M137.F07.TT1.sols

M137.F07.TT1.sols - Time: 4:30 - 6:20 p.m. Family N eme:...

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Unformatted text preview: Time: 4:30 - 6:20 p.m. Family N eme: ID. Number: Faculty of h’latheniatics University of Waterloo hiath 137 Term Test 1 — Fall Term 2007 SOL. uTioUS AIDS: iPINK TIE’ CALCULATORS ONLY Signature: Check the box next to your section: B. J. Marshman (11:30 am.) H. Lee (12:30 p.111.) M. Leifer (1:30 pm.) S. Speziale (2:30 pm.) P. Wood (1:30 11111) C. Subich ( 9:30 am.) I Oancea (10:30 3.111.) Initials: _..._ 008 009 010 011 012 013 ULIULILID' M. Best (10:30 am.) D. Yang (2:30 pm.) P. Roh (12:30 p.111.) 5. Furino (9:30 H.111.) D. Park (8:30 am.) I. Oencea (8:30 am.) Date: October 9, 2007. receive full marks. Your answers must be stated in a clear and logical form in order to Reference any theorems or rules used by their name, or the appropriate abbreviation. Useful abbreviations for this test include LSR (limit sum rule), LPR (limit product rule), LQR (limit quotient rule), LCR (limit composite rule for basic functions). Note: 1. Complete the information section above.: indicating your instructor‘s name by a. Checkmark in the appro priate box. [.0 Place your initials and ID No. at the top right corner of each page. 3. Tear off the blank last page of the test to use for rough work. You may use the reverse of any page if you need extra space. 4. Any 60 marks worth of questions constitutes a complete test. You. will moi-Lie ’flnoi‘ “Ll Solgfilimin s 16 6" 2 ? Mark l Out of i 1 l3 2 3 17 <1 18 Total 60 + 4 all L' H MGR-Lizgilsltjan , CT a cum. no 5 ubsTani‘m ~e 7&ngmmcmt ) ,4 NIATH 137 TERM TEST #1 PAGE 2 1‘ For each qualitative sketch in this problem, explain: clearly how you obtain yo1u‘ graph, and indicate all asymptotes and intercepts with the x and y axes, if they occur. [4] a) Sketch the graph of the function f(2) = |1n(—:::) + 1|, S’rchr @th \oo-olc. gnapln g -= thd g = wch 5 - 0.1705 3231.29... *5 = €n(— x5 + I c M ‘ 93M 61 ‘Sllgplow Tm ‘X'MlS arm—x» WM X I %(—x\ + ll 5: [5] b) Sketch the graph of the function = IE3 + 1 on the upper axes and then Sketch the . . 1 graph of 1ts remprocal — = I3+1 on the lower axes. . . '5 5:134 ISASngrl‘ufhjlrg 15:7‘ . le’lru~ Mctkuocd 3; AT— ,UJLm—Cl -. +£ Ll ')<q’+\:0t53s 11—4 =1> x:-liso. W’Ccd MWTo’Ce 68 j: 215—: «- . on-—\<x¢o°) x5“ is PCS-(ELM * HAUL. =*’ 4“ cs Poswwmwu. ' m—w<x<—( ) K$+\(5MQJm+\nCA. =9!) \ db “- CLLUL ‘ OJ) ._.L——— 0 so £5 “*tcfi : x1“... ’ heme. [4] (3) Sketch the graphs of y = arctan :1: and y = ~rc1/3 and hence find all solutions of the equation arctan 1: + 2:1/3 = 0. ,.. 8 —:. .. 3L“; m hp¥lCle mm M-amis (m 3-6546) 9‘8 A3: 3C”). m M 50W efi mama; 1—38” as M x ¢3+|=\:& x20 =z> MATH 137 TERM TEST #1 :1 At left is the graph of the function J/ll_—'——_- 1 [3] (a f(l‘) : 1+6“. (i) Label the horizontal asymptotes of x y = and the y-intercept. Explain your choices by evaluating suitable 3h: limits. , an}? §ww ‘ x = o x ' x —=; +00 l + e 3 ’ ‘j I 6% _.L__. : .L._. g t LQR 1 'X a- an i i-e l +0 L S R ’ (ii) Find the inverse function f" (:22) and sketch y = f‘l(:i:) on the graph in a) .,_—-\ 7c ’fi - x __ - {5" we} =9 %C\+e)=\ =s> ge=l-H=—’V€-L§ Sol Ding 8m '35 DL={Q/vx(l_—g$.) '25 SUS\VCJf\tl/\C51+Lj" gem {-1 _,. A m L735>O l3 (ad ’8 ls» o<x<l b) Solve the equation 1 + cos 2:: = —35ina:. [ HINT: Recall that cos 29; = 1 — 28in2 ID] Substihfliing Tim t h. (\_’Zs(n2x\= — 38ml =§ 2‘25UY‘27<=‘350Y\9§ :b ’Zsinz'x...350nx_2:o FOLCTMCWQ‘. (Si‘nx—23C1sivxx+0 =0 =x> aw m oimx=—‘/2 sinxf-Vz =a> gar-IVE +2“ or—Ebfli-zlpjl') 5‘61 —\_<ss'nxs\ [7] c) A runner‘s elbow swings back and forth from the sl‘o “SIGHT shoulder as she runs, lf y is the angle shown, then 5 the motion of the elbow is given by I 71' i . y(t) = g cos(37rt) , With t in seconds. <_ 494MB: Ill/[Erna], (i) Sketch a graph of y(t) on the given axes, indicating clearly the amplitude A and period T of the motion. Label the points where y(t) = Al “ W ) fi/ /z We . 1. 4 a Z Pemod. T‘: D‘ng—m. 77 LIV“) (/3 k‘ 3w 3 9 “Mei-it‘wa = 1 t 8 Seton-AS all/is 1r 3 /3 Co 5 3w: (ii) If the distance OP from the shoulder to the elbow is 30 cm, what is the maximum value 3 of the horizontal displacement h of the elbow? 0cm RA will bk. a. moin'mum weuw (.5 & MM. efi 11/8 ix _ - it i. 'ga-Smfl a l‘wa‘ BosM/é) cm- (iii) Assuming the runner takes 2 steps per period of the elbow’s swing, how many steps per minute does she take? 2 steps (m. 3/3 second. =9 3 $295 (unsecoma =P(80 “ pus ma:an (iv) When a person is iii-[ilking instead of running, their elbow swings more slowly. and does not move as far Explain how this would change the amplitude and period of y(L), A mound Ame-unan— . 1‘ ououfid WWW c JZo—ngzn mm nvawé? MATH 137 3‘ [3i ’1 TERM TEST #1 PAGE 4 a) Evaluate each limit, or show that it does not exist. Justify your reasoning. 3 . 0),!22025’; = MM Q/x3*5 _ 0+3 LQR. Lao/ = -5 Luz-£900 x? knive- =: ‘Qi‘m w wwwcfiiefi x—amz Qx_33(yfi mimi+sén:ox+—2 xg—cl (u) 1:11—41:12 3:2 —- II? — 6 :. —2~2 “2‘3. LSJFQ, 4 Qk‘Vszm __ 4 X-ra _. /5 ‘ arctane) [3] (“1) 212311 605632) 9|V:\c1 & Im C a) ' xi”? cos X 2 C°9<8CTQD+123 : C050: i L L { \og LCR. J ‘ WA gig/(MJWU) = ahdan(£i'wx _\_ = W/z lvM ant-3mm“; 1172 _ 1r b L R xjo+ 9‘ ) 7"?0" Cost“) 2 “T— — 7 ‘3 Q ’ [2] b) Find the domain of the function f(17) = V621 - 561 + 4. Domaui/v (:3 Jo; at m 2,0 4:3. of“ “(Woo-o. mud. ezw-‘Se'xir 4 7/0 Coufiéjeyi?oo3 «187 UL ‘ I ll [3] [3] 2w 1- ‘ x e 48") =9 Ce“.4XeX—I3 >,O 6%» a )‘i («3%45 OR e“g\ (150). C) Solve the equation In .1' + ln(z — 1) = 1. Q’WX‘K eviCX-i3 = eluCfi (X-D\ =i =9 xC1_13:e(=e afar-e =0 =I> x = it \ii-HQ t Z ’ gtnuwe. mud 3L>i 8m QMX 093g QMCx—fl To 0‘6“) \Jvu. 011.6 gov: Lu ac: H—JH-‘Qe 2 d) Solve the inequality 6"2 < 62H3 sm‘u a") (s M UnMrng gwc‘hév“: U) '2 ex< 8'27“? :1) x7- 4 2x+3 =3’ Pct—2x45 <0 ="’ (x—3)(x+0 <0 l 3 =9» ~...l < x < 3 grown siliCk 5__ x7-_2x_3 0R -. was: 0.3040 18$ a<o,b>o 0R a>o,io<o ~ 1 - — W} Lon‘ko. x 3, b- :chm- Agikw .44 x<3 no so!"‘— , MATH 137 TERM TEST #1 PAGE 5 5 4. Label each statement as TRUE or FALSE in the blank brovided. (Use the space between questions to justify your answer. Gnesses will not be graded.) [3] a) F If f($) = fl and g(:1:) = arcsinm, then the domain of (g o f)(a:) is {9219: 2 0}. (<3 oglfix) : fiH-(xll = mcstln (07) J? 9 ( :9 x .L. l Domain -. x>o / 3 FaDJD e b) T The equation tans: = C081? has only one real sulnlninn in Line interval (—7r/2‘ 77/2), ' ~ 7. fly; ‘5 17mm =c059< % £23” :cos'x =1> mmx=coszx ‘1 __ 59‘ = l — 5m x FmMSluld") zpfim1x+ow~ 1'0 it x 3=rams< eszmsx x x \ Memo” 999 ‘m<~1T/2:W2) 1') owvx=—\ir\l\+4 l r ’2. F ,‘.’\’<‘u.q 4 d-me: -|+£ (4-5 6 [3] c) The function = ln('2e“) + log3(9”) — arclanfl) has range R, the real 2' Q 2 - numbers. One xfin QCXX=QMZ+€MC€A§ l. Q03 (32") _ 43L 1: (0 NZ) 3 11- 4‘. : e!“ Z -— x + 2 x __ x :Q/WZ .'. g-(x) (5 cmsl—Mk ,4- sz ‘ So FOL-9436 (I) \ If f(;I:) = 2 + .1' + 6” then f“I(—1)= 1+ 6“. SAG“ = \+ e‘l >0 =>§(§"(—13]=—l =§(l+e") ...\ Bud? 611°”le Shire“): 2+e +e - >0 50 \‘l‘ CWol" be —l> Md ms [3] e) T If f is an even function, then f(|$|) = f(:):) for any a: in the domain of f. 93-20% =3> £00155ch £61 at e DCS) gUxU : %§—(X3 Sm 13° ._. S-(fl Sm 30,0 $0.0 {pm x<o MA 541 I<o f .mam ls‘Truu % 51‘“ 3“ “9(5) I') F If f(:17) and g(a:) en‘e positive une—Lu-one functions on lll.‘ then ]L(.’L') = is also one Lo one on H. J‘fi {'00 = ex amo‘ %Cx] = 2 e x m 930:)“ wQuLkLunm f-.(. %Cx) 2 .TCML’S PMQL‘OLU k COME-A "WW I “AM Jaw" Lo FaLse. 3 ...
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This note was uploaded on 01/17/2012 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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M137.F07.TT1.sols - Time: 4:30 - 6:20 p.m. Family N eme:...

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