently
representative as it may give the wrong impression that the digits must
be repeated in pairs. Better illustrations would be the numbers 305503
or 282282. Also it would have been better to include an illustration
of a number which is not a palindrome, e.g. 370037. Anyway, the
concept, if not the name, is familiar to most people through recreational
words, such as ANNA or MALAYALAM or the short sentence ‘Madam,
I’m Adam’ and, last but not least, the famous sentence attributed to
Napoleon ‘Able was I ere I saw Elba’.
Q.7 Let A be a square matrix of real numbers such that A4 = A. Which of
the following is true for every such A?

(A) det (A) 6= −1.

(B) A must be invertible.
(C) A cannot be invertible.
(D) A2 + A + I = 0 where I is the identity matrix.

Answer and Comments: (A). The zero matrix satisfies the hypothe-
sis but fails both (B) and (D). Also taking A to be the identity matrix,
we see that (C) is false. That leaves (A) as the only correct option.
However, to really justify it positively (rather than by elimination),
repeatedly applying the multiplicativity property of determinants to
the hypothesis, we get det (A4 ) = (det (A))4 = det (A). Clearly, this
cannot hold if det (A) = −1.

Whichever way you look at it, the problem is very simple. Option
(D) is probably based on the factorisation of A4 − A as A(A3 − I) and
further as A(A − I)(A2 + A + I). If both A and A − I are non-singular,
then from A4 = A, it would follow that A2 + A + I = 0. Anyway, that
is not applicable here. Perhaps the idea is to penalise those who know
too much but do not know where not to apply what they know!
Lest the problem appears vacuous, we should give a non-trivial
example where the hypothesis holds. One such source is what are

13
 called idempotent matrices. By definition, an idempotent matrix
is a square matrix A for which A2 = A. It then follows that A3 ,
A4 , and, in fact, all powers of A equal A (whence the name). The
zero matrix and the identity matrix are idempotent. But any square
matrix whose all non-diagonal entries are 0 and the diagonal entries
are either
" 0 or 1 is also idempotent. # A less trivial example is the 2 × 2
2
cos α sin α cos α
matrix where α is any angle. (This matrix
sin α cos α sin2 α
represents the projection of the plane IR2 onto the line y = (tan α)x,
which makes it easier to see why it is idempotent, without actually
calculating its square.)
" √ #
−1/2 − 3/2
√
The matrix A = also satisfies the hypothesis. It
3/2 −1/2
represents the complex number ω and satisfies A3 = I.
Q.8 Consider the real valued function h : {0, 1, 2, . . . , 100} −→ IR such that
h(0) = 5, h(100) = 20 and satisfying h(i) = 21 (h(i + 1) + h(i − 1)), for
every i = 1, 2, . . . , 99. Then the value of h(1) is:

(A) 5.15 (B) 5.5 (C) 6 (D) 6.15.

Answer and Comments: (A). This is an example of a problem which

is cunningly formulated. Once you recognise what its hypothesis means,
the problem becomes trivial. The given relation can be rewritten as
2h(i) = h(i − 1) + h(i + 1) (1)
and hence as
h(i + 1) − h(i) = h(i) − h(i − 1) (2)
for every i ≥ 1. From (2), h(0), h(1), h(2), . . . , h(100) is an airthmetic
progression. (Of course, this can also be seen without such a rewriting,
because the given condition means that every term is the arithmetic
mean of the two terms besides it, which is just another way of looking
at an A.P.)
After this there is hardly anything left in the problem. If d is the
common difference of the A.P. then we have
h(100) = h(0) + 100d (3)

14
 From h(0) = 5 and h(100) = 20 we get 100d = 20 − 5 = 15. Hence
d = 0.15. So, h(1) = h(0) + d = 5 + 0.15 = 5.15.

There is not much to comment as mathematically the problem is

trivial. Such questions are more suitable in live interviews to test the
mental alertness of a candidate.
Q.9 An up-right path is a sequence of points a0 = (x0 , y0 ), a1 = (x1 , y1 ), a2 =
(x2 , y2 ), . . . such that ai+1 − ai is either (1, 0) or (0, 1). The number of
up-right paths from (0, 0) to (100, 100) which pass through (1, 2) is:
       
197 100 197 197
(A) 3. 99
(B) 3. 50
(C ) 2. 98
(D) 3. 100
.

Answer and Comments: (A). Yet another problem where the es-
sential idea is very simple but the wording is (perhaps intentionally)
clumsy. It would be much easier to describe an up-right path as a se-
quence of unit steps, where each unit step is either in the direction of
the positive x-axis or in the direction of the positive y-axis. (Custom-
arily, these directions are taken as east and north respectively. So an
up-right path can also be called a north-east path, in which every unit
step is either northward or eastward. If we denote each unit east step
by H (for horizontal) and each unit north step by V (for vertical), then
every up-right path is determined by a unique sequence of these H’s
and V ’s. (Such sequences are called binary sequences.)
y

(5,5) .(8,5)
(2,4) P2 (8,4)
(5,4)
(0,3)
(2,2)
(6,2)
P
1

(0,0) x
(2,0)

In the diagram above we show two up-right paths, P1 and P2 ,

from (0, 0) to (8, 5). Each path has 13 unit steps, of which 8 are H

15
 and 5 are V . The binary sequence corresponding to the path P1 is
(H, H, V, V, H, H, H, H, V, V, H, H, V ) while that corresponding to the
path P2 is (V, V, V, H, H, V, H, H, H, V, H, H, H).
Given a point (a, b) and positive integers m, n it is clear that every
up-right path from (a, b) to (a + m, b + n) corresponds to a unique
binary sequence of length m + n of V ’s and H’s in which m entries are
H and the remaining
 n entries are V . So the
 total number of all such
paths is m+n
m
(which is the same as m+n
n
).
In the given problem, we first count the number of up-right paths
from (0, 0) to (1, 2). We can follow any of these
 paths by any up-right
3
path from (1, 2) to (100, 100). There are 1 = 3 paths of the first type
 
197
and 99
paths of the second type. So the total number of up-right
 
197
paths from (0, 0) to (100, 100) that pass through (1, 2) is 3. 99
.

An easy problem once the essential idea is understood.

Q.10 Let f (x) = 12 x sin x − (1 − cos x). The smallest positive integer k such
f (x)
that lim k 6= 0 is:
x→0 x

(A) 3 (B) 4 (C) 5 (D) 6.

Answer and Comments: (B). Clearly f (x) → 0 as x → 0. Also, for

every positive integer k, xk → 0 as x → 0. So three things are possible
f (x)
for the behaviour of the ratio k as x → 0. For small k, f (x) tends
x
to 0 more rapidly than xk and so the ratio tends to 0. For large k,
xk tends to 0 faster than f (x) and then the limit of the ratio does not
exist. But there is a critical value of k for which xk tends to 0 as rapidly
f (x)
as f (x) and then the ratio k tends to a finite non-zero number as
x
x tends to 0. In such a case we say that near 0, f (x) is comparable
to xk . The problem asks to determine which power of x is comparable
to the given function f (x) as x → 0.
The basic result needed is that as x → 0, sin x is comparable to x.
This follows from the well known result that
sin x
lim =1 (1)
x→0 x

16
Let us write f (x) as g(x) − h(x) where
1
g(x) = x sin x (2)
2
x
and h(x) = 1 − cos x = 2 sin2 ( ) (3)
2
Because of (1), both g(x) and h(x) are comparable to x2 as x → 0. So
it is tempting to think that the answer is k = 2. But there is a catch
g(x)
here. Let L1 and L2 be, respectively, the limits of the ratios 2 and
x
h(x) g(x) − h(x)
as x → 0. Both are non-zero. If L1 6= L2 , then will
x2 x2
tend to L1 − L2 and then k would be 2. But if both these limits happen
f (x)
to be equal then tends to 0 as x → 0. In such a case, f (x) is
x2
2
not comparable to x but to some higher power of x. By not giving 2
as a possible answer, the paper-setters have given an implicit hint that
L1 = L2 . We can, of course, verify this because from (2), both L1 and
L2 equal 21 .
This makes the problem more delicate. We now have to work with
1
x sin x − 2 sin2 ( x2 )
lim 2 and decide for which value of k this limit is
x→0 xk
finite and non-zero. As this limit is of the indeterminate 00 form, one
method is to try l’Hôpital’s rule. Since derivatives will be involved, it is
convenient to put back f (x) in its original form as 21 x sin x − 1 + cos x.
Two applications of the rule give
1
f (x) 2
x sin x − 1 + cos x
lim k
= lim
x→0 x x→0 xk
1
2
sin x + 21 x cos x − sin x
= lim
x→0 kxk−1
1
− cos x − 21 x sin x + 12 cos x
= lim 2 (4)
x→0 k(k − 1)xk−2
− 12 x sin x
= lim (5)
x→0 k(k − 1)xk−2

where, in going from (4) to (5) we have not applied the l’Hôpital’s rule,
but merely simplified the numerator. We can go on applying the rule
again. But that is hardly necessary. We already know that as x → 0,

17
 x sin x is comparable to x2 . So, if k = 4, then the limit in (4) is a finite,
−1/2
non-zero number. The actual value of this limit can be found as .
4×3
But that is not necessary. All that matters is that it is non-zero. Hence
4 is the least value of the integer k for which the given limit is non-zero.
(Actually, that is an understatement, because 4 is the only value of k
for which the limit is non-zero. For k < 4, the limit is 0 while for k > 4
it does not exist.)

A good problem based on the concept of comparable rates of growth.

There is also a solution based on the Taylor series expansions of sin x
and cos x. Using these, we get

x x3 x5 x2 x4
f (x) = (x − + − . . .) − 1 + (1 − + − . . .)
2 3! 5! 2! 4!
x2 x4 x6 x2 x4
= − + + ...− + − ...
2 2.3! 2.5! 2 4!
= A4 x4 + A6 x6 + . . . (6)
1 1 1
where A4 = 24 − 12 = − 24 6= 0. As x → 0, higher powers of x are
negligible and so f (x) is comparable to x4 as x → 0. This solution is
too advanced for the HSC level. But the series for sin x and cos x are
very popular. And, in any case, for a multiple choice format question,
it does not matter how you get the answer, as long as it is correct.

Q.11 Nine students in a class gave a test for 50 marks. Let S1 ≤ S2 ≤ . . . ≤

S5 ≤ . . . ≤ S8 ≤ S9 denote their ordered scores. Given that S1 = 20
9
X
and Si = 250, let m be the smallest value that S5 can take and let
i=1
M be the largest value that S5 can take. Then the pair (m, M) is given
by

(A) (20, 35) (B) (20, 34) (C) (25, 34) (D) (25, 50).

Answer and Comments: (B). From the data, S5 ≥ 20. If S5 = 20,

then S2 , S3 , S4 also equal 20 and so S6 +S7 +S8 +S9 = 250−100 = 150.
This is feasible if we take S8 = S9 = 50 and S6 = S7 = 25. So m = 20.
To determine M, note that if S5 = M then S6 , S7 , S8 , S9 are at least M
each and so S5 + S6 + S7 + S8 + S9 ≥ 5M. But S1 + S2 + S3 + S4 ≥ 80.

18
 So we have 250 ≥ 5M + 80 which gives M ≤ 34. Again this is feasible
if we let Si = 20 for i = 1, 2, 3, 4 and Si = 34 for i = 5, 6, 7, 8, 9. So
M = 34.

A simple problem about inequalities. It is not clear what role the

maximum score in the test (viz. 50) has. Had it been, say 30, the
problem would be more interesting.

Q.12 Let 10 red balls and 10 white balls be arranged in a straight line such
that 10 each are on either side of a central mark. The number of such
symmetrical arrangements about the central mark is
10! 10!
(A) 5!5!
(B) 10! (C) 5!
(D) 2.10!

Answer and Comments: (A). Because of symmetry, on each side

10!
there will be 5 red and 5 white balls. These can be arranged in 5!5!
ways. Once they are arranged, the arrangement on the other side of
the mark is determied uniquely by symmetry.

An absolutely simple problem. The basic idea is that of a palindrome

in Q. 6. But there the subsequent work involved some number theory.
Here there is hardly anything left except arranging 5 red and 5 white
balls in a row, for which there is a formula.
 
Q.13 If z = x + iy is a complex number such that  z−i  < 1, then we must
 
z+i
have

(A) x > 0 (B) x < 0 (C) y > 0 (D) y < 0.

Answer and Comments: (C). One way to tackle the problem would
be to convert the data in terms of x and y. But usually, it is better
to work in terms of complex numbers. The given inequality can be
rewritten as

|z − i| < |z + i| (1)

The L.H.S. is the distance of the point z from the point i while the
R.H.S. is its distance from −i. So, (1) is equivalent to saying that the
point z is closer to i than to −i. Therefore z lies on that side of the

19
 perpendicular bisector of the segment joining i to −i, which contains
i. The perpendicular bisecor here is simply the x-axis. So, (1) holds if
and only if y > 0.

i .
.z
x
O

−i .

An easy problem if approached geometrically. Just in case, you can’t

think of it, we give the approach based on conversion to real numbers.
Then squaring both the sides of (1) we get

x2 + (y − 1)2 < x2 + (y + 1)2 (2)

which is equivalent to

−2y < 2y (3)

i.e. to y > 0. So, this approach, at least in the present problem, is

almost equally easy.

Q.14 Let S = {x − y | x, y are real numbers with x2 + y 2 = 1}. Then the

maximum number in the set S is
√ √ √
(A) 1 (B) 2 (C) 2 2 (D) 1 + 2.

Answer and Comments: (B). This question resembles Q.8 in that

once the clumsy formulation is unveiled, the problem becomes very
familiar. Taking x = cos θ and y = sin θ, the problem asks to maximise
cos θ − sin θ as θ varies over
√ [−π, π]. π Instead of applying calculus,
π
we rewrite cos θ − sin θ as 2(cos θ sin( 4 ) − sin θ cos( 4 ) and further as
√
2 sin( π4 − θ). Its maximum occurs when θ = − π4 . And the maximum
√
value is 2. (If we let θ vary over [0, 2π], which is more customary, the
maximum occurs when θ = 7π 4
. But it gives the same point.)

20
Q.15 In a factory, 20 workers start working on a project of packing consign-
ments. They need exactly 5 hours to pack one consignment. Every
hour 4 new workers join the existing workforce. It is mandatory to
relieve a worker after 10 hours. Then the number of consignments that
would be packed in the initial 113 hours is

(A) 40 (B) 50 (C) 45 (D) 52.

Answer and Comments: (C). There is some ambiguity regarding the

type of work involved. Packages are discrete objects. And one tends
to expect that when 20 workers are working on a particular package,
they will be working as a team and till they finish packaging one con-
signment, it is of no use to add to the team. So the new members who
join will have some work to do only when they reach a strength of 20
and then they will work on a separate consignment.
But if we assume this interpretation, then the new workers who join
will be idle (or ‘sitting on the bench’ in the common parlance of the
IT industry!) for considerable time. The mandatory relief, however,
has to be given to a worker 10 hours after joining, rather than after
ten hours of working. The resulting wastage is considerable and makes
this interpretation unrealistic.
So, we assume that the packaging work is not discrete but continuous
in nature and also that every worker starts contributing independently
immediately upon joining. Each consignment takes 100 man-hours of
work. But these can be contributed by any number of workers. As a
real life example, take the work of painting walls. If 20 workers can
paint a wall in 5 hours, then 4 workers working together at any moment
will do it in 25 hours, with the understanding that they may not be the
same four workers all the time. (A more realistic model will be the job
of filling water tanks of capacity 100 units each, by pumps each of which
pumps 1 unit of water per hour and has to be stopped after operating
for 10 hours. Any number of pumps can simultaneously pump water
into the same tank and as soon as one tank is full, all the pumps filling
into it are instantaneously shifted to start filling another.)
With this interpretation, the problem resembles the umpteeen
number of problems we do in school regarding time, work and rate.
All we need to do is to calculate the total man-hours accumulated in

21
 the first 113 hours and divide this number by 100. The real problem
is with this calculation, because not only the workers keep changing,
their numbers also keep changing.
We begin by keeping an hourly track of how many workers are
working. In the first hour, there are 20 workers. In the second there
are 24, in the third there are 28 and so on. This goes on till the tenth
hour during which we have 56 workers. At the end of the tenth hour,
4 new workers will join but the the original 20 workers will be relieved.
Hence during the 11-th hour, there will be only 40 workers. After this
there will be a steady state, because at the end of each hour, as many
new workers will join as will be relieved. So from the 11-th hour to the
113-th hour, at any time 40 workers will be working.
If we denote the total number of man hours during the first 113
hours by W , we get

W = 20 + 24 + 28 + . . . + 56 + (113 − 10) × 40 (1)

= 380 + 4120 (2)
= 4500 (3)

where in going from (1) to (2) we have used the formula for the sum of
the terms in an A.P. (We can do without it too, by adding terms which
are symmetrically located, like 20 + 56, 24 + 52, till 36 + 40.)
Since each packaging consignment takes 100 man-hours, from (3)
we see that 45 consignments would be completed during the first 113
hours.

As in many earlier problems, the mathematical essence of the problem

is very elementary. Such problems are, therefore, more a test of a
candidate’s ability to reduce a real life problem to a mathematical one.
One only wishes that the formulation were free of ambiguity.

Q.16 Let ABCD be a rectangle with its shorter side a > 0 units and perime-
ter 2s units. Let P QRS be a rectangle such that the vertices A, B, C
and D respectively lie on the lines P Q, QR, RS and SP . Then the
maximum area of such a rectangle P QRS in square units is given by
s2
(A) s2 (B) 2a(s − a) (C) 2
(D) 52 a(s − a).

22
Answer and Comments: (C). The manner in which the data is given
is somewhat puzzling. Normally, the size of a rectangle is specified
in terms of its length and the breadth, say b and a (since usually the
breadth is smaller than the length). Then its semi-perimeter is s = a+b.
Mathematically, if s and a are given one can determine b and so it makes
no difference which two of the three variables a, b, s are specified. The
unusual choice of a and s, rather than of a and b, suggests that the
correct answer is in terms of s. So perhaps (A) or (C) is the right
answer.
Mathematically, this guess has no validity. Nor does it really matter.
With the data as given, the longer side of the rectangle ABCD is s − a.
We take AB as a and AD as s − a. Let θ be the angle between the lines
AD and P Q. Then θ is also the angle between BC and RS, as they
are parallel, respectively to AD and P Q. Moreover, θ is also the angle
between AB and QR since AB ⊥ AD and QR ⊥ P Q. Similarly, θ
equals the angle between CD and P S, as shown in the diagram below.

A θ D
s−a
Q a θ
θ a
S
B s−a θ C

We express the area, say ∆, of the rectangle P QRS in terms of a, s

and θ as follows.
From the figure above, we have

P Q = P A + AQ = AD cos θ + AB sin θ = (s − a) cos θ + a sin θ (1)

and similarly,

P S = P D + DS = (s − a) sin θ + a cos θ (2)

23
 Hence
∆ = P Q.P S
= a(s − a) + (a2 + (s − a)2 ) sin θ cos θ (3)
Since a and s − a are constants, ∆ is a function of θ and it will be
maximum if and only if sin θ cos θ is maximum. Instead of finding
the maximum by calculus, we express sin θ cos θ as 12 sin 2θ which is
maximum when θ = π4 . The maximum value of ∆ is a(s − a) + 12 (a2 +
2 2
(s − a)2 ) = as − a2 + a2 + s2 − as = s2 .

A good problem, involving elementary geometry and trigonometry,

but more importantly, the ability to visualise the data. It could have
been asked as a full length question in Paper 2, along with a straight
edge and compass construction for this rectangle with maximum area.
A still more challenging problem would be to ask a construction for
the rectangle P QRS whose sides contain the vertices of a given rectan-
gle ABCD and whose area equals that of some other given rectangle
A′ B ′ C ′ D ′ . In that case one would first have to construct 2θ (which may
be less than π2 ) using (3). Bisecting it will give θ. After getting hold of
θ, the point P can be determined as the intersection of the semi-circle
with AD as the diameter and a line making angle θ with the side AB.
Q.17 The number of pairs of integers (x, y) satisfying the equation
xy(x + y + 1) = 52018 + 1 is :

(A) 0 (B) 2 (C) 1009 (D) 2018.

Answer and Comments: (A). Equations in several variables where

the variables can take only integral values are called diophantine
equations. Usually, they involve higher powers of these variables
and there is no general method to solve them. Each problem is in
its own class. Even a slight difference can drastically affect the degree
of difficulty. For example all solutions of the Pythagorean equation
x2 + y 2 = z 2 are known. Cancelling common factors, if any, (so that
x, y, z are pairwise coprime), it can be shown that they are of the form
x = u2 − v 2 and y = 2uv (or vice versa) for some integers u, v (so
that z = u2 + v 2 ). But the deceptively similar equation x3 + y 3 = z 3
eluded as great mathematicians as Gauss and others for more than a

24
 century. Fermat conjectured that for n > 2, the diophantine equation
xn + y n = z n has no solutions in positive integers. This conjecture was
settled (in the affirmative) only recently. And the methods needed are
far too advanced.
On this background, when a diophantine equation is given in an
elementary examination, it is a safe bet to say that it will have either
no solutions or only obvious solutions, and further, that the methods
needed will generally not go beyond divisibility. The present problem
is no exception. We claim that it has no solutions. And the key to the
proof is the observation that the R.H.S. 52018 + 1 is divisible by 2 but
not by 4. Divisibility by 2 is obvious since all powers of 5 are odd. For
showing non-divisibility by 4, we note that the successive powers of 5
starting from 52 end in 25. For example, the first few such powers are
25, 125, 625, 3125, . . .. For an inductive proof, if 5k = 100r + 25, then
5k+1 = 5(100r + 25) = (5r + 1)100 + 25.
As a result, 52018 + 1 is of the form 100m + 26 for some integer m.
Therefore it is divisible by 2 but not by 4. So, if xy(x+y+1) = 52018 +1,
then exactly one of the three factors x, y and (x + y + 1) is even. But
this can never happen. Surely, x, y are not both even. If one of them
is even and the other odd, then the factor x + y + 1 would be even, a
contradiction. The only case left is when both x, y are odd. But in that
case x + y + 1 and hence xy(x + y + 1) would be odd, a contradiction.
Hence the given equation has no solutions.

Q.18 Let p(n) be the number of digits when 8n is written in base 6, and let
q(n) be the number of digits when 6n is written in base 4. For example,
p(n)q(n)
82 in base 6 is 144, hence p(2) = 3. Then lim equals :
n−→∞ n2
4 3
(A) 1 (B) 3
(C) 2
(D) 2.

Answer and Comments: (C). A k-digit number with base 6 is at

least 6k−1 and at most 6k − 1. So if 8n has p(n) digits when expressed
with base 6, we have

6p(n)−1 ≤ 8n < 6p(n) (1)

25
Taking logarithms, this means

(p(n) − 1) log 6 ≤ n log 8 < p(n) log 6 (2)

This is valid regardless of the base of the logarithms, as long as the base
is greater than 1. As we shall see, the actual base does not matter for
our purpose. For the sake of definiteness, we assume that all logarithms
are with base 10. Dividing throughout by n log 6, we get
p(n) − 1 log 8 p(n)
≤ < (3)
n log 6 n
which can be rewritten as
log 8 p(n) log 8 1
< ≤ + (4)
log 6 n log 6 n
for every positive integer n. We now apply the Sandwich Theorem for
limits of sequences. The first term is a constant sequence while the
last term is a sequence which tends to this constant (since n1 → 0 as
n → ∞). So, we get
p(n) log 8
lim = (5)
n→∞ n log 6
By an entirely analogous argument, we have
q(n) log 6
lim = (6)
n→∞ n log 4
From (5) and (6),
p(n)q(n) log 8
lim 2
= (7)
n→∞ n log 4

It only remains to evaluate log 8

log 4
. Had we taken the base 2, the numer-
ator and the denominator would be 3 and 2 respectively. Even for any
other base, say 10, by the laws of logarithms, we have log 8 = 3 log 2
log 8
and log 4 = 2 log 2 and so the value of log 4
is 23 .

An excellent and a well designed problem, based on very simple

properties of logarithms. The solution looks longer because we have

26
 given it rigorously. In informal work, instead of (1) we can write that
for large n,

8n ≈ 6p(n) (8)

and hence

n log 8 ≈ p(n) log 6 (9)

This gives a quick derivation of (5). (6) can be derived analogously.

Q.19 For a real number α, let Sα denote the set of those real numbers β that
satisfy α sin(β) = β sin(α). Then which of the following statements is
true?

(A) For any α, Sα is an infinite set.

(B) Sα is a finite set if and only if α is not an integer multiple of π.
(C) There are infinitely many numbers α for which Sα is the set of all
real numbers.
(D) Sα is always finite.

Answer and Comments: (B). Because of the periodicity of the

trigonometric functions, for a given α, there are infinitely many an-
gles β such that sin β = sin α. In fact all such angles are given by

β = nπ + (−1)n α (1)

where n is any integer. But this knowledge is of little help because the
equation in the present problem is different, viz.

α sin β = β sin α (2)

For α, β 6= 0, this can be recast as

sin β sin α
= (3)
β α
or equivalently, as

f (β) = f (α) (4)

27
where the function f (x) is defined by
sin x
(
x
, if x 6= 0
f (x) = (5)
1, if x = 0
Then f is an even function (i.e. f (−x) = f (x) for all x) which is
continuous everywhere. We are more interested in the behaviour of
f (x) as x → ±∞. As the function is even, we confine ourselves only
to x > 0. The numerator sin x is periodic and varies between −1 and
1 in any interval of length 2π. But the denominator x is not periodic.
In fact, it keeps on increasing as x → ∞ and for large x, sin x is
insignificant as compared to x. Put differently,
lim f (x) = 0 (6)
x→∞
| sin x| 1
which follows because we have |f (x)| = x
≤ x
for all x > 0.
As with the sine function, the zeros of f (x) are precisely all integral
multiples of π. Also like the sine function, if n is even then f (x) > 0 for
x ∈ (nπ, (n + 1)π) and if n is odd, then f (x) < 0 if x ∈ (nπ, (n + 1)π).
So, the graph of f (x) is also wavelike, like that of the sine function, but
because of the factor x in the denominator, the ‘amplitudes’ of these
waves go on decreasing. In fact, for x ∈ [nπ, (n + 1)π], we have
1
|f (x)| ≤ (7)
nπ
which gives a ceiling on the amplitudes.
The graph of y = f (x) is shown below.
y

sin x
1.00 y= x

0.75

0.50

0.25

x
−20 −15 −10 −5 5 10 15 20

−0.25

−0.50

28
 In terms of the function f , for every α, Sα = {β : f (β) = f (α)}. If
α = nπ for some n ∈ ZZ, then f (α) = 0 and there are infinitely many
points in the set Sα since f (nπ) = 0 for all n ∈ ZZ. On the other hand,
if α is not an integral multiple of π, then f (α) 6= 0. Let ǫ = |f (α)|.
Then ǫ > 0 and so by (6), there exists some R > 0 such that for all
|x| > R, |f (x)| < ǫ. We can take R to be kπ for some sufficiently large
integer k. Then the possible values of β for which f (β) = f (α) all lie
in the interval [−kπ, kπ]. Moreover β cannot be of the form mπ for
any m ∈ ZZ. So every element of the set Sα lies in an open interval
(mπ, (m + 1)π) for some m, −k ≤ m < k.
If we could show that each of these 2k open intervals contains at
most three elements of Sα , it would follow that the set Sα is finite. This
is obvious from the graph. (In fact, from the graph, each such interval
contains at most two elements of Sα .) However, for an analytical proof,
suppose there exist four elements, say, β1 < β2 < β3 < β4 of Sα in
(mπ, (m + 1)π), then each f (βi ) equals f (α) and so by Rolle’s theorem,
there would be points γ1 ∈ (β1 , β2 ), γ2 ∈ (β2 , β3 ) and γ3 ∈ (β3 , β4 ) such
that

f ′ (γ1 ) = f ′ (γ2 ) = f ′ (γ3 ) = 0 (8)

Now note that

x cos x − sin x
f ′ (x) = (9)
x2
So, the zeros of f ′ are also the zeros of g(x) = tan x − x which is unde-
fined at x = 2m+1
2
π. But on each of the two subintervals [mπ, 2m+1 2
π)
2m+1
and ( 2 π, (m+1)π], g(x) is a strictly increasing function on as we see
from its derivative sec2 x−1. So each of these two subintervals contains
at most one zero of tan x − x and hence of f ′ (x). But that contradicts
(8) since at least two γ’s must be in one of these two subintervals.
Summing up, we have shown that the set Sα is infinite if and only
if sin α = 0, i.e. α is an integer multiple of π. This is precisely (B), in
its contrapositive form.

sin x
A good problem requiring study of the graph of y = x
. More
suitable for Paper 2 where reasoning could be tested.

29
 ! !
1 1 2018 a b
Q.20 If A = and A = , then a + d equals:
0 i c d

(A) 1 + i (B) 0 (C) 2 (D) 22018 .

Answer and Comments: (B). The matrix A is an example of what is

called an upper triangular matrix, i.e. a matrix in which all entries
below the diagonal are 0. By a direct calculation, for any real (or
complex) numbers x, y, z, x′ , y,′ z ′ , we have
! ! !
x y x′ y ′ xx′ xy ′ + yz ′
= (1)
0 z 0 z′ 0 yy ′

Verbally, the product of two upper triangular matrices (of order 2) is

an upper triangular matrix whose diagonal elements are the products
of the respective diagonal elements of the two matrices.
Applying this fact repeatedly, we see that
!
2018 12018 λ
A = (2)
0 i2018
!
a b
for some complex number λ. Equating this with , we get
c d
a = 12018 = 1, d = i2018 = i2 = −1. So a + d = 1 + (−1) = 0.

A simple problem for those who know (1). Those who don’t, can
begin by calculating A2 and A3 by direct multiplication. Then they
will guess that the diagonal entries of An are the n-th powers of the
corresponding diagonal entries of A. A scrupulous student will foolishly
waste time in proving this guess by induction on n. But ultimately, all
will get the answer.

Q.21 Let f : IR −→ IR and g : IR −→ IR be two functions. Consider the

following two statements.

P(1) : If lim f (x) exists and lim f (x)g(x) exists, then lim g(x) must
x→0 x→0 x→0
exist.
P(2) : If f, g are differentiable with f (x) < g(x) for every real number
x, then f ′ (x) < g ′ (x) for all x.

30
 Then, which of the following is a correct statement?

(A) Both P(1) and P(2) are true. (B) Both P(1) and P(2) are false.
(C) P(1) is true and P(2) false. (D) P(1) is false and P(2) true.

Answer and Comments: (B). For P(1), take f (x) = x and g(x) =
sin( x1 ) for a counterexample. For P(2), take f (x) = sin x and g(x) =
100.

P(1) would have been true if lim f (x) were non-zero, because in
x→0
that case f (x) 6= 0 in some sufficiently small deleted neighbourhood
of 0 and we can write g(x) as the ratio of the functions f (x)g(x) and
f (x). The essential idea in dealing with P(2) is that a function can be
made arbitrarily large by adding a constant to it, without increasing
its derivative.

Q.22 The number of solutions of the equation sin 7x + sin 3x = 0 with 0 ≤

x ≤ 2π is

(A) 9 (B) 12 (C) 15 (D) 18.

Answer and Comments: (C). The given equation can be rewritten

as

2 sin(5x) cos(2x) = 0 (1)

We find the zeros of sin 5x and of cos 2x separately.

If sin 5x = 0, then 5x = kπ for some integer k. Since 0 ≤ x ≤ 2π,
the possible values of k are 0 to 10. Each such value gives a solution
and these 11 solutions are all distinct. The second equation cos 2x = 0
has only four solutions in [0, 2π], viz. π4 , 3π
4
, 5π
4
and 7π
4
. So, in all, (1)
has 11 + 4 i.e. 15 solutions.

An absolutely simple problem. The last problem also involved

solutions of sin x = 0. But there is no comparison between their levels
of difficulty. 14 should have been included as a fake answer to punish
those who hastily conclude that sin 5x = 0 has only 10 solutions.

31
Q.23 A bag contains some candies, 25 of them are made of white chocolate
and the remaining 53 are made of dark chocolate. Out of the white
chocolate candies, 1/3 are wrapped in red paper, the rest are wrapped
in blue paper. Out of the dark chocolate candies, 32 are wrapped in
red paper, the rest are wrapped in blue paper. If a randomly selected
candy from the bag is found to be wrapped in red paper, then what is
the probability that it is made up of dark chocolate?
2 3 3 1
(A) 3
(B) 4
(C) 5
(D) 4

Answer and Comments: (B). If the narrative in Q.3 (where Roger

had 5 red tennis balls and 4 green tennis balls), was primary schoolish,
the one in the present problem belongs to the kindergarten level!.
Anyway, coming to the problem itself, this is a problem about
conditional probability. The following tree diagram summarises the
data. The labels on the nodes and on the arrows are self explanatory.

2/5 3/5

W D

2/3 1/3
1/3 2/3

R B R B

Clearly,
2 1 3 2
P (R) = × + ×
5 3 5 3
2 6 8
= + = (1)
15 15 15
while
3 2 6
P (R ∩ D) = × = (2)
5 3 15

32
 The problem asks for the conditional probability P (D|R), i.e. the
probability of D, given R. By Bayes theorem,

P (R ∩ D) 6/15 3
P (D|R) = = = (3)
P (R) 8/15 4

So, this is the probability that a chocolate with a red wrapper happened
to be dark.

It takes more time to read the problem than to solve it! In fact,
after getting (1), the calculations can be done mentally.
Such problems definitely do not enhance the prestige of a test in
academic circles. Far better problems on conditional probability have
been asked in JEE. See Comment No. 11 of Chapter 22, and especially
the 1994 JEE problem there. In those days the candidates had to show
their work. And sometimes it was a torture for the examiner to go
through pages and pages of hastily scribbled, and often inconclusive,
calculations with hardly any words of explanation. (Perhaps this was
one of the reasons that led to the adoption of a totally objective type
JEE). On this background, this particular problem was an examiner’s
193
dream. The correct answer to it was 792 . As soon as an examiner saw
this magic figure displayed prominently at the end of a solution, he
could safely skip the justification, because there is no way one can get
this answer by fluke!

Q.24 A party is attended by twenty people. In any subset of four people,

there is at least one person who knows the other three (we assume that
if X knows Y , then Y knows X). Suppose there are three people in
the party who do not know each other. How many people in the party
know everyone?

(A) 16 (B) 17 (C) 18

(D) Cannot be determined from the given data.

Answer and Comments: (B). Forget about the use of the old fash-
ioned ‘people’. We shall call them as persons. Name them from P1 to
P20 . Suppose that P1 , P2 , P3 are the ones who do not know each other.
(It would be better to say, ‘no two of whom know each other’.) Then,

33
 for every i = 4, 5, . . . , 20, out of the four persons P1 , P2 , P3 and Pi ,
someone knows all the other three from the first part of the data. This
person has to be Pi since P1 , P2 , P3 fail to know at least two persons in
this foursome.
Let S = {P4 , P5 , . . . , P20 }. What we have shown so far is that
every Pi ∈ S knows P1 , P2 , P3 . But we now claim that for any two (dis-
tinct) Pi and Pj in this set, Pi and Pj know each other. For otherwise
in the foursome {P1 , P2 , Pi , Pj }, no one will know all the other three,
contradicting the hypothesis.
Thus we have shown that all members of S know each other. Since
they also know P1 , P2 , P3 , at least these 17 persons meet the condition
of knowing everybody else. Since P1 , P2 , P3 obviously fail to satisfy this
condition, the answer is 17.

A good problem, requiring simple deductive reasoning.

Q.25 The sum of all natural numbers a such that a2 − 16a + 67 is a perfect
square is:

(A) 10 (B) 12 (C) 16 (D) 22.

Answer and Comments: (C). Completing squares, the condition

means
b2 + 3 = c2 (1)
where b = |a − 8| and c is some positive integer. This can be further
rewritten as
(c − b)(c + b) = 3 (2)
Since 3 is a prime, its only factorisation is 1 × 3. Equating the smaller
factor 1 with c − b and the larger one 3 with c + b and solving the
resulting system, we get c = 2 and b = 1. We are interested only in b.
Having got |a − 8| = 1, a is either 7 or 9. So these are the only integral
values of a which satisfy the given condition. Their sum is 16.

Another problem which can be solved without much writing. And

unlike the chocolate problem, here the time to read it is also negli-
gible. So, although a trivial problem by itself, once in a while it is

34
 welcome, especially in an entrance test which is not solely based on
such problems.

Q.26 The sides of a regular hexagon ABCDEF are extended by doubling

them (for example, BA extends to BA′ with BA′ = 2BA) to form a
bigger regular hexagon A′ B ′ C ′ D ′ E ′ F ′ as in the figure.

A’

1
B’ 60 F’
A 1 F 1

B E

C D
C’ E’

D’

Then the ratio of the areas of the bigger to the smaller hexagon is:
√
(A) 2 (B) 3 (C) 2 3 (D) π.

Answer and Comments: (B). As both the hexagons are regular, the
ratio of their areas will be the square of the ratio of their sides. Let
us take the sides of the smaller hexagon 1 unit each. Then AA′ = 1
and AF ′ = AF + F F ′ = 2. Also 6 A′ AF ′ equals 60◦ , being an external
angle of a regular hexagon. So, applying the cosine rule to the triangle
A′ AF ′ , we get

A′ F ′2 = AA′2 + AF ′2 − 2AA′ .AF ′ cos 60◦ = 5 − 2 = 3 (1)

√
So A′ F ′ = 3. Therefore
√ the ratio of the sides of the bigger to the
smaller hexagons is 3. Hence the ratio of their areas is 3.

A good, simple problem in geometry. By supplying the diagram,

the paper-setters have ensured that it can be done within a reasonable
time. A single application of the cosine rule is the only computation
involved. So this problem is well suited for Paper 1.

35
Q.27 Between 12 noon and 1 PM, there are two instants, when the hour hand
and the minute hand of a clock are at right angles. The difference in
minutes between these two instants is:
8 8 5 5
(A) 3211 (B) 3011 (C) 3211 (D) 3011 .

Answer and Comments: (A). A typical high school problem. Let

the two hands be at right angles for the first time x minutes after 12
noon. During this time the minute hand travels x minutes. But the
x
hour hand travels only 12 minutes. Being at right angles means an
angular distance of 15 minutes. This gives an equation for x, viz.
x
x− = 15 (1)
12
180
solving which, we get x = 11
.

12 12

3 9 3
9
y

Now the two hands will again be at right angles after, say, y minutes
past 12 noon. This time the minute hand has travelled y minutes and
y
the hour hand only 12 minutes. The angular distance between them is
still 15 minutes. But now it is as if the minute hand is lagging behind
the hour hand. So the equation for y is
y
y + 15 = 60 + (2)
12
540 360 8
Solving which we get y = 11
. Hence y − x = 11
= 3211 .

Another very simple problem. Its commendable feature is that it

tests the ability to write down equations from a real life data. But the
setting of a clock is very common in recreational puzzles in newspapers.
A different setting based on, say, the orbital motion of celestial bodies
would have been welcome.

36
Q.28 For which values of θ, with 0 < θ < π/2, does the quadratic polynomial
in t given by t2 + 4t cos θ + cot θ have repeated roots?
π 5π π 5π π 5π π 5π
(A) 6
or 18
(B) 6
or 12
(C) 12
or 18
(D) 12
or 12

Answer and Comments: (D). An ideal problem for those who hate
word problems. You only have to know that a quadratic has repeated
roots if and only if its discriminant vanishes. In the present problem,
this means

16 cos2 θ = 4 cot θ (1)

This resolves into two cases. (i) cos θ = 0 and (ii) 4 sin θ cos θ = 1. The
first possibility gives nothing because cos θ > 0 for 0 < θ < /pi/2. (ii)
simplifies to sin(2θ) = 12 . Since 2θ ∈ (0, π), there are two possibilities,
either 2θ = π6 or 2θ = π − π6 = 5π 6
. Correspondingly, θ = 12π
or 5π
12
.

Another absolutely straightforward problem. The relationship with

quadratic equations is superficial. Once it is removed, the problem vies
with Q.22 for triviality. One really fails to know the purpose of such
questions, both based on the same narrow area.

Q.29 Let α, β, γ be complex numbers which are the vertices of an equilateral

triangle. Then we must have:
(A) α + β + γ = 0 (B) α2 + β 2 + γ 2 = 0
(C) α2 + β 2 + γ 2 + αβ + βγ + γα = 0
(D) (α − β)2 + (β − γ)2 + (γ − α)2 = 0

Answer and Comments: (D). Obviously, whatever be the right an-

swer, it has to be cyclically symmetric in α, β, γ. As this is the case
with all the four options, we cannot rule out any of them on this prelim-
inary ground. But the first three options can be ruled out on a different
ground. We can choose an equilateral triangle whose vertices are very
close to each other but far away from 0. Then α + β + γ will be close to
3α but not 0. Similarly for α2 +β 2 +γ 2 and α2 +β 2 +γ 2 +αβ +βγ +γα.
Hence, by elimination, (D) is the correct option.
But instead of this back door entry, we give a proof that (D)
holds if the complex numbers α, β, γ correspond to the vertices A, B, C

37
respectively of an equilateral triangle. Then the directed line segments
−→ −→
AC and AB are represented, respectively, by the complex numbers
γ − α and β − α. But the former segment is obtained by rotating
the latter around the point A through an angle 60◦ . (In the diagram
below this rotation is counterclockwise. The argument can be modified
suitably, if it is clockwise.)

Cγ

60
A
α βB

This observation gives us

γ − α = eπi/3 (β − α) (1)
and hence by squaring,
(γ − α)2 = ω(α − β)2 (2)
where ω = e2πi/3 = cos( 2π
3
) + i sin( 2π
3
).
By a similar reasoning (or by cyclical symmetry) and using (1)
(β − γ)2 = ω(γ − α)2 = ω 2(α − β)2 (3)
Adding,
(α − β)2 + (β − γ)2 + (γ − α)2 = (1 + ω + ω 2 )(α − β)2 = 0 (4)
where the last step follows from the fact that ω 3 = 1 but ω 6= 1.
So, if α, β, γ represent the vertices of an equilateral triangle, then (D)
holds.

There are easier ways to derive (D). The simplest probably is to

|α − β|2
write α − β as . Write β − γ and γ − α similarly and add to
α−β
get 0. As all numerators are equal (because of equilaterality), we get
1 1 1
+ + =0 (5)
α−β β−γ γ−α

38
 which, on simplification gives

α2 + β 2 + γ 2 = αβ + βγ + γα (6)

(D) follows by multiplying both the sides by 2 and rearranging. The

converse is also true. That is, if (D) holds, then the triangle is equilat-
eral. In fact, this characterisation of equilaterality is fairly well known
(see Comment No. 12 of Chapter 8). Those who know it will get the
answer instantaneously. To avoid the consequent inequity, this ques-
tion should have been in Paper 2. If at all it is to be asked in the
multiple choice format, care should have been taken to see that all the
fake options cannot be eliminated without doing any work.

Q.30 Assume that n copies of the unit cube are glued together side by side
to form a rectangular solid block. If the number of unit cubes that are
completely invisible is 30, then the minimum possible value of n is:

(A) 204 (B) 180 (C) 140 (D) 84.

Answer and Comments: (C). Let the size of the rectangular block
be x × y × z. Then we have

n = xyz (1)

For a unit cube in this block to be completely invisible, it must not

lie on any of the six surfaces. Put differently, it must lie in the ‘core’
of the block which is obtained by removing one layer of unit cubes on
each surface. Since we are removing two layers of cubes parallel to each
surface, the size of the core is (x − 2) × (y − 2) × (z − 2). We are given
that

(x − 2)(y − 2)(z − 2) = 30 (2)

The problem asks to minimise (1) subject to the constraint (2). Here
both the objective function (i.e. the function to be minimised) and the
constraint are functions of three variables. If these variables can assume
all possible positive real values, then we need calculus or some tricks
such as the A.M.-G.M. inequality. But in the present problem, the
dimensions of the block and hence of the core are integers. As 30 is a
small number, it can be factored into a product of three positive integers

39
in only a few ways. We can try all such factorisations, one by one, and
calculate the value of n for each. For example the factorisation 6×5×1
gives x − 2 = 6, y − 2 = 5 and z − 2 = 1. Hence x = 8, y = 7, z = 3. So
n = 168. But intuitively, it is clear that to minimise n, the rectangular
block should be as close to a cube as possible. 30 is not a perfect cube.
But the factorisation 5 × 3 × 2 is the closest we can come. For this
factorisation, n = 7 × 5 × 4 = 140.

A good problem, based on a realistic situation. Sometimes, we

want to enclose some precious cubes made of gold, say, by a collection
of relatively cheaper cubes of the same size, to protect against theft.
The problem tells you how many cheaper cubes will be needed.

40
 ISI BStat-BMath-UGA-2018 Paper 2 with Comments

There are EIGHT questions. Time allowed is 2 hours. Each question

carries 10 marks.

Notations: In the following, IN = {1, 2, 3, . . .} denotes the set of all

natural numbers, IR denotes the set of real numbers.

Q. 1 Find all pairs x, y with x, y real satisfying the equations:

x+y
 
sin = 0, |x| + |y| = 1.
2

Solution and Comments: The best method is by drawing a diagram

in which the solution set of each equation is shown. Let A be the
solution set of the first equation. That is
x+y
 
2
A = {(x, y) ∈ IR : sin = 0} (1)
2
Then (x, y) ∈ A if and only if x + y = 2nπ for some integer n. This is a
disjoint union of an infinite family of parallel lines Ln for every integer
n, where Ln = {(x, y) ∈ IR2 : x + y = 2nπ}. Only one such line, viz.
L0 = {(x, y) : x + y = 0} is shown as a dotted thick line in the figure
below.
y

(0,1)

B2
(−1/2, 1/2) B1
P

x
(−1,0) O (1,0)

B Q (−1/2, −1/2)
3
B4
L0
(0,−1)

41
 The solution set, say B, of the second equation, viz. |x| + |y| = 1.
consists of four line segments B1 , B2 , B3 and B4 depending upon the
quadrant in which (x, y) lies. They are all shown in the figure.
It is cleat that L0 intersects only B2 and B4 at the points P = (−1/2, 1/2)
and Q = (1/2, −1/2). As for the other lines Ln , in the set A, they all lie
outside the square B. (That is why they are not shown in the figure.)
So, there are only two solutions to the system of equations, viz. (− 12 , 21 )
and ( 12 , − 12 ).

A fairly simple problem once the idea of drawing the solution sets of
the two equations strikes.
Q. 2 Suppose P Q and RS are two chords of a circle intersecting at a point
O. It is given that P O = 3 cm and SO = 4 cm. Moreover the area of
the triangle P OR is 7 cm2 . Find the area of the triangle QOS.

Solution and Comments: Since the quadrilateral P RQS is cyclic,

6 P RO = 6 SQO = θ (say) (1)
Hence the triangles P OR and SOQ are similar.

4 O
S θ R
3
P

The ratio of the sides opposite to the angles θ is 43 . Hence the ratio of
9
their areas is 16 . Therefore
16 16 112
△QOS = ∆P OR = ×7= (2)
9 9 9
cm2 .

A very simple problem. Should have been asked in Paper 1.

42
Q. 3 Let f : IR −→ IR be a continuous function such that for all x and for
all t ≥ 0,
f (x) = f (et x).
Show that f is a constant function.

Solution and Comments: We first show that if x, y are of the same

sign (i.e. both positive or both negative) then f (x) = f (y). In this
case both the ratios xy and xy are positive and one of them, say the first
one, is at least 1. So taking t = log( xy ) we have y = et x and t ≥ 0.
Therefore from the hypothesis f (x) = f (et x) = f (y).
Thus f is some constant, say C1 , on the positive x-axis and some
constant C2 on the negative x-axis. By continuity of f at 0, C1 = C2 .
But then f is constant on IR.

Another very simple problem.

Q. 4 Let f : (0, ∞) −→ IR be a continuous function such that for all
x ∈ (0, ∞),
f (2x) = f (x).
Show that the function g defined by the equation
Z 2x dt
g(x) = f (t) for x > 0
x t
is a constant function.

Solution and Comments: The function g(x) is a function defined by

an integral. As the integrand is a continuous function of t and both the
upper and lower limits of the integral (viz. 2x and x respectively), are
differentiable functions of x, g(x) is differentiable on (0, ∞). Moreover,
by Leibnitz rule (which is a combination of the Fundamental Theorem
of Calculus and the Chain Rule), we have
f (2x) d f (x) d
g ′ (x) = (2x) − (x)
2x dx x dx
f (2x) f (x)
= −
x x
f (2x) − f (x)
= =0 (1)
x
43
 for all x > 0 from the hypothesis. So g ′ vanishes identically and hence
g is constant.

The proof is very similar to that of the assertion that if f is con-

Z a+T
tinuous and periodic with period T , then the integral f (t)dt is
a
independent of a, i.e. is a constant function of a. The result is still
true even if f is not continuous, as long as it is integrable. But this
proof becomes inapplicable.
Q. 5 Let f : IR −→ IR be a differentiable function such that its derivative f ′
is a continuous function. Moreover, assume that for all x ∈ IR,
1
0 ≤ |f ′(x)| ≤ .
2

Define a sequence of real numbers {an }n∈IN by:

a1 = 1

an+1 = f (an ) for all n ∈ IN.

Prove that there exists a positive real number M such that for all
n ∈ IN,
|an | ≤ M.

Solution and Comments: In more common terminology, the prob-

lem asks to show that the sequence {an }n∈IN is bounded. In fact, the
formulation would have been less clumsy had this term been used. It
is avoided apparently to do justice to those candidates who may not
know it. While such an intention, if at all present, is commendable,
sometimes it has the very opposite effect. The concept of a bounded
sequence is familiar to most candidates even at the HSC level and often
used in statements such as ‘Every convergent sequence is bounded, but
not conversely’. When some commonly used term is replaced by its
definition, one tends to get confused as to whether the paper setters
have something else in mind. If you say ‘the woman who gave birth to
me’, the first query would be ‘You mean your mother?’ and if you are
not there to answer it, then the first suspicion is that although your
mother gave birth to you, she probably died or abandoned you soon!

44
 Let us leave aside this verbal objection and get down to the problem.
We are given a1 = 1 but not a2 = f (a1 ) since the function f is not given.
If a2 = a1 , then a3 = f (a2 ) = f (a1 ) and by induction on n, it will follow
that an = a1 for all n ∈ IN. Then the sequence {an }n∈IN would be the
constant sequence {1} and we can take M = 1.
So we assume a2 = f (a1 ) 6= a1 . In that case, by Lagrange’s Mean
Value Theorem (LMVT), there exists some c1 in the interval (a1 , a2 )
(or the interval (a2 , a1 ) as the case may be depending upon whether
a1 < a2 ot a2 < a1 ), such that

a3 − a2 = f (a2 ) − f (a1 ) = f ′ (c1 )(a2 − a1 ) (1)

Taking absolute values of both the sides and using the hypothesis about
|f ′(x)|, we get
1
|a3 − a2 | = |f ′ (c1 )||a2 − a1 | ≤ |a2 − a1 | (2)
2

We now repeat the argument for a2 and a3 to get some c2 (in case
a2 6= a3 ) such that

|a4 − a3 | = |f ′ (c2 )||a3 − a2 |

1
≤ |a3 − a2 |
2
1
≤ |a2 − a1 | (3)
4
where in the last step we have used (2).
By induction on n, it follows that
1
|an − an−1 | ≤ |a2 − a1 | (4)
2n−2
for all n ≥ 2. (This includes the possibility that for some r, ar+1 =
f (ar ) = ar . If this happens then the L.H.S. of (4) is 0 for all n ≥ r + 1.)
As |a2 −a1 | is a fixed positive number, (4) shows that the consecutive
terms of the sequence {an }n∈IN are getting closer and closer. This
does not automatically prove that the sequence is bounded. A classic
counterexample is the sequence of harmonic numbers, defined by,

45
Hn = 1 + 12 + 13 + . . . + n1 . Here |Hn − Hn−1 | = n1 which gets smaller
and smaller as n increases. Still, it can be shown that the sequence
{Hn }n∈IN is not bounded.
But the situation in our problem is more pleasant. Because of
the factor 2n−2 in the denominator of the R.H.S. of (4), the difference
between the consecutive terms of the sequence {an }n∈IN tends to 0
much more rapidly than n1 . In fact, using the triangle inequality and
the formula for the sum of a geometric progression, we get

|an − a1 | = |(an − an−1 ) + (an−1 − an−2 ) + . . . + (a3 − a2 ) + (a2 − a1 )|

n
X
≤ |ak − ak−1 |
k=2
n
X 1
≤ |a2 − a1 | (5)
k=2 2k−2
1
1− 2n−1
= 1 |a2 − a1 | (6)
2
≤ 2(|a2 − a1 |) (7)

where in (5) we have used (4).

To finish the solution, we apply the triangle inequality again to get
that for all n ≥ 2,

|an | ≤ |an − a2 | + |a2 | ≤ 2|a2 − a1 | + |a2 | ≤ 3|a2 | + 2|a1 | (8)

As this also holds for n = 1, we can take M as 3|a2 | + 2|a1 |.

The essence of the solution was that by applying LMVT and the
hypothesis we showed that for all a, b ∈ IR
1
|f (b) − f (a)| ≤ |b − a| (9)
2
and then to conclude that {an }n∈IN is bounded we crucially used that
1
2
< 1.
More generally, a function for which there exists some λ ∈ [0, 1)
such that for all a, b ∈ IR

|f (b) − f (a)| ≤ λ|b − a| (10)

46
 is called a contraction, because it brings the points of the domain
closer. So, we have used LMVT and the hypothesis to show that f is
a contraction. The subsequent part of the solution is applicable to any
contraction. That is, if f : IR −→ IR is a contraction, then starting with
any a1 ∈ IR, the recursively defined sequence an = f (an−1 ) for n ≥ 2, is
bounded. Actually, a stronger result is true. This sequence of iterates,
as it is called, is not only bounded, but is convergent and its limit is the
unique fixed point of f . This result is called the contraction fixed
point theorem. The proof is also essentially the same, based on the
∞
λn is convergent for |λ| < 1.
X
fact that the geometric series
n=0
The contraction fixed point theorem also holds for functions f from
IRn to IRn . The only change is that in the the definition of contraction,
(1) is replaced by
||f(b) − f(a)|| ≤ λ||b − a|| (11)
for all a, b ∈ IRn . That is, we replace the absolute value by the eu-
clidean distance. In fact, the theorem holds more generally, for what
are called complete metric spaces and has important applications
in proving the existencxe of solutions of certain differential equations.
We note in passing that the contraction fixed point theorem may
fail if (10) is replaced by the weaker hypothesis
|f (b) − f (a)| < |b − a| (12)
for all a, b in the domain. For example, the function f : [2, ∞) −→
[2, ∞) defined by f (x) = x + x1 satisfies (12) as we see by applying
LMVT and the fact that f ′ (x) = 1 − x12 < 1 for all x ∈ [0, ∞). A
function which satisfies (12) is called a weak contraction.
Q. 6 Let a ≥ b ≥ c > 0 be real numbers such that for all n ∈ IN, there
exist triangles of side lengths an , bn , cn . Prove that the triangles are
isosceles.

Solution and Comments: It suffices to show that a = b. For every

n ∈ IN, we have an ≥ bn ≥ cn > 0. They will form a triangle if and
only if bn + cn > an which translates as
un + v n > 1 (1)

47
 b
where u = a
and v = ac .
Now, if a > b then a > c and so both u, v will lie in the interval
(0, 1). But then un as well as v n tends to 0 as n → ∞. That will
be a contradition if we take limits of both the sides of (1) as n → ∞.
So a = b. Therefore for every n, the triangle with sides an , bn , cn is
isosceles.

A simple problem based on an interesting application of the fact

that xn → 0 if |x| < 1. Note that we cannot assert that the triangles
are equilateral. In fact, for any a > c > 0, we can form a triangle with
sides an , an and cn .

Q. 7 Let a, b, c ∈ IN be such that

a2 + b2 = c2 and c − b = 1.

Prove that

(i) a is odd,
(ii) b is divisible by 4,
(iii) ab + ba is divisible by c.

Solution and Comments: Since c = b + 1, a direct substitution into

the equation given gives

a2 = 2b + 1 (1)

which shows that a2 and hence a is odd. Also from (1) we get

2b = (a − 1)(a + 1) (2)

The integers a − 1 and a + 1 is a pair of consecutive even integers. So

one of them is divisible by 4. Hence 2b is divisible by 8, whence b is
divisible by 4.
It only remains to prove (iii). Since b = c − 1, we can rewrite (1) as

a2 = 2c − 1 (3)

48
 We already showed that b = 4m for some positive integer m. So, from
(3),
ab = a4m = (a2 )2m = (2c − 1)2m (4)
When the last term is expanded by the binomial theorem, all terms
except the constant term will be divisible by c (in fact by c2 ). The
constant term is (−1)2m which is simply 1. So we have
ab = pc + 1 (5)
for some integer p. As for ba , we write b as c − 1. Then
ba = (c − 1)a (6)
once again all the terms in the binomial expansion of the R.H.S., except
the constant term are divisible by c. But this time the constant term
is (−1)a which equals −1 since a is odd. Therefore
ba = qc − 1 (7)
for some integer q.
From (5) and (7),
ab + ba = (p + q)c (8)
which proves (iii).

A simple problem about Pythagorean triplets. Examples of such

triples are (3, 4, 5) and (5, 12, 13). But because of the restriction that
c − b = 1, the general form of the Pythagorean triplets given in the
comments to Q.17 of Paper 1 is not needed here and that makes the
problem elementary. A more sophisticated formulation of the solution
to (iii) is possible if we use the concept of congruency modulo c. In
essence, the proof of (iii) boils down to showing that ab ≡ 1 modulo
c and ba ≡ −1 modulo c. But we have avoided this terminology and
worked directly using the binomial theorem.
Q. 8 Let n ≥ 3. Let A = ((aij ))1≤i,j≤n be an n × n matrix such that
aij ∈ {1, −1} for all 1 ≤ i, j ≤ n. Suppose that
ak1 = 1 for all 1 ≤ k ≤ n and

49
 n
X
aki akj = 0 for all i 6= j.
k=1

Show that n is a multiple of 4.

Solution and Comments: We are given that the first column of A has
all 1’s in it. The second part of the hypothesis is popularly expressed
by saying that every two distinct columns of A are orthogonal to each
other, because their inner product (regarded as n-dimensional column
vectors) is 0. Since every column of A other than the first column
is orthogonal to the first column and has only 1 and −1 as possible
entries, exactly half of them are 1 and the other half must be −1. In
particular, this shows that n is even, say n = 2m. Our task is to show
that m is itself even.
For this we normalise the second column of the matrix A. We
know that exactly m entries of the second column of A are 1 and
the remaining are −1. Since any reshuffling of rows does not affect the
orthogonality condition, we may assume, without loss of generality that
the first m entries of the second column are 1 and the last m entries
are −1. Thus, if n = 4, the submatrix of A consisting of the first two
columns is
1 1
 
 1 1 
(1)
 
 1 −1 
 

1 −1
Now consider the third column of A, which exists because it is given
that n ≥ 3. Because of orthogonality with the first column, m of its
entries are 1 and the remaining m entries are −1. This we already know.
Let r be the number of the first m entries in the third column which
equal 1. Then the remaining m − r entries among the first entries equal
−1 each. Also in the last m entries of the third column, there must be
m−r entries that equal 1 and r entries that equal −1 (because the total
of each type is m). We already know the entries of the (normalised)
second column. But now the inner product of the third column and
the second column can be split into an ‘upper’ and a ‘lower’ part as
n
X 2m
X m
X 2m
X
ak2 ak3 = ak2 ak3 = ak2 ak3 + ak2 ak3 (2)
k=1 k=1 k=1 k=m+1

50
In the first summation in the last term, ak2 = 1 for all 1 ≤ k ≤ m. But
ak3 equals 1 for r values of k and equals −1 for the remaining m − r
values of k. We don’t know exactly which ones are 1 and which ones
are −1. But regardless of that, the ‘upper’ part of the inner product
equals
m
X
ak2 ak3 = r − (m − r) = 2r − m (3)
k=1

On the other hand, for the second summation of the last term in (2),
ak2 = −1 for all m + 1 ≤ k ≤ m. So by an analogous reasoning we have
2m
X
ak2 ak3 = r − (m − r) = 2r − m (4)
k=m+1

As the leftmost term of (2) is given to be 0, we get

0 = 2r − m + 2r − m = 4r − 2m (5)

which shows that m = 2r and hence n = 4r. Thus n is divisible by 4.

We can complete (1) to a 4 × 4 matrix A

1 1 1 1
 
 1 1 −1 −1 
(6)
 
1 −1 1 −1 
 

1 −1 −1 1

which satisfies all the conditions in the data.

Matrices which satisfy the conditions in the problem are called
Hadamard matrices. We just gave an example, of a ‘smallest’ Hadamard
matrix. There is a way to build larger Hadamard matrices starting
from smaller ones, by taking what is called their tensor product. So
Hadamard matrices of arbitrarily large size are possible to construct.
Interested readers will find the details on p.704 of the author’s Applied
Discrete Structures, New Age International (P) Ltd., New Delhi.

51
 CONCLUDING REMARKS
By and large, most problems this year are simpler as compared with those
in the ISI entrance tests of recent years. In fact, many of them are triv-
ial. This includes Q.2 (about teams of officers), Q.3 (about Roger and the
tennis balls), Q.12 (about symmetric arrangement of balls), Q.23 (about con-
ditional probability) and Q.25 (about a quadratic assuming integral values).
Both the problems (Q.22 and Q.28) on trigonometric equations are absolutely
straightforward.
A few questions in Paper 1 are trivial, once the essential idea is cor-
rectly grasped. The masterpiece is Q.8 where the condition for an arithmetic
progression is given in an intentionally clumsy form. Q. 13 and Q.14 also
falls under this category. Q.15 has an ambiguous formulation. Once it is
understood, it is a high school problem about time, work and rate. The non-
existence of any solutions to the diophantine equation in Q.17 also hinges
simply on parity considerations.
As it happens every year, there are a few questions in Paper 1, where the
correct answer can be identified simply by eliminating the wrong options.
These include Q.5 (about the possible numbers of roots and local extrema
of a fourth degree polynomial). Q.7 (about matrices satisfying A4 = A)
and, most prominently, Q.29 (about characterisation of equilateral triangles
in terms of complex numbers).
Among the better questions in Paper 1, Q.18 is probably the best. It
combines the representation of an integer in a given base and logarithms.
Q.10 is also good but has an unintended easy solution using power series
for sin x and cos x. Q.4 (about permutations which move every point by at
most 1), Q.16 (about maximising the area of a rectangle whose sides contain
the vertices of a given rectangle), Q.19 (about finiteness of the number of
solutions of sin x/x = c) and Q.27 (about the two hands of a clock being at
right angles) are more suitable in Paper 2, with the modifications suggested.
Q.24 gives the impression that graph theory will be useful. But the solution
requires only pure deductive logic. Q.30 is a nice problem of constrained
integral optimisation in a real life situation.
Over-all, Paper 1 is a mixed bag. Once again one wishes that by removing
some absolutely easy questions and retaining only 15 to 20 relatively good
ones, the test would have been a more valid measure of a candidate’s abilities.
It is Paper 2 that is really disappointing in comparison with the past
years. There is not even one question which can be called unusual and really
makes you think. The first four questions belong more to Paper 1. The

52
Pythagorean triplets in Q.7 are too restricted and do not need their general
structure. Instead, the problem reduces to a divisibility problem. Q.5 and
Q.8 are forerunners of the contraction fixed point theorem and Hadamard
matrices respectively. The former is based on the fact that xn → 0 as x → ∞
if |x| < 1. Q.6 is also a cute application of this fact. This duplication could
have been avoided. Instead, a question about inequalities would have been
welcome as this topic is not even paid a lip service this year.

53