cmi preparation

7/21/2019 Cmi Preparation

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Chennai Mathematical Institute

BSc (Honours) Mathematics and Computer Science

Topics covered in entrance examination

The entrance examination for the B.Sc. (Hons) Maths and Computer Science program, is atest of aptitude to do Mathematics. We expect the students to be familiar with the followingtopics.

Algebra: Arithmetic and geometric progressions, arithmetic mean, geometric mean,harmonic mean and related inequalities, polynomial equations, roots of polynomials,matrices, determinants, linear equations, solvability of equations, binomial theorem andmultinomial theorem, permutations and combinations, mathematical induction. primenumbers and divisibility, GCD, LCM, modular arithmetic logarithms, probability.

Geometry: Vectors, triangles, two dimensional geometry of Conics - straight lines,parabola, hyperbola, ellipses and circles, tangents, measurement of area and volume,co-ordinate geometry.

Trigonometry: addition, subtraction formulas, double-angle formulas.

Calculus: Limits, continuity, derivatives, integrals, indefinite and definite integrals,maxima and minima of functions in a single variable, series and sequences, convergencecriterion.

Complex numbers, roots of unity.

Suggested reading material

The following books are suggested, in addition to the Class XI and XII Mathematics Text-books of the National Council of Educational Research and Training, New Delhi.

1. V. Krishnamoorthy, C.R. Pranesachar, K. N. Ranganathan, B.J.Venkatachala: Chal-lenge and Thrill of Pre-College Mathematics, New Age International Publishers.

2. M.R. Modak, S.A. Katre, V.V. Acharya: An Excursion in Mathematics, BhaskaracharyaPratishtan (Pune).

3. D. Fomin, S. Genkin, I. Itenberg: Mathematical Circles: Russian experience, Universi-ties Press (Hyderabad) 1998.

4. H.S. Hall, S.R. Knight: Higher Algebra.


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CMI BSc (Hons) Math. and C.S., Sample Examination 1

PART A

Instructions:

There are 13 questions in this part. Each question carries4 marks. Answer all questions.

1. For a polynomialf(x), letf(n)(x) denote thenthderivativefor n 1 and f(0)(x) =f(x). Is the following true or false? Give brief reasons.

f(n)(a) = 0, f o r n= 0, 1, . . . , k (xa)(k+1) divides f(x)

2. (a) Find the value of

nk=1 k k! as a function ofn.

(b) True or false ? Give brief reasons.

1000

k=2

1

logkN =

1

log1000!

N

3. Letp(x) be a polynomial such that when divided by (x1)it leaves the remainder 2 and when divided by (x 2) itleaves the remainder 1. What is the remainder when it isdivided by (x 1)(x 2)?

4. Prove that for any fixed n > 2, among the two integers2n 1 and 2n + 1, at the most one of them can be a prime.

5. Show that n! (n+1

2 )n

,n 1.6. Show that the sum

n

k=1

k! =m2

for any integer m, for n 4.


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7. Give an example of a subset Sof the plane such that the

following property is satisfied:

There exist infinitely many lines in the plane such that the

intersection S is a line segment of any length howeverlarge.

8. True or false ? Give brief reasons. There exist non-zeroploynomialsf(x)with integer coefficients of any degreed 5 which vanish atx= 2, 1 +

3 and 1 +5.9. If the sum of 113 terms of an arithmetic progression is equal

to 6780, then find the 57th term of the arithmetic progres-sion. Give an example of such an arithmetic progression.

10. Show that ifp and p+ 2 with p 5, are both primes thenthe number p+ 1 is always divisible by 6.

11. Show that for all real numbers a, b, c,

(a+b+c)2

3(a2

+b2

+c2

)Further show that 3 is the smallest real number with thisproperty.

12. True or false ? Give brief reasons. Ifa, b are positive inte-gers then the number

(a+

b)n + (a

b)n

is an integer for all values ofn.

13. True or false ? Give brief reasons. The number of commontangents to the circlesx2 + y2 = 1 andx2 + y2 4x + 3 = 0is 1.


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Part B

Instructions:

There are 6 questions in this part. Each question carries 8marks. Answer all questions.

1. Let A(x1, y1), B(x2, y2) and C(x3, y3) form a triangle withcircumcentre (0, 0). Show thatABC is an equilateral tri-

angle if and only if

x1+x2+x3= y1+y2+y3= 0

2. Show the following:

(a) limx0sin(x) log(x) = 0

(b) limn xn

n! = 0

(c) limx

(cos(x

))x = 1, where is a constant.

3. In a jail with 100 rooms, all locked initially, 100 riotersbreak in and disturb the rooms in the following way. Firstone stops at all rooms and opens them all. Second riotersstops at rooms numbered 2, 4, 6, . . .and locks open rooms,and leaving the other rooms as they were. The third ri-oter stops at rooms numbered 3, 6, 9, . . .and again opens alocked room and locks an open room, leaving others undis-turbed. And this process continues. After all the 100 riotershave left which rooms would be open?

4. (a) If two sides of a triangle are given, then show thatthe area of this triangle is maximum if the sides areperpendicular to each other.

USE SERIES EXPANSION

1infinity


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(b) Construction: For any given positive real number ,

find a point D on a line AB such that the differenceAD2 BD2 =. (Note that the point D need not bewithin the segment AB).

5. Show that if [x] denote the greatest integer x, thenn

3

[n

3]

is a natural number divisible by 3 for all values ofn.6. Is it possible to remove one square from a 5 5 board so

that the remaining 24 squares can be covered by eight 31rectangles? If yes, find all such squares.

(Hint: A domino is a 2 1 rectangle. As you may know,if two diagonally opposite squares of an ordinary 8 8chessboard are removed, the remaining 62 squares cannotbe covered by 31 non-overlapping dominos. The reason

being, after removing the two corners 32 squares of onecolour and 30 of the other are left. No matter how you placea domino it will cover one white and one black square.)


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Chennai Mathematical Institute

Entrance Examination for B.Sc. (Mathematics & Computer Science) May 2010

Duration: 3 hours Maximum Score: 100

PART A

Instructions:

There are 13 questions in this part. Each question carries 4 marks. Answer all questions.

1. Find all x [, ] such that cos 3x+ cos x= 0.2. A polynomial f(x) has integer coefficients such that f(0) and f(1) are both odd num-

bers. Prove that f(x) = 0 has no integer solutions.

3. Evaluate:

(a) limx1

n nk=1xk1 x

(b) limx0

e1/x

x

4. Show that there is no infinite arithmetic progression consisting of distinct integers allof which are squares.

5. Find the remainder given by 389 786 when divided by 17.6. Prove that

2

0! + 1! + 2!+

3

1! + 2! + 3!+ + n

(n 2)! + (n 1)! +n! = 1 1

n!

7. Ifa, b, care real numbers> 1, then show that

1

1 + loga2bca

+ 1

1 + logb2cab

+ 1

1 + logc2abc

= 3

8. If 8 points in a plane are chosen to lie on or inside a circle of diameter 2cm then showthat the distance between some two points will be less than 1cm.

9. Iff(x) = xn

n!+

xn1

(n 1)!+ +x+ 1, then show that f(x) = 0 has no repeated roots.

10. Given cos x+ cos y+ cos z = 3

3

2 and sin x+ sin y+ sin z =

3

2 then show that x =

6+ 2k, y=

6+ 2, z=

6+ 2m for some k,, m Z.

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11. Using the fact that

n is an irrational number whenever n is not a perfect square,show that

3 +

7 +

21 is irrational.

12. In an isocelesABC with A at the apex the height and the base are both equal to1cm. Points D, E and F are chosen one from each side such that BDEF is a rhombus.Find the length of the side of this rhombus.

13. Ifbis a real number satisfying b4 + 1

b4= 6, find the value of

b +

i

b

16wherei =

1.

PART B

Instructions:

There are seven questions in this part. Each question carries 8 marks.

Answer any six questions.

1. Let a1, a2,...,a100 be 100 positive integers. Show that for somem, nwith 1 m n 100,

ni=mai is divisible by 100.

2. In ABC, BE is a median, and O the mid-point of BE. The line joining A and Omeets BC at D. Find the ratio AO : OD (Hint: Draw a line through E parallel to AD.)

3. (a) A computer program prints out all integers from 0 to ten thousand in base 6 usingthe numerals 0,1,2,3,4 and 5. How many numerals it would have printed?

(b) A 3-digt number abc in base 6 is equal to the 3-digit number cba in base 9. Findthe digits.

4. (a) Show that the area of a right-angled triangle with all side lengths integers is aninteger divisible by 6.

(b) If all the sides and area of a triangle were rational numbers then show that thetriangle is got by pasting two right-angled triangles having the same property.

5. Prove that b1

alogbx dx >ln bwhere a,b >0, b = 1.

6. Let C1, C2 be two circles of equal radii R. IfC1 passes through the centre ofC2 prove

that the area of the region common to them is R2

6 (4 27).7. Leta1, a2, . . . , anand b1, b2, . . . , bnbe two arithmetic progressions. Prove that the points

(a1, b1), (a2, b2), . . . , (an, bn) are collinear.

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Entrance Examination for CMI BSc (Mathematics & Computer Science) May 2011

Attempt all problems from parts A and C. Attempt any 7 problems from part B.

Part A. Choose the correct option and explain your reasoning briefly. Each problem is

worth 3 points.

1. The word MATHEMATICS consists of 11 letters. The number of distinct ways torearrange these letters is(A) 11! (B) 11!

3 (C) 11!

6 (D) 11!

8

2. In a rectangle ABCD, the length BC is twice the width AB. Pick a point P on side BCsuch that the lengths of AP and BC are equal. The measure of angle CPD is(A) 75 (B) 60 (C) 45 (D) none of the above

3. The number of with 0


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Part B. Attempt any 7 problems. Explain your reasoning. Each problem is worth 7points.

1. In a business meeting, each person shakes hands with each other person, with the

exception of Mr. L. Since Mr. L arrives after some people have left, he shakes hands onlywith those present. If the total number of handshakes is exactly 100, how many peopleleft the meeting before Mr. L arrived? (Nobody shakes hands with the same person morethan once.)

2. Show that the power ofxwith the largest coefficient in the polynomial (1 + 2x3

)20 is 8,i.e., if we write the given polynomial as

iaix

i then the largest coefficient ai is a8.

3. Show that there are infinitely many perfect squares that can be written as a sum of sixconsecutive natural numbers. Find the smallest such square.

4. Let S be the set of all 5-digit numbers that contain the digits 1,3,5,7 and 9 exactly once(in usual base 10 representation). Show that the sum of all elements of S is divisible by11111. Find this sum.

5. It is given that the complex number i 3 is a root of the polynomial 3x4 + 10x3 +Ax2 +Bx 30, where Aand B are unknown real numbers. Find the other roots.

6. Show that there is no solid figure with exactly 11 faces such that each face is a polygon

having an odd number of sides.

7. To find the volume of a cave, we fit X, Y and Z axes such that the base of the cave isin the XY-plane and the vertical direction is parallel to the Z-axis. The base is the regionin the XY-plane bounded by the parabola y2 = 1 x and the Y-axis. Each cross-sectionof the cave perpendicular to the X-axis is a square.

(a) Show how to write a definite integral that will calculate the volume of this cave.(b) Evaluate this definite integral. Is it possible to evaluate it without using a formula forindefinite integrals?

8. f(x) =x3 +x2 +cx+d, where c and d are real numbers. Prove that ifc > 13

, then fhas exactly one real root.

9. A real-valued function f defined on a closed interval [a, b] has the properties thatf(a) = f(b) = 0 and f(x) =f(x) + f(x) for all x in [a, b]. Show that f(x) = 0 for all xin [a, b].

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Part C. Explain your reasoning. Each problem is worth 10 points.

1. Show that there are exactly 16 pairs of integers (x, y) such that 11x+ 8y+ 17 = xy.You need not list the solutions.

2. A function g from a set X to itself satisfies gm =gn for positive integers mand nwithm > n. Here gn stands for g g g (ntimes). Show that gis one-to-one if and only ifg is onto. (Some of you may have seen the term one-one function instead of one-to-onefunction. Both mean the same.)

3. In a quadrilateral ABCD, angles at vertices B and D are right angles. AM and CN arerespectively altitudes of the triangles ABD and CBD. See the figure below. Show that BN= DM.

In this figure the angles ABC, ADC, AMD and CNB are right angles.

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Entrance Examination for BSc Programmes at CMI, May 2012

Attempt all 5 problems in part A, each worth 6 points. Attempt 7 out of the 9 problems in part B, each

worth 10 points.

Part A. (5 problems

6 points = 30 points.) Clearly explain your entire reasoning.

1. Find the number of real solutions to the equation x= 99 sin(x).

2. A differentiable function f : R R satisfies f(1) = 2, f(2) = 3 and f(3) = 1. Show that f(x) = 0for some x.

3. Show that ln(12)ln(18) is irrational.

4. Show that

limx

x100 ln(x)

ex tan1(3 + sin x)= 0 .

5. (a) n identical chocolates are to be distributed among the k students in Tinkus class. Find theprobability that Tinku gets at least one chocolate, assuming that the nchocolates are handed out oneby one in n independent steps. At each step, one chocolate is given to a randomly chosen student, witheach student having equal chance to receive it.

(b) Solve the same problem assuming instead that all distributions are equally likely. You are giventhat the number of such distributions is

n+k1k1

. (Here all chocolates are considered interchangeable

but students are considered different.)

Part B. (9 problems10 points = 90 points.) Clearly explain your entire reasoning.

Attempt at least 7 problems. You may solve only part of a problem and get partial credit. If you cannotsolve an earlier part, you may assume it and proceed to the next part. For al l such partial answers,

clearly mention what you are solving and what you are assuming.

1. a) Find a polynomial p(x) with realcoefficients such that p(

2 +i) = 0.

b) Find a polynomial q(x) with rationalcoefficients and having the smallest possible degree such thatq(

2 +i) = 0. Show that any other polynomial with rational coefficients and having

2 +i as a roothas q(x) as a factor.

2. a) Let E, F, G and H respectively be the midpoints of the sides AB, BC, CD and DA of a convexquadrilateral ABCD. Show that EFGH is a parallelogram whose area is half that of ABCD.

b) Let E = (0, 0), F = (0,1), G = (1,1), H = (1, 0). Find all points A = (p, q) in the first quadrantsuch that E, F, G and H respectively are the midpoints of the sides AB, BC, CD and DA of a convexquadrilateral ABCD.

3. a) We want to choose subsets A1, A2, . . . , Ak of{1, 2, . . . , n} such that any two of the chosen subsetshave nonempty intersection. Show that the size k of any such collection of subsets is at most 2n1.

b) For n >2 show that we can always find a collection of 2n1 subsetsA1, A2, . . . of{1, 2, . . . , n} suchthat any two of the Ai intersect, but the intersection of all Ai is empty.

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4. Define

x=10

i=1

1

10

3

1

1 + ( i103

)2 and y=

9

i=0

1

10

3

1

1 + ( i103

)2.

Show that a) x < 6 < y and b) x+y

2 < 6 . (Hint: Relate these sums to an integral.)

5. Using the steps below, find the value ofx2012 +x2012, where x+x1 =5+12 .

a) For any realr, show that|r+r1| 2. What does this tell you about the given x?b) Show that cos(5 ) =

5+14 , e.g. compare sin(

25 ) and sin(

35 ).

c) Combine conclusions of parts a and b to express x and therefore the desired quantity in a suitableform.

6. For n >1, a configurationconsists of 2n distinct points in a plane, n of them red, the remaining nblue, with no three points collinear. A pairingconsists ofn line segments, each with one blue and onered endpoint, such that each of the given 2n points is an endpoint of exactly one segment. Prove the

following.a) For any configuration, there is a pairing in which no two of the n segments intersect. (Hint: considertotal length of segments.)

b) Given n red points (no three collinear), we can place n blue points such that any pairing in theresulting configuration will have two segments that do not intersect. (Hint: First consider the casen= 2.)

7. A sequence of integers cn starts with c0 = 0 and satisfies cn+2= acn+1+ bcn forn 0, where a andb are integers. For any positive integer kwithgcd(k, b) = 1, show that cn is divisible byk for infinitelymany n.

8. Letf(x) be a polynomial with integer coefficients such that for each nonnegative integern, f(n) = aperfect power of a prime number, i.e., of the form pk, where p is prime and k a positive integer. (pandkcan vary with n.) Show thatfmust be a constant polynomial using the following steps or otherwise.

a) If such a polynomial f(x) exists, then there is a polynomial g(x) with integer coefficients such thatfor each nonnegative integer n, g(n) = a perfect power of a fixedprime number.

b) Show that a polynomial g(x) as in part a must be constant.

9. LetNbe the set of non-negative integers. Suppose f :N Nis a function such that f(f(f(n))) f(1) and f(2) > f(3)). Sothe maximum must be at an interior point x and then f(x) = 0. OR By the meanvalue theorem, f(y) = 1 > 0 for some y(1, 2) and f(z) =2 cfor anyx, e.g. c= tan1(0.04)

will work since > 3.12, sin(x) 1 and tan1 is an increasing function. Moreoverln(x) < x for x >0. So the given ratio is sandwiched between 0 andx101/cex. Now useLHospitals rule repeatedly.

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Consider the diagonals AC and BD. By the basic proportionality theorem in triangle ABC,we get that EF and AC are parallel and AC = 2 EF. Moreover, ABC and EBF are similar.Using triangles ADC and HDG, we similarly get that AC is parallel to HG, AC = 2 HG.Thus EF and HG are parallel. Likewise FG and EH are parallel (both parallel to BD), soEFGH is a parallelogram. Also by similarity, Area(ABC) = 4 Area(EBF), Area(ADC) =

4 Area(HDG), Area(BAD) = 4 Area(EAH) and Area(BCD) = 4 Area(FCG). (Note. Sofar convexity of ABCD is unnecessary. But the next steps need it, draw pictures and see.)

Area(EFGH) = Area(ABCD) [Area(EBF) + Area(FCG) + Area(GDH) + Area(HAE)]= Area(ABCD) 14 [Area(ABC)+ Area(BCD) + Area(CDA) + Area(DAB)]= Area(ABCD) 12Area(ABCD) =

12Area(ABCD).

b) Let E = (0, 0), F = (0, 1), G = (1, 1), H = (1, 0). Find all points A = (p, q) in thefirst quadrant such that E, F, G and H respectively are the midpoints of the sides AB,BC, CD and DA of a convex quadrilateral ABCD.

If A = (p, q) is such a point, then E = (0,0) being the midpoint of AB is equivalent tohaving B = (p, q). Similarly we get C = (p, q 2), D = (2 p, q). In particular AC =BD = 2, AC is vertical and BD horizontal. By the reasoning in part a), these facts implythat the quadrilateral constructed from the midpoints of the sides of ABCD is a square ofside 1. So we just need to ensure that the listed coordinates make ABCD into a convexquadrilateral. This happens if and only ifp, qare both positive (which is given) and 1 then A will be to the right ofH and so D to the left of H. Ifq >1, then B will be below F and so C will be above F. Ifp or q= 1, then three of the points A, B, C, D become collinear. In all cases ABCD will

not be a convex quadrilateral. If both p, q >1, ABCD will even be self-intersecting.

B3. a)We want to choose subsets A1, A2, . . . , Ak of{1, 2, . . . , n}such that any two of thechosen subsets have nonempty intersection. Show that the size k of any such collection ofsubsets is at most 2n1.

If a set A is in such a collectionC, then the complement of A cannot be inC. Therefore|C| 12 (total number of subsets of{1, 2, . . . , n}) = 122n = 2n1.

b) For n > 2 show that we can always find a collection of 2n1 subsets A1, A2, . . . of

{1, 2, . . . , n} such that any two of the Ai intersect, but the intersection of all Ai is empty.There are many ways to build such a collection, e.g., take all 2n1 subsets of{1, 2, . . . , n}containing 1, remove the singleton set{1} and instead include its complement. ORNote that for n = 3, the four sets{1, 2}, {2, 3}, {1, 3}, {1, 2, 3} give a (unique) solution.For n >3 take the union of each of these 4 sets with all 2n3 subsets of{4, . . . , n}. ORForn = 2k + 1, take all subsets of{1, 2, . . . , n}of size> k. Any two of these will intersect.Now use

ni

= nni

. For n= 2k, take all subsets of size > k along with half the subsets

of size k, namely those containing a fixed number. (Check the details.)

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B4. Define

x=10

i=1

1

10

3

1

1 + ( i103

)2 and y =

9

i=0

1

10

3

1

1 + ( i103

)2.

Show that a) x <

6

< y and b) x+y

2

<

6

. (Hint: Relate these sums to an integral.)

a) Let f(t) = 1/(1 +t2). Then y and x are respectively the left and right hand Riemannsums for fover the interval [0, 1

3] using 10 equal parts, each of width 1/10

3. Sincef is

a positive decreasing function,y overestimates the area under fover the given interval and

xunderestimates it. The area underfover [0, 13

] is1/30

f(t)dt= tan1(t)|1/3

0 = 6 , so

x < 6 < y. Note. Different normalizations are possible for f, e.g., the more simplemindedchoicef(t) = 1

103

11+( t

103)2

considered over the interval [0,10] will work too.

b) x+y2

can be interpreted as the sum of areas of 10 trapezoids as follows. Dividing

[0, 13 ] into 10 equal parts, let the i-th subinterval be [ti1, ti] with i= 0, 1, . . . , 10. Then

the i-th trapezoid has base [ti1, ti] and it has two vertical sides, the left one of heightf(ti1) and the right one of height f(ti) (draw a picture and see). So we have to provethat the total area of trapezoids is less than the area under f. For this we should checkconcavity off(draw pictures and see why). Check that over the interval (0, 1

3), we have

f(t) = 6t22

(1+t2)3 0 by replacingrwith rif necessary.Now use AM-GM inequality or the fact that (

r1/r)2 0. Sincex +x1 =

5+12 0, we get cos = 5+14 .

c) Combine conclusions of parts a and b to express x and therefore the desired quantityin a suitable form.

Let x = dei = d(cos + i sin ). Then x1 = d1ei = d1(cos i sin ). Addingand using that x+x1 =

5+12 = 2 cos(

5 ), we get d = 1 and =. So x = e

i

5 and

x2012 +x2012 = 2 cos(20125 ) = 2 cos(402+ 25 ) = 2 cos(

25 ) = 2 cos

2(5 ) 1 =512 .

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B6. Forn >1, aconfigurationconsists of 2ndistinct points in a plane,n of them red, theremainingnblue, with no three points collinear. Apairingconsists ofnline segments, eachwith one blue and one red endpoint, such that each of the given 2npoints is an endpointof exactly one segment. Prove the following.

a) For any configuration, there is a pairing in which no two of the n segments intersect.(Hint: consider total length of segments.)

For any configuration, there are only finitely many pairings. Choose one with least possibletotal length of segments. Here no two of the n segments can interest, because ifRB andRB intersect in point Xthen we get a contradiction as follows. Using triangle inequalityin triangles RXB and RXB, we get RB +RB < RB +RB (draw a picture). Soreplacing RB and RB withRB and RB would give a pairing with smaller total length.

b)Givennred points (no three collinear), we can place nblue points such that any pairingin the resulting configuration will have two segments that do not intersect. (Hint: First

consider the case n= 2.)

Forn = 2, place the two blue points on opposite sides of the line passing through the giventwo red points. There are two possible pairings and the two segments in either one do notintersect. We use a similar idea in general. Given nred points, find a triangle ABC suchthat A is a red point and all other red points are inside triangle ABC. (This is alwayspossible. Why?) Place one blue point at B and all other blue points in the region oppositeto triangleABCat vertexC. (More precisely, letCbe betweenA and Aand also betweenB and B. Place the remaining blue points inside triangle ACB.) Now in any pairing, ifAandB are connected, then AB will not intersect any other segment. Otherwise the two

segments having A and B as vertices will not intersect. Draw a picture to see this.

B7. A sequence of integers cn starts with c0 = 0 and satisfies cn+2 = acn+1+bcn forn0, wherea and b are integers. For any positive integer k withgcd(k, b) = 1, show thatcn is divisible by k for infinitely many n.

Consider pairs of consecutive entries of the sequence modulo k, i.e., (cn, cn+1), where adenotesa modulok. Since there are only finitely many possibilities (namelyk2), some pairof consecutive residues will repeat. Suppose (ci, ci+1) = (ci+p, ci+p+1) for some i. We willshow that in fact the previous equation holds for alli, i.e., whole sequence of consecutive

pairs is periodic. This will prove in particular that (c0, c1) = (cp, cp+1) = (c2p, c2p+1) = .Since c0 = 0 is divisible by k, so is cip for all i.

The equationcn+2=acn+1 + bcn shows that bcn= cn+2 acn+1. Nowgcd(k, b) = 1 meansb is invertible modulo k, i.e., there is a b with bb =1. Therefore cn = b(cn+2 acn+1).Thus knowing a pair of consecutive residues uniquely determines the previous residue(this is why we considered pairs of residues). Therefore (ci, ci+1) = (ci+p, ci+p+1) implies(ci1, ci) = ( ci+p1, ci+p) and (by the given recurrence) (ci+1, ci+2) = ( ci+p+1, ci+p+2).Thus the whole sequence (cn, cn+1) becomes periodic as soon as a single such pair repeats.

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B8. Let f(x) be a polynomial with integer coefficients such that for each nonnegativeintegern,f(n) = a perfect power of a prime number, i.e., of the form pk, wherep is primeand k a positive integer. (p and k can vary with n.) Show that f must be a constantpolynomial using the following steps or otherwise.

a)If such a polynomialf(x) exists, then there is a polynomial g(x) with integer coefficientssuch that for each nonnegative integer n, g(n) = a perfect power of a fixedprime number.

Write f(x) = anxn +an1xn1 + +a1x+a0. Thena0 = f(0) = pk for some prime

p and integer k > 0. Defineg(x) =f(px). Theng(x) is a polynomial such that for eachnonnegative integer n, g(n) = f(pn) = a perfect power of a prime number. This primenumber has to be p, because by evaluating we see that g(n) =f(pn) is divisible by p.

b)Show that a polynomial g(x) as in part a must be constant.

Let g(x) =bn

xn + bn1

xn1 +

+ b1

x + b0

. Then b0

=g(0) =pk. Consider g(mpk+1) =bn(mp

k+1)n+bn1(mpk+1)n1+ +b1(mpk+1)+pk. Clearly for each non-negative integerm, this expression is divisible by pk, but not by pk+1 (since it is pk modulo pk+1). Thisforcesg(mpk+1) =pk for allm, since it must be a perfect power ofp. Thus the polynomialg takes the value pk infinitely often, so it must be identically equal to pk. (Otherwise thepolynomial g(x) pk would have infinitely many roots.) To finish the problem, note thatsince g(x) =f(px) is constant, f(x) must be constant by the same logic.

B9. LetNbe the set ofnon-negativeintegers. Supposef :NNis a function such thatf(f(f(n)))< f(n+ 1) for every nN. Prove that f(n) =n for all nusing the followingsteps or otherwise.a)Iff(n) = 0, then n= 0.

Let f(n) = 0. Ifn >0, then n 1 is in the domain off and f(f(f(n 1))) < f(n) = 0,which is a contradiction, since 0 is the smallest possible value off. (Note that this doesNOT prove that f(0) = 0, only that iff(some n) = 0, then that n = 0. In fact provingf(0) = 0 along with part a would essentially solve the problem, see below.)

b)Iff(x)< n, then x < n. (Start by considering n= 1.)

Induction onn. Ifn = 1, then this is just part a. Assuming the statement up to n we needto prove that iff(x) < n+ 1, then x < n+ 1. Iff(x) < n, then by induction x < n, sox < n+ 1. So let f(x) =n. Ifx= 0, we are done. Otherwise f(f(f(x 1)))< f(x) =nand by using induction thrice we get in succession f(f(x 1))< n, then f(x 1)< nandthen x 1< n, i.e., x < n+ 1 as desired.

c) f(n)< f(n+ 1) and n < f(n+ 1) for alln.

Apply part b to f(f(f(m))) < f(m+ 1) (with x = f(f(m)) and n = f(m+ 1)) to get

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f(f(m)) < f(m+ 1). Apply part b to this with x = f(m) and n = f(m+ 1) to getf(m)< f(m+ 1). Again apply part b to get m < f(m+ 1).

d)f(n) =n for all n.

By part c, f is increasing and f(n) n. If f(n) > n, then f(f(n)) > f(n) (since fis increasing) and so f(f(n)) > n, i.e., f(f(n)) n+ 1. Again, since f is increasing,f(f(f(n)))f(n+ 1), a contradiction.

Alternative solution after part a. Let us prove f(0) = 0. We know that f(n) = 0implies n = 0, so n > 0 implies f(n) > 0. Applying this to any positive f(k), we getf(f(k))> 0. Denoting f(f(k)) =x, we therefore get f(f(f(x 1)))< f(x) =f(f(f(k))).This means that for k such that f(f(f(k))) is the smallest number in{f(f(f(n)))|n0},we must have f(k) = 0. In particular 0 is in the range off, so by part a f(0) = 0.

Since f(n) = 0 for no other n, we may restrict the function f by deleting 0 from the

domain and the range. The resulting function would satisfy f(f(f(n))) < f(n+ 1) forevery n >0. Repeat the reasoning substituting 1 (the new lowest element of the domainand the range) for 0 and conclude f(1) = 1. Then restrict ton > 1 and show f(2) = 2and so on.

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2013 Entrance Examination for BSc Programmes at CMI

Part A. (10 problems 5 points = 50 points.) Attempt all questions in this partbefore going to part B. Carefully read the details of marking scheme given

below. Note that wrong answers will get negative marks!

In each problem you have to fill in 4 blanks as directed. Points will be given based only onthe filled answer, so you need not explain your answer. Each correct answer gets 1 pointand having all 4 answers correct will get 1 extra point for a total of 5 points per problem.But each wrong/illegible/unclear answer will get minus 1 point. Negative points from anyproblem will be counted in your total score, so it is better not to guess! If you are unsureabout a part, you may leave it blank without any penalty. If you write something and thenwant it not to count, cross it out and clearly write no attempt next to the relevant part.

1. For sets A and B, let f :A B andg: B A be functions such that f(g(x)) = x foreach x. For each statement below, write whether it is TRUE or FALSE.

a) The function fmust be one-to-one.

Answer:

b) The function fmust be onto.

Answer:

c) The function g must be one-to-one.

Answer:

d) The function g must be onto.

Answer:

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2. Let f : R R be a function, where R is the set of real numbers. For each statementbelow, write whether it is TRUE or FALSE.

a) If|f(x) f(y)| 39|x y| for all x, y then fmust be continuous everywhere.

Answer:

b) If|f(x) f(y)| 39|x y| for all x, y then fmust be differentiable everywhere.

Answer:

c) If|f(x) f(y)| 39|x y|2 for all x, y then fmust be differentiable everywhere.

Answer:

d) If|f(x) f(y)| 39|x y|2 for all x, y then f must be constant.

Answer:

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3. LetSbe a circle with centerO. SupposeA, Bare points on the circumference ofSwithAOB = 120. For triangle AOB, let Cbe its circumcenter and D its orthocenter (i.e.,the point of intersection of the three lines containing the altitudes). For each statementbelow, write whether it is TRUE or FALSE.

a) The triangle AOC is equilateral.

Answer:

b) The triangle ABD is equilateral.

Answer:

c) The point Clies on the circle S.

Answer:

d) The point D lies on the circle S.

Answer:

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4. A polynomial f(x) with real coefficients is said to be a sum of squares if we can writef(x) =p1(x)

2 + +pk(x)2, wherep1(x), . . . , pk(x) are polynomials with real coefficients.For each statement below, write whether it is TRUE or FALSE.

a) If a polynomial f(x) is a sum of squares, then the coefficient of every odd power ofxinf(x) must be 0.

Answer:

b) Iff(x) =x2 +px+qhas a non-real root, then f(x) is a sum of squares.

Answer:

c) Iff(x) =x3 +px2 +qx+r has a non-real root, then f(x) is a sum of squares.

Answer:

d) If a polynomial f(x)> 0 for all real values ofx, then f(x) is a sum of squares.

Answer:

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5. There are 8 boys and 7 girls in a group. For each of the tasks specified below, write anexpression for the number of ways of doing it. Do NOT try to simplify your answers.

a) Sitting in a row so that all boys sit contiguously and all girls sit contiguously, i.e., no

girl sits between any two boys and no boy sits between any two girls

Answer:

b) Sitting in a row so that between any two boys there is a girl and between any two girlsthere is a boy

Answer:

c) Choosing a team of six people from the group

Answer:

d) Choosing a team of six people consisting ofunequalnumber of boys and girls

Answer:

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6. Calculate the following integrals whenever possible. If a given integral does not exist,state so. Note that [x] denotes the integer part ofx, i.e., the unique integer n such thatn x < n+ 1.

a)41

x2dx

Answer:

b)31

[x]2dx

Answer:

c)21

[x2]dx

Answer:

d) 1

1

1

x2 dx

Answer:

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7. LetA, B , C be angles such that eiA, eiB , eiC form an equilateral triangle in the complexplane. Find values of the given expressions.

a) eiA +eiB +eiC

Answer:

b) cos A+ cos B+ cos C

Answer:

c) cos2A+ cos 2B+ cos 2C

Answer:

d) cos2 A+ cos2 B+ cos2 C

Answer:

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8. Consider the quadratic equation x2 +bx+c = 0, where b and c are chosen randomlyfrom the interval [0,1] with the probability uniformly distributed over all pairs (b, c). Letp(b) = the probability that the given equation has a real solution for given(fixed) value ofb. Answer the following questions by filling in the blanks.

a) The equation x2 +bx+c = 0 has a real solution if and only ifb2 4c is

Answer:

b) The value ofp(12

), i.e., the probability that x2 + x2

+c= 0 has a real solution is

Answer:

c) As a function ofb, is p(b) increasing, decreasing or constant?

Answer:

d) As b and c both vary, what is the probability that x2 +bx+c = 0 has a real solution?

Answer:

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9. Let R= the set of real numbers. A continuous function f : R R satisfies f(1) = 1,f(2) = 4, f(3) = 9 and f(4) = 16. Answer the independent questions below by choosingthe correct option from the given ones.

a) Which of the following values must be in the range off ?Options: 5 25 both neither

Answer:

b) Supposefis differentiable. Then which of the follwing intervals must contain an x suchthat f(x) = 2x ? Options: (1,2) (2,4) both neither

Answer:

c) Suppose f is twice differentiable. Which of the following intervals must contain an xsuch that f(x) = 2 ? Options: (1,2) (2,4) both neither

Answer:

d) Supposef is a polynomial, then which of the following are possible values of its degree?Options: 3 4 both neither

Answer:

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10. Let

f(x) = x4

(x 1)(x 2) (x n)

where the denominator is a product ofn factors,n being a positive integer. It is also given

that the X-axis is a horizontal asymptote for the graph of f. Answer the independentquestions below by choosing the correct option from the given ones.

a) How many vertical asymptotes does the graph off have?Options: n less thann more thann impossible to decide

Answer:

b) What can you deduce about the value ofn ?Options: n 4 impossible to decide

Answer:

c) As one travels along the graph offfrom left to right, at which of the following points

is the sign off(x) guaranteed to change from positive to negative?Options: x= 0 x= 1 x= n 1 x= n

Answer:

d) How many inflection points does the graph offhave in the region x


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Part B. (Problems 14 15 points + problems 56 20 points = 100 points.) Solvethese problems in the space provided for each problem after this page. You may solve onlypart of a problem and get partial credit. Clearly explain your entire reasoning. Nocredit will be given without reasoning.

1. In triangle ABC, the bisector of angle A meets side BC in point D and the bisector ofangle B meets side AC in point E. Given that DE is parallel to AB, show that AE = BDand that the triangle ABC is isosceles.

2. A curve Chas the property that the slope of the tangent at any given point ( x, y) on

C is x2+y2

2xy .

a) Find the general equation for such a curve. Possible hint: let z= yx

.

b) Specify all possible shapes of the curves in this family. (For example, does the family

include an ellipse?)

3. A positive integerNhas its first, third and fifth digits equal and its second, fourth andsixth digits equal. In other words, when written in the usual decimal system it has theform xyxyxy, where x and y are the digits. Show that Ncannot be a perfect power, i.e.,N cannot equal ab, where a and b are positive integers with b >1.

4. Supposef(x) is a function from Rto Rsuch that f(f(x)) = f(x)2013. Show that thereare infinitely many such functions, of which exactly four are polynomials. (Here R= theset of real numbers.)

5. Consider the function f(x) = ax+ 1x+1

, where a is a positive constant. Let L = the

largest value off(x) andS= the smallest value off(x) forx [0, 1]. Show thatL S > 112

for any a >0.

6. Define fk(n) to be the sum of all possible products of k distinct integers chosen fromthe set {1, 2, . . . , n}, i.e.,

fk(n) =

1i

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Solutions for 2013 Entrance Examination for BSc Programmes at CMI

Part A. (10 problems 5 points = 50 points.) Attempt all questions in this partbefore going to part B. Carefully read the details of marking scheme given

below. Note that wrong answers will get negative marks!

In each problem you have to fill in 4 blanks as directed. Points will be given based only onthe filled answer, so you need not explain your answer. Each correct answer gets 1 pointand having all 4 answers correct will get 1 extra point for a total of 5 points per problem.But each wrong/illegible/unclear answer will get minus 1 point. Negative points from anyproblem will be counted in your total score, so it is better not to guess! If you are unsureabout a part, you may leave it blank without any penalty. If you write something and thenwant it not to count, cross it out and clearly write no attempt next to the relevant part.

1. For setsAand B , let f :A B and g :B A be functions such that f(g(x)) =x foreach x. For each statement below, write whether it is TRUE or FALSE.

a) The function fmust be one-to-one.b) The function fmust be onto.c) The function g must be one-to-one.d) The function g must be onto.

Answer: FTTF.

Ifg(x1) = g(x2), then x1 =f(g(x1)) =f(g(x2)) = x2, so g is one-to-one. Also f is onto

because each xB is in the image off, namely x= f(g(x)). The other two statementsare false, e.g. by constructing an example in which A is a larger finite set than B.

2. Let f : R R be a function, where R is the set of real numbers. For each statementbelow, write whether it is TRUE or FALSE.

a) If|f(x) f(y)| 39|x y| for all x, y then fmust be continuous everywhere.b) If|f(x) f(y)| 39|x y| for all x, y then fmust be differentiable everywhere.c) If|f(x) f(y)| 39|x y|2 for all x, y then fmust be differentiable everywhere.d) If|f(x) f(y)| 39|x y|2 for all x, y then fmust be constant.Answer: TFTT

In parts a and b, we have |f(x)f(a)| sandwiched between39|xa|. As x a,39|x a| 0 and hence f(x) f(a) 0, so f is continuous. But it need not bedifferentiable, e.g. f(x) = |x| satisfies f(x) f(y) = |x| |y| |x y| 39|x y|. But fis not differentiable at 0.

In parts c and d, we have| f(x)f(a)xa | 39|x a|, so by reasoning as for part a, we have

limxaf(x)f(a)

xa = 0, i.e.,f(a) = 0 for all a, so f is a constant function.

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3. LetSbe a circle with center O. SupposeA, Bare points on the circumference ofSwithAOB = 120. For triangle AOB, let Cbe its circumcenter and D its orthocenter (i.e.,the point of intersection of the three lines containing the altitudes). For each statementbelow, write whether it is TRUE or FALSE.

a) The triangle AOCis equilateral.b) The triangle ABD is equilateral.c) The point C lies on the circle S.d) The point D lies on the circle S.

Answer: TTTT

Draw a picture and see that the bisector of AOB splits this angle into two angles of60 each and meets the circle, say in point C. Now the triangles OAC and OBC areboth equilateral, so AC = OC = BC, making C = C, the cirumcenter of triangleAOB. Similarly, letting CD be a diameter of the circle S, it is easy to deduce thatAOD = BOD = 120 and that triangle ABD is also equilateral with O as itscentroid. Hence CD AB, line BO AD and line AO BD, making D = D, theorthocenter of triangle AOB.

4. A polynomial f(x) with real coefficients is said to be a sum of squares if we can writef(x) =p1(x)

2 + +pk(x)2, wherep1(x), . . . , pk(x) are polynomials with real coefficients.For each statement below, write whether it is TRUE or FALSE.

a) If a polynomial f(x) is a sum of squares, then the coefficient of every odd power ofx inf(x) must be 0.b) Iff(x) =x2 +px+qhas a non-real root, then f(x) is a sum of squares.

c) Iff(x) =x3 +px2 +qx+r has a non-real root, then f(x) is a sum of squares.d) If a polynomial f(x)> 0 for all real values ofx, then f(x) is a sum of squares.

Answer: FTFT

For part b, complete the square to getf(x) =x2 +px + q= (x + p2 )2 + (4qp

2

4 ), which is asum of squares since 4qp2 >0 due to the roots being non-real. Since pneed not be 0, thisdisproves part a. For part d, since all roots offare non-real and occur in conjugate pairs,f(x) = a product of quadratic polynomials each of which is a sum of squares by part b.For part c, note that f(x) as x , so in particular f(x) takes negative valuesand hence can never be a sum of squares. (This applies to any odd degree polynomial.)

5. There are 8 boys and 7 girls in a group. For each of the tasks specified below, write anexpression for the number of ways of doing it. Do NOT try to simplify your answers.

a) Sitting in a row so that all boys sit contiguously and all girls sit contiguously, i.e., nogirl sits between any two boys and no boy sits between any two girls.

Answer: 2 8! 7! (The factor of 2 arises because the two blocks of boys and girls canswitch positions.)

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b) Sitting in a row so that between any two boys there is a girl and between any two girlsthere is a boy

Answer: 8! 7! (There is no factor of 2 because there must be a boy at each end.)c) Choosing a team of six people from the group Answer:

15

6

d) Choosing a team of six people consisting ofunequalnumber of boys and girls

Answer:156

83

73

=86

+85

71

+84

72

+82

74

+81

75

+76

6. Calculate the following integrals whenever possible. If a given integral does not exist,state so. Note that [x] denotes the integer part ofx, i.e., the unique integer n such thatn x < n+ 1.a)41

x2dx= x3

3|41 = 21 using the fundamental theorem of calculus.

b)31

[x]2dx= 1(12) + 1(22) = 5 = area under the piecewise constant function [x]2

c)21

[x2]dx= 1(

2 1) + 2(3 2) + 3(2 3) = 5 2 3 since the function [x]2is constant on intervals [1,

2), [

2,

3), [

3, 2), taking values 1, 2, 3 respectively.

d)11

1x2

dx = 2limt0+1t

1x2

dx = 2limt0+(1 + 1t ) =. The fundamental theoremdoes not apply over the interval [1, 1] because 1

x2goes to in the interval. It is also ok

to answer that the integral does not exist (as a real number).

7. LetA, B,Cbe angles such that eiA, eiB, eiC form an equilateral triangle in the complexplane. Find values of the given expressions.

a) eiA +eiB +eiC = 0 by taking the vector sum of the three points on the unit circle.

b) cos A+ cos B+ cos C= 0 = real part ofeiA +eiB +eiC, which is 0 by part a.

c) cos 2A+cos2B+cos2C= 0 because the pointse2iA, e2iB, e2iC on the unit circle also forman equilateral triangle in the complex plane, since taking B =A + (2/3), C=A + (4/3),we get 2B= 2A + (4/3) and 2C= 2A + (8/3) = 2A + (2/3 ) + 2 and the last term 2does not change the position of the point.

d) cos2 A+ cos2 B+ cos2 C = 32 because, using the formula for cos 2 in part c, we getcos2 A+ cos2 B+ cos2 C= sin2 A+ sin2 B+ sin2 Cand the sum of the LHS and the RHSin this equation is 3.

8. Consider the quadratic equation x2 +bx+c = 0, where b and c are chosen randomlyfrom the interval [0,1] with the probability uniformly distributed over all pairs (b, c). Letp(b) = the probability that the given equation has a real solution for given(fixed) value ofb. Answer the following questions by filling in the blanks.

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a) The equation x2 +bx+c= 0 has a real solution if and only ifb2 4c is 0.b) The value ofp(1

2), i.e., the probability that x2 + x

2+c= 0 has a real solution is

Answer: 116 since a real solution occurs precisely whenb24c= 14 4c 0, i.e., 0 c 116 ,

which is 116

thfraction of the interval [0, 1] over which c ranges.

c) As a function ofb, is p(b) increasing, decreasing or constant?

Answer: increasing, because b2 4c0 if and only if 0c b24, so p(b) = b2

4, which is

increasing for 0 b 1.d) Asband c both vary, what is the probability that x2 +bx+c= 0 has a real solution?

Answer: This is the fraction of the area of the unit square [0, 1] [0, 1] that is occupied bythe region b2 4c0, i.e., it is the area underthe parabola c= b24 from b = 0 to b= 1,which is

10

b2

4db = 112 .

9. Let R = the set of real numbers. A continuous functionf : R R satisfies f(1) = 1,f(2) = 4, f(3) = 9 and f(4) = 16. Answer the independent questions below by choosingthe correct option from the given ones.

a) Which of the following values must be in the range off ?Options: 5 25 both neither

Answer: 5, by the intermediate value theorem, e.g., over the interval [2,3]. Alsof(x) neednot take the value 25, e.g., take f(x) =x2 for x


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g(x) = 2 sin(x), which is 0 only when x is an integer, in particular g(x) = 0 for anyx (1, 2).For part d, note that g(x), now being a polynomial vanishing at 1, 2, 3 and 4, must bedivisible by (x 1)(x 2)(x 3)(x 4). So g(x), if non-zero, must have degree at least 4.Thus f(x) =x

2

or a polynomial of degree at least 4.

10. Let

f(x) = x4

(x 1)(x 2) (x n)where the denominator is a product ofn factors,n being a positive integer. It is also giventhat the X-axis is a horizontal asymptote for the graph of f. Answer the independentquestions below by choosing the correct option from the given ones.

a) How many vertical asymptotes does the graph off have?Options: n less thann more thann impossible to decide

Answer: n, at x= 1, 2, . . . , n.

b) What can you deduce about the value ofn ?Options: n 4 impossible to decide

Answer: n > 4, because limx f(x) = 0 and for this to happen, the degree of thedenominator off(x) must be greater than that of the numerator.

c) As one travels along the graph off from left to right, at which of the following pointsis the sign off(x) guaranteed to change from positive to negative?Options: x= 0 x= 1 x= n 1 x= nAnswer: x= n1, becausef(x) is positive forx > nand f(x) changes sign precisely whenit passes through x = 1, 2 . . . , n. Note that the sign off(x) for x < 0 and for x (0, 1)depends on the parity ofn.

d) How many inflection points does the graph offhave in the region x 0 for x


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Part B. (Problems 14 15 points + problems 56 20 points = 100 points.) Solvethese problems in the space provided for each problem after this page. You may solve onlypart of a problem and get partial credit. Clearly explain your entire reasoning. Nocredit will be given without reasoning.

1. In triangle ABC, the bisector of angle A meets side BC in point D and the bisector ofangle B meets side AC in point E. Given that DE is parallel to AB, show that AE = BDand that the triangle ABC is isosceles.

Answer: EAD = DAB = EDA, the first equality because AD bisects EAB andthe second because alternate angles made by line AD intersecting parallel lines DE andAB are equal. ThusEAD is isosceles with EA = ED. Similarly ED =DB using thefact that BE bisects DBA also intersects parallel lines DE and AB. ThereforeEA =ED =DB. Now by the basic proportionality theorem, CE

EA = CD

DB. As the denominators

EA and DB are equal, the numerators must be equal as well, i.e., CE = CD. Finally,

CA= C E+EA= C D+DB =C B, soABCis isosceles.

2. A curve Chas the property that the slope of the tangent at any given point (x, y) on

C is x2+y2

2xy .

a) Find the general equation for such a curve. Possible hint: let z = yx

.

b) Specify all possible shapes of the curves in this family. (For example, does the familyinclude an ellipse?)

Answer: The defining property of the curve C is equivalent to the differential equationdydx

= x2+y2

2xy = 12 (

xy

+ yx

). It is convenient to let z = y/x, so the equation becomesdydx

= 12 (1z

+ z). To get this in terms of only x and z , differentiatez =y/xwith respect tox

to get dzdx

= 1xdydx

yx2

= 1x

( dydx

z) = 1x

(12

( 1z

+ z) z) = 1x1z22z

, where we have substituted

for dydx

using the differential equation and then simplified. Separating the variables and

integrating, we get dxx

= 2zdz1z2 , which gives log |x| = log |1 z2|+ a constant, i.e.,

log |1 z2| = log |x| +K= log |x|1 +K. Exponentiating, we get 1 z2 =eKx

= cx

,

where c is anonzeroconstant. Substitutingz = y/x, we get 1 y2x2

= cx

, i.e.,x2 y2 =cx.To be precise, we have to delete the points (0 , 0) and (c, 0) from this solution, because

for the given equation dy

dx = x2+y2

2xy to make sense, bothx and y must be nonzero. If theequation were given as 2xy dy

dx=x2 +y2, then this issue would not arise.

To see the shape of the curve, complete the square to get (x c2 )2 y2 = c2

4, which isa hyperbola when c= 0. (Note: By differentiating x2 y2 = cx, it is easy to see thatdydx

= 2xcy

= x2+y2

2xy and that this holds even whenc = 0. Thus we get the two straight lines

y = xalso as solutions. The reason the above answer missed this possibility was becausewe put 1 z2 in the denominator while separating variables, which precludes z = 1, i.e.,y = x. To be precise, even here we have to delete the origin from the two lines.)

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3. A positive integer Nhas its first, third and fifth digits equal and its second, fourth andsixth digits equal. In other words, when written in the usual decimal system it has theformxyxyxy, where x and y are the digits. Show that Ncannot be a perfect power, i.e.,Ncannot equal ab, where a and b are positive integers with b >1.

Answer: We have N = (105

+ 103

+ 10)x + (104

+ 102

+ 1)y = 10101(10x+ y) =3 7 13 37 (10x+y). Therefore for Nto be a perfect power, the primes 3,7,13,37must all occur (and in fact with equal power) as factors in the prime factorization of10x+y. In particular 10x+y 10101. But since x and y are digits, each is between 0and 9, so 10x+y 99. So Ncannot be a perfect power.

4. Supposef(x) is a function from R to Rsuch thatf(f(x)) =f(x)2013. Show that thereare infinitely many such functions, of which exactly four are polynomials. (Here R = theset of real numbers.)

Answer: Iff is a polynomial, then we make two cases. (i) Iff(x) = a constant c, thenthe given condition is equivalent to c = c2013, which happens precisely for three valuesof c, namely c = 0, 1, 1 (since we have c(c2012 1) = 0, so c = 0 or c2012 = 1). Thusthere are three constant functions with the given property. (ii) Iff(x) is a non-constantpolynomial, then consider its range set A= {f(x)|x R}. Now for all a A, we have bythe given property f(a) =a2013. So the polynomial f(x) x2013 has all elements ofA asits roots. Since there are infinitely many values in A (e.g. applying the intermediate valuetheorem because f is continuous), the polynomial f(x) x2013 has infinitely many rootsand thus must be the zero polynomial, i.e., f(x) =x2013 for all real number x.

Note: One can also deduce that the degree offmust be 0 or 2013 by equating the degrees

off(f(x)) and f(x)2013

. Then, in the non-constant case, it is possible to argue first thatthe leading coefficient is 1 and then that all other coefficients must be 0.

To find infinitely many function with the given property, define f(0) = 0, f(1) = 1 andf(1) = 1. For every other real number x, arbitrarily define f(x) to be 0, 1 or1. It iseasy to see that any such function satisfies the given property. (Other answers are possible,e.g., more systematically, observe that f(a) = a2013 for at least one real number a (e.g.,

any number in the range off) and then this forcesf(x) =x2013 for allx S= {a2013i |i=0, 1, 2, . . .}. We use this as follows. Fix a real number a. Then definef(x) =x2013 for allx S= {a2013i |i= 0, 1, 2, . . .}. For all x S, simply define f(x) = any element of the setS, e.g., a itself will do.)

5. Consider the function f(x) = ax+ 1x+1

, where a is a positive constant. LetL = the

largest value off(x) andS= the smallest value off(x) forx [0, 1]. Show thatLS > 112for any a >0.

Answer: Let f(x) = ax+ 1x+1

. We wish to understand the minimum and maximum of

this function in the interval [0, 1]. Now f(0) = 1, f(1) = a+ 12

and f(x) = a 1(x+1)2 .Over the interval [0, 1], the value off(x) increases froma 1 at x= 0 to a 14 at x= 1.

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We should consider what happens to the sign off(x). For this we consider the followingcases.

(1) Suppose a 1/4. Because 1/(x+ 1)2 1/4 on the interval [0, 1], f(x) 0, so themaximum is at 0 and the minimum is at x = 1. So the difference is 1

(1/2 +a) =

1/2 a 1/4 1/12.(2) Supposea 1. Thenf(x) 0 on the interval [0, 1], so maximum is at 1 and minimumat 0. We get a+ 1/2 1 =a 1/2 1/2 1/12.(3) Suppose 1/4 a 1. Now f(x) = 0 at x = 1

a 1. For this range ofa, x [0, 1].

In the interval [0,x], f(x) 0 and in the interval [x, 1], f(x) 0. Now we make twosub-cases depending on at which endpoint the maximum occurs.

(3i) Suppose 1/4 a 1/2. Then f(0) f(1). So minimum is at x, maximum isat x = 0. f(x) =

a

a+

a = 2

a

a. So the difference between maximum and

minimum is 1 +a 2a = (1 a)2. This is smallest whena is closest to 1 and so(1 a)2 (1 1/2)2 = 3/2 2. This is bigger than 1/12 since ( 32 112 ) = 17/12 and172 = 289 2 122.(3ii) Suppose 1/2 a 1. Now f(1) f(0). Max is at 1 and minimum is at x. Thedifference is a+ 1/2 a+a a = 2a 2a+ 1/2 = (2a 1

2)2. By a calculation

similar to the above it is bigger than 1/12.

6. Define fk(n) to be the sum of all possible products ofk distinct integers chosen from

the set{1, 2, . . . , n}, i.e.,fk(n) =

1i1


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b) We prove the statement by induction onk. Firstf1(n) =n

i=1i= n(n+1)

2 , a polynomial

of degree 2 as desired. For k > 1, we have by part a the equation fk(n) fk(n 1) =nfk1(n 1). The right hand side is a polynomial of degree 1 + 2(k 1) = 2k 1, where2(k 1) is the degree offk1(n 1) by induction and the added 1 comes from the factorn. Since successive differences in the values of fk are given by a polynomial of degree

2k 1, the functionfk on positive integers is given by a polynomial of degree 1 more, i.e.,of degree 2k.

Note: The previous statement is a standard fact, which can be explained as follows. (1)If we assume that fk(n) is a polynomial, then its degree is easily found, because for anypolynomialfof degreem, its successive difference functionf(x)f(x1) is a polynomialof degree m 1. (Reason: If the leading term off(x) is axm, then the leading term inf(x) f(x 1) is amxm1, as seen by expanding the power ofx 1 in axm a(x 1)m.The remaining terms in f(x) f(x 1) do not matter because by expanding powers ofx 1 in them and simplifying, we only get monomials of degree < m 1.) (2) In fact,based on the difference equation, fk(n) must be a polynomial in the variable n. This is aconsequence of the following well-known fact.

Claim: given a polynomial h(x) of degree d, there is a polynomial g(x) of degree d+ 1such that g(x) g(x 1) = h(x). Proof: Induction on d, the degree ofh. Ifh(x) = c,a constant, then g(x) =cx works. Now ford >1, it is enough to find a polynomial g(x)such that g(x) g(x 1) =xd (because ifh(x) =cxd +h(x), whereh has degree < d, byinduction we find g forh and then cg(x) + g(x) works for h(x)). To find such g(x), noticethat forg1(x) =x

d+1, we haveh1(x) =g1(x) g1(x 1) = (d + 1)xd + h2(x), whereh2(x)is a polynomial of degree d 1. By induction h2(x) =g2(x) g2(x 1) for a polynomialg2(x) of degree d. Nowg(x) =

1d+1

(g1(x) g2(x)) works.

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Test Codes: UGA (Multiple-choice Type) andUGB (Short Answer Type) 2013

Questions will be set on the following and related topics.Algebra: Sets, operations on sets. Prime numbers, factorization of integersand divisibility. Rational and irrational numbers. Permutations and com-binations, Binomial Theorem. Logarithms. Polynomials: relations betweenroots and coefficients, Remainder Theorem, Theory of quadratic equationsand expressions. Arithmetic and geometric progressions. Inequalities involv-ing arithmetic, geometric & harmonic means. Complex numbers.

Geometry: Plane geometry. Geometry of 2 dimensions with Cartesian andpolar coordinates. Equation of a line, angle between two lines, distance froma point to a line. Concept of a Locus. Area of a triangle. Equations of circle,parabola, ellipse and hyperbola and equations of their tangents and normals.

Mensuration.Trigonometry: Measures of angles. Trigonometric and inverse trigonomet-ric functions. Trigonometric identities including addition formulae, solutionsof trigonometric equations. Properties of triangles. Heights and distances.

Calculus: Sequences - bounded sequences, monotone sequences, limit of asequence. Functions, one-one functions, onto functions. Limits and continu-ity. Derivatives and methods of differentiation. Slope of a curve. Tangentsand normals. Maxima and minima. Using calculus to sketch graphs of func-tions. Methods of integration, definite and indefinite integrals, evaluation ofarea using integrals.

Reference (For more sample questions)Test of Mathematics at the 10 + 2 level, Indian Statistical Institute. Pub-lished by Affiliated East-West Press Pvt. Ltd., 105, Nirmal Tower, 26Barakhamba Road, New Delhi 110001.

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Sample Questions for UGA

Instructions. UGA is a multiple choice examination. In each of the fol-lowing questions, exactly one of the choices is correct. You get four marksfor each correct answer, one mark for each unanswered question, and zeromarks for each incorrect answer.

1 Define an = (12 + 22 +. . .+n2)n and bn = n

n(n!)2. Recalln! is theproduct of the first n natural numbers. Then,

(A) an< bn for all n >1 (B) an> bn for all n >1

(C) an= bn for infinitely many n (D) None of the above

2 The sum of all distinct four digit numbers that can be formed usingthe digits 1, 2, 3, 4, and 5, each digit appearing at most once, is

(A) 399900 (B) 399960 (C) 390000 (D) 360000

3 The last digit of (2004)5 is

(A) 4 (B) 8 (C) 6 (D) 2

4 The coefficient ofa3b4c5 in the expansion of (bc + ca + ab)6 is

(A) 12!3!4!5!

(B)

63

3! (C) 33 (D) 3

63

5 Let ABCD be a unit square. Four pointsE, F, G and H are chosen

on the sides AB, BC, CD and DA respectively. The lengths of thesides of the quadrilateral EFGH are , , and . Which of thefollowing is always true?

(A) 1 2 + 2 + 2 + 2 22(B) 2

2 2 + 2 + 2 + 2 42

(C) 2 2 + 2 + 2 + 2 4(D)

2

2 + 2 + 2 + 2

2 +

2

6 If log10 x= 10log1004 then x equals

(A) 410 (B) 100 (C) log104 (D) none of the above

7 z1,z2are two complex numbers withz2= 0 andz1=z2and satisfyingz1+ z2z1 z2 = 1. Then z1z2 is

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(A) real and negative

(B) real and positive

(C) purely imaginary

(D) none of the above need to be true always

8 The set of all real numbersxsatisfying the inequalityx3(x+1)(x2) 0 is

(A) the interval [2, ) (B) the interval [0, )(C) the interval [1, ) (D) none of the above

9 Let z be a non-zero complex number such that z1+z is purely imagi-nary. Then

(A) z is neither real nor purely imaginary (B)z is real(C) z is purely imaginary (D) none of the above

10 Let A be the fixed point (0, 4) and B be a moving point (2t, 0). LetMbe the mid-point ofAB and let the perpendicular bisector ofABmeet the y-axis atR. The locus of the mid-pointP ofM R is

(A) y + x2 = 2 (B) x2 + (y 2)2 = 1/4(C) (y 2)2 x2 = 1/4 (D) none of the above

11 The sides of a triangle are given to be x2 +x+ 1, 2x+ 1 andx2 1.Then the largest of the three angles of the triangle is

(A) 75 (B)

xx + 1

radians (C) 120 (D) 135

12 Two poles, AB of length two metres and CD of length twenty me-tres are erected vertically with bases at B and D. The two polesare at a distance not less than twenty metres. It is observed thattan ACB = 2/77. The distance between the two poles is

(A) 72m (B) 68m (C) 24m (D) 24.27m

13 If A,B, C are the angles of a triangle and sin2 A+ sin2 B = sin2 C,then C is equal to

(A) 30 (B) 90 (C) 45 (D) none of the above

14 In the interval (2, 0), the function f(x) = sin

1

x3

2


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(A) never changes sign

(B) changes sign only once

(C) changes sign more than once, but finitely many times

(D) changes sign infinitely many times

15 The limit

limx0

(ex 1)tan2 xx3

(A) does not exist (B) exists and equals 0

(C) exists and equals 2/3 (D) exists and equals 1

16 Let f1(x) = ex, f2(x) = e

f1(x) and generally fn+1(x) = efn(x) for all

n 1. For any fixed n, the value of ddx

fn(x) is equal to

(A) fn(x) (B) fn(x)fn1(x)

(C) fn(x)fn1(x) f1(x) (D)fn+1(x)fn(x) f1(x)ex

17 If the function

f(x) =

x22x+A

sinx ifx = 0B ifx = 0

is continuous atx= 0, then

(A) A = 0, B = 0 (B)A = 0, B =

2

(C) A = 1, B = 1 (D) A = 1, B = 0

18 A truck is to be driven 300 kilometres (kms.) on a highway at a con-stant speed of x kms. per hour. Speed rules of the highway requirethat 30 x 60. The fuel costs ten rupees per litre and is consumedat the rate 2 + (x2/600) litres per hour. The wages of the driver are200 rupees per hour. The most economical speed (in kms. per hour)to drive the truck is

(A) 30 (B) 60 (C) 30

3.3 (D) 20

33

19 Ifb = 10

et

t + 1 dt then aa1

et

t a 1 dt is(A) bea (B) bea (C)bea (D)bea

20 In the triangle ABC, the angle BACis a root of the equation3cos x + sin x= 1/2.

Then the triangle AB C is3


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(A) obtuse angled (B) right angled

(C) acute angled but not equilateral (D) equilateral

21 Let n be a positive integer. Consider a square Sof side 2n units. Di-vide S into 4n2 unit squares by drawing 2n 1 horizontal and 2n 1vertical lines one unit apart. A circle of diameter 2n 1 is drawn withits centre at the intersection of the two diagonals of the squareS. Howmany of these unit squares contain a portion of the circumference ofthe circle?

(A) 4n 2 (B) 4n (C) 8n 4 (D) 8n 2

22 A lantern is placed on the ground 100 feet away from a wall. A mansix feet tall is walking at a speed of 10 feet/second from the lanternto the nearest point on the wall. When he is midway between thelantern and the wall, the rate of change (in ft./sec.) in the length ofhis shadow is

(A) 2.4 (B) 3 (C) 3.6 (D) 12

23 An isosceles triangle with base 6 cms. and base angles 30 each isinscribed in a circle. A second circle touches the first circle and alsotouches the base of the triangle at its midpoint. If the second circle issituated outside the triangle, then its radius (in cms.) is

(A) 3

3/2 (B)

3/2 (C)

3 (D) 4/

3

24 Let n be a positive integer. Definef(x) = min{|x 1|, |x 2|, . . . , |x n|}.

Then

n+10

f(x)dx equals

(A) (n + 4)

4 (B)

(n + 3)

4 (C)

(n + 2)

2 (D)

(n + 2)

4

25 LetS= {1, 2, . . . , n}. The number of possible pairs of the form (A, B)with A B for subsets A and B ofS is(A) 2n (B) 3n (C)

n

k=0

n

k n

n

k (D) n!26 The number of maps f from the set{1, 2, 3}into the set{1, 2, 3, 4, 5}

such that f(i) f(j) whenever i < j is(A) 60 (B) 50 (C) 35 (D) 30

27 Consider three boxes, each containing 10 balls labelled 1, 2, . . . , 10.Suppose one ball is drawn from each of the boxes. Denote by ni, the

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label of the ball drawn from the i-th box,i = 1, 2, 3. Then the numberof ways in which the balls can be chosen such that n1< n2< n3 is

(A) 120 (B) 130 (C) 150 (D) 160

28 Leta be a real number. The number of distinct solutions (x, y) of thesystem of equations (x a)2 + y2 = 1 and x2 =y2, can only be

(A) 0, 1, 2, 3, 4 or 5 (B) 0, 1 or 3

(C) 0, 1, 2 or 4 (D) 0, 2, 3, or 4

29 The maximum of the areas of the isosceles triangles with base on thepositive x-axis and which lie below the curve y = ex is:

(A) 1/e (B) 1 (C) 1/2 (D)e

30 Supposea,band nare positive integers, all greater than one. Ifan+bnis prime, what can you say about n?(A) The integer n must be 2(B) The integer n need not be 2, but must be a power of 2(C) The integer n need not be a power of 2, but must be even(D) None of the above is necessarily true

31 Water falls from a tap of circular cross section at the rate of 2 me-tres/sec and fills up a hemispherical bowl of inner diameter 0.9 metres.If the inner diameter of the tap is 0.01 metres, then the time neededto fill the bowl is

(A) 40.5 minutes (B) 81 minutes

(C) 60.75 minutes (D) 20.25 minutes

32 The value of the integral 5/2/2

etan1(sinx)

etan1(sinx) + etan1(cosx) dx

equals (A) 1 (B) (C)e (D) none of these

33 The set of all solutions of the equation cos 2 = sin + cos is givenby(A) = 0

(B) = n+ 2 , where n is any integer(C) = 2n or = 2n 2 or = n 4 , where n is any integer(D) = 2n or = n+ 4 , where n is any integer

34 For k 1, the value ofn

0

+

n + 1

1

+

n + 2

2

+ +

n + k

k

equals

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(A)

n + k+ 1

n + k (B) (n + k+ 1)

n + k

n + 1(C)

n + k+ 1

n + 1

(D)

n + k+ 1

n

35 The value of

sin1 cot

sin1

1

2

1

5

6

+ cos1

2

3+ sec1

8

3

is

(A) 0 (B) /6 (C) /4 (D)/2

36 Which of the following graphs represents the function

f(x) =

x0

eu2/xdu, for x >0 and f(0) = 0?

(A) (B)

(C) (D)

37 Ifan=

1 +

1

n2

1 +

22

n2

21 +

32

n2

3

1 +n2

n2

n, then

limn

a1/n2

n

is(A) 0 (B) 1 (C)e (D)

e/2

38 The function x( x) is strictly increasing on the interval 0< x


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40 A box contains 10 red cards numbered 1, . . . , 10 and 10 black cardsnumbered 1, . . . , 10. In how many ways can we choose 10 out of the

20 cards so that there are exactly 3 matches, where a matchmeans ared card and a black card with the same number?

(A)

10

3

7

4

24 (B)

10

3

7

4

(C)

10

3

27 (D)

10

3

14

4

41 LetPbe a point on the ellipse x2 + 4y2 = 4 which does not lie on theaxes. If the normal at the pointPintersects the major and minor axesat C and D respectively, then the ratio P C :P D equals

(A) 2 (B) 1/2 (C) 4 (D) 1/4

42 The set of complex numbers z satisfying the equation

(3 + 7i)z+ (10 2i)z+ 100 = 0represents, in the complex plane,(A) a straight line(B) a pair of intersecting straight lines(C) a pair of distinct parallel straight lines(D) a point

43 The number of triplets (a,b,c) of integers such that a < b < c anda,b,care sides of a triangle with perimeter 21 is

(A) 7 (B) 8 (C) 11 (D) 12.

44 Suppose a, b and c are three numbers in G.P. If the equationsax2 + 2bx+c = 0 and dx2 + 2ex+f= 0 have a common root, thend

a, e

b and

f

c are in

(A) A.P. (B) G.P. (C) H.P. (D) none of the above.

45 The number of solutions of the equation sin1 x= 2 tan1 x is

(A) 1 (B) 2 (C) 3 (D) 5.

46 Suppose ABCD is a quadrilateral such that BAC= 50, CAD =60, CBD = 30 and BDC= 25. IfEis the point of intersectionofAC and B D, then the value ofAEB is

(A) 75 (B) 85 (C) 95 (D) 110.

47 Let R be the set of all real numbers. The function f : R R definedby f(x) =x3 3x2 + 6x 5 is(A) one-to-one, but not onto(B) one-to-one and onto

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(C) onto, but not one-to-one(D) neither one-to-one nor onto.

48 LetL be the point (t, 2) andMbe a point on the y-axis such thatLMhas slopet. Then the locus of the midpoint ofLM, as t varies overall real values, is

(A) y = 2 + 2x2 (B)y = 1 + x2

(C) y = 2 2x2 (D) y = 1 x2.

49 Suppose x, y (0, /2) and x=y . Which of the following statementis true?(A) 2 sin(x + y)< sin 2x + sin 2y for all x, y.(B) 2 sin(x + y)> sin 2x + sin 2y for all x, y.

(C) There existx, y such that 2 sin(x + y) = sin 2x + sin 2y.(D) None of the above.

50 A triangle ABChas a fixed base BC. IfAB : AC= 1 : 2, then thelocus of the vertex A is(A) a circle whose centre is the midpoint ofB C(B) a circle whose centre is on the line BCbut not the midpoint of

BC(C) a straight line(D) none of the above.

51 LetPbe a variable point on a circle Cand Q be a fixed point outsideC. IfR is the mid-point of the line segment P Q, then the locus ofRis

(A) a circle (B) an ellipse

(C) a line segment (D) segment of a parabola

52 N is a 50 digit number. All the digits except the 26th from the rightare 1. IfNis divisible by 13, then the unknown digit is

(A) 1 (B) 3 (C) 7 (D) 9.

53 Suppose a < b. The maximum value of the integral

ba

34 x x2

dx

over all possible values ofa and b is

(A) 3

4 (B)

4

3 (C)

3

2 (D)

2

3.

54 For any n 5, the value of 1 +12

+1

3+ + 1

2n 1 lies between8


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(A) r cosecn (B)r(1 + cosec2n)

(C) r(1 + cosec 2n) (D) r(1 + cosecn)

61 Ifn is a positive integer such that 8n + 1 is a perfect square, then

(A) n must be odd

(B) n cannot be a perfect square

(C) 2n cannot be a perfect square

(D) none of the above

62 Let C denote the set of all complex numbers. Define

A= {(z, w)|z, w C and|z| = |w|}

B = {(z, w)|z, w C, and z2

=w2

}.Then,

(A) A = B (B) A B and A =B(C) B A and B=A (D) none of the above

63 Letf(x) =a0+a1|x|+a2|x|2+a3|x|3,wherea0, a1, a2, a3are constants.Then

(A) f(x) is differentiable at x = 0 whatever be a0, a1, a2, a3(B) f(x) is not differentiable at x = 0 whatever be a0, a1, a2, a3(C) f(x) is differentiable at x = 0 only ifa1= 0

(D) f(x) is differentiable at x = 0 only ifa1= 0, a3= 0

64 Iff(x) = cos(x) 1 + x22, then(A) f(x) is an increasing function on the real line

(B) f(x) is a decreasing function on the real line

(C) f(x) is increasing on < x 0 and decreasing on 0 x < (D) f(x) is decreasing on < x 0 and increasing on 0 x <

65 The number of roots of the equation x2 + sin2 x = 1 in the closedinterval [0, 2 ] is

(A) 0 (B) 1 (C) 2 (D) 3

66 The set of values ofmfor whichmx26mx+5m+1 > 0 for all realxis(A) m < 14 (B) m 0(C) 0 m 14 (D) 0 m < 14

67 The digit in the units place of the number 1! + 2! + 3! + . . . + 99! is10


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(A) 3 (B) 0 (C) 1 (D) 7

68 The value of limn

13+23+...+n3

n4 is:

(A) 34 (B) 14 (C) 1 (D) 4

69 For any integern 1, define an= 1000nn! . Then the sequence{an}(A) does not have a maximum

(B) attains maximum at exactly one value ofn

(C) attains maximum at exactly two values ofn

(D) attains maximum for infinitely many values ofn

70 The equation x3y+ xy3 + xy = 0 represents

(A) a circle (B) a circle and a pair of straight lines

(C) a rectangular hyperbola (D) a pair of straight lines

71 For each positive integer n, define a function fn on [0, 1] as follows:

fn(x) =

0 if x= 0

sin

2n if 0< x 1

n

sin2

2n if

1

n < x 2

n

sin32n

if 2n

< x 3n

... ...

...

sinn

2n if

n 1n

< x 1.

Then, the value of limn

10

fn(x) dx is

(A) (B) 1 (C) 1

(D)

2

.

72 Let d1, d2, . . . , dk be all the factors of a positive integer n including 1and n. Ifd1+ d2+ . . . + dk = 72, then 1d1

+ 1d2 + + 1dk is:(A) k

2

72 (B) 72k (C)

72n (D) none of the above

73 A subset Wof the set of real numbers is called a ring if it contains 1and if for all a, b W, the numbers a b and ab are also in W. LetS=

m2n| m, n integers

and T =

pq| p, qintegers, qodd

. Then

11


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(A) neither Snor T is a ring (B) Sis a ring T is not a ring

(C) T is a ring S is not a ring (D) both Sand Tare rings

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Hints and Answers to selected problems.

There are also other ways to solve the problems apart from the ones sketchedin the hints. Indeed, a student should feel encouraged upon finding a differentway to solve some of these problems.

Hints and Answers to selected UGA Sample Questions.

1 (B). Take the nth root ofan and bn and use A.M. G.M.3 (A). As 2004 = 2000 +4, the last digits of (2004)5 and 45 are equal.

4 (D) Use binomial expansion of (bc + a (b + c))6.6 (B) Let y = log10 x. Then log10 y = log1004. Hencey = 2.8 (D) Check for test points.9 (A) Check (B) and (C) are false, and then that (A) is true.

14 (D) sin 1x3

changes sign at the points (n)1

3 for all n 1.15 (D) Observe that (e

x1)tan2 xx3

= (ex1)x sin

2 xx2

1cos2 x

.16 (C) Use induction and chain rule of differentiation.22 (B) Show that the height function is 60t .26 (C) Compute the number of maps such that f(3) = 5, f(3) = 4 etc..Alternatively, define g :{1, 2, 3} {1, 2, . . . , 7} by g (i) = f(i) + (i 1).Then,g is a strictly increasing function and its image is a subset of size 3 of{1, 2, . . . 7}.28 (D) Draw graphs of (x + y)(x y) = 0 and (x a)2 + y2 = 1.38 (A) Differentiate.51 (A) Compute for C= x2 + y2 = 1and Q = (a, 0) for some a >1.57 (C) Compute the integral

21/2

2xdx 21/2

log xdx.

60 (D) Let s be distance between the centre of the big circle and the centreof (any) one of the small circles. Then there exists a right angle trianglewith hypoteneuse s, side r and angle n .

61(C)If 8n + 1 =m2, then 2n is a product of two consecutive integers.62 (C) z2 =w2 z= w B A. But|i| = 1 and i2 = 1.63 (C) Amongst 1, |x|, |x|2, |x|3, only|x| is not differentiable at 0.64 (D) Look at the derivative off.65 (B) Draw graphs ofy = cos xand y = xand find the number of pointsof intersections.

66 (D) Calculate the discriminant (b2 4ac) of the given quadratic.67 (A) The unit digit of all numbers n! with n 5 is 0.68 (B) Use the formula for

ni=1

i3.

69 (C) Find out the first values ofn for which an+1an becomes < 1.

70 (D) The equation is xy (x2 + y2 + 1) = 0.72 (C) Multiply the given sum by n.73 (D) Verify using the given definition of a ring.

13


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Sample Questions for UGB

Instructions UGB consists of questions that will require you to provide

answers with appropriate justification.1 Find the sum of all distinct four digit numbers that can be formed

using the digits 1, 2, 3, 4, 5, each digit appearing at most once.

2 How many natural numbers less than 108 are there, with sum of digitsequal to 7?

3 Consider the squares of an 8 8 chessboard filled with the numbers 1to 64 as in the figure below. If we choose 8 squares with the propertythat there is exactly one from each row and exactly one from eachcolumn, and add up the numbers in the chosen squares, show that thesum obtained is always 260.

1 2 3 4 5 6 7 89 10 11 12 13 14 15 16

17 18 19 20 21 22 23 2425 26 27 28 29 30 31 3233 34 35 36 37 38 39 4041 42 43 44 45 46 47 4849 50 51 52 53 54 55 5657 58 59 60 61 62 63 64

4 Consider the function

f(x) = limn

loge(2 + x) x2n sin x

1 + x2n

defined for x > 0. Is f(x) continuous at x = 1? Justify your answer.Show that f(x) does not vanish anywhere in the interval 0x 2 .Indicate the points where f(x) changes sign.

5 An isosceles triangle with base 6 cms. and base angles 30o each isinscribed in a circle. A second circle, which is situated outside thetriangle, touches the first circle and also touches the base of the triangleat its midpoint. Find its radius.

6 Suppose a is a complex number such that

a2 + a +1

a

+ 1

a2

+ 1 = 0.

Ifm is a positive integer, find the value of

a2m + am + 1

am+

1

a2m.

7 Let an= 1 . . . 1 with 3n digits. Prove that an is divisible by 3an1.

8 Let f(u) be a continuous function and, for any real number u, let [u]denote the greatest integer less than or equal to u. Show that for any

1


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x >1,

x1

[u]([u] + 1)f(u)du= 2

[x]i=1

i xi

f(u)du.

9 If a circle intersects the hyperbola y = 1/x at four distinct points(xi, yi), i= 1, 2, 3, 4, then prove that x1x2= y3y4.

10 Two intersecting circles are said to be orthogonal to each other if thetangents to the two circles at any point of intersection are perpendicu-lar to each other. Show that every circle through the points (2, 0) and(2, 0) is orthogonal to the circle x2 + y2 5x + 4 = 0.

11 Show that the function f(x) defined below attains a unique minimum

for x > 0. What is the minimum value of the function? What is thevalue ofx at which the minimum is attained?

f(x) =x2 + x +1

x+

1

x2 for x= 0.

Sketch on plain paper the graph of this function.

12 Show that there is exactly one value ofx which satisfies the equation

2cos2(x3 + x) = 2x + 2x.

13 Let S={1, 2, . . . , n}. Find the number of unordered pairs{A, B} ofsubsets of S such that A and B are disjoint, where A or B or both

may be empty.

14 An oil-pipe has to connect the oil-well O and the factory F, betweenwhich there is a river whose banks are parallel. The pipe must crossthe river perpendicular to the banks. Find the position and nature ofthe shortest such pipe and justify your answer.

15 Find the maximum value ofx2 +y2 in the bounded region, includingthe boundary, enclosed by y = x2 , y=x2 and x = y2 + 1.

16 Letx = (x1, . . . , xn) and y = (y1, . . . , yn) where x1, , xn, y1, , ynare real numbers. We write x > y if either x1 > y1 or for some k,with 1

k

n

1, we have x1 =y1, . . . , xk =yk, but xk+1 > yk+1.

Show that foru = (u1, . . . , un),v = (v1, . . . , vn),w = (w1, . . . , wn) andz = (z1, . . . , zn), ifu > v and w > z, then u + w > v+ z.

17 How many real roots does x4 + 12x 5 have?18 For any positive integer n, let f(n) be the remainder obtained on

dividing n by 9. For example, f(263) = 2.(a) Let n be a three-digit number and m be the sum of its digits.

Show that f(m) =f(n).2


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(b) Show that f(n1n2) = f(f(n1)f(n2)) where n1, n2 are any twopositive three-digit integers.

19 Find the maximum among 1, 21/2, 31/3, 41/4, . . ..

20 Show that it is not possible to have a triangle with sides a, b and cwhose medians have lengths 23a,

23b and

45c.

21 For real numbers x, y and z , show that

|x| + |y| + |z| |x + y z| + |y+ z x| + |z+ x y|.

22 Let

P(x) =xn + an1xn1 + an2xn2 + + a1x + a0be a polynomial with integer coefficients, such thatP(0) andP(1) are

odd integers. Show that:(a) P(x) does not have any even integer as root.(b) P(x) does not have any odd integer as root.

23 Let N ={1, 2, . . . , n} be a set of elements called voters. LetC ={S : S N} be the set of all subsets ofN. Members ofC are calledcoalitions. Let f be a function fromC to{0, 1}. A coalition S Nis said to be winning if f(S) = 1; it is said to be a losing coalitionif f(S) = 0. A pairN, f as above is called a voting game if thefollowing conditions hold.(a) Nis a winning coalition.(b) The empty set

is a losing coalition.

(c) IfSis a winning coalition and SS, then S is also winning.(d) If both S and S are winning coalitions, then S S=, i.e., S

and S have a common voter.Show that the maximum number of winning coalitions of a votinggame is 2n1. Find a voting game for which the number of winningcoalitions is 2n1.

24 Supposefis a real-valued differentiable function defined on [1, ) withf(1) = 1. Suppose, moreover, that f satisfies f(x) = 1/(x2 +f2(x)).Show that f(x)1 + /4 for every x1.

25 If the normal to the curve x2/3 +y2/3 =a2/3 at some point makes an

angle with the X-axis, show that the equation of the normal isy cos x sin = a cos2.

26 Suppose that a is an irrational number.(a) If there is a real numberbsuch that both (a+b) andabare rational

numbers, show that a is a quadratic surd. (a is a quadratic surdif it is of the form r+

s or r s for some rationals r and s,

where s is not the square of a rational number).3


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(b) Show that there are two real numbers b1 and b2 such that(i) a + b1 is rational but ab1 is irrational.

(ii) a + b2 is irrational but ab2 is rational.(Hint: Consider the two cases, where a is a quadratic surdand a is not a quadratic surd, separately).

27 Let A, B , and Cbe three points on a circle of radius 1.(a) Show that the area of the triangle ABC equals

1

2(sin(2ABC) + sin(2BC A) + sin(2CAB)) .

(b) Suppose that the magnitude ofABC is fixed. Then show thatthe area of the triangle ABCis maximized when BC A= CAB.

(c) Hence or otherwise show that the area of the triangle ABC is

maximum when the triangle is equilateral.28 In the given figure, E is the midpoint of the arc

ABEC and ED is perpendicular to the chord BCat D. If the length of the chord AB isl1, and that ofBD is l2, determine the length ofDC in terms of l1and l2

A

B

E

C

D

29 (a) Let f(x) = xxe1/x, x > 0. Show that f(x) is an increasingfunction on (0, ), and limx f(x) = 1.

(b) Using part (a) and calculus, sketch the graphs ofy = x1,y = x,y =x + 1, and y = xe1/|x| for < x


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Show thatf(n)

n! =

n

k=0

cmi preparation

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