Cmi Preparation

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 a test of aptitude to do Mathematics. We expect the students to be familiar with the following topics. • 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 and multinomial theorem, permutations and combinations, mathematical induction. prime numbers 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, convergence criterion. • Complex numbers, roots of unity. Suggested reading material The following books are suggested, in addition to the Class XI and XII Mathematics Textbooks of the National Council of Educational Research and Training, New Delhi. 1. V. Krishnamoorthy, C.R. Pranesachar, K. N. Ranganathan, B.J.Venkatachala: Challenge and Thrill of Pre-College Mathematics, New Age International Publishers. 2. M.R. Modak, S.A. Katre, V.V. Acharya: An Excursion in Mathematics, Bhaskaracharya Pratishtan (Pune). 3. D. Fomin, S. Genkin, I. Itenberg: Mathematical Circles: Russian experience, Universities Press (Hyderabad) 1998. 4. H.S. Hall, S.R. Knight: Higher Algebra.

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

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PART A Instructions: • There are 13 questions in this part. Each question carries 4 marks. Answer all questions. 1. For a polynomial f (x), let f (n) (x) denote the nth –derivative for n ≥ 1 and f (0) (x) = f (x). Is the following true or false ? Give brief reasons. f (n) (a) = 0, f or n = 0, 1, . . . , k ⇐⇒ (x−a)(k+1) divides f (x) 2. (a) Find the value of

Pn k=1 k

· k! as a function of n.

(b) True or false ? Give brief reasons. 1000 X

1 1 = log1000! N k=2 logk N

3. Let p(x) be a polynomial such that when divided by (x − 1) it leaves the remainder 2 and when divided by (x − 2) it leaves the remainder 1. What is the remainder when it is divided by (x − 1)(x − 2)? 4. Prove that for any fixed n > 2, among the two integers 2n − 1 and 2n + 1, at the most one of them can be a prime. )n , ∀n ≥ 1. 5. Show that n! ≤ ( n+1 2 6. Show that the sum

n X

k=1

k! 6= m2

for any integer m, for n ≥ 4.

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

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7. Give an example of a subset S of the plane such that the following property is satisfied: There exist infinitely many lines ℓ in the plane such that the intersection ℓ ∩ S is a line segment of any length however large. 8. True or false ? Give brief reasons. There exist non-zero ploynomials f (x) with integer√coefficients of√any degree d ≥ 5 which vanish at x = 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 progression. Give an example of such an arithmetic progression. 10. Show that if p and p + 2 with p ≥ 5, are both primes then the 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 this property. 12. True or false ? Give brief reasons. If a, b are positive integers then the number √ √ (a + b)n + (a − b)n is an integer for all values of n. 13. True or false ? Give brief reasons. The number of common tangents to the circles x2 + y 2 = 1 and x2 + y 2 − 4x + 3 = 0 is 1.

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

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Part B Instructions: • There are 6 questions in this part. Each question carries 8 marks. Answer all questions. 1. Let A(x1, y1), B(x2, y2) and C(x3, y3 ) form a triangle with circumcentre (0, 0). Show that ABC is an equilateral triangle if and only if x1 + x2 + x3 = y1 + y2 + y3 = 0 2. Show the following: (a) limx→0 sin(x) log(x) = 0

USE SERIES EXPANSION

(b) limn→∞ xn! = 0 n

(c) limx→∞ (cos( αx ))x = 1, where α is a constant. 1infinity 3. In a jail with 100 rooms, all locked initially, 100 rioters break in and disturb the rooms in the following way. First one stops at all rooms and opens them all. Second rioters stops at rooms numbered 2, 4, 6, . . . and locks open rooms, and leaving the other rooms as they were. The third rioter stops at rooms numbered 3, 6, 9, . . . and again opens a locked room and locks an open room, leaving others undisturbed. And this process continues. After all the 100 rioters have left which rooms would be open? 4. (a) If two sides of a triangle are given, then show that the area of this triangle is maximum if the sides are perpendicular to each other.

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

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(b) Construction: For any given positive real number λ, find a point D on a line AB such that the difference AD2 − BD2 = λ. (Note that the point D need not be within the segment AB). 5. Show that if [x] denote the greatest integer ≤ x, then  

n n −[ ] 3 3

 

is a natural number divisible by 3 for all values of n. 6. Is it possible to remove one square from a 5 × 5 board so that the remaining 24 squares can be covered by eight 3 × 1 rectangles? 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 × 8– chessboard are removed, the remaining 62 squares cannot be covered by 31 non-overlapping dominos. The reason being, after removing the two corners 32 squares of one colour and 30 of the other are left. No matter how you place a domino it will cover one white and one black square.)

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 ∈ [−π, π] 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 numbers. Prove that f (x) = 0 has no integer solutions. 3. Evaluate: n − nk=1 xk (a) lim x→1 1−x −1/x e (b) lim x→0 x P

4. Show that there is no infinite arithmetic progression consisting of distinct integers all of which are squares. 5. Find the remainder given by 389 × 786 when divided by 17. 6. Prove that 2 3 n 1 + + ··· + =1− 0! + 1! + 2! 1! + 2! + 3! (n − 2)! + (n − 1)! + n! n! 7. If a, b, c are real numbers > 1, then show that 1 1 + loga2 b

c a

+

1 1 + logb2 c

a b

+

1 1 + logc2 a

b c

=3

8. If 8 points in a plane are chosen to lie on or inside a circle of diameter 2cm then show that the distance between some two points will be less than 1cm. xn xn−1 + + · · · + x + 1, then show that f (x) = 0 has no repeated roots. n! (n − 1)! √ 3 3 3 and sin x + sin y + sin z = then show that x = 10. Given cos x + cos y + cos z = 2 2 π π π + 2kπ, y = + 2`π, z = + 2mπ for some k, `, m ∈ Z. 6 6 6 9. If f (x) =

1

√ 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 isoceles 4ABC with A at the apex the height and the base are both equal to 1cm. 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. If b is a real number satisfying b4 +

 √ 1 i 16 −1. = 6, find the value of b + where i = b4 b

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 some m, n with 1 ≤ m ≤ n ≤ P 100, ni=m ai is divisible by 100. 2. In 4 ABC, BE is a median, and O the mid-point of BE. The line joining A and O meets 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 using the 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. Find the digits. 4. (a) Show that the area of a right-angled triangle with all side lengths integers is an integer divisible by 6. (b) If all the sides and area of a triangle were rational numbers then show that the triangle is got by ‘pasting’ two right-angled triangles having the same property. 5. Prove that

Z b

alogb x dx > ln b where a, b > 0, b 6= 1.

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6. Let C1 , C2 be two circles of equal radii R. If C1 passes through the centre of C2 prove √ R2 that the area of the region common to them is (4π − 27). 6 7. Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be 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 to rearrange these letters is (A) 11! (B) 11! (C) 11! (D) 11! 3 6 8 2. In a rectangle ABCD, the length BC is twice the width AB. Pick a point P on side BC such 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 θ with 0 ≤ θ < 2π such that 4 sin(3θ + 2) = 1 is (A) 2 (B) 3 (C) 6 (D) none of the above 4. Given positive real numbers a1 , a2 , . . . , a2011 whose product a1 a2 · · · a2011 is 1, what can you say about their sum S = a1 + a2 + · · · + a2011 ? (A) S can be any positive number. (B) 1 ≤ S ≤ 2011. (C) 2011 ≤ S and S is unbounded above. (D) 2011 ≤ S and S is bounded above. 5. A function f is defined by f (x) = ex if x < 1 and f (x) = loge (x) + ax2 + bx if x ≥ 1. Here a and b are unknown real numbers. Can f be differentiable at x = 1? (A) f is not differentiable at x = 1 for any a and b. (B) There exist unique numbers a and b for which f is differentiable at x = 1. (C) f is differentiable at x = 1 whenever a + b = e. (D) f is differentiable at x = 1 regardless of the values of a and b. 6. The equation x2 + bx + c = 0 has nonzero real coefficients satisfying b2 > 4c. Moreover, exactly one of b and c is irrational. Consider the solutions p and q of this equation. (A) Both p and q must be rational. (B) Both p and q must be irrational. (C) One of p and q is rational and the other irrational. (D) We cannot conclude anything about rationality of p and q unless we know b and c. 7. When does the polynomial 1 + x + · · · + xn have x − a as a factor? Here n is a positive integer greater than 1000 and a is a real number. (A) if and only if a = −1 (B) if and only if a = −1 and n is odd (C) if and only if a = −1 and n is even (D) We cannot decide unless n is known. 1

Part B. Attempt any 7 problems. Explain your reasoning. Each problem is worth 7 points. 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 only with those present. If the total number of handshakes is exactly 100, how many people left the meeting before Mr. L arrived? (Nobody shakes hands with the same person more than once.) 20 2. Show that the power of x with the largest in the polynomial (1 + 2x is 8, 3 ) P coefficient i i.e., if we write the given polynomial as i ai x then the largest coefficient ai is a8 .

3. Show that there are infinitely many perfect squares that can be written as a sum of six consecutive 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 by 11111. 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 A and 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 is in the XY-plane and the vertical direction is parallel to the Z-axis. The base is the region in the XY-plane bounded by the parabola y 2 = 1 − x and the Y-axis. Each cross-section of 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 for indefinite integrals? 8. f (x) = x3 + x2 + cx + d, where c and d are real numbers. Prove that if c > 31 , then f has exactly one real root. 9. A real-valued function f defined on a closed interval [a, b] has the properties that f (a) = f (b) = 0 and f (x) = f ′ (x) + f ′′ (x) for all x in [a, b]. Show that f (x) = 0 for all x in [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 g m = g n for positive integers m and n with m > n. Here g n stands for g ◦ g ◦ · · · ◦ g (n times). Show that g is one-to-one if and only if g is onto. (Some of you may have seen the term “one-one function” instead of “one-to-one function”. Both mean the same.) 3. In a quadrilateral ABCD, angles at vertices B and D are right angles. AM and CN are respectively 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(πx). 2. A differentiable function f : R → R satisfies f (1) = 2, f (2) = 3 and f (3) = 1. Show that f 0 (x) = 0 for some x. 3. Show that

ln(12) ln(18)

is irrational.

4. Show that lim

x→∞

x100 ln(x) = 0. ex tan−1 ( π3 + sin x)

5. (a) n identical chocolates are to be distributed among the k students in Tinku’s class. Find the probability that Tinku gets at least one chocolate, assuming that the n chocolates are handed out one by one in n independent steps. At each step, one chocolate is given to a randomly chosen student, with each student having equal chance to receive it. (b) Solve the same problem assuming instead that
all distributions are equally likely. You are given that the number of such distributions is n+k−1 k−1 . (Here all chocolates are considered interchangeable but students are considered different.) Part B. (9 problems × 10 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 cannot solve an earlier part, you may assume it and proceed to the next part. For all such partial answers, clearly mention what you are solving and what you are assuming. √ 1. a) Find a polynomial p(x) with real coefficients such that p( 2 + i) = 0. b)√Find a polynomial q(x) with rational coefficients and having the smallest possible degree such that √ q( 2 + i) = 0. Show that any other polynomial with rational coefficients and having 2 + i as a root has 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 convex quadrilateral 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 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. 3. a) We want to choose subsets A1 , A2 , . . . , Ak of {1, 2, . . . , n} such that any two of the chosen subsets have nonempty intersection. Show that the size k of any such collection of subsets is at most 2n−1 . b) For n > 2 show that we can always find a collection of 2n−1 subsets A1 , A2 , . . . of {1, 2, . . . , n} such that any two of the Ai intersect, but the intersection of all Ai is empty. 1

4. Define x=

10 X i=1

Show that a) x <

π 6

1 1 √ 10 3 1 + ( 10i√3 )2

< y and b)

x+y 2

<

π 6

and

y=

9 X i=0

1 1 √ . 10 3 1 + ( 10i√3 )2

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

5. Using the steps below, find the value of x2012 + x−2012 , where x + x−1 =

√

5+1 2 .

a) For any real r, show that |r + r−1 | ≥ 2. What does this tell you about the given x? b) Show that cos( π5 ) =

√

5+1 4 ,

3π e.g. compare sin( 2π 5 ) and sin( 5 ).

c) Combine conclusions of parts a and b to express x and therefore the desired quantity in a suitable form. 6. For n > 1, a configuration consists of 2n distinct points in a plane, n of them red, the remaining n blue, with no three points collinear. A pairing consists of n line segments, each with one blue and one red 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: consider total length of segments.) b) Given n red points (no three collinear), we can place n blue points such that any pairing in the resulting configuration will have two segments that do not intersect. (Hint: First consider the case n = 2.) 7. A sequence of integers cn starts with c0 = 0 and satisfies cn+2 = acn+1 + bcn for n ≥ 0, where a and b are integers. For any positive integer k with gcd(k, b) = 1, show that cn is divisible by k for infinitely many n. 8. Let f (x) be a polynomial with integer coefficients such that for each nonnegative integer n, f (n) = a perfect power of a prime number, i.e., of the form pk , where p is prime and k a positive integer. (p and k can vary with n.) Show that f must 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 that for each nonnegative integer n, g(n) = a perfect power of a fixed prime number. b) Show that a polynomial g(x) as in part a must be constant. 9. Let N be the set of non-negative integers. Suppose f : N → N is a function such that f (f (f (n))) < f (n + 1) for every n ∈ N . Prove that f (n) = n for all n using the following steps or otherwise. a) If f (n) = 0, then n = 0. b) If f (x) < n, then x < n. (Start by considering n = 1.) c) f (n) < f (n + 1) and n < f (n + 1) for all n. d) f (n) = n for all n.

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Solutions to the 2012 CMI BSc Entrance Examination Part A: 5 problems × 6 marks.

Part B: 7 out of 9 problems × 10 marks.

A1. Find the number of real solutions to the equation x = 99 sin(πx). x The number of solutions is the number of times the line y = 99 meets the graph of y = sin(πx). This can occur only for x ∈ [−99, 99] because sin(πx) has range [−1, 1]. Also sin(πx) is periodic with period 2. For x ≥ 0, the two graphs meet twice in each cycle of sin(πx), both intersections occurring in the first half of the cycle. There are 50 such half-cycles from x = 0 to x = 99, over intervals [0, 1], [2, 3], . . . , [98, 99]. So there are 100 non-negative solutions. Similarly there are 100 solutions ≤ 0 because both graphs are odd. Since x = 0 is counted twice, the total number of solutions is 100 + 100 − 1 = 199.

A2. A differentiable function f : R → R satisfies f (1) = 2, f (2) = 3 and f (3) = 1. Show that f 0 (x) = 0 for some x. Since f is differentiable, it is continuous. By the intermediate value theorem, there is a ∈ (2, 3) with f (a) = 2 = f (1). Now by Rolle’s theorem there is x ∈ (1, a) with f 0 (x) = 0. –OR– The continuous function f over the closed interval [1, 2] must attain its absolute maximum, which cannot be at either endpoint (since f (2) > f (1) and f (2) > f (3)). So the maximum must be at an interior point x and then f 0 (x) = 0. –OR– By the mean value theorem, f 0 (y) = 1 > 0 for some y ∈ (1, 2) and f 0 (z) = −2 < 0 for some z ∈ (2, 3). So f 0 (x) = 0 for some x ∈ (y, z) since for a differentiable f , the function f 0 satisfies the intermediate value property by Darboux’s theorem. (This is important to say because f 0 need not be continuous.) A3. Show that

ln(12) ln(18)

is irrational.

ln(12) ln(18)

= log18 (12). Suppose this is rational, say = ab where a, b are integers with b 6= 0. a Then 18 b = 12, so 18a = 12b . By factoring into primes this gives 32a 2a = 3b 22b , which by unique factorization can happen only if 2a = b and a = 2b. But this gives a = b = 0, ln(2) a contradiction. (Alternatively and similarly, prove that r = ln(3) is irrational and show that rationality of

ln(12) ln(18)

=

ln 3+2 ln 2 2 ln 3+ln a

A4. Show that lim

x→∞

=

1+2r 2+r

would force r to be rational as well.)

x100 ln(x) = 0. ex tan−1 ( π3 + sin x)

There is a positive constant c such that tan−1 ( π3 +sin x) > c for any x, e.g. c = tan−1 (0.04) will work since π > 3.12, sin(x) ≥ −1 and tan−1 is an increasing function. Moreover ln(x) < x for x > 0. So the given ratio is sandwiched between 0 and x101 /cex . Now use L’Hospital’s rule repeatedly. 1

A5. a) n identical chocolates are to be distributed among the k students in Tinku’s class. Find the probability that Tinku gets at least one chocolate, assuming that the n chocolates are handed out one by one in n independent steps. At each step, one chocolate is given to a randomly chosen student, with each student having equal chance to receive it. P(Tinku gets at least one chocolate) = 1 – P(Tinku gets none) = 1 − (1 − k1 )n , because in each of the independent steps the probability of Tinku not getting a chocolate is 1 − k1 . b) Solve the same problem assuming instead that all distributions are equally likely. You
are given that the number of such distributions is n+k−1 . (Here all chocolates are conk−1 sidered interchangeable but students are considered different.)
There are (n−1)+k−1 distributions in which Tinku gets at least a chocolate: give Tinku a k−1 chocolate and then use the given formula to find number of distributions
n+k−1
of then remaining (n−1)+k−1 n − 1 chocolates among k students. So the answer is / k−1 = n+k−1 . –OR– k−1 The number of distributions in which Tinku gets no chocolate = number of distributions
n+k−2 of n chocolates among the remaining k − 1 students = k−2 . So the desired probability
n+k−1
n = n+k−1 is 1 − n+k−2 . k−2 / k−1 √ B1. a) Find a polynomial p(x) with real coefficients such that p( 2 + i) = 0. Non-real polynomial with √ real coefficients occur in conjugate pairs. p(x) = √ roots of a √ 2 (x − ( 2 + i))(x − ( 2 − i)) = x − 2 2x + 3 works. b) Find a polynomial q(x) with rational coefficients and having the smallest possible degree √ such that q( 2 + i) = 0. Show that any other polynomial with rational coefficients and √ having 2 + i as a root has q(x) as a factor. √

√ √ 2 + i satisfies x2 − 2 2x + 3 = 0, i.e., x2 + 3 = 2 2x and so satisfies (x2 + 3)2 = 8x2 . So q(x) = (x2 + 3)2 − 8x2 works. A cubic with rational coefficients will not work because, after √ dividing by the necessarily rational leading coefficient, √it must be of the 2 form (x − 2 2x + 3)(x − r). This forces the coefficients −3r and −2 2 − r to be both rational, which is impossible. √ Let f (x) be a polynomial with rational coefficients such that f ( 2 + i) = 0. Divide f (x) by q(x) using long division to get √quotient a(x) and√remainder b(x), both polynomials with rational coefficients. √ Using f ( 2 + i) = 0 and q( 2 + i) = 0 in the equation f (x) = q(x)a(x) + b(x) gives b( 2 + i) = 0. Now if the remainder b(x)√is a nonzero polynomial, then it would have rational coefficients, degree less than 4 and 2 + i as a root. But we just proved that this is impossible. Hence b(x) = 0, i.e., f (x) is a multiple of q(x). B2. a) Let E, F, G and H respectively be the midpoints of the sides AB, BC, CD and DA of a convex quadrilateral ABCD. Show that EFGH is a parallelogram whose area is half that of ABCD. 2

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), so EFGH 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. So far 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) – 41 [Area(ABC)+ Area(BCD) + Area(CDA) + Area(DAB)] = Area(ABCD) – 12 Area(ABCD) = 12 Area(ABCD). b) Let E = (0, 0), F = (0, −1), G = (1, −1), H = (1, 0). Find all points A = (p, q) in the first 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 to having 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 imply that the quadrilateral constructed from the midpoints of the sides of ABCD is a square of side 1. So we just need to ensure that the listed coordinates make ABCD into a convex quadrilateral. This happens if and only if p, q are both positive (which is given) and < 1. It is easy to see that these conditions are sufficient to make ABCD a convex quadrilateral. For necessity see the following (pictures will help). If p > 1 then A will be to the right of H and so D to the left of H. If q > 1, then B will be below F and so C will be above F. If p 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 the chosen subsets have nonempty intersection. Show that the size k of any such collection of subsets is at most 2n−1 . If a set A is in such a collection C, then the complement of A cannot be in C. Therefore |C| ≤ 21 (total number of subsets of {1, 2, . . . , n}) = 21 2n = 2n−1 . b) For n > 2 show that we can always find a collection of 2n−1 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 2n−1 subsets of {1, 2, . . . , n} containing 1, remove the singleton set {1} and instead include its complement. –OR– Note 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 2n−3 subsets of {4, . . . , n}. –OR– For n = 2k +
1, take
all subsets of {1, 2, . . . , n} of size > k. Any two of these will intersect. n Now use ni = n−i . 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.) 3

B4. Define x=

10 X i=1

Show that a) x <

π 6

1 1 √ 10 3 1 + ( 10i√3 )2 < y and b)

x+y 2

<

and

y=

9 X i=0

π 6

1 1 √ . 10 3 1 + ( 10i√3 )2

. (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 √ Riemann 1 sums for f over the interval [0, √3 ] using 10 equal parts, each of width 1/10 3. Since f is a positive decreasing function, y overestimates the area under f over the given interval and √ R 1/√3 1/ 3 1 π −1 x underestimates it. The area under f over [0, √3 ] is 0 f (t)dt = tan (t)|0 = 6 , so π x < 6 < y. Note. Different normalizations are possible for f , e.g., the more simpleminded choice f (t) = 101√3 1+( 1t√ )2 considered over the interval [0,10] will work too. 10

3

b) x+y can be interpreted as the sum of areas of 10 trapezoids as follows. Dividing 2 [0, √13 ] into 10 equal parts, let the i-th subinterval be [ti−1 , ti ] with i = 0, 1, . . . , 10. Then the i-th trapezoid has base [ti−1 , ti ] and it has two vertical sides, the left one of height f (ti−1 ) and the right one of height f (ti ) (draw a picture and see). So we have to prove that the total area of trapezoids is less than the area under f . For this we should check concavity of f (draw pictures and see why). Check that over the interval (0, √13 ), we have 2

6t −2 f 00 (t) = (1+t 2 )3 < 0, so f is concave down and hence each trapezoid lies completely below the graph of f .

B5. Using the steps below, find the value of x2012 + x−2012 , where x + x−1 =

√

5+1 2 .

a) For any real r, show that |r + r−1 | ≥ 2. What does this tell you about the given x? Because of the absolute value we may assume that rp> 0 by replacing r with −r if √ necessary. √ 2 −1 Now use AM-GM inequality or the fact that ( r − 1/r) ≥ 0. Since x+x = 5+1 < 2, 2 given x must be a non-real (complex) number. b) Show that cos( π5 ) =

√

5+1 4 ,

3π e.g. compare sin( 2π 5 ) and sin( 5 ).

Let θ = π5 . Then sin(2θ) = sin(π − 2θ) = sin(3θ). Using the formulas for sin(2θ) and sin(3θ), canceling sin θ (it is nonzero) and substituting sin2 θ = 1 − cos2√θ, gives the quadratic equation 4 cos2 θ − 2 cos θ − 1 = 0. Since cos θ > 0, we get cos θ = 5+1 4 . c) Combine conclusions of parts a and b to express x and therefore the desired quantity in a suitable form. Let x = deiα = d(cos α + i√sin α). Then x−1 = d−1 e−iα = d−1 (cos α − i sin α). Adding iπ and using that x + x−1 = 5+1 = 2 cos( π5 ), we get d = 1 and α = ±θ. So x = e± 5 and 2 √ 2π 2π 5−1 2 π x2012 + x−2012 = 2 cos( 2012π ) = 2 cos(402π + ) = 2 cos( ) = 2 cos ( ) − 1 = 5 5 5 5 2 . 4

B6. For n > 1, a configuration consists of 2n distinct points in a plane, n of them red, the remaining n blue, with no three points collinear. A pairing consists of n line segments, each with one blue and one red 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: consider total length of segments.) For any configuration, there are only finitely many pairings. Choose one with least possible total length of segments. Here no two of the n segments can interest, because if RB and R0 B 0 intersect in point X then we get a contradiction as follows. Using triangle inequality in triangles RXB 0 and R0 XB, we get RB 0 + R0 B < RB + R0 B 0 (draw a picture). So replacing RB and R0 B 0 with R0 B and RB 0 would give a pairing with smaller total length. b) Given n red points (no three collinear), we can place n blue points such that any pairing in the resulting configuration will have two segments that do not intersect. (Hint: First consider the case n = 2.) For n = 2, place the two blue points on opposite sides of the line passing through the given two red points. There are two possible pairings and the two segments in either one do not intersect. We use a similar idea in general. Given n red points, find a triangle ABC such that A is a red point and all other red points are inside triangle ABC. (This is always possible. Why?) Place one blue point at B and all other blue points in the region opposite to triangle ABC at vertex C. (More precisely, let C be between A and A0 and also between B and B 0 . Place the remaining blue points inside triangle A0 CB 0 .) Now in any pairing, if A and B 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 for n ≥ 0, where a and b are integers. For any positive integer k with gcd(k, b) = 1, show that cn is divisible by k for infinitely many n. Consider pairs of consecutive entries of the sequence modulo k, i.e., (¯ cn , c¯n+1 ), where a ¯ 2 denotes a modulo k. Since there are only finitely many possibilities (namely k ), some pair of consecutive residues will repeat. Suppose (¯ ci , c¯i+1 ) = (¯ ci+p , c¯i+p+1 ) for some i. We will show that in fact the previous equation holds for all i, i.e., whole sequence of consecutive pairs is periodic. This will prove in particular that (¯ c0 , c¯1 ) = (¯ cp , c¯p+1 ) = (¯ c2p , c¯2p+1 ) = · · ·. Since c0 = 0 is divisible by k, so is cip for all i. The equation cn+2 = acn+1 + bcn shows that ¯b¯ cn = c¯n+2 − a ¯c¯n+1 . Now gcd(k, b) = 1 means 0 0 ¯ ¯ ¯ b is invertible modulo k, i.e., there is a b with bb = 1. Therefore c¯n = ¯b0 (¯ cn+2 − a ¯c¯n+1 ). Thus knowing a pair of consecutive residues uniquely determines the previous residue (this is why we considered pairs of residues). Therefore (¯ ci , c¯i+1 ) = (¯ ci+p , c¯i+p+1 ) implies (¯ ci−1 , c¯i ) = (¯ ci+p−1 , c¯i+p ) and (by the given recurrence) (¯ ci+1 , c¯i+2 ) = (¯ ci+p+1 , c¯i+p+2 ). Thus the whole sequence (¯ cn , c¯n+1 ) becomes periodic as soon as a single such pair repeats. 5

B8. Let f (x) be a polynomial with integer coefficients such that for each nonnegative integer n, f (n) = a perfect power of a prime number, i.e., of the form pk , where p is prime and k a positive integer. (p and k can vary with n.) Show that f must 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 that for each nonnegative integer n, g(n) = a perfect power of a fixed prime number. Write f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 . Then a0 = f (0) = pk for some prime p and integer k > 0. Define g(x) = f (px). Then g(x) is a polynomial such that for each nonnegative integer n, g(n) = f (pn) = a perfect power of a prime number. This prime number 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 + bn−1 xn−1 + · · · + b1 x + b0 . Then b0 = g(0) = pk . Consider g(mpk+1 ) = bn (mpk+1 )n +bn−1 (mpk+1 )n−1 +· · ·+b1 (mpk+1 )+pk . Clearly for each non-negative integer m, this expression is divisible by pk , but not by pk+1 (since it is pk modulo pk+1 ). This forces g(mpk+1 ) = pk for all m, since it must be a perfect power of p. Thus the polynomial g takes the value pk infinitely often, so it must be identically equal to pk . (Otherwise the polynomial g(x) − pk would have infinitely many roots.) To finish the problem, note that since g(x) = f (px) is constant, f (x) must be constant by the same logic. B9. Let N be the set of non-negative integers. Suppose f : N → N is a function such that f (f (f (n))) < f (n + 1) for every n ∈ N . Prove that f (n) = n for all n using the following steps or otherwise. a) If f (n) = 0, then n = 0. Let f (n) = 0. If n > 0, then n − 1 is in the domain of f and f (f (f (n − 1))) < f (n) = 0, which is a contradiction, since 0 is the smallest possible value of f . (Note that this does NOT prove that f (0) = 0, only that if f (some n) = 0, then that n = 0. In fact proving f (0) = 0 along with part a would essentially solve the problem, see below.) b) If f (x) < n, then x < n. (Start by considering n = 1.) Induction on n. If n = 1, then this is just part a. Assuming the statement up to n we need to prove that if f (x) < n + 1, then x < n + 1. If f (x) < n, then by induction x < n, so x < n + 1. So let f (x) = n. If x = 0, we are done. Otherwise f (f (f (x − 1))) < f (x) = n and by using induction thrice we get in succession f (f (x − 1)) < n, then f (x − 1) < n and then x − 1 < n, i.e., x < n + 1 as desired. c) f (n) < f (n + 1) and n < f (n + 1) for all n. Apply part b to f (f (f (m))) < f (m + 1) (with x = f (f (m)) and n = f (m + 1)) to get 6

f (f (m)) < f (m + 1). Apply part b to this with x = f (m) and n = f (m + 1) to get f (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 f is 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) = 0 implies n = 0, so n > 0 implies f (n) > 0. Applying this to any positive f (k), we get f (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)))|n ≥ 0}, we must have f (k) = 0. In particular 0 is in the range of f , 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) for every n > 0. Repeat the reasoning substituting 1 (the new lowest element of the domain and the range) for 0 and conclude f (1) = 1. Then restrict to n > 1 and show f (2) = 2 and so on.

7

2013 Entrance Examination for BSc Programmes at CMI

Part A. (10 problems × 5 points = 50 points.) Attempt all questions in this part before 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 on the filled answer, so you need not explain your answer. Each correct answer gets 1 point and 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 any problem will be counted in your total score, so it is better not to guess! If you are unsure about a part, you may leave it blank without any penalty. If you write something and then want 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 and g : B → A be functions such that f (g(x)) = x for each x. For each statement below, write whether it is TRUE or FALSE. a) The function f must be one-to-one.

Answer:

b) The function f must be onto.

Answer:

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

Answer:

d) The function g must be onto.

Answer:

1

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

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

Answer:

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

Answer:

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

Answer:

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

Answer:

2

3. Let S be a circle with center O. Suppose A, B are points on the circumference of S with ∠AOB = 120◦ . For triangle AOB, let C be its circumcenter and D its orthocenter (i.e., the point of intersection of the three lines containing the altitudes). For each statement below, write whether it is TRUE or FALSE.

a) The triangle AOC is equilateral.

Answer:

b) The triangle ABD is equilateral.

Answer:

c) The point C lies on the circle S.

Answer:

d) The point D lies on the circle S.

Answer:

3

4. A polynomial f (x) with real coefficients is said to be a sum of squares if we can write f (x) = p1 (x)2 + · · · + pk (x)2 , where p1 (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 of x in f (x) must be 0.

Answer:

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

Answer:

c) If f (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 of x, then f (x) is a sum of squares.

Answer:

4

5. There are 8 boys and 7 girls in a group. For each of the tasks specified below, write an expression 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 girls there is a boy

Answer:

c) Choosing a team of six people from the group

Answer:

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

Answer:

5

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

a)

R4 1

x2 dx

Answer:

b)

R3 1

[x]2 dx

Answer:

c)

R2 1

[x2 ]dx

Answer:

d)

R1

1 dx −1 x2

Answer:

6

7. Let A, B, C be angles such that eiA , eiB , eiC form an equilateral triangle in the complex plane. Find values of the given expressions.

a) eiA + eiB + eiC

Answer:

b) cos A + cos B + cos C

Answer:

c) cos 2A + cos 2B + cos 2C

Answer:

d) cos2 A + cos2 B + cos2 C

Answer:

7

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

a) The equation x2 + bx + c = 0 has a real solution if and only if b2 − 4c is

Answer:

b) The value of p( 12 ), i.e., the probability that x2 +

x 2

+ c = 0 has a real solution is

Answer:

c) As a function of b, 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:

8

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 choosing the correct option from the given ones.

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

Answer:

b) Suppose f is differentiable. Then which of the follwing intervals must contain an x such that f 0 (x) = 2x ? Options: (1,2) (2,4) both neither

Answer:

c) Suppose f is twice differentiable. Which of the following intervals must contain an x such that f 00 (x) = 2 ? Options: (1,2) (2,4) both neither

Answer:

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

Answer:

9

10. Let f (x) =

x4 (x − 1)(x − 2) · · · (x − n)

where the denominator is a product of n 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 independent questions below by choosing the correct option from the given ones.

a) How many vertical asymptotes does the graph of f have? Options: n less than n more than n impossible to decide

Answer:

b) What can you deduce about the value of n ? Options: n < 4 n=4 n>4 impossible to decide

Answer:

c) As one travels along the graph of f from left to right, at which of the following points is the sign of f (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 of f have in the region x < 0 ? Options: none 1 more than 1 impossible to decide (Hint: Sketching is better than calculating.)

Answer:

10

Part B. (Problems 1–4 × 15 points + problems 5–6 × 20 points = 100 points.) Solve these problems in the space provided for each problem after this page. You may solve only part of a problem and get partial credit. Clearly explain your entire reasoning. No credit will be given without reasoning. 1. In triangle ABC, the bisector of angle A meets side BC in point D and the bisector of angle B meets side AC in point E. Given that DE is parallel to AB, show that AE = BD and that the triangle ABC is isosceles. 2. A curve C has the property that the slope of the tangent at any given point (x, y) on 2 +y 2 . C is x 2xy a) Find the general equation for such a curve. Possible hint: let z = xy . b) Specify all possible shapes of the curves in this family. (For example, does the family include an ellipse?) 3. A positive integer N has its first, third and fifth digits equal and its second, fourth and sixth digits equal. In other words, when written in the usual decimal system it has the form xyxyxy, where x and y are the digits. Show that N cannot be a perfect power, i.e., N cannot equal ab , where a and b are positive integers with b > 1. 4. Suppose f (x) is a function from R to R such that f (f (x)) = f (x)2013 . Show that there are infinitely many such functions, of which exactly four are polynomials. (Here R = the set of real numbers.) 1 5. Consider the function f (x) = ax + x+1 , where a is a positive constant. Let L = the 1 largest value of f (x) and S = the smallest value of f (x) for x ∈ [0, 1]. Show that L−S > 12 for any a > 0.

6. Define fk (n) to be the sum of all possible products of k distinct integers chosen from the set {1, 2, . . . , n}, i.e., X fk (n) = i1 i2 . . . ik . 1≤i1 0 due to the roots being non-real. Since p need not be 0, this disproves part a. For part d, since all roots of f are 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 values and 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 an expression 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: 2 × 8! × 7! (The factor of 2 arises because the two blocks of boys and girls can switch positions.) 2

b) Sitting in a row so that between any two boys there is a girl and between any two girls there 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 of unequal number of boys and girls












8 7 8 8 7 8 7 8 7 8 7 7 Answer: 15 − = + + + + + 6 3 3 6 5 1 4 2 2 4 1 5 6 6. Calculate the following integrals whenever possible. If a given integral does not exist, state so. Note that [x] denotes the integer part of x, i.e., the unique integer n such that n ≤ x < n + 1. R4 3 a) 1 x2 dx = x3 |41 = 21 using the fundamental theorem of calculus. R3 b) 1 [x]2 dx = 1(12 ) + 1(22 ) = 5 = area under the piecewise constant function [x]2 √ √ √ √ √ √ R2 c) 1 [x2 ]dx = 1( 2 − 1) + 2( 3 − 2) + 3(2 − 3) = 5 − 2 − 3 since the function [x]2 √ √ √ √ is constant on intervals [1, 2), [ 2, 3), [ 3, 2), taking values 1, 2, 3 respectively. R1 R1 d) −1 x12 dx = 2 limt→0+ t x12 dx = 2 limt→0+ (−1 + 1t ) = ∞. The fundamental theorem does not apply over the interval [−1, 1] because x12 goes to ∞ in the interval. It is also ok to answer that the integral does not exist (as a real number).

7. Let A, B, C be angles such that eiA , eiB , eiC form an equilateral triangle in the complex plane. 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 of eiA + eiB + eiC , which is 0 by part a. c) cos 2A+cos 2B+cos 2C = 0 because the points e2iA , e2iB , e2iC on the unit circle also form an equilateral triangle in the complex plane, since taking B = A + (2π/3), C = A + (4π/3), we get 2B = 2A + (4π/3) and 2C = 2A + (8π/3) = 2A + (2π/3) + 2π and the last term 2π does not change the position of the point. d) cos2 A + cos2 B + cos2 C = 32 because, using the formula for cos 2θ in part c, we get cos2 A + cos2 B + cos2 C = sin2 A + sin2 B + sin2 C and the sum of the LHS and the RHS in this equation is 3. 8. Consider the quadratic equation x2 + bx + c = 0, where b and c are chosen randomly from the interval [0,1] with the probability uniformly distributed over all pairs (b, c). Let p(b) = the probability that the given equation has a real solution for given (fixed) value of b. Answer the following questions by filling in the blanks. 3

a) The equation x2 + bx + c = 0 has a real solution if and only if b2 − 4c is ≥ 0. b) The value of p( 21 ), i.e., the probability that x2 + Answer: which is

x 2

+ c = 0 has a real solution is

1 1 2 16 since a real solution occurs precisely when b −4c = 4 −4c 1 th fraction of the interval [0, 1] over which c ranges. 16

≥ 0, i.e., 0 ≤ c ≤

1 16 ,

c) As a function of b, is p(b) increasing, decreasing or constant? Answer: increasing, because b2 − 4c ≥ 0 if and only if 0 ≤ c ≤ increasing for 0 ≤ b ≤ 1.

b2 4 ,

so p(b) =

b2 4 ,

which is

d) As b and 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 by 2 the region b2 − 4c ≥ 0, i.e., it is the area under the parabola c = b4 from b = 0 to b = 1, R1 2 1 . which is 0 b4 db = 12 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 choosing the correct option from the given ones. a) Which of the following values must be in the range of f ? Options: 5 25 both neither Answer: 5, by the intermediate value theorem, e.g., over the interval [2,3]. Also f (x) need not take the value 25, e.g., take f (x) = x2 for x < 4 and f (x) = 16 for x ≥ 4. b) Suppose f is differentiable. Then which of the follwing intervals must contain an x such that f 0 (x) = 2x ? Options: (1,2) (2,4) both neither Answer: both c) Suppose f is twice differentiable. Which of the following intervals must contain an x such that f 00 (x) = 2 ? Options: (1,2) (2,4) both neither Answer: (2,4) d) Suppose f is a polynomial, then which of the following are possible values of its degree? Options: 3 4 both neither Answer: 4 For parts b,c and d, let g(x) = f (x) − x2 . We have g(1) = g(2) = g(3) = g(4) = 0. For part b, applying Rolle’s theorem to g(x) gives g 0 (x) = 0 for some values of x in each of the intervals (1, 2), (2, 3), (3, 4). For these values of x, we have f 0 (x) = g 0 (x) + 2x = 2x. Far part c, take from part b values r ∈ (2, 3) and s ∈ (3, 4) with g 0 (r) = 0 = g 0 (s). Applying Rolle’s theorem to g 0 (x) in the interval [r, s], we get for some x ∈ (r, s) ⊂ (2, 4) the equality g 00 (x) = 0 and so f 00 (x) = g 00 (x) + 2 = 2. There need not be an x ∈ (1, 2) with f 00 (x) = 2, i.e., g 00 (x) = 0. There are many ways to arrange this, for example let g(x) = sin(πx). Then 4

g 00 (x) = −π 2 sin(πx), which is 0 only when x is an integer, in particular g 00 (x) 6= 0 for any x ∈ (1, 2). For part d, note that g(x), now being a polynomial vanishing at 1, 2, 3 and 4, must be divisible by (x − 1)(x − 2)(x − 3)(x − 4). So g(x), if non-zero, must have degree at least 4. Thus f (x) = x2 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 of n 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 independent questions below by choosing the correct option from the given ones. a) How many vertical asymptotes does the graph of f have? Options: n less than n more than n impossible to decide Answer: n, at x = 1, 2, . . . , n. b) What can you deduce about the value of n ? Options: n < 4 n=4 n>4 impossible to decide Answer: n > 4, because limx→±∞ f (x) = 0 and for this to happen, the degree of the denominator of f (x) must be greater than that of the numerator. c) As one travels along the graph of f from left to right, at which of the following points is the sign of f (x) guaranteed to change from positive to negative? Options: x = 0 x=1 x=n−1 x=n Answer: x = n−1, because f (x) is positive for x > n and f (x) changes sign precisely when it passes through x = 1, 2 . . . , n. Note that the sign of f (x) for x < 0 and for x ∈ (0, 1) depends on the parity of n. d) How many inflection points does the graph of f have in the region x < 0 ? Options: none 1 more than 1 impossible to decide (Hint: Sketching is better than calculating.) Answer: more than 1. Note that f (x) = 0 only at x = 0, with multiplicity 4. Without loss of generality, let n be even. (If n is odd, the reasoning is completely parallel, see note at the end.) Now f (x) > 0 for x < 1 except at x = 0 and f has all derivatives for x < 1. Due to the multiple root at x = 0, the graph of f must be concave up (i.e. f 00 (x) > 0) near x = 0. Further, as x → −∞, the values of f (x) stay positive and → 0. Therefore, as one traces the graph leftward from the origin, it must become concave down at least once and eventually concave up again so as to approach the X-axis from above. (Note: If n is odd, f (x) < 0 for x < 1 except at x = 0. As one traces the graph leftward from the origin, the function is initially as well as eventually concave down and must be concave up at least once in-between so as to approach the X-axis from below.) 5

Part B. (Problems 1–4 × 15 points + problems 5–6 × 20 points = 100 points.) Solve these problems in the space provided for each problem after this page. You may solve only part of a problem and get partial credit. Clearly explain your entire reasoning. No credit will be given without reasoning. 1. In triangle ABC, the bisector of angle A meets side BC in point D and the bisector of angle B meets side AC in point E. Given that DE is parallel to AB, show that AE = BD and that the triangle ABC is isosceles. Answer: ∠EAD = ∠DAB = ∠EDA, the first equality because AD bisects ∠EAB and the second because alternate angles made by line AD intersecting parallel lines DE and AB are equal. Thus 4EAD is isosceles with EA = ED. Similarly ED = DB using the fact that BE bisects ∠DBA also intersects parallel lines DE and AB. Therefore EA = CD ED = DB. Now by the basic proportionality theorem, CE EA = DB . As the denominators EA and DB are equal, the numerators must be equal as well, i.e., CE = CD. Finally, CA = CE + EA = CD + DB = CB, so 4ABC is isosceles. 2. A curve C has the property that the slope of the tangent at any given point (x, y) on 2 +y 2 C is x 2xy . a) Find the general equation for such a curve. Possible hint: let z = xy . b) Specify all possible shapes of the curves in this family. (For example, does the family include an ellipse?) Answer: The defining property of the curve C is equivalent to the differential equation dy x2 +y 2 = = 12 ( xy + xy ). It is convenient to let z = y/x, so the equation becomes dx 2xy dy dx

= 21 ( z1 + z). To get this in terms of only x and z, differentiate z = y/x with respect to x 2 dy dy dz = x1 dx − xy2 = x1 ( dx − z) = x1 ( 12 ( z1 + z) − z) = x1 1−z to get dx 2z , where we have substituted dy for dx using the differential equation and then simplified. Separating the variables and R R 2zdz 2 integrating, we get dx = x 1−z 2 , which gives log |x| = − log |1 − z |+ a constant, i.e., K

log |1 − z 2 | = − log |x| + K = log |x|−1 + K. Exponentiating, we get 1 − z 2 = ± ex = 2

c x,

where c is a nonzero constant. Substituting z = y/x, we get 1 − xy 2 = xc , i.e., x2 − y 2 = cx. To be precise, we have to delete the points (0, 0) and (c, 0) from this solution, because 2 dy +y 2 for the given equation dx = x 2xy to make sense, both x and y must be nonzero. If the dy equation were given as 2xy dx = x2 + y 2 , then this issue would not arise. 2

To see the shape of the curve, complete the square to get (x − 2c )2 − y 2 = c4 , which is a hyperbola when c 6= 0. (Note: By differentiating x2 − y 2 = cx, it is easy to see that 2 dy +y 2 2x−c = x 2xy and that this holds even when c = 0. Thus we get the two straight lines dx = y y = ±x also as solutions. The reason the above answer missed this possibility was because we put 1 − z 2 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.) 6

3. A positive integer N has its first, third and fifth digits equal and its second, fourth and sixth digits equal. In other words, when written in the usual decimal system it has the form xyxyxy, where x and y are the digits. Show that N cannot be a perfect power, i.e., N cannot 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 N to be a perfect power, the primes 3,7,13,37 must all occur (and in fact with equal power) as factors in the prime factorization of 10x + y. In particular 10x + y ≥ 10101. But since x and y are digits, each is between 0 and 9, so 10x + y ≤ 99. So N cannot be a perfect power. 4. Suppose f (x) is a function from R to R such that f (f (x)) = f (x)2013 . Show that there are infinitely many such functions, of which exactly four are polynomials. (Here R = the set of real numbers.) Answer: If f is a polynomial, then we make two cases. (i) If f (x) = a constant c, then the given condition is equivalent to c = c2013 , which happens precisely for three values of c, namely c = 0, 1, −1 (since we have c(c2012 − 1) = 0, so c = 0 or c2012 = 1). Thus there are three constant functions with the given property. (ii) If f (x) is a non-constant polynomial, then consider its range set A = {f (x)|x ∈ R}. Now for all a ∈ A, we have by the given property f (a) = a2013 . So the polynomial f (x) − x2013 has all elements of A as its roots. Since there are infinitely many values in A (e.g. applying the intermediate value theorem because f is continuous), the polynomial f (x) − x2013 has infinitely many roots and thus must be the zero polynomial, i.e., f (x) = x2013 for all real number x. Note: One can also deduce that the degree of f must be 0 or 2013 by equating the degrees of f (f (x)) and f (x)2013 . Then, in the non-constant case, it is possible to argue first that the 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 and f (−1) = −1. For every other real number x, arbitrarily define f (x) to be 0, 1 or −1. It is easy 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., i any number in the range of f ) and then this forces f (x) = x2013 for all x ∈ S = {a2013 |i = 0, 1, 2, . . .}. We use this as follows. Fix a real number a. Then define f (x) = x2013 for all i x ∈ S = {a2013 |i = 0, 1, 2, . . .}. For all x 6∈ S, simply define f (x) = any element of the set S, e.g., a itself will do.) 1 , where a is a positive constant. Let L = the 5. Consider the function f (x) = ax + x+1 1 largest value of f (x) and S = the smallest value of f (x) for x ∈ [0, 1]. Show that L−S > 12 for any a > 0. 1 Answer: Let f (x) = ax + x+1 . We wish to understand the minimum and maximum of 1 this function in the interval [0, 1]. Now f (0) = 1, f (1) = a + 21 and f 0 (x) = a − (x+1) 2. Over the interval [0, 1], the value of f 0 (x) increases from a − 1 at x = 0 to a − 14 at x = 1.

7

We should consider what happens to the sign of f 0 (x). For this we consider the following cases. (1) Suppose a ≤ 1/4. Because 1/(x + 1)2 ≥ 1/4 on the interval [0, 1], f 0 (x) ≤ 0, so the maximum 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) Suppose a ≥ 1. Then f 0 (x) ≥ 0 on the interval [0, 1], so maximum is at 1 and minimum at 0. We get a + 1/2 − 1 = a − 1/2 ≥ 1/2 ≥ 1/12. (3) Suppose 1/4 ≤ a ≤ 1. Now f 0 (x) = 0 at x ˜ = √1a − 1. For this range of a, x ˜ ∈ [0, 1]. 0 0 In the interval [0, x ˜], f (x) ≤ 0 and in the interval [˜ x, 1], f (x) ≥ 0. Now we make two sub-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 is at x = 0. f (˜ x) = a√− a + a √ = 2 a − a. So the difference√between maximum and 2 minimum is 1 + a − 2 a = (1 − to 1 and so √ 2 √ a) . This is smallest when a is3 closest √ 2 1 (1 − a) ≥ (1 − 1/ 2) = 3/2 − 2. This is bigger than 1/12 since ( 2 − 12 ) = 17/12 and 172 = 289 ≥ 2 × 122 . (3ii) Suppose 1/2 ≤ a √ ≤ 1. Now√f (1) ≥ f (0). 1 and minimum is at x ˜. The √ Max is at √ 1 2 √ difference is a + 1/2 − a + a − a = 2a − 2 a + 1/2 = ( 2a − 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 of k distinct integers chosen from the set {1, 2, . . . , n}, i.e., fk (n) =

X

i1 i2 . . . ik .

1≤i1 1, it is enough to find a polynomial g(x) ˜ ˜ has degree < d, by such that g(x) − g(x − 1) = xd (because if h(x) = cxd + h(x), where h ˜ and then cg(x) + g˜(x) works for h(x)). To find such g(x), notice induction we find g˜ for h d+1 that for g1 (x) = x , we have h1 (x) = g1 (x) − g1 (x − 1) = (d + 1)xd + h2 (x), where h2 (x) is a polynomial of degree d − 1. By induction h2 (x) = g2 (x) − g2 (x − 1) for a polynomial 1 (g1 (x) − g2 (x)) works. g2 (x) of degree d. Now g(x) = d+1

9

Test Codes: UGA (Multiple-choice Type) and UGB (Short Answer Type) 2013 Questions will be set on the following and related topics. Algebra: Sets, operations on sets. Prime numbers, factorization of integers and divisibility. Rational and irrational numbers. Permutations and combinations, Binomial Theorem. Logarithms. Polynomials: relations between roots and coefficients, Remainder Theorem, Theory of quadratic equations and expressions. Arithmetic and geometric progressions. Inequalities involving arithmetic, geometric & harmonic means. Complex numbers. Geometry: Plane geometry. Geometry of 2 dimensions with Cartesian and polar coordinates. Equation of a line, angle between two lines, distance from a 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 trigonometric functions. Trigonometric identities including addition formulae, solutions of trigonometric equations. Properties of triangles. Heights and distances. Calculus: Sequences - bounded sequences, monotone sequences, limit of a sequence. Functions, one-one functions, onto functions. Limits and continuity. Derivatives and methods of differentiation. Slope of a curve. Tangents and normals. Maxima and minima. Using calculus to sketch graphs of functions. Methods of integration, definite and indefinite integrals, evaluation of area using integrals.

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

1

Sample Questions for UGA Instructions. UGA is a multiple choice examination. In each of the following questions, exactly one of the choices is correct. You get four marks for each correct answer, one mark for each unanswered question, and zero marks for each incorrect answer.

1 Define an = (12 + 22 + . . . + n2 )n and bn = nn (n!)2 . Recall n! is the product of the first n natural numbers. Then, (A) an < bn for all n > 1 (C) an = bn for infinitely many n

(B) an > bn for all n > 1 (D) None of the above

2 The sum of all distinct four digit numbers that can be formed using the 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 of a3 b4 c5 in the expansion of (bc + ca + ab)6 is     6 6 12! 3! (C) 33 (D) 3 (B) (A) 3 3 3!4!5! 5 Let ABCD be a unit square. Four points E, F , G and H are chosen on the sides AB, BC, CD and DA respectively. The lengths of the sides of the quadrilateral EF GH are α, β, γ and δ. Which of the following is always true? √ (A) 1 ≤ α2 + β 2 + γ 2 + δ 2 ≤ 2 2 √ √ (B) 2 2 ≤ α2 + β 2 + γ 2 + δ 2 ≤ 4 2 (C) 2 ≤ α2 + β 2 + γ 2 + δ 2 ≤ 4 √ √ (D) 2 ≤ α2 + β 2 + γ 2 + δ 2 ≤ 2 + 2 6 If log10 x = 10log100 4 then x equals (A) 410

(B) 100

(C) log10 4

(D) none of the above

7 z 1 , z2 are two complex numbers with z2 6= 0 and z1 6= z2 and satisfying z1 + z2 z1 z1 − z2 = 1. Then z2 is 1

(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 numbers x satisfying the inequality x3 (x+1)(x−2) ≥ 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 nary. Then (A) z is neither real nor purely imaginary (C) z is purely imaginary

z 1+z

is purely imagi-

(B) z is real (D) none of the above

10 Let A be the fixed point (0, 4) and B be a moving point (2t, 0). Let M be the mid-point of AB and let the perpendicular bisector of AB meet the y-axis at R. The locus of the mid-point P of M R is (A) y + x2 = 2 (C) (y − 2)2 − x2 = 1/4

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

11 The sides of a triangle are given to be x2 + x + 1, 2x + 1 and x2 − 1. Then the largest of the three angles of the triangle is   x ◦ (A) 75 (B) π radians (C) 120◦ (D) 135◦ x+1 12 Two poles, AB of length two metres and CD of length twenty metres are erected vertically with bases at B and D. The two poles are at a distance not less than twenty metres. It is observed that tan ∠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 (−2π, 0), the function f (x) = sin

2



1 x3



(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

(ex − 1) tan2 x x→0 x3 lim

(A) does not exist (C) exists and equals 2/3

(B) exists and equals 0 (D) exists and equals 1

16 Let f1 (x) = ex , f2 (x) = ef1 (x) and generally fn+1 (x) = efn (x) for all d n ≥ 1. For any fixed n, the value of fn (x) is equal to dx (A) fn (x) (B) fn (x)fn−1 (x) (C) fn (x)fn−1 (x) · · · f1 (x) (D) fn+1 (x)fn (x) · · · f1 (x)ex 17 If the function f (x) =

(

x2 −2x+A sin x

B

if x 6= 0 if x = 0

is continuous at x = 0, then (A) A = 0, B = 0 (C) A = 1, B = 1

(B) A = 0, B = −2 (D) A = 1, B = 0

18 A truck is to be driven 300 kilometres (kms.) on a highway at a constant speed of x kms. per hour. Speed rules of the highway require that 30 ≤ x ≤ 60. The fuel costs ten rupees per litre and is consumed at the rate 2 + (x2 /600) litres per hour. The wages of the driver are 200 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 If b = (A)

Z

0 a be

1

et dt then t+1 (B)

Z

a

a−1 −a be

e−t dt is t−a−1

(C) −be−a

(D) −bea

20 In the triangle ABC, the angle ∠BAC is a root of the equation √ 3 cos x + sin x = 1/2. Then the triangle ABC is 3

(A) obtuse angled (C) acute angled but not equilateral

(B) right angled (D) equilateral

21 Let n be a positive integer. Consider a square S of side 2n units. Divide S into 4n2 unit squares by drawing 2n − 1 horizontal and 2n − 1 vertical lines one unit apart. A circle of diameter 2n − 1 is drawn with its centre at the intersection of the two diagonals of the square S. How many of these unit squares contain a portion of the circumference of the 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 man six feet tall is walking at a speed of 10 feet/second from the lantern to the nearest point on the wall. When he is midway between the lantern and the wall, the rate of change (in ft./sec.) in the length of his 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 is inscribed in a circle. A second circle touches the first circle and also touches the base of the triangle at its midpoint. If the second circle is situated outside the triangle, then its radius (in cms.) is √ √ √ √ (B) 3/2 (C) 3 (D) 4/ 3 (A) 3 3/2 24 Let n be a positive integer. Define

Then (A)

Z

n+1

f (x) = min{|x − 1|, |x − 2|, . . . , |x − n|}. f (x)dx equals

0

(n + 4) 4

(B)

(n + 3) 4

(C)

(n + 2) 2

(D)

(n + 2) 4

25 Let S = {1, 2, . . . , n}. The number of possible pairs of the form (A, B) with A ⊆ B for subsets A and B of S is  n   X n n n n (A) 2 (B) 3 (C) (D) n! k n−k k=0

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 4

label of the ball drawn from the i-th box, i = 1, 2, 3. Then the number of ways in which the balls can be chosen such that n1 < n2 < n3 is (A) 120 (B) 130 (C) 150 (D) 160 28 Let a be a real number. The number of distinct solutions (x, y) of the system of equations (x − a)2 + y 2 = 1 and x2 = y 2 , can only be (A) 0, 1, 2, 3, 4 or 5 (C) 0, 1, 2 or 4

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

29 The maximum of the areas of the isosceles triangles with base on the positive x-axis and which lie below the curve y = e−x is: (A) 1/e (B) 1 (C) 1/2 (D) e 30 Suppose a, b and n are positive integers, all greater than one. If an +bn is 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 metres/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 needed to 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 Z 5π/2 π/2

equals (A) 1

etan

etan

−1 (sin x)

−1 (sin x)

(B) π

+ etan

−1 (cos x)

(C) e

dx (D) none of these

33 The set of all solutions of the equation cos 2θ = sin θ + cos θ is given by (A) θ = 0 (B) θ = nπ + π2 , where n is any integer (C) θ = 2nπ or θ = 2nπ − π2 or θ = nπ − π4 , where n is any integer (D) θ = 2nπ or θ = nπ + π4 , where n is any integer 34 For k ≥ 1, the value of         n n+1 n+2 n+k + + + ··· + 0 1 2 k equals

5

 n+k+1 n+k   n+k+1 (C) n+1 (A)



35 The value of "

sin−1 cot sin−1

(

1 2



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

1−

r

5 6

!)

+ cos−1

r

2 + sec−1 3

r

8 3



#

is (A) 0

(B) π/6

(C) π/4

(D) π/2

36 Which of the following graphs represents the function Z √x 2 f (x) = e−u /x du, for x > 0 and f (0) = 0? 0

(A)

(B)

(C)

(D)

  2  3 n  1 22 32 n2 37 If an = 1 + 2 1+ 2 1 + 2 · · · 1 + 2 , then n n n n lim a−1/n n

2

n→∞

is (A) 0

(B) 1

(C) e

(D)

√

e/2

38 The function x(α − x) is strictly increasing on the interval 0 < x < 1 if and only if (A) α ≥ 2 (C) α < −1

(B) α < 2 (D) α > 2

39 Consider a circle with centre O. Two chords AB and CD extended intersect at a point P outside the circle. If ∠AOC = 43◦ and ∠BP D = 18◦ , then the value of ∠BOD is (A) 36◦ (B) 29◦ (C) 7◦ (D) 25◦ 6

40 A box contains 10 red cards numbered 1, . . . , 10 and 10 black cards numbered 1, . . . , 10. In how many ways can we choose 10 out of the 20 cards so that there are exactly 3 matches, where a match means a red card and a black card with the same number?       10 7 10 7 4 2 (B) (A) 4 3 4 3      10 14 10 7 2 (D) (C) 4 3 3 41 Let P be a point on the ellipse x2 + 4y 2 = 4 which does not lie on the axes. If the normal at the point P intersects the major and minor axes at 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 = 0 represents, 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 and a, b, c are 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 equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root, then f d e , and are in a b c (A) A.P. (B) G.P. (C) H.P. (D) none of the above. 45 The number of solutions of the equation sin−1 x = 2 tan−1 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◦ . If E is the point of intersection of AC and BD, then the value of ∠AEB is (A) 75◦ (B) 85◦ (C) 95◦ (D) 110◦ . 47 Let R be the set of all real numbers. The function f : R → R defined by f (x) = x3 − 3x2 + 6x − 5 is (A) one-to-one, but not onto (B) one-to-one and onto 7

(C) onto, but not one-to-one (D) neither one-to-one nor onto. 48 Let L be the point (t, 2) and M be a point on the y-axis such that LM has slope −t. Then the locus of the midpoint of LM , as t varies over all real values, is (A) y = 2 + 2x2

(B) y = 1 + x2

(C) y = 2 − 2x2

(D) y = 1 − x2 .

49 Suppose x, y ∈ (0, π/2) and x 6= y. Which of the following statement is 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 exist x, y such that 2 sin(x + y) = sin 2x + sin 2y. (D) None of the above. 50 A triangle ABC has a fixed base BC. If AB : AC = 1 : 2, then the locus of the vertex A is (A) a circle whose centre is the midpoint of BC (B) a circle whose centre is on the line BC but not the midpoint of BC (C) a straight line (D) none of the above. 51 Let P be a variable point on a circle C and Q be a fixed point outside C. If R is the mid-point of the line segment P Q, then the locus of R is (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 right are 1. If N is 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  Z b 3 2 −x−x dx 4 a over all possible values of a and b is 3 4 (A) (B) 4 3 54 For any n ≥ 5, the value of 1 +

(C)

3 2

(D)

1 1 1 + + ··· + n lies between 2 3 2 −1

8

2 . 3

n and n 2 (D) none of the above.

n 2 (C) n and 2n

(B)

(A) 0 and

55 Let ω denote a cube root of unity which is not equal to 1. Then the number of distinct elements in the set  (1 + ω + ω 2 + · · · + ω n )m : m, n = 1, 2, 3, · · · is (A) 4

(B) 5

(C) 7

(D) infinite.

56 The value of the integral

(A) is less than 2

Z

3 2

dx loge x (B) is equal to 2

(C) lies in the interval (2, 3)

(D) is greater than 3.

57 The area of the region bounded by the straight lines x = 12 and x = 2, and the curves given by the equations y = loge x and y = 2x is √ √ (A) log1 2 (4 + 2) − 25 loge 2 + 32 (B) log1 2 (4 − 2) − 25 loge 2 e e √ (D) none of the above (C) log1 2 (4 − 2) − 25 loge 2 + 32 e

58 In a win-or-lose game, the winner gets 2 points whereas the loser gets 0. Six players A, B, C, D, E and F play each other in a preliminary round from which the top three players move to the final round. After each player has played four games, A has 6 points, B has 8 points and C has 4 points. It is also known that E won against F. In the next set of games D, E and F win their games against A, B and C respectively. If A, B and D move to the final round, the final scores of E and F are, respectively, (A) 4 and 2 (B) 2 and 4 (C) 2 and 2 (D) 4 and 4. 59 The number of ways in which one can select six distinct integers from the set {1, 2, 3, · · · , 49}, such that no two consecutive integers are selected, is       49 48 43 (A) −5 (B) 6 5 6     44 25 . (D) (C) 6 6 60 Let n ≥ 3 be an integer. Assume that inside a big circle, exactly n small circles of radius r can be drawn so that each small circle touches the big circle and also touches both its adjacent small circles. Then, the radius of the big circle is 9

(A) r cosec πn π (C) r(1 + cosec 2n )

(B) r(1 + cosec 2π n ) (D) r(1 + cosec πn )

61 If n is a positive integer such that 8n + 1 is a perfect square, then (A) (B) (C) (D)

n must be odd n cannot be a perfect square 2n cannot be a perfect square 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 z 2 = w2 }. Then, (A) A = B (C) B ⊂ A and B 6= A

(B) A ⊂ B and A 6= B (D) none of the above

63 Let f (x) = a0 +a1 |x|+a2 |x|2 +a3 |x|3 , where a0 , a1 , a2 , a3 are 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 if a1 = 0 (D) f (x) is differentiable at x = 0 only if a1 = 0, a3 = 0 64 If f (x) = cos(x) − 1 +

x2 2 ,

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 closed interval [0, π2 ] is (A) 0

(B) 1

(C) 2

(D) 3

66 The set of values of m for which mx2 −6mx+5m+1 > 0 for all real x is (A) m < 41 (B) m ≥ 0 1 (D) 0 ≤ m < 14 (C) 0 ≤ m ≤ 4 67 The digit in the unit’s place of the number 1! + 2! + 3! + . . . + 99! is 10

(A) 3

(B) 0 13 +23 +...+n3 n4 n→∞ (B) 14

68 The value of lim (A)

3 4

(D) 7

(C) 1

(D) 4

is:

69 For any integer n ≥ 1, define an = (A) (B) (C) (D)

(C) 1

1000n n! .

Then the sequence {an }

does not have a maximum attains maximum at exactly one value of n attains maximum at exactly two values of n attains maximum for infinitely many values of n

70 The equation x3 y + xy 3 + 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:  0 if x=0    π 1    sin if 0 1, Z

x

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

[x] X i=1

i

Z

x

f (u)du. i

9 If a circle intersects the hyperbola y = 1/x at four distinct points (xi , yi ), i = 1, 2, 3, 4, then prove that x1 x2 = y3 y4 . 10 Two intersecting circles are said to be orthogonal to each other if the tangents to the two circles at any point of intersection are perpendicular to each other. Show that every circle through the points (2, 0) and (−2, 0) is orthogonal to the circle x2 + y 2 − 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 the value of x at which the minimum is attained? 1 1 f (x) = x2 + x + + 2 for x 6= 0. x x Sketch on plain paper the graph of this function. 12 Show that there is exactly one value of x which satisfies the equation 2 cos2 (x3 + x) = 2x + 2−x . 13 Let S = {1, 2, . . . , n}. Find the number of unordered pairs {A, B} of subsets 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 , between which there is a river whose banks are parallel. The pipe must cross the river perpendicular to the banks. Find the position and nature of the shortest such pipe and justify your answer. 15 Find the maximum value of x2 + y 2 in the bounded region, including the boundary, enclosed by y = x2 , y = − x2 and x = y 2 + 1. 16 Let x = (x1 , . . . , xn ) and y = (y1 , . . . , yn ) where x1 , · · · , xn , y1 , · · · , yn are 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 for u = (u1 , . . . , un ), v = (v1 , . . . , vn ), w = (w1 , . . . , wn ) and z = (z1 , . . . , zn ), if u > 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

(b) Show that f (n1 n2 ) = f (f (n1 ) · f (n2 )) where n1 , n2 are any two positive 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 c whose medians have lengths 23 a, 32 b and 45 c. 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 + an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0

be a polynomial with integer coefficients, such that P (0) and P (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. Let C = {S : S ⊆ N } be the set of all subsets of N . Members of C are called coalitions. Let f be a function from C to {0, 1}. A coalition S ⊆ N is said to be winning if f (S) = 1; it is said to be a losing coalition if f (S) = 0. A pair hN, f i as above is called a voting game if the following conditions hold. (a) N is a winning coalition. (b) The empty set ∅ is a losing coalition. (c) If S is a winning coalition and S ⊆ S ′ , then S ′ is also winning. (d) If both S and S ′ are winning coalitions, then S ∩ S ′ 6= ∅, i.e., S and S ′ have a common voter. Show that the maximum number of winning coalitions of a voting game is 2n−1 . Find a voting game for which the number of winning coalitions is 2n−1 . 24 Suppose f is a real-valued differentiable function defined on [1, ∞) with f (1) = 1. Suppose, moreover, that f satisfies f ′ (x) = 1/(x2 + f 2 (x)). Show that f (x) ≤ 1 + π/4 for every x ≥ 1. 25 If the normal to the curve x2/3 + y 2/3 = a2/3 at some point makes an angle θ with the X-axis, show that the equation of the normal is y cos θ − x sin θ = a cos 2θ. 26 Suppose that a is an irrational number. (a) If there is a real number b such that both (a+b) and ab are rational numbers, show that a is√a quadratic √ surd. (a is a quadratic surd if 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

(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 surd and a is not a quadratic surd, separately). 27 Let A, B, and C be three points on a circle of radius 1. (a) Show that the area of the triangle ABC equals 1 (sin(2∠ABC) + sin(2∠BCA) + sin(2∠CAB)) . 2 (b) Suppose that the magnitude of ∠ABC is fixed. Then show that the area of the triangle ABC is maximized when ∠BCA = ∠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 BC at D. If the length of the chord AB is l1 , and that of BD is l2 , determine the length of DC in terms of l1 and l2

E C B

D A

29 (a) Let f (x) = x − xe−1/x , x > 0. Show that f (x) is an increasing function on (0, ∞), and limx→∞ f (x) = 1. (b) Using part (a) and calculus, sketch the graphs of y = x − 1, y = x, y = x + 1, and y = xe−1/|x| for −∞ < x < ∞ using the same X and Y axes. 30 For any integer n greater than 1, show that   2n 2n n 2 < < n−1 . n Y i (1 − ) n i=0

31 Show that there exists a positive real number x 6= 2 such that log2 x = x 2 . Hence obtain the set of real numbers c such that log2 x =c x has only one real solution. 32 Find a four digit number M such that the number N = 4 × M has the following properties. (a) N is also a four digit number. (b) N has the same digits as in M but in the reverse order. 33 Consider a function f on nonnegative integers such that f (0) = 1, f (1) = 0 and f (n) + f (n − 1) = nf (n − 1) + (n − 1)f (n − 2) for n ≥ 2. 4

Show that

n

f (n) X (−1)k = . n! k! k=0

34 Of all triangles with a given perimeter, find the triangle with the maximum area. Justify your answer. 35 A 40 feet high screen is put on a vertical wall 10 feet above your eyelevel. How far should you stand to maximize the angle subtended by the screen (from top to bottom) at your eye? 36 Study the derivatives of the function p y = x3 − 4x and sketch its graph on the real line.

37 Suppose P and Q are the centres of two disjoint circles C1 and C2 respectively, such that P lies outside C2 and Q lies outside C1 . Two tangents are drawn from the point P to the circle C2 , which intersect the circle C1 at points A and B. Similarly, two tangents are drawn from the point Q to the circle C1 , which intersect the circle C2 at points M and N . Show that AB = M N .   2n 1 . 38 Evaluate: lim log n→∞ 2n n 39 Consider the equation x5 + x = 10. Show that (a) the equation has only one real root; (b) this root lies between 1 and 2; (c) this root must be irrational. 40 In how many ways can you divide the set of eight numbers {2, 3, . . . , 9} into 4 pairs such that no pair of numbers has g.c.d. equal to 2? 41 Suppose S is the set of all positive integers. For a, b ∈ S, define l.c.m(a, b) a∗b= g.c.d(a, b) For example, 8 ∗ 12 = 6. Show that exactly two of the following three properties are satisfied : (a) If a, b ∈ S then a ∗ b ∈ S. (b) (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ S. (c) There exists an element i ∈ S such that a ∗ i = a for all a ∈ S.

5

Hints and Answers to selected problems. There are also other ways to solve the problems apart from the ones sketched in the hints. Indeed, a student should feel encouraged upon finding a different way to solve some of these problems. Hints and Answers to selected UGB Sample Questions. 1. The answer is 399960. For each x ∈ {1, 2, 3, 4, 5}, there are 4! such numbers whose last digit is x. Thus the digits in the unit place of all the 120 numbers add up to 4! (1 + 2 + 3 + 4 + 5). Similarly the numbers at ten’s place add up to 360 and so on. Thus the sum is 360 (1 + 10 + 100 + 1000). 3. Let the chosen entries be in the positions (i, ai ), 1 ≤ i ≤ 8. Thus a1 , . . . , a8 is a permutation of {1, . . . , 8}. The entry in the square corresponding to 8 P (i + 8 (aj − 1)). (i, j)th place is i + 8 (j − 1). Hence the required sum is i=1

5. Radius is

√ 3 3 2 .

Use trigonometry.


n 7. Observe that an = an−1 1 + t + t2 where t = 103

9. Substitute y = x1 in the equation of a circle and clear denominator to get a degree 4 equation in x. The product of its roots is the constant term, which is 1.

11. The function f (x) − 4 is a sum of squares and hence non-negative. So the minimum is 4 which is attained at x = 1. n

13. The number is 3 2+1 . An ordered pair (A, B) of disjoint subsets of S is determined by 3 choices for every element of S (either it is in A, or in B or in neither of them). Hence such pairs are 3n in number. An unordered pair will be counted twice in this way, except for the case A and B both empty. n Hence the number is 1 + 3 2−1 . 15. Answer is 5. The maximum is attained at points (2, 1) and (2, −1). 17. Answer is 2. Let f be the given polynomial. Then f (0) is negative and f is positive as x tends to ±∞. √ Hence it has at least 2 real roots. Since the derivative of f is zero only at 3 −3, it cannot have more than two real roots. √ 1 3 19. Maximum is 3. Either check the maximum of the function x x , or √ √ compare 3 3 with n n. 21. Rewrite the given inequality in terms of the new variables α = x + y − z, β = y + z − x, γ = x + z − y, and use the triangle inequality.

6

A Model Question Paper for B.Math/B.Stat

Test Code : UGA

Booklet No.

Forenoon

Questions : 30

Time : 2 hours

Write your Name, Registration Number, Test Centre, Test Code and the Number of this Booklet in the appropriate places on the Answersheet.

This test contains 30 questions in all. For each of the 30 questions, there are four suggested answers. Only one of the suggested answers is correct. You will have to identify the correct answer in order to get full credit for that question. Indicate your choice of the correct answer by darkening the appropriate oval completely on the answersheet.

You will get 4 marks for each correctly answered question, 0 marks for each incorrectly answered question and 1 mark for each unattempted question.

All rough work must be done on this booklet only. You are not allowed to use calculator.

WAIT FOR THE SIGNAL TO START.

Each of the following questions has exactly one correct answer among the given options and you have to identify it. 1. A rod AB of length 3 rests on a wall as follows:

P is a point on AB such that AP : P B = 1 : 2. If the rod slides along the wall, then the locus of P lies on (A) 2x + y + xy = 2 (B) 4x2 + y 2 = 4 (C) 4x2 + xy + y 2 = 4 (D) x2 + y 2 − x − 2y = 0. 2. Consider the equation x2 + y 2 = 2007. How many solutions (x, y) exist such that x and y are positive integers? (A) None (B) Exactly two (C) More than two but finitely many (D) Infinitely many. 3. Consider the functions f1 (x) = x, f2 (x) = 2 + loge x, x > 0 (where e is the base of natural logarithm). The graphs of the functions intersect (A) once in (0, 1) and never in (1, ∞) (B) once in (0, 1) and once in (e2 , ∞) (C) once in (0, 1) and once in (e, e2 ) (D) more than twice in (0, ∞).

1

4. Consider the sequence un =

n X r , n ≥ 1. 2r r=1

Then the limit of un as n → ∞ is (A) 1

(B) 2

(C) e

(D) 1/2.

5. Suppose that z is any complex number which is not equal to any of {3, 3ω, 3ω 2 } where ω is a complex cube root of unity. Then 1 1 1 + + z − 3 z − 3ω z − 3ω 2 equals (A)

3z 2 +3z (z−3)3

(B)

3z 2 +3ωz z 3 −27

(C)

3z 2 z 3 −3z 2 +9z−27

(D)

3z 2 . z 3 −27

6. Consider all functions f : {1, 2, 3, 4} → {1, 2, 3, 4} which are one-one, onto and satisfy the following property: if f (k) is odd then f (k + 1) is even, k = 1, 2, 3. The number of such functions is (A) 4

(B) 8

(C) 12

(D) 16.

7. A function f : R → R is defined by ( 1 e− x , x > 0 f (x) = 0 x ≤ 0. Then (A) f is not continuous (B) f is differentiable but f 0 is not continuous (C) f is continuous but f 0 (0) does not exist (D) f is differentiable and f 0 is continuous. 8. The last digit of 9! + 39966 is (A) 3

(B) 9

(C) 7 2

(D) 1.

9. Consider the function f (x) =

2x2 + 3x + 1 , 2 ≤ x ≤ 3. 2x − 1

Then (A) maximum of f is attained inside the interval (2, 3) (B) minimum of f is 28/5 (C) maximum of f is 28/5 (D) f is a decreasing function in (2, 3). 10. A particle P moves in the plane in such a way that the angle between the two tangents drawn from P to the curve y 2 = 4ax is always 90◦ . The locus of P is (A) a parabola

(B) a circle

(C) an ellipse

(D) a straight line.

11. Let f : R → R be given by f (x) = |x2 − 1|, x ∈ R. Then (A) f has a local minima at x = ±1 but no local maximum (B) f has a local maximum at x = 0 but no local minima (C) f has a local minima at x = ±1 and a local maximum at x = 0 (D) none of the above is true. 12. The number of triples (a, b, c) of positive integers satisfying 2a − 5b 7c = 1 is (A) infinite

(B) 2

(C) 1

(D) 0.

13. Let a be a fixed real number greater than −1. The locus of z ∈ C satisfying |z − ia| = Im(z) + 1 is (A) parabola

(B) ellipse

(C) hyperbola 3

(D) not a conic.

14. Which of the following is closest to the graph of tan(sin x), x > 0?

15. Consider the function f : R \ {1} → R \ {2} given by f (x) =

2x . x−1

Then (A) f is one-one but not onto (B) f is onto but not one-one (C) f is neither one-one nor onto (D) f is both one-one and onto. 16. Consider a real valued continuous function f satisfying f (x+1) = f (x) for all x ∈ R. Let Z

t

f (x) dx, t ∈ R.

g(t) = 0

Define h(t) = limn→∞

g(t+n) n ,

provided the limit exists. Then

(A) h(t) is defined only for t = 0 (B) h(t) is defined only when t is an integer (C) h(t) is defined for all t ∈ R and is independent of t (D) none of the above is true. 4

17. Consider the sequence a1 = 241/3 , an+1 = (an + 24)1/3 , n ≥ 1. Then the integer part of a100 equals (A) 2

(B) 10

(C) 100

(D) 24.

18. Let x, y ∈ (−2, 2) and xy = −1. Then the minimum value of 4 9 + 4 − x2 9 − y 2 is (A) 8/5

(B) 12/5

19. What is the limit of 

(C) 12/7

1 n2 + 1+ 2 n +n

√

(D) 15/7.

n

as n → ∞? (A) e

(B) 1

(D) ∞.

(C) 0

20. Consider the function f (x) = x4 + x2 + x − 1, x ∈ (−∞, ∞). The function (A) is zero at x = −1, but is increasing near x = −1 (B) has a zero in (−∞, −1) (C) has two zeros in (−1, 0) (D) has exactly one local minimum in (−1, 0). 21. Consider a sequence of 10 A’s and 8 B’s placed in a row. By a run we mean one or more letters of the same type placed side by side. Here is an arrangement of 10 A’s and 8 B’s which contains 4 runs of A and 4 runs of B: AAAB B AB B B AAB AAAAB B In how many ways can 10 A’s and 8 B’s be arranged in a row so that there are 4 runs of A and 4 runs of B?



(A) 2 93 73 (B) 93 73 (C)

5

10 4


8
4

(D)

10 5


8
5

.

22. Suppose n ≥ 2 is a fixed positive integer and f (x) = xn |x|, x ∈ R. Then (A) f is differentiable everywhere only when n is even (B) f is differentiable everywhere except at 0 if n is odd (C) f is differentiable everywhere (D) none of the above is true. 23. The line 2x + 3y − k = 0 with k > 0 cuts the x axis and y axis at points A and B respectively. Then the equation of the circle having AB as diameter is (A) x2 + y 2 − k2 x − k3 y = k 2 (B) x2 + y 2 − k3 x − k2 y = k 2 (C) x2 + y 2 − k2 x − k3 y = 0 (D) x2 + y 2 − k3 x − k2 y = 0. 24. Let α > 0 and consider the sequence xn =

(α + 1)n + (α − 1)n , n = 1, 2, . . . . (2α)n

Then limn→∞ xn is (A) 0 for any α > 0 (B) 1 for any α > 0 (C) 0 or 1 depending on what α > 0 is (D) 0, 1 or ∞ depending on what α > 0 is. 25. If 0 < θ < π/2 then (A) θ < sin θ (B) cos(sin θ) < cos θ (C) sin(cos θ) < cos(sin θ) (D) cos θ < sin(cos θ). 6

26. Consider a cardboard box in the shape of a prism as shown below. The length of the prism is 5. The two triangular faces ABC and A0 B 0 C 0 are congruent and isosceles with side lengths 2,2,3. The shortest distance between B and A0 along the surface of the prism is

(A)

√

29

(B)

√

28

(C)

p

29 −

√

5

(D)

p √ 29 − 3

27. Assume the following inequalities for positive integer k: √ √ 1 1 √ < k+1− k < √ . 2 k+1 2 k The integer part of 9999 X k=2

1 √ k

equals (A) 198

(B) 197

(C) 196

(D) 195.

28. Consider the sets defined by the inequalities A = {(x, y) ∈ R2 : x4 + y 2 ≤ 1}, B = {(x, y) ∈ R2 : x6 + y 4 ≤ 1}. Then (A) B ⊆ A (B) A ⊆ B (C) each of the sets A − B, B − A and A ∩ B is non-empty (D) none of the above is true.

7

29. The number

 210 11 11

is




10 2 10 2 10 2 10 2 10 1 2 3 4 5



10 2 10 2 10 2 10 2 strictly larger than 1 but 3 4




2 10 2 10 2 10 2 10 2 10 1 2 3 4 5
2 10
2 10
2 10
2 less than or equal to 10 1 2 3 4




2 2 2 2 10 10 10 10 equal to 10 1 2 3 4 5 .

(A) strictly larger than (B)

(C) (D)

strictly smaller than

30. If the roots of the equation x4 + ax3 + bx2 + cx + d = 0 are in geometric progression then (A) b2 = ac

(B) a2 = b

(C) a2 b2 = c2

8

(D) c2 = a2 d.

A Model Question Paper for B.Math/B.Stat BOOKLET No.

TEST CODE : UGB Afternoon Session

There are 3 pages in this booklet. The exam has 8 questions. Answer as many as you can.

Time : 2 hours

Write your Name, Registration number, Test Centre, Test Code and the Number of this booklet in the appropriate places on the answer-booklet.

ALL ROUGH WORK IS TO BE DONE ON THIS BOOKLET AND/OR THE ANSWER-BOOKLET. CALCULATORS ARE NOT ALLOWED.

STOP! WAIT FOR THE SIGNAL TO START.

P.T.O. 1

1. Let X, Y , Z be the angles of a triangle. (i) Prove that tan

X Y X Z Z Y tan + tan tan + tan tan = 1. 2 2 2 2 2 2

(ii) Using (i) or otherwise prove that tan

X Y Z 1 tan tan ≤ √ . 2 2 2 3 3

2. Let α be a real number. Consider the function 2

g(x) = (α + | x |)2 e(5−|x|) , −∞ < x < ∞. (i) Determine the values of α for which g is continuous at all x. (ii) Determine the values of α for which g is differentiable at all x.

3. Write the set of all positive integers in a triangular array as 1 2 4 7 11

3 5 8 12 .

6 9 13 . .

10 14 . . .

15 . . . .

. . . . .

. . . . .

Find the row number and column number where 20096 occurs. For example 8 appears in the third row and second column.

4. Show that the polynomial x8 − x7 + x2 − x + 15 has no real root. 5. Let m be a natural number with digits consisting entirely of 6’s and 0’s. Prove that m is not the square of a natural number. P.T.O. 2

6. Let 0 < a < b. (i) Show that amongst the triangles with base a and perimeter a + b, the maximum area is obtained when the other two sides have equal length 2b . (ii) Using the result of (i) or otherwise show that amongst the quadrilateral of given perimeter the square has maximum area.

7. Let 0 < a < b. Consider two circles with radii a and b and centers (a, 0) and (b, 0) respectively with 0 < a < b. Let c be the center of any circle in the crescent shaped region M between the two circles and tangent to both (See figure below). Determine the locus of c as its circle traverses through region M maintaining tangency.

c01 0

00 11 1 0 00 0 11 1

a b

M

8. Let n ≥ 1, S = {1, 2, . . . , n}. For a function f : S → S, a subset D ⊂ S is said to be invariant under f , if f (x) ∈ D for all x ∈ D. Note that the empty set and S are invariant for all f . Let deg(f ) be the number of subsets of S invariant under f . (i) Show that there is a function f : S → S such that deg(f ) = 2.

(ii) Further show that for any k such that 1 ≤ k ≤ n there is a function f : S → S such that deg(f ) = 2k .

3