INMO-2018 problems and solutions
|
|
- Gary Melton
- 5 years ago
- Views:
Transcription
1 Pioeer Educatio The Best Way To Success INMO-018 proles ad solutios NTSE Olypiad AIPMT JEE - Mais & Advaced 1. Let ABC e a o-equilateral triagle with iteger sides. Let D ad E e respectively the id-poits BC ad CA : let G e the cetroid of triagle ABC. Suppose D, C, E, G are cocyclic. Fid the least possile perieter of triagle ABC. Solutio: BD.BC = BG.BE a. a. 3 a 3...(1) 4 By appoloius theore a + c = 4 a c () 4 Fro (1) ad () a c 3 a (Fro (1) ad ()) 4 4 c a (3) a + ust e eve a, ust e of sae partiy. Now, c = W. L. O. G let a a a a a a x N, y N (as a, of sae parity) c = x + y For y = 0, c = x a = ad c = a c = a = equilateral which is ot the case. y > 0 Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page 1 of 6
2 Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced Now c, x, y are sides of a right agle triagle ad sallest pythagorea triple is 5, 4, 3 ; secod sallest 5, 1, 13 For 5, 4, 3 we have c = 5, a 4, a 3 a = 7, = 1 Not possile For 13, 1, 5 we have c = 13, a 1, a 5 c = 13 ad a = 17, = 7 as 17 < least perieter of ABC will e = 37. For ay atural uer, cosider a 1 rectagular oard ade up of uit squares. This is covered y three types of tiles 1 1 red tile, 1 1 gree tile ad 1 lue doio. (For exaple, we ca have 5 types of tilig whe = : red-red; red-gree; gree-red; gree-gree ad lue.) Let t deote the uer of ways of coverig 1 rectagular oard y these three types of tiles. Prove that t divides t+1. Solutio: Let r, g, respectively e the uer of 1 tiles that ed with a red, gree ad lue tiles. Clearly, t = r + g +. To get a 1 ( + 1) tile edig i a red tile, we ca apped a 1 1 red tile to ay of the aove three. Hece r+1= r + g +. Siilarly, g+1 = r + g +. To get +1, we eed to apped a lue tile to a 1 ( 1) tile. Thus +1 = r 1 + g Thus t+1 = r+1 + g = (r + g + ) + (r + g + ) + (r 1 + g 1+ 1) = t + t 1 Thus we have recurrece relatio t+1 t. t 1= 0 whose characteristic equatio is Thus has characteristic roots 1±. Thus α 1 ad β1. Sice t1 = ad t = 5, we get A = t α 1 1 β t A 1 B 1 Aα ββ, where α β ad B. Thus λ y 1 0. Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page of 6
3 Pioeer Educatio The Best Way To Success Now, NTSE Olypiad AIPMT JEE - Mais & Advaced t 1 α β α β α β = 1 1 α β 1 1 = α β = 1 t 1 α β Note that α β is a iteger ad t + 1 is 4 divisile y t. 3. Let Γ1 ad Γe two circles with respective cetres O1 ad O itersectig i two distict poits A ad B such that O1AO is a otuse agle. Let the circucircle of triagle O1AO itersect Γ1 ad Γ respectively i poits C( A) ad D( A). Let the lie CB itersect i Γ i E; let the lie DB itersect Γ 1 i F. Prove that the poits C, D, E, F are cocyclic. Solutio: Clai : CB passes through O ad DB through O1 1 Proof : For circle Γ 1, AO1O AO1B ACB (1) Also for circle Γ3 AO1O ACO () Fro (1) ad () we set ACB = ACO CB CO CB passes through O Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page 3 of 6
4 Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced Siilarly BD, passes through O1 Now BAE = 90 (as BF diaeter of Γ 1 ) ad BAF = 90 (as BF diaeter of Γ 1 ) FAE are colliear ad to O1O Let FEC = α O1OB = α (as O1O FE) or OOC 1 α O1D C α (o Γ 3) or FDC α FEC α FDC CDEF are cocyclic. 4. Fid all polyoials with real coefficiets P(x) such that P(x + x + 1) divides P(x 3 1). Solutio: Possiility (1) : P(x) is costat = c the P(x 3 1) = c ad P(x + x + 1) = c ad we are doe. Let P(x) e o cotat polyoial. As P(x + x +1) P(x 3 1) P(x 3 1) = P(x + x + 1) Q(x) where Q(x) i soe polyoial i x. P(x 1)(x + x + 1) = P(x + x + 1) Q(x) Wheever x + x + 1 i a root of P(x), (x 1) (x + x + 1) i also a root... (1) Let α e a root of P(x) such that α e axiu. Now tae x + x + 1 = α x = x1x = (say), roots with x1 + x = 1, Atleast oe root out of x1, x will have distace ore tha 1 (fro '1'). Let x1 1 x 1 x1 x 1 1 x 1 3 x 1 3 x = x () Fro oe we have (x 1) ( x x 1 ) = (x 1) α β(say) is aother root of P(x) = 0. Here B (x1)α x1 α α Which is a cotradictio α 0 α 0 All root of o cotat polyoial ust e '0'. P(x) = a. x, a R, N. A other solutio p(x) = c, c R. 5. There are 3 girls i a class sittig aroud a circular tale, each havig soe apples with her. Every tie the teacher otices a girl havig ore apples tha oth of her eighors coied, the Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page 4 of 6
5 Pioeer Educatio The Best Way To Success NTSE Olypiad AIPMT JEE - Mais & Advaced teacher taes away oe apple fro that girl ad gives oe apple each to her eighors. Prove that this process stops after a fiite uer of steps. (Assue that the teacher has a audat supply of apples.) Solutio: Let Iitially i th girl have ai apples Let as defie di = ai a i1 i = 1,, 3,,. here a+1 = a1 Cosider : Su d1 + d + d d = S1 (say) Let a > a 1 + a+1...(1) the after first step a a 1, a a 1, a a ' d a 1 a 1 a a d...() ' ad d a 1 a 1 a a d (3) 1 1 Equality i () ad (3) ca t hold siultaeously otherwise it will violate (1) ' ' 1 1 d d d d ' ' 1 1 S = d1 + d + d3. + d d... d S as Si 0ad Si is a decreasig sequece of o-egative itegers, it ca't e eep o decreasig forever. After soe fiite steps process will stop. 6. Let N deote the set of all atural uers ad let f : N N e a fuctio such that (a) f() = f()f() all, i N ; () + divides f() + f() for all, i N. Prove that there exists a odd atural uer such that f() = for all i N. Solutio: P(, ) : f() = f().f(); Q(, ) : + (f() + f()) P(1, 1) : f(11) = f(1). f(1) f(1) = 1 as f N Q(, ) : + (f() + f()) f() f() =.q, q soe odd uer N. If possile let q > 1 the there will exist a prie p such that p q p = odd prie. Also we set p f () p 1 p 1 p 1 P. : f p 1 f. f.f Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page 5 of 6
6 Pioeer Educatio The Best Way To Success P f p 1 NTSE Olypiad AIPMT JEE - Mais & Advaced Q(1, p 1) : 1 + (p 1) (f(1) + f (p 1) p (1 f p1 p 1 Which is a cotradictio q = 1 f() = (, 1) : ( + 1) (f() + f(1)) ( + 1) (od 3) or ( 1) (od 3) = odd. Also fro f() = f().f() f(...) f(). f()...f().... ties ties ties f( ) = () = Now Q(, ): ( + ) (f() f( )) i.e. f.(1) as (x + y) x y for = odd () Fro (1) ad () we get f N f f N f(). = 0 has ifiite divisors f() = for soe odd N Pioeer Educatio SCO 30, Sector 40 D, Chadigarh , Page 6 of 6
Jacobi symbols. p 1. Note: The Jacobi symbol does not necessarily distinguish between quadratic residues and nonresidues. That is, we could have ( a
Jacobi sybols efiitio Let be a odd positive iteger If 1, the Jacobi sybol : Z C is the costat fuctio 1 1 If > 1, it has a decopositio ( as ) a product of (ot ecessarily distict) pries p 1 p r The Jacobi
More information+ {JEE Advace 03} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If A (α, β) = (a) A( α, β) = A( α, β) (c) Adj (A ( α, β)) = Sol : We
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationFirst selection test, May 1 st, 2008
First selectio test, May st, 2008 Problem. Let p be a prime umber, p 3, ad let a, b be iteger umbers so that p a + b ad p 2 a 3 + b 3. Show that p 2 a + b or p 3 a 3 + b 3. Problem 2. Prove that for ay
More informationDifferent kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationChapter 2. Asymptotic Notation
Asyptotic Notatio 3 Chapter Asyptotic Notatio Goal : To siplify the aalysis of ruig tie by gettig rid of details which ay be affected by specific ipleetatio ad hardware. [1] The Big Oh (O-Notatio) : It
More informationSEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
More informationMathematics Extension 1
016 Bored of Studies Trial Eamiatios Mathematics Etesio 1 3 rd ctober 016 Geeral Istructios Total Marks 70 Readig time 5 miutes Workig time hours Write usig black or blue pe Black pe is preferred Board-approved
More informationProblem. Consider the sequence a j for j N defined by the recurrence a j+1 = 2a j + j for j > 0
GENERATING FUNCTIONS Give a ifiite sequece a 0,a,a,, its ordiary geeratig fuctio is A : a Geeratig fuctios are ofte useful for fidig a closed forula for the eleets of a sequece, fidig a recurrece forula,
More informationPARTIAL DIFFERENTIAL EQUATIONS SEPARATION OF VARIABLES
Diola Bagayoko (0 PARTAL DFFERENTAL EQUATONS SEPARATON OF ARABLES. troductio As discussed i previous lectures, partial differetial equatios arise whe the depedet variale, i.e., the fuctio, varies with
More informationREVIEW OF CALCULUS Herman J. Bierens Pennsylvania State University (January 28, 2004) x 2., or x 1. x j. ' ' n i'1 x i well.,y 2
REVIEW OF CALCULUS Hera J. Bieres Pesylvaia State Uiversity (Jauary 28, 2004) 1. Suatio Let x 1,x 2,...,x e a sequece of uers. The su of these uers is usually deoted y x 1 % x 2 %...% x ' j x j, or x 1
More informationUSA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year
/3/27. Fill i each space of the grid with either a or a so that all sixtee strigs of four cosecutive umbers across ad dow are distict. You do ot eed to prove that your aswer is the oly oe possible; you
More information2011 Problems with Solutions
1. Argumet duplicatio The Uiversity of Wester Australia SCHOOL OF MATHEMATICS AND STATISTICS BLAKERS MATHEMATICS COMPETITION 2011 Problems with Solutios Determie all real polyomials P (x) such that P (2x)
More informationSolutions for May. 3 x + 7 = 4 x x +
Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits
More informationBITSAT MATHEMATICS PAPER III. For the followig liear programmig problem : miimize z = + y subject to the costraits + y, + y 8, y, 0, the solutio is (0, ) ad (, ) (0, ) ad ( /, ) (0, ) ad (, ) (d) (0, )
More informationCHAPTER 1 SEQUENCES AND INFINITE SERIES
CHAPTER SEQUENCES AND INFINITE SERIES SEQUENCES AND INFINITE SERIES (0 meetigs) Sequeces ad limit of a sequece Mootoic ad bouded sequece Ifiite series of costat terms Ifiite series of positive terms Alteratig
More information3sin A 1 2sin B. 3π x is a solution. 1. If A and B are acute positive angles satisfying the equation 3sin A 2sin B 1 and 3sin 2A 2sin 2B 0, then A 2B
1. If A ad B are acute positive agles satisfyig the equatio 3si A si B 1 ad 3si A si B 0, the A B (a) (b) (c) (d) 6. 3 si A + si B = 1 3si A 1 si B 3 si A = cosb Also 3 si A si B = 0 si B = 3 si A Now,
More informationJEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018)
JEE(Advaced) 08 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 0 th MAY, 08) PART- : JEE(Advaced) 08/Paper- SECTION. For ay positive iteger, defie ƒ : (0, ) as ƒ () j ta j j for all (0, ). (Here, the iverse
More informationUNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007
UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =
More informationExam 2 CMSC 203 Fall 2009 Name SOLUTION KEY Show All Work! 1. (16 points) Circle T if the corresponding statement is True or F if it is False.
1 (1 poits) Circle T if the correspodig statemet is True or F if it is False T F For ay positive iteger,, GCD(, 1) = 1 T F Every positive iteger is either prime or composite T F If a b mod p, the (a/p)
More informationSolutions to Final Exam Review Problems
. Let f(x) 4+x. Solutios to Fial Exam Review Problems Math 5C, Witer 2007 (a) Fid the Maclauri series for f(x), ad compute its radius of covergece. Solutio. f(x) 4( ( x/4)) ( x/4) ( ) 4 4 + x. Sice the
More information3 Show in each case that there is a root of the given equation in the given interval. a x 3 = 12 4
C Worksheet A Show i each case that there is a root of the equatio f() = 0 i the give iterval a f() = + 7 (, ) f() = 5 cos (05, ) c f() = e + + 5 ( 6, 5) d f() = 4 5 + (, ) e f() = l (4 ) + (04, 05) f
More information2) 3 π. EAMCET Maths Practice Questions Examples with hints and short cuts from few important chapters
EAMCET Maths Practice Questios Examples with hits ad short cuts from few importat chapters. If the vectors pi j + 5k, i qj + 5k are colliear the (p,q) ) 0 ) 3) 4) Hit : p 5 p, q q 5.If the vectors i j
More informationObjective Mathematics
. If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic
More informationBertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.
Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,,
More informationMTH Assignment 1 : Real Numbers, Sequences
MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a
More informationWBJEE Answer Keys by Aakash Institute, Kolkata Centre
WBJEE - 7 Aswer Keys by, Kolkata Cetre MATHEMATICS Q.No. B A C B A C A B 3 D C B B 4 B C D D 5 D A B B 6 C D B B 7 B C C A 8 B B A A 9 A * B D C C B B D A A D B B C B 3 A D D D 4 C B A A 5 C B B B 6 C
More informationSubstitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get
Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics.
More informationTutorial F n F n 1
(CS 207) Discrete Structures July 30, 203 Tutorial. Prove the followig properties of Fiboacci umbers usig iductio, where Fiboacci umbers are defied as follows: F 0 =0,F =adf = F + F 2. (a) Prove that P
More informationEXERCISE - 01 CHECK YOUR GRASP
J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). The maximum value of the sum of the A.P. 0, 8, 6,,... is - 68 60 6. Let T r be the r th term of a A.P. for r =,,,...
More informationM17 MAT25-21 HOMEWORK 5 SOLUTIONS
M17 MAT5-1 HOMEWORK 5 SOLUTIONS 1. To Had I Cauchy Codesatio Test. Exercise 1: Applicatio of the Cauchy Codesatio Test Use the Cauchy Codesatio Test to prove that 1 diverges. Solutio 1. Give the series
More informationTEMASEK JUNIOR COLLEGE, SINGAPORE JC One Promotion Examination 2014 Higher 2
TEMASEK JUNIOR COLLEGE, SINGAPORE JC Oe Promotio Eamiatio 04 Higher MATHEMATICS 9740 9 Septemer 04 Additioal Materials: Aswer paper 3 hours List of Formulae (MF5) READ THESE INSTRUCTIONS FIRST Write your
More informationQ.11 If S be the sum, P the product & R the sum of the reciprocals of a GP, find the value of
Brai Teasures Progressio ad Series By Abhijit kumar Jha EXERCISE I Q If the 0th term of a HP is & st term of the same HP is 0, the fid the 0 th term Q ( ) Show that l (4 36 08 up to terms) = l + l 3 Q3
More informationBernoulli Polynomials Talks given at LSBU, October and November 2015 Tony Forbes
Beroulli Polyoials Tals give at LSBU, October ad Noveber 5 Toy Forbes Beroulli Polyoials The Beroulli polyoials B (x) are defied by B (x), Thus B (x) B (x) ad B (x) x, B (x) x x + 6, B (x) dx,. () B 3
More informationECE Spring Prof. David R. Jackson ECE Dept. Notes 20
ECE 6341 Sprig 016 Prof. David R. Jackso ECE Dept. Notes 0 1 Spherical Wave Fuctios Cosider solvig ψ + k ψ = 0 i spherical coordiates z φ θ r y x Spherical Wave Fuctios (cot.) I spherical coordiates we
More informationJoe Holbrook Memorial Math Competition
Joe Holbrook Memorial Math Competitio 8th Grade Solutios October 5, 07. Sice additio ad subtractio come before divisio ad mutiplicatio, 5 5 ( 5 ( 5. Now, sice operatios are performed right to left, ( 5
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More informationSAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS
. If their mea positios coicide with each other, maimum separatio will be A. Now from phasor diagram, we ca clearly see the phase differece. SAFE HANDS & IIT-ia's PACE ad Aswer : Optio (4) 5. Aswer : Optio
More informationMATHEMATICS 9740 (HIGHER 2)
VICTORIA JUNIOR COLLEGE PROMOTIONAL EXAMINATION MATHEMATICS 970 (HIGHER ) Frida 6 Sept 0 8am -am hours Additioal materials: Aswer Paper List of Formulae (MF5) READ THESE INSTRUCTIONS FIRST Write our ame
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationSummer MA Lesson 13 Section 1.6, Section 1.7 (part 1)
Suer MA 1500 Lesso 1 Sectio 1.6, Sectio 1.7 (part 1) I Solvig Polyoial Equatios Liear equatio ad quadratic equatios of 1 variable are specific types of polyoial equatios. Soe polyoial equatios of a higher
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More information07 - SEQUENCES AND SERIES Page 1 ( Answers at he end of all questions ) b, z = n
07 - SEQUENCES AND SERIES Page ( Aswers at he ed of all questios ) ( ) If = a, y = b, z = c, where a, b, c are i A.P. ad = 0 = 0 = 0 l a l
More information[ 47 ] then T ( m ) is true for all n a. 2. The greatest integer function : [ ] is defined by selling [ x]
[ 47 ] Number System 1. Itroductio Pricile : Let { T ( ) : N} be a set of statemets, oe for each atural umber. If (i), T ( a ) is true for some a N ad (ii) T ( k ) is true imlies T ( k 1) is true for all
More informationBertrand s postulate Chapter 2
Bertrad s postulate Chapter We have see that the sequece of prie ubers, 3, 5, 7,... is ifiite. To see that the size of its gaps is ot bouded, let N := 3 5 p deote the product of all prie ubers that are
More information(a) (b) All real numbers. (c) All real numbers. (d) None. to show the. (a) 3. (b) [ 7, 1) (c) ( 7, 1) (d) At x = 7. (a) (b)
Chapter 0 Review 597. E; a ( + )( + ) + + S S + S + + + + + + S lim + l. D; a diverges by the Itegral l k Test sice d lim [(l ) ], so k l ( ) does ot coverge absolutely. But it coverges by the Alteratig
More informationBRAIN TEASURES TRIGONOMETRICAL RATIOS BY ABHIJIT KUMAR JHA EXERCISE I. or tan &, lie between 0 &, then find the value of tan 2.
EXERCISE I Q Prove that cos² + cos² (+ ) cos cos cos (+ ) ² Q Prove that cos ² + cos (+ ) + cos (+ ) Q Prove that, ta + ta + ta + cot cot Q Prove that : (a) ta 0 ta 0 ta 60 ta 0 (b) ta 9 ta 7 ta 6 + ta
More informationMath 4707 Spring 2018 (Darij Grinberg): midterm 2 page 1. Math 4707 Spring 2018 (Darij Grinberg): midterm 2 with solutions [preliminary version]
Math 4707 Sprig 08 Darij Griberg: idter page Math 4707 Sprig 08 Darij Griberg: idter with solutios [preliiary versio] Cotets 0.. Coutig first-eve tuples......................... 3 0.. Coutig legal paths
More informationCOMP 2804 Solutions Assignment 1
COMP 2804 Solutios Assiget 1 Questio 1: O the first page of your assiget, write your ae ad studet uber Solutio: Nae: Jaes Bod Studet uber: 007 Questio 2: I Tic-Tac-Toe, we are give a 3 3 grid, cosistig
More informationAssignment ( ) Class-XI. = iii. v. A B= A B '
Assigmet (8-9) Class-XI. Proe that: ( A B)' = A' B ' i A ( BAC) = ( A B) ( A C) ii A ( B C) = ( A B) ( A C) iv. A B= A B= φ v. A B= A B ' v A B B ' A'. A relatio R is dified o the set z of itegers as:
More informationdistinct distinct n k n k n! n n k k n 1 if k n, identical identical p j (k) p 0 if k > n n (k)
THE TWELVEFOLD WAY FOLLOWING GIAN-CARLO ROTA How ay ways ca we distribute objects to recipiets? Equivaletly, we wat to euerate equivalece classes of fuctios f : X Y where X = ad Y = The fuctios are subject
More informationAssignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1
Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationSequence A sequence is a function whose domain of definition is the set of natural numbers.
Chapter Sequeces Course Title: Real Aalysis Course Code: MTH3 Course istructor: Dr Atiq ur Rehma Class: MSc-I Course URL: wwwmathcityorg/atiq/fa8-mth3 Sequeces form a importat compoet of Mathematical Aalysis
More informationSolving equations (incl. radical equations) involving these skills, but ultimately solvable by factoring/quadratic formula (no complex roots)
Evet A: Fuctios ad Algebraic Maipulatio Factorig Square of a sum: ( a + b) = a + ab + b Square of a differece: ( a b) = a ab + b Differece of squares: a b = ( a b )(a + b ) Differece of cubes: a 3 b 3
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationJEE ADVANCED 2013 PAPER 1 MATHEMATICS
Oly Oe Optio Correct Type JEE ADVANCED 0 PAPER MATHEMATICS This sectio cotais TEN questios. Each has FOUR optios (A), (B), (C) ad (D) out of which ONLY ONE is correct.. The value of (A) 5 (C) 4 cot cot
More information2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.
CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.
More informationCALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=
More informationA Recurrence Formula for Packing Hyper-Spheres
A Recurrece Formula for Packig Hyper-Spheres DokeyFt. Itroductio We cosider packig of -D hyper-spheres of uit diameter aroud a similar sphere. The kissig spheres ad the kerel sphere form cells of equilateral
More informationChapter 1. Complex Numbers. Dr. Pulak Sahoo
Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia E-mail : sahoopulak1@gmail.com 1 Module-2: Stereographic Projectio 1 Euler
More informationAN ALMOST LINEAR RECURRENCE. Donald E. Knuth Calif. Institute of Technology, Pasadena, Calif.
AN ALMOST LINEAR RECURRENCE Doald E. Kuth Calif. Istitute of Techology, Pasadea, Calif. form A geeral liear recurrece with costat coefficiets has the U 0 = a l* U l = a 2 " ' " U r - l = a r ; u = b, u,
More information42 Dependence and Bases
42 Depedece ad Bases The spa s(a) of a subset A i vector space V is a subspace of V. This spa ay be the whole vector space V (we say the A spas V). I this paragraph we study subsets A of V which spa V
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More information18.S34 (FALL, 2007) GREATEST INTEGER PROBLEMS. n + n + 1 = 4n + 2.
18.S34 (FALL, 007) GREATEST INTEGER PROBLEMS Note: We use the otatio x for the greatest iteger x, eve if the origial source used the older otatio [x]. 1. (48P) If is a positive iteger, prove that + + 1
More informationLecture 10: Bounded Linear Operators and Orthogonality in Hilbert Spaces
Lecture : Bouded Liear Operators ad Orthogoality i Hilbert Spaces 34 Bouded Liear Operator Let ( X, ), ( Y, ) i i be ored liear vector spaces ad { } X Y The, T is said to be bouded if a real uber c such
More informationUnit 6: Sequences and Series
AMHS Hoors Algebra 2 - Uit 6 Uit 6: Sequeces ad Series 26 Sequeces Defiitio: A sequece is a ordered list of umbers ad is formally defied as a fuctio whose domai is the set of positive itegers. It is commo
More informationIYGB. Special Extension Paper E. Time: 3 hours 30 minutes. Created by T. Madas. Created by T. Madas
YGB Special Extesio Paper E Time: 3 hours 30 miutes Cadidates may NOT use ay calculator. formatio for Cadidates This practice paper follows the Advaced Level Mathematics Core ad the Advaced Level Further
More informationTHE GREATEST ORDER OF THE DIVISOR FUNCTION WITH INCREASING DIMENSION
MATHEMATICA MONTISNIGRI Vol XXVIII (013) 17-5 THE GREATEST ORDER OF THE DIVISOR FUNCTION WITH INCREASING DIMENSION GLEB V. FEDOROV * * Mechaics ad Matheatics Faculty Moscow State Uiversity Moscow, Russia
More informationBinomial transform of products
Jauary 02 207 Bioial trasfor of products Khristo N Boyadzhiev Departet of Matheatics ad Statistics Ohio Norther Uiversity Ada OH 4580 USA -boyadzhiev@ouedu Abstract Give the bioial trasfors { b } ad {
More informationIf the escalator stayed stationary, Billy would be able to ascend or descend in = 30 seconds. Thus, Billy can climb = 8 steps in one second.
BMT 01 INDIVIDUAL SOLUTIONS March 01 1. Billy the kid likes to play o escalators! Movig at a costat speed, he maages to climb up oe escalator i 4 secods ad climb back dow the same escalator i 40 secods.
More informationOn the transcendence of infinite sums of values of rational functions
O the trascedece of ifiite sus of values of ratioal fuctios N. Saradha ad R. Tijdea Abstract P () = We ivestigate coverget sus T = Q() ad U = P (X), Q(X) Q[X], ad Q(X) has oly siple ratioal roots. = (
More informationMATH spring 2008 lecture 3 Answers to selected problems. 0 sin14 xdx = x dx. ; (iv) x +
MATH - sprig 008 lecture Aswers to selected problems INTEGRALS. f =? For atiderivatives i geeral see the itegrals website at http://itegrals.wolfram.com. (5-vi (0 i ( ( i ( π ; (v π a. This is example
More informationPUTNAM TRAINING, 2008 COMPLEX NUMBERS
PUTNAM TRAINING, 008 COMPLEX NUMBERS (Last updated: December 11, 017) Remark. This is a list of exercises o Complex Numbers Miguel A. Lerma Exercises 1. Let m ad two itegers such that each ca be expressed
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More informationStanford Math Circle January 21, Complex Numbers
Staford Math Circle Jauary, 007 Some History Tatiaa Shubi (shubi@mathsjsuedu) Complex Numbers Let us try to solve the equatio x = 5x + x = is a obvious solutio Also, x 5x = ( x )( x + x + ) = 0 yields
More informationTHE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: 2 hours
THE 06-07 KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: hours Let x, y, ad A all be positive itegers with x y a) Prove that there are
More informationBINOMIAL COEFFICIENT AND THE GAUSSIAN
BINOMIAL COEFFICIENT AND THE GAUSSIAN The biomial coefficiet is defied as-! k!(! ad ca be writte out i the form of a Pascal Triagle startig at the zeroth row with elemet 0,0) ad followed by the two umbers,
More informationCourse : Algebraic Combinatorics
Course 8.32: Algebraic Combiatorics Lecture Notes # Addedum by Gregg Musier February 4th - 6th, 2009 Recurrece Relatios ad Geeratig Fuctios Give a ifiite sequece of umbers, a geeratig fuctio is a compact
More informationARITHMETIC PROGRESSION
CHAPTER 5 ARITHMETIC PROGRESSION Poits to Remember :. A sequece is a arragemet of umbers or objects i a defiite order.. A sequece a, a, a 3,..., a,... is called a Arithmetic Progressio (A.P) if there exists
More informationDefine a Markov chain on {1,..., 6} with transition probability matrix P =
Pla Group Work 0. The title says it all Next Tie: MCMC ad Geeral-state Markov Chais Midter Exa: Tuesday 8 March i class Hoework 4 due Thursday Uless otherwise oted, let X be a irreducible, aperiodic Markov
More informationIIT JAM Mathematical Statistics (MS) 2006 SECTION A
IIT JAM Mathematical Statistics (MS) 6 SECTION A. If a > for ad lim a / L >, the which of the followig series is ot coverget? (a) (b) (c) (d) (d) = = a = a = a a + / a lim a a / + = lim a / a / + = lim
More informationHomework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.
Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger. 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that
More informationMath 132, Fall 2009 Exam 2: Solutions
Math 3, Fall 009 Exam : Solutios () a) ( poits) Determie for which positive real umbers p, is the followig improper itegral coverget, ad for which it is diverget. Evaluate the itegral for each value of
More informationSOME FINITE SIMPLE GROUPS OF LIE TYPE C n ( q) ARE UNIQUELY DETERMINED BY THEIR ELEMENT ORDERS AND THEIR ORDER
Joural of Algebra, Nuber Theory: Advaces ad Applicatios Volue, Nuber, 010, Pages 57-69 SOME FINITE SIMPLE GROUPS OF LIE TYPE C ( q) ARE UNIQUELY DETERMINED BY THEIR ELEMENT ORDERS AND THEIR ORDER School
More informationAxioms of Measure Theory
MATH 532 Axioms of Measure Theory Dr. Neal, WKU I. The Space Throughout the course, we shall let X deote a geeric o-empty set. I geeral, we shall ot assume that ay algebraic structure exists o X so that
More informationReal Variables II Homework Set #5
Real Variables II Homework Set #5 Name: Due Friday /0 by 4pm (at GOS-4) Istructios: () Attach this page to the frot of your homework assigmet you tur i (or write each problem before your solutio). () Please
More information(s)h(s) = K( s + 8 ) = 5 and one finite zero is located at z 1
ROOT LOCUS TECHNIQUE 93 should be desiged differetly to eet differet specificatios depedig o its area of applicatio. We have observed i Sectio 6.4 of Chapter 6, how the variatio of a sigle paraeter like
More informationUnit 4: Polynomial and Rational Functions
48 Uit 4: Polyomial ad Ratioal Fuctios Polyomial Fuctios A polyomial fuctio y px ( ) is a fuctio of the form p( x) ax + a x + a x +... + ax + ax+ a 1 1 1 0 where a, a 1,..., a, a1, a0are real costats ad
More informationTopic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or
Topic : Sequeces ad Series A sequece is a ordered list of umbers, e.g.,,, 8, 6, or,,,.... A series is a sum of the terms of a sequece, e.g. + + + 8 + 6 + or... Sigma Notatio b The otatio f ( k) is shorthad
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationFACTORS OF SUMS OF POWERS OF BINOMIAL COEFFICIENTS. Dedicated to the memory of Paul Erdős. 1. Introduction. n k. f n,a =
FACTORS OF SUMS OF POWERS OF BINOMIAL COEFFICIENTS NEIL J. CALKIN Abstract. We prove divisibility properties for sus of powers of bioial coefficiets ad of -bioial coefficiets. Dedicated to the eory of
More informationU8L1: Sec Equations of Lines in R 2
MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie
More information1 lim. f(x) sin(nx)dx = 0. n sin(nx)dx
Problem A. Calculate ta(.) to 4 decimal places. Solutio: The power series for si(x)/ cos(x) is x + x 3 /3 + (2/5)x 5 +. Puttig x =. gives ta(.) =.3. Problem 2A. Let f : R R be a cotiuous fuctio. Show that
More informationRegn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: ,
. Sectio-A cotais 30 Multiple Choice Questios (MCQ). Each questio has 4 choices (a), (b), (c) ad (d), for its aswer, out of which ONLY ONE is correct. From Q. to Q.0 carries Marks ad Q. to Q.30 carries
More information