# Theorems Solutions. Multiple Choice Solutions

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1 Solutions We hve intentionlly included more mteril thn cn be covered in most Student Study Sessions to ccount for groups tht re ble to nswer the questions t fster rte. Use your own judgment, bsed on the group of students, to determine the order nd selection of questions to work in the session. Be sure to include vriety of types of questions (multiple choice, free response, clcultor, nd non-clcultor) in the time llotted. Multiple Choice Solutions 1. B (1998 AB4) B is flse becuse this specil cse of the MVT clled Rolle s Theorem lso requires tht f( ) = f( b). A is true by MVT; C nd D re true by EVT, E is true by the definition of definite integrl.. C (1988 AB) Rolle s Theorem gurntees t lest one vlue of x between nd b such tht f ( x) =.. B (1985 BC1) f( b) f( ) 7 7 Show f () c = using MVT: x 6x= ; xx ( ) = when x = nd b x = ; however, only is eligible since there is n endpoint t x =. 4. A (1998 AB6) Any vlue of k less thn 1 will require the function to ssume the vlue of 1 t lest twice becuse of the Intermedite Vlue Theorem on the intervls [, 1] nd [1, ], so k = is the only option. 5. D (197 BC18 pproprite for AB) D could be flse, consider gx ( ) = 1 xon [, 1]. A is true by the Extreme Vlue Theorem. B is true becuse g is function. C is true by the Intermedite Vlue Theorem. E is true becuse g is continuous. 6. D (199 AB18) π π π π f 1 c f sin sin f ( c) = cos ; f () c = = = = ; π π π π π 1 c cos = ; c = π Copyright 1 Ntionl Mth + Science Inititive, Dlls, TX. All rights reserved. Visit us online t

2 7. E (199 BC44 pproprite for AB) By the Intermedite Vlue Theorem, there is c stisfying c b verge vlue of f on the intervl [, b ]. The verge vlue is lso given by Equting the two gives option E. t < < such tht f() c is equl to the 1 b ( ) b f x dx. Alterntively, let F() t = f ( x) dx. By the Men Vlue Theorem, there is c stisfying < c< b Fb ( ) F ( ) b such tht = F () c. F( b) F( ) f ( x) dx b = nd F () c = f() c by the Fundmentl Theorem of Clculus. This gin gives option E s the nswer. This result is clled the Men Vlue Theorem for Integrls. 8. B (1969 BC pproprite for AB) 1 y = x, so y =. By the Men Vlue Theorem, x 1 =, so c = 1. The point is (1, 1). x 4 9. E (1998 AB91) I nd III re true by IVT; II is true by MVT. 1. E (8 AB89/BC89) Since there is no c for which f ( c) =, Rolle s Theorem is violted nd f ( k) does not exist on (, ). 11. B ( AB8) B could be flse since this is specil cse of MVT (Rolle s Theorem) which lso requires tht f( ) = f( b). A nd C re true by IVT; D is true by MVT; E is true by EVT. 1. D (1997 BC81 pproprite for AB) f ssumes every vlue between 1 nd on the intervl (, 6), so f() c = 1t lest once. Copyright 1 Ntionl Mth + Science Inititive, Dlls, TX. All rights reserved. Visit us online t

3 AB/BCb Yes; Since, the Men Vlue Theorem gurntees tht there is t, t 4, such tht nswer MVT or equivlent 14. 9B ABbc f(6) f( ) = when f( ) = f(6). There 6 re two vlues of for which this is true. (c) Yes, =. The function f is differentible on the intervl < x < 6 nd continous on f(6) f() 1 1 x 6. Also, = =. 6 6 By the Men Vlue Theorem, there is 1 vlue c, < x < 6, such tht f () c =. expression for verge rte of chnge nswer with reson nswers yes nd identifies = justifiction 15. 8B AB5/BC5d (d) No, the MVT does not gurntee the existence of vlue c with the stted properties becuse is not differentible for t lest one point in verge vlue of nswer No with reson Copyright 1 Ntionl Mth + Science Inititive, Dlls, TX. All rights reserved. Visit us online t

4 16. 7 AB b ( ) ( ) () h( ) f g( ) f h( ) f ( g( )) f ( ) Since h( ) 5 h( 1) 1 = 1 6= 6= 9 6= = 6= 4 6= 1 6= 7 < < nd h is continuous, by the Intermedite Vlue Theorem, there exists vlue r, 1< r <, such tht h( r ) = 5. h( 1) nd h ( ) conclusion, using IVT ( ) h( ) h 1 7 = = Since h is continuous nd differentible, by the Men Vlue Theorem, there exists vlue c, 1< c <, such tht h c =. ( ) 5 ( ) h( 1) h 1 conclusion, using MVT 17. 7B AB 6bd () The Men Vlue Theorem gurntees tht there is vlue c, with so tht conclusion, using MVT uses MVT with Since f is twice-differentible, is differentible everywhere, so the Men Vlue Theorem pplied to on [, 5] gurntees there is vlue k, with such tht (d) conclusion, using IVT Since the Intermedite Vlue Theorem gurntees tht there is vlue r, with such tht Copyright 1 Ntionl Mth + Science Inititive, Dlls, TX. All rights reserved. Visit us online t

5 18. AB 6c (c) By the Men Vlue Theorem there is c with < c <.5 such tht f (.5) f ( ) f ( c) = = = 6 = r.5.5 reference to MVT for f (or differentibility of f ) vlue of r for intervl x.5 Copyright 1 Ntionl Mth + Science Inititive, Dlls, TX. All rights reserved. Visit us online t

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### 4181H Problem Set 11 Selected Solutions. Chapter 19. n(log x) n 1 1 x x dx, 48H Problem Set Selected Solutions Chpter 9 # () Tke f(x) = x n, g (x) = e x, nd use integrtion by prts; this gives reduction formul: x n e x dx = x n e x n x n e x dx. (b) Tke f(x) = (log x) n, g (x)

### sec x over the interval (, ). x ) dx dx x 14. Use a graphing utility to generate some representative integral curves of the function Curve on 5 Curve on Clcultor eperience Fin n ownlo (or type in) progrm on your clcultor tht will fin the re uner curve using given number of rectngles. Mke sure tht the progrm fins LRAM, RRAM, n MRAM. (You nee to

### Higher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector

### Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

### 5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship 5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is

### Bob Brown Math 251 Calculus 1 Chapter 5, Section 4 1 CCBC Dundalk Bo Brown Mth Clculus Chpter, Section CCBC Dundlk The Fundmentl Theorem of Clculus Informlly, the Fundmentl Theorem of Clculus (FTC) sttes tht differentition nd definite integrtion re inverse opertions

### A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

### 11 An introduction to Riemann Integration 11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

### SUMMER KNOWHOW STUDY AND LEARNING CENTRE SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

### 0.1 THE REAL NUMBER LINE AND ORDER 6000_000.qd //0 :6 AM Pge 0-0- CHAPTER 0 A Preclculus Review 0. THE REAL NUMBER LINE AND ORDER Represent, clssify, nd order rel numers. Use inequlities to represent sets of rel numers. Solve inequlities.

### Chapters 4 & 5 Integrals & Applications Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO - Ares Under Functions............................................ 3.2 VIDEO - Applictions Test 3 Review Jiwen He Test 3 Test 3: Dec. 4-6 in CASA Mteril - Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 14-17 in CASA You Might Be Interested November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.