3sin 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

Size: px
Start display at page:

Download "3sin 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"

Transcription

1 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, cos (A + B) = cosa cos B sia sib = cos A (3 si A) si A ( 3 si A) = 3 si A cos A 3 si A cos A = 0 A + B =. The umber of solutios of cos x 1si x,0 x 3, is (a) 3 (b) (c) (d) oe of these. As (a) Clearly 1 + si x > 0 equatio becomes cos x 1 si x cos x si x = 1 Clearly x = 0 is a solutio x = is a solutio 3π x is a solutio No other solutio is possible

2 3. The sum of all the solutios of the equatio x 0, 6 1 cos x.cos x.cos x, 3 3 is (a) 5 (b) 30 (c) 0 (d) oe of these. As (b) We have 1 cos xcos si x cos x (1 cos x) 3 1 cos xcos x cos x ( cos x 3) = 1 cos 3x = 1 3x = x π 3 Where = 0, 1,, 3,, 5, 6, 7, 8, 9. the reqd. sum π ( ) = 3 π 9(9 1). 30 π. 3. If ta m ta, ( m ) the the differet values of are i (a) A.P. (c) G.P. (b) H.P. (d) o particular sequece. ta m ta m K m K K where K I m These values of are i A.P. with commo differece = m Hece (a) is the correct aswer

3 5. The set of value of x for which 0, (B) 0, 3 3 si x.cos x cos x.si x, 0 x, is, (D) Noe of these 3 3 si xcos x cos xsi x si x cos x cos x si x 0 1 si x cos x 0 si x 0 x ; I 0 x (give i the optios) 6. The umber of values of K for which the system of equatios K 1 x 8y K ad Kx K 3 y 3K 1 has ifiitely may solutios is 0 (B) 1 (D) Ifiite For ifiitely may solutios, we have k 1 8 k k k 3 3k 1 from first two k k 3 8k k k 3 0 k 1,3 from first ad last, k k 1 0 k 1 0 k 1 3k k 1 k takig the commo solutio we get k = 1 Hece oly oe solutio is possible

4 7. The umber of differet permutatios of all the letters of the word 'PERMUTATION' such that ay two cosecutive letters i the arragemet are either both vowels or both idetical is 63 6! 5! (B) 57 5! 5! 33 6! 5! (D) 7 7! 5! The letters other tha vowels are : PRMTTN Number of permutatios with o two vowels together is 6!! 7 C 5 5! Further amog these permutatios the umber of cases i which T's are together is 5! 6 C 5 5! So the required umber = 6!! 7 C 5 5! - 5! 6 C 5 5! = 57 (5!) 8. The umber of differet words that ca be formed usig all the letters of the word 'SHASHANK' such that i ay word the vowels are separated by atleast two cosoats, is 700 (B) (D) 600 The letters other tha vowels are SHSHNK which ca be arraged i 6!!! ways Now i its each case, let the first A be placed i the r th gap the the umber of ways to place the d A will be (7 - r - 1). So, the total umber of ways = 5 6! (6 r)!! r 1 = 6! ( ) = 700.!!

5 The sum of the series upto ifiity (B) (D) 5 r 1 Hit : Tr r, r 5 rr ( 1) 5 r( r1) 1 1 r r1 r 5 r( r 1) 5 ( r 1) 5 r r T r If x x x x x x x x 7 the 16 x is 15 (B) 16 x is less tha x greater tha 15 (D) Nothig ca be said about 16 x Hit : x x x x x x x x 8 x( x 1)( x 1)( x 1) 16 8 x 1 ( x 1)( x 1)( x 1)( x 1) 7 1 ( x 1) 7 x 1 x x 16 x 15

6 11. If a, b, c, d are distict itegers i AP such that 0 (B) 1 d a b c the a + b + c + d is (D) Noe Hit : d a b c a 3 t ( a t) a ( a t) 5t 3(a 1) t 3a a 0 D 0 a 16a a 3 3 a 1,0 3 a0, t 0, 5 a 1, t 1, 5 t 1 a b c d 1. The least value of sec + cosec + sec + cosec i (0, /) is + (B) - 5 (D) 5 - Sice sec + cosec ad sec + cosec has miimum value at = /, we have miimum value of sec + cosec + sec + cosec = +

7 13. If cosa = cosb + cos 3 B, sia = sib - si 3 B. The si(a-b) = 1/ (B) 1/3 /3 (D) 1/5 si(a - B) = sia cosb - cosa sib... (i) Substitutig the values of cosa ad sia from the give equatios i (i), he have si(a - B) = si Bcos B... (ii) squarig ad addig give equatio we get cosb = 1/3 si(a - B) = 1/3 1. If x + y + z + w = 5, the the least value of x cot9 + y cot7 + z cot63 + w cot81 is 5/ (B) 5 5 5/ (D) 5 5 We have x y z w ta9 ta 7 ta 63 ta The umber of solutios of [si x] + cos x = 1, i x, [where [x] = greatest iteger fuctio], is 3 (B) 5 (D) 6 [si x] = - 1, 0, 1 If [si x] = - 1 the cosx ot possible If si x 0 the cos x 1 x =,, 3, If si x 1 The cosx 0 5 x Number or solutios = 5 for x,.

8 16. Let f(x) = cosec x + sec x + cosec x, the miimum value of f(x), for x (0, /), is 1 1 (B) (D) 1 (B) f(x) = si x cos x = si x si x 1si x cos x si x = 1 si x cos x = si x cos x 1 si x cos x 1 f (x) cos x si x si x cos x 1 f(x) is decreasig for 0 < x < / ad icrease for / < x < / miimum is at x = f mi (x) = f lim, r r1 where [] deotes the G.I.F, is 1 (B) 0 does ot exist (D) oe of these (B) Clearly r r 1 lim 0 r r1

9 x x 18. The maximum value of f(x) = x 1, (xr), is - (B) -3 - (D) -1 f(x) = 3 - ( x 1) ( x 1) 3 - = The umber of all the odd divisor of 3600 is 5 (B) 18 (D) 9 Solutio : (D) 3600 = 3 5 so umber of odd divisors = ( + 1) ( + 1) = 9 0. Let A be the set of -digit umbers a 1 a a 3 a where a 1 > a > a 3 > a, the is equal to 16 (B) 8 10 (D) oe of these Solutio: Ay selectio of four digits from the te digits 0, 1,, 3,..., 9 gives oe such umber. So, the required umber of umbers = 10 C = 10. Hece is the correct aswer.

10 1. Total umber of ways of selectig two umbers from the set {1,, 3,,., 3} so that their sum is divisible by 3 is equal to: (B) 3 Solutio : (D) 3 Give umbers ca be rearraged as type type type That meas we must take two umbers from last row or oe umber each from first ad secod row. Total ways = C + C 1. C 1 = 1 3 =. A polygo has 65 diagoals. The umber of its sides is 8 (B) (D) 13 Solutio : (D) Let umber of sides be, the C = 65 ( 3) = 130 = 13

11 3. I a chess touramet, all participats were to play oe game with the other. Two players fell ill after havig played 3 games each. If total umber of games played i the touramet is equal to 8, the total umber of participats i the begiig was equal to: Solutio : 10 (B) 15 1 (D) 1 Let there were players i the begiig. Total umber of games to be played was played to C ad each player would have played ( 1) games. Thus C (( 1) + ( 1) 1) + 6 = = 0 = 15. If objects are arraged i a row, the the umber of ways of selectig three objects so that o two of them are ext to each other is 3 (B) 6 - C 3-3 C C (D) oe of these. Solutio: Let x 0 be the umber of objects to the left of the first object chose, x 1 the umber of objects betwee the first ad the secod, x the umber of objects betwee the secod ad the third ad x 3 the umber of objects to the right of the third object. We have x 0, x 3 0, x 1, x 1 ad x 0 + x 1 +x + x 3 = 3... (1) The umber of solutios of (1) = coefficiet of y -3 i (1+ y+ y +...)(1+ y + y +...)(y + y + y )(y + y + y ) = coefficiet of y -3 i y ( 1+ y + y +y ) = coefficiet of y -5 i (1- y) - = coefficiet of y -5 i ( 1+ C 1 y + 5 C y + 6 C 3 y ) = C -5 = - C 3 3 = 6 also -3 C C = - C 3. Hece, (B) ad are the correct aswers.

12 I order to solve the equatio of the form Put a rcos ; b rsi b Paragraph for Questios Nos. 5 to 7 a cos bsi c... 1 r a b, ta equatio (1) becomes a rcos cos rsi si c cos a c b If c a b, the the equatio acos bsi c has o solutio If c a b, the put a c b cos t cos cos t t t 5. The solutio of si x 3 cos x is 5, (B) 1 1 5, 1 1, (D) Noe of these si x cos x cosx cos 6 x 6 x 6 5 x or x 1 1

13 6. kcos x 3si x k 1 is solvable oly if k (, ] (B) k (, ) k cos x 3si x k 1 k 3 k 1 cos x si x k 9 k 9 k 9 k 1, (D) Noe of these k 1 k the cos x ; where cos k 9 k 9 ad si 3 k 9 1 cos x 1 ; k1 1 1 k 9 k1 1 k 1 k 9 k 9 7. If 0 x, ; k 3 3 the umber of solutios of equatio 3 si x cos x si x cos x 8 is 0 (B) 1 (D) 3 si x cos x si x cos x si x cos x si x cos x 8 si x cos x 3 1si x cos x 8 si x cos x1 si x cos x 8 si x cos x 3 8 ; si x cos x six 1, Not possible The give equatio has o. solutio

14 Paragraph for Questio Nos. 8 to 0 If a sequece or series is ot a direct form of a AP, GP, etc. The its th term ca ot be determied. I such cases, we use the followig steps to fid the th term T of the give sequece. Step I : Fid the differeces betwee the successive terms of the give sequece. If these differeces are i AP, the take T a b c, where a,b,c are costats. Step II : If the successive differeces foud i step I are i GP with commo ratio r, the 1 take T a b cr, where a, b, c are costats. Step III : If the secod successive differeces (Differeces of the differeces) i step I are i 3 AP, the take T a b c d, where a, b, c, d are costats. Step IV : If the secod successive differeces (Differeces of the differeces) i step I are i 1 GP, the take T a b c dr, where a, b, c, d are costats. Now let sequeces : A : 1, 6, 18, 0, 75, 16,.. B : 1, 1, 6, 6, 91, 91,. C : l l, l 3, l If the th term of the sequece A is T a b c d the the value 6a b d is l (B) l 8 (D) Hit : (D) 3 T a b c d T1 a b c d 1 T 8a b c d 6 6a b d 9. For the sequece 1, 1, 6, 6, 91, 91,.. Fid the S 50 where S T r r (B)

15 (D) Noe of these 8 50 Hit : T S The sum of the series 1 6 (B) (D) Hit : r r 1 1 r r r1 r1 r = 6 1 6

SEQUENCE AND SERIES NCERT

SEQUENCE 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 information

2) 3 π. EAMCET Maths Practice Questions Examples with hints and short cuts from few important chapters

2) 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 information

Objective Mathematics

Objective 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 information

11 C 90. Through what angle has it turned in 10 seconds? cos24 +cos55 +cos125 +cos204 +cos300 =½. Prove that :

11 C 90. Through what angle has it turned in 10 seconds? cos24 +cos55 +cos125 +cos204 +cos300 =½. Prove that : Class - XI Maths Assigmet 0-07 Topic : Trigoometry Q If the agular diameter of the moo by 0, how far from the eye a coi of diameter cm be kept to hide the moo? cm Q Fid the agle betwee the miute had of

More information

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK) VERY SHORT ANSWER TYPE QUESTIONS ( MARK). If th term of a A.P. is 6 7 the write its 50 th term.. If S = +, the write a. Which term of the sequece,, 0, 7,... is 6? 4. If i a A.P. 7 th term is 9 ad 9 th

More information

07 - SEQUENCES AND SERIES Page 1 ( Answers at he end of all questions ) b, z = n

07 - 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

Q.11 If S be the sum, P the product & R the sum of the reciprocals of a GP, find the value of

Q.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 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 ] 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 information

BRAIN TEASURES TRIGONOMETRICAL RATIOS BY ABHIJIT KUMAR JHA EXERCISE I. or tan &, lie between 0 &, then find the value of tan 2.

BRAIN 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 information

Created by T. Madas SERIES. Created by T. Madas

Created by T. Madas SERIES. Created by T. Madas SERIES SUMMATIONS BY STANDARD RESULTS Questio (**) Use stadard results o summatios to fid the value of 48 ( r )( 3r ). 36 FP-B, 66638 Questio (**+) Fid, i fully simplified factorized form, a expressio

More information

Topic 1 2: Sequences and Series. A sequence is an ordered list of numbers, e.g. 1, 2, 4, 8, 16, or

Topic 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 information

LIMITS AND DERIVATIVES NCERT

LIMITS AND DERIVATIVES NCERT . Overview.. Limits of a fuctio Let f be a fuctio defied i a domai wic we take to be a iterval, say, I. We sall study te cocept of it of f at a poit a i I. We say f ( ) is te epected value of f at a give

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES 9 SEQUENCES AND SERIES INTRODUCTION Sequeces have may importat applicatios i several spheres of huma activities Whe a collectio of objects is arraged i a defiite order such that it has a idetified first

More information

LIMITS AND DERIVATIVES

LIMITS AND DERIVATIVES Capter LIMITS AND DERIVATIVES. Overview.. Limits of a fuctio Let f be a fuctio defied i a domai wic we take to be a iterval, say, I. We sall study te cocept of it of f at a poit a i I. We say f ( ) is

More information

Putnam Training Exercise Counting, Probability, Pigeonhole Principle (Answers)

Putnam Training Exercise Counting, Probability, Pigeonhole Principle (Answers) Putam Traiig Exercise Coutig, Probability, Pigeohole Pricile (Aswers) November 24th, 2015 1. Fid the umber of iteger o-egative solutios to the followig Diohatie equatio: x 1 + x 2 + x 3 + x 4 + x 5 = 17.

More information

EXERCISE - 01 CHECK YOUR GRASP

EXERCISE - 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 information

Objective Mathematics

Objective Mathematics 6. If si () + cos () =, the is equal to :. If <

More information

ARITHMETIC PROGRESSION

ARITHMETIC 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 information

CHAPTER - 9 SEQUENCES AND SERIES KEY POINTS A sequece is a fuctio whose domai is the set N of atural umbers. A sequece whose rage is a subset of R is called a real sequece. Geeral A.P. is, a, a + d, a

More information

SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2

SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2 Class-Jr.X_E-E SIMPLE HOLIDAY PACKAGE CLASS-IX MATHEMATICS SUB BATCH : E-E SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY. siθ+cosθ + siθ cosθ = ) ) ). If a cos q, y bsi q, the a y b ) ) ). The value

More information

MISCELLANEOUS SEQUENCES & SERIES QUESTIONS

MISCELLANEOUS SEQUENCES & SERIES QUESTIONS MISCELLANEOUS SEQUENCES & SERIES QUESTIONS Questio (***+) Evaluate the followig sum 30 r ( 2) 4r 78. 3 MP2-V, 75,822,200 Questio 2 (***+) Three umbers, A, B, C i that order, are i geometric progressio

More information

Solutions to Math 347 Practice Problems for the final

Solutions 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 information

ANSWERSHEET (TOPIC = ALGEBRA) COLLECTION #2

ANSWERSHEET (TOPIC = ALGEBRA) COLLECTION #2 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC ALGEBRA) COLLECTION # Questio Type A.Sigle Correct Type Q. (B) Sol ( 5 7 ) ( 5 7 9 )!!!! C

More information

WBJEE MATHEMATICS

WBJEE MATHEMATICS WBJEE - 06 MATHEMATICS Q.No. 0 A C B B 0 B B A B 0 C A C C 0 A B C C 05 A A B C 06 B C B C 07 B C A D 08 C C C A 09 D D C C 0 A C A B B C B A A C A B D A A A B B D C 5 B C C C 6 C A B B 7 C A A B 8 C B

More information

a= x+1=4 Q. No. 2 Let T r be the r th term of an A.P., for r = 1,2,3,. If for some positive integers m, n. we 1 1 Option 2 1 1

a= x+1=4 Q. No. 2 Let T r be the r th term of an A.P., for r = 1,2,3,. If for some positive integers m, n. we 1 1 Option 2 1 1 Q. No. th term of the sequece, + d, + d,.. is Optio + d Optio + (- ) d Optio + ( + ) d Optio Noe of these Correct Aswer Expltio t =, c.d. = d t = + (h- )d optio (b) Q. No. Let T r be the r th term of A.P.,

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces

More information

JEE ADVANCED 2013 PAPER 1 MATHEMATICS

JEE 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 information

Section 5.1 The Basics of Counting

Section 5.1 The Basics of Counting 1 Sectio 5.1 The Basics of Coutig Combiatorics, the study of arragemets of objects, is a importat part of discrete mathematics. I this chapter, we will lear basic techiques of coutig which has a lot of

More information

SEQUENCE AND SERIES. Contents. Theory Exercise Exercise Exercise Exercise

SEQUENCE AND SERIES. Contents. Theory Exercise Exercise Exercise Exercise SEQUENCE AND SERIES Cotets Topic Page No. Theory 0-0 Exercise - 05-09 Exercise - 0-3 Exercise - 3-7 Exercise - 8-9 Aswer Key 0 - Syllabus Arithmetic, geometric ad harmoic progressios, arithmetic, geometric

More information

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.

and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime. MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

Find a formula for the exponential function whose graph is given , 1 2,16 1, 6

Find 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 information

Complex Numbers Solutions

Complex 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 information

Sequences. A Sequence is a list of numbers written in order.

Sequences. A Sequence is a list of numbers written in order. Sequeces A Sequece is a list of umbers writte i order. {a, a 2, a 3,... } The sequece may be ifiite. The th term of the sequece is the th umber o the list. O the list above a = st term, a 2 = 2 d term,

More information

[ 47 ] then T ( m ) is true for all n a. 2. The greatest integer function : [ ] is defined by selling [ x]

[ 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 information

Solutions to Final Exam Review Problems

Solutions 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 information

MATH 304: MIDTERM EXAM SOLUTIONS

MATH 304: MIDTERM EXAM SOLUTIONS MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest

More information

XT - MATHS Grade 12. Date: 2010/06/29. Subject: Series and Sequences 1: Arithmetic Total Marks: 84 = 2 = 2 1. FALSE 10.

XT - MATHS Grade 12. Date: 2010/06/29. Subject: Series and Sequences 1: Arithmetic Total Marks: 84 = 2 = 2 1. FALSE 10. ubject: eries ad equeces 1: Arithmetic otal Mars: 8 X - MAH Grade 1 Date: 010/0/ 1. FALE 10 Explaatio: his series is arithmetic as d 1 ad d 15 1 he sum of a arithmetic series is give by [ a ( ] a represets

More information

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is

Homework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of

More information

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n =

sin(n) + 2 cos(2n) n 3/2 3 sin(n) 2cos(2n) n 3/2 a n = 60. Ratio ad root tests 60.1. Absolutely coverget series. Defiitio 13. (Absolute covergece) A series a is called absolutely coverget if the series of absolute values a is coverget. The absolute covergece

More information

Intermediate Math Circles November 4, 2009 Counting II

Intermediate Math Circles November 4, 2009 Counting II Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we

More information

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions

MA541 : Real Analysis. Tutorial and Practice Problems - 1 Hints and Solutions MA54 : Real Aalysis Tutorial ad Practice Problems - Hits ad Solutios. Suppose that S is a oempty subset of real umbers that is bouded (i.e. bouded above as well as below). Prove that if S sup S. What ca

More information

MATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006

MATH 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 information

LIMITS MULTIPLE CHOICE QUESTIONS. (a) 1 (b) 0 (c) 1 (d) does not exist. (a) 0 (b) 1/4 (c) 1/2 (d) 1/8. (a) 1 (b) e (c) 0 (d) none of these

LIMITS MULTIPLE CHOICE QUESTIONS. (a) 1 (b) 0 (c) 1 (d) does not exist. (a) 0 (b) 1/4 (c) 1/2 (d) 1/8. (a) 1 (b) e (c) 0 (d) none of these DSHA CLASSES Guidig you to Success LMTS MULTPLE CHOCE QUESTONS.... 5. Te value of + LEVEL (Objective Questios) (a) e (b) e (c) e 5 (d) e 5 (a) (b) (c) (d) does ot et (a) (b) / (c) / (d) /8 (a) (b) (c)

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

Section 7 Fundamentals of Sequences and Series

Section 7 Fundamentals of Sequences and Series ectio Fudametals of equeces ad eries. Defiitio ad examples of sequeces A sequece ca be thought of as a ifiite list of umbers. 0, -, -0, -, -0...,,,,,,. (iii),,,,... Defiitio: A sequece is a fuctio which

More information

MTH112 Trigonometry 2 2 2, 2. 5π 6. cscθ = 1 sinθ = r y. secθ = 1 cosθ = r x. cotθ = 1 tanθ = cosθ. central angle time. = θ t.

MTH112 Trigonometry 2 2 2, 2. 5π 6. cscθ = 1 sinθ = r y. secθ = 1 cosθ = r x. cotθ = 1 tanθ = cosθ. central angle time. = θ t. MTH Trigoometry,, 5, 50 5 0 y 90 0, 5 0,, 80 0 0 0 (, 0) x, 7, 0 5 5 0, 00 5 5 0 7,,, Defiitios: siθ = opp. hyp. = y r cosθ = adj. hyp. = x r taθ = opp. adj. = siθ cosθ = y x cscθ = siθ = r y secθ = cosθ

More information

Chapter 10: Power Series

Chapter 10: Power Series Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because

More information

Chapter 4. Fourier Series

Chapter 4. Fourier Series Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,

More information

BITSAT 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 information

ARITHMETIC PROGRESSIONS

ARITHMETIC PROGRESSIONS CHAPTER 5 ARITHMETIC PROGRESSIONS (A) Mai Cocepts ad Results A arithmetic progressio (AP) is a list of umbers i which each term is obtaied by addig a fixed umber d to the precedig term, except the first

More information

Infinite Sequences and Series

Infinite Sequences and Series Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet

More information

Sets. Sets. Operations on Sets Laws of Algebra of Sets Cardinal Number of a Finite and Infinite Set. Representation of Sets Power Set Venn Diagram

Sets. Sets. Operations on Sets Laws of Algebra of Sets Cardinal Number of a Finite and Infinite Set. Representation of Sets Power Set Venn Diagram Sets MILESTONE Sets Represetatio of Sets Power Set Ve Diagram Operatios o Sets Laws of lgebra of Sets ardial Number of a Fiite ad Ifiite Set I Mathematical laguage all livig ad o-livig thigs i uiverse

More information

Math 155 (Lecture 3)

Math 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 information

Sequences and Series of Functions

Sequences and Series of Functions Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges

More information

On a Smarandache problem concerning the prime gaps

On a Smarandache problem concerning the prime gaps O a Smaradache problem cocerig the prime gaps Felice Russo Via A. Ifate 7 6705 Avezzao (Aq) Italy felice.russo@katamail.com Abstract I this paper, a problem posed i [] by Smaradache cocerig the prime gaps

More information

( ) GENERATING FUNCTIONS

( ) GENERATING FUNCTIONS GENERATING FUNCTIONS Solve a ifiite umber of related problems i oe swoop. *Code the problems, maipulate the code, the decode the aswer! Really a algebraic cocept but ca be eteded to aalytic basis for iterestig

More information

Unit 6: Sequences and Series

Unit 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 information

ARRANGEMENTS IN A CIRCLE

ARRANGEMENTS IN A CIRCLE ARRANGEMENTS IN A CIRCLE Whe objects are arraged i a circle, the total umber of arragemets is reduced. The arragemet of (say) four people i a lie is easy ad o problem (if they liste of course!!). With

More information

Objective Mathematics

Objective Mathematics -0 {Mais & Advace} B.E.(CIVIL), MNIT,JAIPUR(Rajastha) Copyright L.K.Sharma 0. Er. L.K.Sharma a egieerig graduate from NIT, Jaipur (Rajastha), {Gold medalist, Uiversity of Rajastha} is a well kow ame amog

More information

Math 128A: Homework 1 Solutions

Math 128A: Homework 1 Solutions Math 8A: Homework Solutios Due: Jue. Determie the limits of the followig sequeces as. a) a = +. lim a + = lim =. b) a = + ). c) a = si4 +6) +. lim a = lim = lim + ) [ + ) ] = [ e ] = e 6. Observe that

More information

Poornima University, For any query, contact us at: ,18

Poornima University, For any query, contact us at: ,18 AIEEE/1/MAHS 1 S. No Questios Solutios Q.1 he circle passig through (1, ) ad touchig the axis of x at (, ) also passes through the poit (a) (, ) (b) (, ) (c) (, ) (d) (, ) Q. ABCD is a trapezium such that

More information

Zeros of Polynomials

Zeros 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 information

6.3 Testing Series With Positive Terms

6.3 Testing Series With Positive Terms 6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial

More information

Math 475, Problem Set #12: Answers

Math 475, Problem Set #12: Answers Math 475, Problem Set #12: Aswers A. Chapter 8, problem 12, parts (b) ad (d). (b) S # (, 2) = 2 2, sice, from amog the 2 ways of puttig elemets ito 2 distiguishable boxes, exactly 2 of them result i oe

More information

First selection test, May 1 st, 2008

First 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 information

05 - PERMUTATIONS AND COMBINATIONS Page 1 ( Answers at the end of all questions )

05 - PERMUTATIONS AND COMBINATIONS Page 1 ( Answers at the end of all questions ) 05 - PERMUTATIONS AND COMBINATIONS Page 1 ( Aswers at the ed of all questios ) ( 1 ) If the letters of the word SACHIN are arraged i all possible ways ad these words are writte out as i dictioary, the

More information

10.6 ALTERNATING SERIES

10.6 ALTERNATING SERIES 0.6 Alteratig Series Cotemporary Calculus 0.6 ALTERNATING SERIES I the last two sectios we cosidered tests for the covergece of series whose terms were all positive. I this sectio we examie series whose

More information

Chapter 6. Progressions

Chapter 6. Progressions Chapter 6 Progressios Evidece is foud that Babyloias some 400 years ago, kew of arithmetic ad geometric progressios. Amog the Idia mathematicias, Aryabhata (470 AD) was the first to give formula for the

More information

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

Substitute 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 information

Solutions for May. 3 x + 7 = 4 x x +

Solutions 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 information

Chapter 7 COMBINATIONS AND PERMUTATIONS. where we have the specific formula for the binomial coefficients:

Chapter 7 COMBINATIONS AND PERMUTATIONS. where we have the specific formula for the binomial coefficients: Chapter 7 COMBINATIONS AND PERMUTATIONS We have see i the previous chapter that (a + b) ca be writte as 0 a % a & b%þ% a & b %þ% b where we have the specific formula for the biomial coefficiets: '!!(&)!

More information

e to approximate (using 4

e to approximate (using 4 Review: Taylor Polyomials ad Power Series Fid the iterval of covergece for the series Fid a series for f ( ) d ad fid its iterval of covergece Let f( ) Let f arcta a) Fid the rd degree Maclauri polyomial

More information

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as

More information

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence

Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet

More information

Eton Education Centre JC 1 (2010) Consolidation quiz on Normal distribution By Wee WS (wenshih.wordpress.com) [ For SAJC group of students ]

Eton Education Centre JC 1 (2010) Consolidation quiz on Normal distribution By Wee WS (wenshih.wordpress.com) [ For SAJC group of students ] JC (00) Cosolidatio quiz o Normal distributio By Wee WS (weshih.wordpress.com) [ For SAJC group of studets ] Sped miutes o this questio. Q [ TJC 0/JC ] Mr Fruiti is the ower of a fruit stall sellig a variety

More information

AIEEE 2004 (MATHEMATICS)

AIEEE 2004 (MATHEMATICS) AIEEE 00 (MATHEMATICS) Importat Istructios: i) The test is of hours duratio. ii) The test cosists of 75 questios. iii) The maimum marks are 5. iv) For each correct aswer you will get marks ad for a wrog

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

(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 information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9. Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say that

More information

CSE 21 Mathematics for

CSE 21 Mathematics for CSE 2 Mathematics for Algorithm ad System Aalysis Summer, 2005 Outlie What a geeratig fuctio is How to create a geeratig fuctio to model a problem Fidig the desired coefficiet Partitios Expoetial geeratig

More information

Permutations, Combinations, and the Binomial Theorem

Permutations, Combinations, and the Binomial Theorem Permutatios, ombiatios, ad the Biomial Theorem Sectio Permutatios outig methods are used to determie the umber of members of a specific set as well as outcomes of a evet. There are may differet ways to

More information

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0,

x x x Using a second Taylor polynomial with remainder, find the best constant C so that for x 0, Math Activity 9( Due with Fial Eam) Usig first ad secod Taylor polyomials with remaider, show that for, 8 Usig a secod Taylor polyomial with remaider, fid the best costat C so that for, C 9 The th Derivative

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)

(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 information

Math 113 Exam 3 Practice

Math 113 Exam 3 Practice Math Exam Practice Exam will cover.-.9. This sheet has three sectios. The first sectio will remid you about techiques ad formulas that you should kow. The secod gives a umber of practice questios for you

More information

Mathematics Extension 1

Mathematics 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 information

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at

More information

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of

MATHEMATICS. 61. The differential equation representing the family of curves where c is a positive parameter, is of MATHEMATICS 6 The differetial equatio represetig the family of curves where c is a positive parameter, is of Order Order Degree (d) Degree (a,c) Give curve is y c ( c) Differetiate wrt, y c c y Hece differetial

More information

MODEL TEST PAPER II Time : hours Maximum Marks : 00 Geeral Istructios : (i) (iii) (iv) All questios are compulsory. The questio paper cosists of 9 questios divided ito three Sectios A, B ad C. Sectio A

More information

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and

MATH301 Real Analysis (2008 Fall) Tutorial Note #7. k=1 f k (x) converges pointwise to S(x) on E if and MATH01 Real Aalysis (2008 Fall) Tutorial Note #7 Sequece ad Series of fuctio 1: Poitwise Covergece ad Uiform Covergece Part I: Poitwise Covergece Defiitio of poitwise covergece: A sequece of fuctios f

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

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018)

JEE(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 information

MATH2007* Partial Answers to Review Exercises Fall 2004

MATH2007* Partial Answers to Review Exercises Fall 2004 MATH27* Partial Aswers to Review Eercises Fall 24 Evaluate each of the followig itegrals:. Let u cos. The du si ad Hece si ( cos 2 )(si ) (u 2 ) du. si u 2 cos 7 u 7 du Please fiish this. 2. We use itegratio

More information

, 4 is the second term U 2

, 4 is the second term U 2 Balliteer Istitute 995-00 wwwleavigcertsolutioscom Leavig Cert Higher Maths Sequeces ad Series A sequece is a array of elemets seperated by commas E,,7,0,, The elemets are called the terms of the sequece

More information

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics:

Chapter 6 Overview: Sequences and Numerical Series. For the purposes of AP, this topic is broken into four basic subtopics: Chapter 6 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets

More information

Joe Holbrook Memorial Math Competition

Joe 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

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel)

For use only in Badminton School November 2011 C2 Note. C2 Notes (Edexcel) For use oly i Badmito School November 0 C Note C Notes (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk

More information

International Contest-Game MATH KANGAROO Canada, Grade 11 and 12

International Contest-Game MATH KANGAROO Canada, Grade 11 and 12 Part A: Each correct aswer is worth 3 poits. Iteratioal Cotest-Game MATH KANGAROO Caada, 007 Grade ad. Mike is buildig a race track. He wats the cars to start the race i the order preseted o the left,

More information

Coffee Hour Problems of the Week (solutions)

Coffee Hour Problems of the Week (solutions) Coffee Hour Problems of the Week (solutios) Edited by Matthew McMulle Otterbei Uiversity Fall 0 Week. Proposed by Matthew McMulle. A regular hexago with area 3 is iscribed i a circle. Fid the area of a

More information

Assignment ( ) Class-XI. = iii. v. A B= A B '

Assignment ( ) 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 information

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016 18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam

More information