International Practice Exam MCQ Answer Key

Name: АРе Calculuв ВС Anвwer Sheet for :Мultiple-Choice Section No. 1 ,д 2 с 76 4 '('\ . .д.. 6 1Л 81 В2 83 з б 7 8 9 10 11 .- No. Answer 12 13 14 15 16 17 18 19 20 21 22 23 ;-;:,, с.. А3 ,1..1. 84 85 86 87 88 89 90 1) с [) r.z) Г) R 7 л (__ с с 28 f?� 30 - ?, ) ) 'f) 29 во А- 24 25 26 27 77 78 LL с i2, ., - Answer с__ i1 С' п, � [) а (.,1... '< Е с.. !?. f,. � ·р, АР ® CALCULUS ВС 2017 SCORING GUIDELINES Question 1 (а) ,/(х'(2))2 + (у'(2))2 = 3.272461 2 : { 1 : expression for specd 1 : answer with units The speed ofthe particle at tiine t = 2 seconds is 3.272 meters per seco11d. 2 2 2 2 (Ь) s(t) = ,/(х'(1)) + (у'(1)) = ,/(2cos(2!)) + (2t- 1) 2 ·{ 1 : conside1-s s'(4) • 1 : answer with reaso11 2 _ { 1 : iпtegraJ • 1 : ans�1er s'( 4} = 2.16265 Since s'( 4) > О, the speed ofthe particle is incгeasiпg at time 1 = 4. (с) s: 2 2 ,/(x'(r)) + (y'(t)) dt = 22.381767 Tl1e total distat1ce ihe pa11icle travels ove,· the tiшe interval О ;;; t ;;; 5 seconds is 22.3 82 (or 22.381) inete1·s. (d) x(IO) =х(8)+х'(8)·2 =sin 16+х'(8)-2 = -4.118541 2 у(\О) = у(8)+ у'(8)-2 = (8 -8)+ у'(8}·2 = 86 The position of the particle at time t = \ О seconds is (-4.119, 86) (01· (-4.118, 86)). 3: { 1 : uses position at 1 = 8 1 : uses velocity at t = 8 J : position at 1 = 10 АР Ф CALCULUS вс 2017 SCORING GUIDELINES Ouestion 2 (а) Jr4.s a(t) dt = 66.532128 o 2: j 1 : integral 1 : ans\ver At time t = 4.5, tank А contains 66.532 liters of,vater. (Ь) a(k) = 20.5 � k = 0.892040 J: (20.5 - a(r)) dt = 10.599191 3 : At tinie 1 = k, the difference in the amounts ofwate,· in the tanks is 10.599 liters. [2.4\б (с) Jo s: 12.416 b(I) dt = Jfk o Ь(!) dt + k b(I) dt 2: { 1 : sets a(k) = 20.5 1 : integral 1 : answer { 1 : J 2.416 k a(t) dr 1 : a11swer b(t) dt = 20.5 • k = 18.286826 • 2.416 On k < t < 2.416, tank А receives 1k a(t) dt = 44.497051 lite1,; of waler, wblch is 14.470 mo,·e lite1·s ofwater than tank В. J2-416 Тherefore, k 2.416 b(t)dt+ k Jr; o J 2.4l6 b(t) dt = ; J a(r) dt -14.470 = 30.027051. b(1)d1=48.3l3876 Attime 1 = 2.416, tank В contains48.3l4(or48.ЗJЗ) litersof,vaieг. (d) и/(3.5)- а'(З.5) = -1.14298 < О The diffe1,ance w(r) - a(t) is decreasing at r = 3.5. 2· • { 1 : w'(З.5) - а'(З.5) < О 1 : conclt,sion AP CALCULUS ВС 2017 SCORING GUIDELINES Question З (а) g(5) - g(-5) = 12 - (,r + 7) = 5 - ,.. 5 - (-5) 10 10 3 (Ь) g'(x) = f(x) l : answer : { 1 : difference quotient 2: answer g'(З) = 1(3) = 4 The instan!aneous i-ate of change of g at х = 3 is 4. (с) The graph of g is concave up оп -5 < х < -2 and О < х < 3, because g'(x) = f(x) is increasing оп these intervals. 2 : intervals with justification (d) g'(x) = /(х) is defined at all х with -5 < х < 5. g'(x) = f(x) = О at х = -2 and х = \. Therefore, g has critical points at х = -2 and х = \. g has neither а local inaxiinum по,· а local шiniшum at х = -2 because g' does not change sign there. g has а Iooal minimшn at х = l because g' changes from negative to positive there. 3 : 1 1 : considers /(х) = О 1 : critica\ poin!s at х = -2 and х = 1 1 : answers with justifications АР ® CALCULUS ВС 2017 SCORING GUIDELINES Question 4 (а) f f'(x)dx"" 2·3.5+2•0.8+2·5.8 = 20.2 /(6) - /(О)= J: f'(x) 6 f'(x) dx"' 20 + 20.2 = 40.2 j l : шidpoint sшn 1 : Fundamental Theorem 3 ofCalcL1lus 1 : answer J: f'(x) dx S: б • 7 = 42. 2 0 /(6) = /(О)+ s: (Ь) Since f'(x) s 7, dx 1(6) - f(0) s 42 � /(6) s: 20 + 42 = 62 : . ( 1 : integral bound • 1 : answer with reasoning Thei·efore, the actual value of f( 6) could not Ье 70. (с) J: f''(x) dx = f'(4)- f'(2) = 1.7 - 2 = -0.3 (d) lim (f(x) - 20е') = О .х➔О lim (0.5/(х) - 10) = О ,➔О Using L'Hospiia]'s Rule, х lim f(x) - 2Ое·' = lim f'(x) - 20е = 4 - 20 = _8 0.5(4) х ➔оО.5/(х)-10 х ➔ о 0.5f'(x) 2: { 1 : Ft1ndamentaJ Theore111 of Calculus 1 : answer 2· • \ l : L'Hospital's Rule 1 : aнswer AP @ CALCULUSBC 2017 SCORING GUIDELINES Question 5 2 = 2/ - у -2.,.у 2 х (Ь) dy 1 = -1 + _± = О 4 dx (,,у)=(4, 2) 2: 2: d1y = 2, 8 - 4 - 16 = _ _!_ < 0 16 4 dx2 1 (х,у)=(4, 2) Ву the Second De,·ivative Test, g has а гelative 111axi111u111 at х = 4. (с) :Z:lr_,,.,.)=(1,2) = -1 + Т =3 3: { I : quotient rule d1 J. ---2. • dx2 { 1 : considers dy \ dx (х,у)=(4, 2) 1 : ans1ver withjustification ! 1 : uses dyl dx (х. y)=(I. 2) 1 and d yl dx1 (х, y)=(I. 2) 2 : Taylor polynomial 11,е second-deg1·ee Taylo1· polynomiaJ for h about х = l is ' ' 8 Т2 {х) = 2 + З(х -1) + (х - 1)- = 2 + З(х -1) + 4(х -1)-. 21 (d)lh(l.l)-Als max lMx �1. 1 "' lh (x)I 11 . 1 - 11 З! 3 < 60 __ 1_=_!0,__=__l_ - 6 1000 1000 100 2· • { J : form ofthe eiтor boнnd J : analysis j АР ® CALCULUS ВС 2017 SCORING GUIDELINES Question 6 (а) r00 [- ь _ _l_ dx = lim r _,_I_ dx = lim 1 tan- 1 Ь➔оо з Ь➔00Jз ,.- + 9 Jз х-, + 9 ь (-зх)lз ) = lim (l1an- 1 (/J_) _ _l_tan- 1 (1)) = tr__ - -"- = -"---ь➔оо З 3 3 6 12 12 (Ь) The function f is continuo11s, positive, and deci-easing оп [3, оо). Ву the integ1--al test_, since 1 : antide,;vative 3 : { 1 : limit expression 1 : answer 2 : conclusion ,vith conditions 00 f3О()f(x) dx converges, Lf(n) converges. 11,=] -OR1 1 r1or п > о _ :,. 0 < -2 п +9 <2 п L- 00 1 converges, the series "' 00 1 Since the series L L f( 11) = 2 2-11=З n 11=Э 11=3 11 + 9 converges Ьу the compш·ison test. оо оО 2 9 1 (с) Considcr the series L = L 11 � . ,, ) ( ,,:1 е , f(n) »:J е lim 11➔СО е" <Х) 1 Ву the ratio test, L converges. n:I ( е,, • J(n) ) 00 ( 1)" The1·efo1·e, L converges absolutely. n•I (е,, • f(n) ) 4: ! 1 1 1 1 : sets up ratio : computes limit ofratio : conc]usion ofratю test : coпverges absolutely