SFE - SOLUTIONS TO FINAL EXAM PART I (5 marks each) 1. Let...

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Unformatted text preview: SOLUTIONS TO FINAL EXAM PART I (5 marks each) 1. Let f 2 y when :r 2 arcsiny. List all of the following intervals that are domains on which f is a function. [071] [07 2i [07 77] [—7T7 77] 7r 7T Solution Recall a: 2 arcsiny means y 2 sinrr with at E [—5, 5]. Note I < g < 2. Hence [0,1] C [—g, and [0, I] is a suitable domain for the function f. However, none of the other three intervals are subsets of [—g, and are therefore not suitable domains of a fucnton f. That is, there are numbers at in each of these three intervals for which f is not defined. 1 2. Find the range of the function 2 arc an cc Solution The range of arctanat is 'I‘lqn v-An;v\v-r\nnlr\ nl- V111W\|r\/\v-r\ 1n +l’1n n+nv- ml I A {H nv-A {Jan V111W\|r\/\v-r\ nl- +l’1n n+nv-nnl I An El Lilb LbbllJLUK/(LLD UL lilJLLLlUL/LD 11L Ulib llilJK/L Volll \ r) s U, (Lib Uilb lilJLLLlUL/LD UL Ulib llilJK/L Volll \ W, ~ ,- 7 A ,. / The recrprocal of 0 is not defined m . 1 n 1 . .1 . 1 / 7r\ 0 /2 \ 1-13 recrpbcars or “unites in the interval |\U7 Z ,1 are The mmoers or The IDTEI‘VBI G7 00) m p n a / 2\ . . /2 \ r-_erer-re Le ranve or T is l—oo ——I U l— ow O J \ 7 7T / \ 7|" 7 / 3. Let g(:r) 2 I + 3:2 with domain (—00. 0). Assume that g is oneitoione. Find the formula for 971(27). Solution Let y 2 971(36). Then a: 2 g(y) 2 1 + yg. Hence 3/2 2 a: — 1 and y 2 —\/$ — I because the range of 9’1 equals the domain of g which consists of negative numbers. Thus g’l($) 2 —\/:c — I. 671‘ 4. Evaluate arctan tan 7. Solution Recall that the domain of the tangenet function which is used to define the arctangent is 7T 7T ( 2, 2). Thus 67? 671' 7r 7r arctan tan — 2 arctantan — — 7r 2 arctan tan —— 2 ——. 7 7 7 7 5 (3111.311 ( \ H 121” a value m? X “minim ohmm that 11m Kim J. i‘\ : 7 . 1n / , him I “m ,i men n, i” in WW . I , .. 1 1, m—u Solution We need to find 6 so that if 0 < [:13 7 2| < 6 then + 1)7 7] < 6, i.e. I32: 7 6| 2 312: i 2! < 6. 3 ' . 1m W . a 113; 3 92 g p Solution Since —1 S sin9 S +17 we have 0(—l) < 0sin6 < 6(+1) 7 v 592 T 7 v 592 T 7 v 502 # 0 < 0 sin 6 < 6 7 v 592 T 7 v 592 T 7 v 592 We know 1' — 6 2 0 d l' 6 2 0 9L”; 7 + 562 7 a“ 9L“; 7 + 562 7 By the Pinching Theorem, lim 08m 6 2 0 0—00 7 + 592 7. Let f be a continuous function with domain [a, b]. Which theorems are needed to conclude that the range of f is a closed interval? J USTIFY YOUR ANSWER. Solution By the Maximum Value Theorem, the continuous function f has a maximum value fUVI) and and minimum value f (m) on the closed interval [(1,1)]. By the Intermediate Value Theorem, every number between and is in the range of f. Hence the range of f equals f(]t1)]. 8. Let 2 W for 36 7S —1, 1. Let f(—1) 2 #1 and f(1)21. m _ Determine where f is continuous, where f has a removable discontinuity and where f has an essential discontinuity. J USTIFY YOUR ANSWER. Solution Observe that for w 2 1, we have f(x) = ling—“(wan : W 1X12 ‘I w = 1)| m i 1M2 a 1 1) I l($—1)(x2+::rl)l(x2—I+l) :(x2 a a: : 1)(Jc2 x.1)l::il since $2 + :1: + 1 2 (at + 1/2)2 + 3/4 2 0 for all x. Thus f is continuous at every x other than m 2 *1 91nd c‘r m — _\_'| RV ‘r wuu um .A/ I 1. uy u x 2 —1. Observe that Hence f has an essential discontinuity at a: 2 1. 9. Use derivative formulas to find the derivative of y 2 arctan \/ tan Solution By the chain rule: dy 1 d 1 1 d _ : —_ x/t : ——_ t dm 1+(«/tan$)2 d$( angs) 1+tan172x/tan$d$ allm) 1 2 sech 2 —sec :3 2 2x/tanx(1+tanx) 2x/tanx(1 +tanx) 10. Let g(:1:) 2 tanassec2 x for 0 S :1: < Assume that g is oneitoione. Note 9 2 2. Find D (g’l) 1 Solution The derivative of the inverse function g’ is given by: _ 1 _ 1 — Dani—1(2)) T Bait/4) since 9(71' / 4) 2 2. By the product and chain rules: Dig—1M2) g'(a:) 2 D(tan .91:)(sec2 it) + (tan a:)D(sec2 at) 2 (sec2 31:)(sec2 at) + (tan 3:) (2 sec 33D(sec 2 sec4 x + 2 tanx sec :I:(sec :1: tan x) 2 sec4 at + 2 tan2 :1: sec2 at Hence air/4) : (fir ~ 2mm 2 4 + 4 z 8 and D (9-1) <2) : a $2 11. Find%ify= 2 2. 96 ~74 Solution Use implicit differentiation and the quotient rule. dz] d ;L2 MCI}? + y?) — 1:2 (21: + 2$y2 i 2x2y% E _ E (352 + y2> _ (x2 + y2)2 _ ($2 + y2>2 dy l + 22234 _ 21-1/2 dm ($2 + y2)2 _ ($2 + y2)2 dy _ 2333/2 _ 2$y2 _ .. 2 2 2 2 da: (352 +y2)2 (1+ (33312;?) (:16 +1} ) +213 y 12. Let f be a differentiable function with domain 3? that has xiintercepts at it : —2, LB 2 0 and a: = 1. Show that f has at least two critical points. J USTIFY YOUR ANSWER. Solution Since f(—2) = f(0) = 07 apply Rolle’s Theorem to the differentiable (hence continuous) function f on the closed interval [—2, 0]: there is at least one number 0 E (—2, 0) with f’(c) : 0. Since f (0) = f (1) : 07 apply Rolle’s Theorem t0 the differentiable (hence continuous) function f on the closed interval [0, 1]: there is at least one number 6 6 (0,1) with f’(e) = 0. Thus 0 and e are two distinct critical points of f. 13. Assume that f”(:1:)= 12x,f’(1): 0 and f(1) : 2. Find Solution Note D(6a:2) = 123:. By a corollary of the MVT, there is a number C with f’(a:) = 62:2 + C. Then 0 = f’(1)= 6(1)2 + C = 6+0. Hence 0 = *6 and f’(;z:) = 6$2 * 6. Note D(2x3 * 6x) = 6x2 7 6. By a corollary of the MVT, there is a number B with : 22:3 — 63: + B. Then 2:f(1) =2(1)3 76(1)+B=B74. Hence B = 6 and = 23:3 — 62: + 6. 14. Find and identify the crictical points of = 332/3 — 2211/3. Solution 1/3 1/3 f,(m):2m_1/3#21m_2/3: 2 2 _2:r. 2 22m —1) 3 3 3ml/3 3m2/3 3322/3 3332/3 Bag/3 Thus f has a critical point at x = 1 where f’(a:) = 07 and f has a critical point at cc = 0 Where f’(w) DNE. Note that the denominator of f’(x) is always greater than or equal to zero. Hence f’($) > 0 for m > 1 while f’ < 0 for :1: < 1. By the First Derivative Test, f does not have a local extremum at a: = 0 while f has a local minimum at a: : 1. 15. Find all the inflection points of f if : — 1)2/3. JUSTIFY YOUR ANSWER. Solution By the chain rule and the quotient rule: 222 (333 _ 01/3 f”(x) : §<w3 — 1>*1/3<3x2) : The numerator of f” is always greater than or equal to zero. Also 3:3 — 1 changes from negative to positive at .r = 1. Hence f changes from concave down to concave up at $ = 1 which is the only inflection point of f. (Note a: = O is not an inflection point of f because f is concave down to the left and to the right of :1: = 0.) 16. Find all the asymptotes of the function 2 2x 1. x _ Solution f has vertical asymptotes at at = —1 and at :1: = 1 where its denominator is zero. Since x3 2 12(332 — 1) + 3:7 _ fix? — 1) :r _ . ' a: f($)— 302—1 +$2—1 _1 V 1:2—1' Note lim ‘ m : 0 and lim $ : 0. Hence f has the oblique asymptote y : :L' on both the left arfioo $2 —]_ wfiioc $2 — and the right. . CC 17. Use any method to find hm ‘ 2 . xflO 8111 :5 Solution 1 1 1 hm ‘ 2 — hm . hm . — (1) hm _ : hm , . xfiO 3111 1‘ x~>0 SlIliL' .rfiO SlIliL‘ 17~>0 sm :1: WHO SIIIIL' Observe that 1 1 I hm . : —M and hm . : +00. x—>0— sma: x—>O+ smx . 1 . 26 Hence hm . DNE7 and hm _ 2 DNE. JCHO smx wHO sm :3 18. The area of a circular puddle is increasing at 2 cm2 per minute. How fast is the radius of the puddle increasing when its area is 9 cm2? Solution Let R be the radius of the puddle in cm, and let A be the area of the puddle in 01112. Let t be time measured in minutes. \Mn mm +nm +Lm+ L114 _ I0 VVK/ (Lib UUlu ULLCLU — IAI- M/n mm nnImA +n {Wm JR m M. /I _ , VVK/ (Lib “OAK/U UU 1111u Vllbli .11. — LI- Slnce A : 7TB2. it follows by implicit differentiation that dA d dB 2:—:—( R2):27rR— dt dt dt Hence dB 1 dt 7rR \MLmn n _ /l _ Nu? "m kn... L) _ 3 A+ +kn+ me JR _ 1 _ 1 nnnnn warmlm vv ilk/11 .J — 11 — Nib . WU LLLWL, .LL — C 11v uuow “111k,q ,u — .j_/ E — .7 C L111 ppj. running. V/I Lbb uJ )Vn UV l ’ 19. Find the minimum and maximum values (if any) of the function f = 7 1| on the interval (’21 Solution Note = 3:3 — 1 for 1 S x S 2 and = —(a:3 — 1) for —2 < a: < 1. Hence f’(;z:) = 3:172 for *2 < at < 1 and f’(m) = 73x2 for 1 < :c < 2. Thus f has critical points at :I: = 0 Where f’(;13) is zero and at w = 1 where f’ DNE. To find the minimum and maximum values of we Ammn‘nn II n vvv:‘-L\ 11/ \ ’7 n Am/JWA:V\" I‘m/J 12m fl~.\ i CUJJJLJGLC J \\ U K) J \\ (I I OJ Cllqulllb Gallu 111.114» J \le — M «72 u‘r Thus f has its minmum value of 0 at m = 1 and has no maximum value. 1):) 4. Lkn .0:ka b buc 113:1“; Q3 n 1am. nmmnzva. 11C 1610 Unpuuib. 100 20. Evaluate :09 — 416 + 10). k=0 Solution 100 100 100 100 Z(k2—4k+10) = 218—421“th k=0 k=0 k=0 k=0 (100)(101)201) (100)(101) — 6 4 2 : (10)(101) [I (50)(101)(67) 2 (200)(101) + (10)(101) 2 (101)(10) [(5)(67) 2 20 + 1] = (1010)(316) 2 319,160 21. Let f be a continuous function with domain [0,1]. Let P7,, be the regular partition of [0,1] into n subintervals. Assume 2113 — 4712 + 1 2n3 + 412.2 + 1 L(me) : T and U(me) : Estimate fol f dm with error loss than .001. Write your answer in terms of fractions. Do not simplify the arithmetic. J USTIFY YOUR ANSWER. Solution Recall that I l 1 . , 1 272324712121 271324772421 1 4n3+2 2 6713 ' 6713 2 6713 _ 2713 +1 _ 2 I 1 _ 1 1 ‘ 6713 ‘ 6 ‘ 6113 ‘ 3 6713 approximates f div with error less than 1MP f) MP W 1 2n3+4712+1 2713—4712—1—1 1 8n? 2 — n — n : : : 2 7 l 2 6713 6713 2 6713 311 “In mm"; 1m “IMAM M "4111 i / 1 ,m m \ 2000 _ 1212122 whim M _ 525217 'T‘an 1‘1 1'/m\ ,1". N l | 1 WI: want bu LllUUDC lb Wlbll .h \ 1mm U1 lb / s — UUUq- 1 1x1: I — UUI. 111171 In [Nu] U N o T wig/3mg QH/ LUUU U Q UU " \ ’ I) U\UU‘) with error less than .001. 22. Let f be a continuous function on the interval [(1,1)]. Let P be a parition of the interval [(1,1)]. Consider the three numbers 1'37 I n/ \ -. 7211—. n\ -.— 11—. n I 1*erer (/1le Llf’T) J \ / 7 \ 7-1 l’ \ 7-1 Ja Which is the smallest of these numbers? Which is the largest of these numbers? n 1 Pb I-/ \ 1 /11/h I-\ 11 1/h /-\- .1 11 1 (‘11 1 171/11 I-\ 301111 on , j I 11331 x 1111’. 11, nence 1111’. Ti 18 the smanesri or these numbers ano 1/11”. I) ()(7, v/ \ / —‘ \ 7v/ / \ 7v/ / \ 7I/ / 23. Find the average value of = a: cosa: on the interval [271371]. JUSTIFY YOUR ANSWER. Solution By definition of the average value H of mcos .12: 1 7r 1 W H:—/ xcosxd$:—/ xcosw (117:0 712(271) Hr 277 7” because : :ccosa: is an odd function. 1 24. Evaluate / f’(g(x))g’(x) dm when 2 cos 77:3 and 9(110) : x2. 0 Solution By the Second Fundamental Theorem of Calculus: 1 1 / f’<g<x>>g’<x> doc 2 f % mm») due 2 f(g(w))lt 2 mun—119(0)) 2 cow—case 2 21— 1 2 22 0 0 .— -—,. - I A" kind — g \ Solution By the First Fundamental Theorem of Calculus and the chain rule: 1 , no \ . / m/l- \ . A . ,— rI / I \ r! I I v \ r1 1 arncln , /:c W I I _, , i ,LLl _ W I I _, , i ,u I _ A, /,,,_\ W I /,,,_\ _ , /_ _ w W v e I I rtlbhllll/ (LII I — — I — I aimint (llll I — —Iru(,sin \/.I,I— I\/.I,I — —ru(,sin v.1, _ — — _ Am\ I _ I Am l I I \ v I,” \' / v r) m I) m cm \\.l,V/;r ,/ mu \\ .IV /I we uvw uvw PART II 1. [15 Marks] Use any method to evaluate each of these limits. JUSTIFY YOUR ANSWERS. . arctan :c (a) hm — :c—>oo l + arctanm Solution Recall that y = arctanx has the line 3/ : 7T / 2 as a horizontal asymptote on the right. Take the limit as 1' approaches 00: , arctan at 7T / 2 7T hm aHoo l+arctanzc : 1+7r/2 : 2+7? (b) hm a: — arctana: 90—>0 a: — sin$ Oiarctan 0 Solution Note : %. Hence we can apply L’Hopital’s Rule: ()—sin 0 hm a: — arctana: I hm D(sc — arctan :13) _ hm l — l/(l + :02). m—>0 a: — sma: arr—>0 D(x — sm x) m—>0 l — cosx Substitute cc : 0 into the latter fraction: % : %. Therefore, we can apply L’Hopital’s Rule again: , x—arctanx , D[1—1/(l+$2)] _ 2a:/(l+:c2)2 hm —. = hm — = hm .—. wfi0 a: * smm aceO D(l * cos :5) 1H0 sma: Substitute a: : 0 into the latter fraction: : %. Hence we apply L’Hopital’s Rule a third time: , a: — arctana: _ D [217/(1 + x2)2] _ [2(1 + 1:2)2 — 2w(2)(l + :02)(2:c))] /(1 + :02)4 2/1 hm _ — hm . — hm — — 2 .77—>0 a: — s1nx m—>0 D(sma:) m—m cos LE 1 2. [10 Marks] An object moves along the 3: axis with position :1: : 8(15) : 3135 i 5133 + 4 meters at time 75 seconds for all t E 3%. Describe the motion of this object for all t E 3%. In particular, state where the object moves right, moves left, stops, slows down and speeds up. You may give your answer in terms of times only, i.e. you do not need to compute the positions of the object at these times. Solution Differentiating: not) = 8’(t) : 1564 7 15t2 = 151%? 7 1) = 15152057 1)(t +1). a(t) _ 1/(25) _ 60753 302: a 6025(252 1/2) s 601505 1/\/§)(t + 1/\/§). The signs of 2105) and a(t) are given in the diagram below. - - - - - - - -- 0+++0—---—0++++++ signum 4—4—4—n , 7 7 7 7 7 , ,,I1 7 , , 7 7 7 7 ,,H—l——l——l— CIITI’IIY/lfi u “5” m, ’ t right _1 left, 1 left 0 left 1 left I right slows speeds *2 slows speeds “2 slows speeds nO-nun .... n‘nun a... n‘Aun H swps up stops up nLulJo up It follows that the object moves right, slowing down, until it stops at t = —1. Then it moves left, speeding up, until t : 71/ Then it continues moving left, slowing down, until it stops at t = 0. Then it continues moving left, speeding up, until t : 1/ Then it continues moving left, slowing down, until it stops at t = 1. Then it moves right, speeding up. See the diagram below. speeds up > > F] Stops ‘ speeds up I : 7— : - I «2 slows > stops ’X 3. [10 Marks] A straight east west railroad track crosses a straight north south highway on a level plain at a crossing. A passenger in a car driving north at 100 km/ hr towards this crossing watches a train moving west at 150 km/ hr away from this crossing. How fast is the head of the passenger turning when both the car and the train are 500 meters from the crossing? Solution Place the origin of our coordinate system where the railroad tracks cross the highway. As usual, the xiaxis points west while the yiaxis points north both with units measured in km. Let x denote the position of train, and let 3/ denote the position of the car. Let t denote time measured in hours. We are given: dm dy — 2 —150 — 2 +100 dt dt Let 9, measured in radians, denote the counterclockwise angle from the yiaxis to the passenger’s line of n-‘INL" XXL-x nv-n (\nlrnr] +n Hvu—J +lnn "nlnn ml: i6 “anm m _ n1 _ n K l\lr\‘>r\ l'v-rwv‘ +Ln 1*]:nrr'wnvv1 Lnlnn that 015111,” VVC 011C aDnCU DU 11 1U U116 VCLIUC UL VVllCil ab ‘_ y ‘_ U.U. LVUUC LLULLL ULLC Ula/E‘la/lll UCLUVV Ullalj ‘y y tan¢9 : — : —. —m :r: Differentiate this equation with respect to t: 1(tan6?) : d(y) dt E E d , sec, 6cm _ md—g — 14% : $(100)— y(—150) dt 2:2 r2 When at : y : #05, we have tané’ : : 1. Hence (9 : % radians. Then secg 7r d6 _ (0.5)(100) + (0.5)(150) 4 dt (0.5)2 ‘ d6 50 + 75 125 E _ 0.25 _ fl _ 500 d—0 — 250 d' /h dt — ra rans r y <— train l C A. l ’X \\ '1‘ 3 \\ J \\ . \ ’yi \ \ l \\ .i l 1 \9l 4. [10 Marks] Find the lower Riemann sum L(Pn, f) Where f = 277333 and P7,, is the regular partition of the interval [0, 3] into n subintervals. Write your answer in sigma notation. DO NOT SIMPLIFY YOUR ANSWER. c 0 th unction f is decreasing for 0 S a: S 3. Hence on each subinterval [:ck_1, 3%], n . . . . . . S . m t. t . . . Anr l < I?- < (n (1 Tpr i " 'mi'm 1 xmi1 p T¢ 51+ hp rich pndnmnr 7-1, T’i’n \ w .t _ v _ t, M W- M.-. J c M, “w “MMMMMM. WWW wk cw My “0-- ouurw-” WM mums t" — _k k. — . 7L n1 . 1 . 1 ~ 1.1 r‘ 1 1 ~ . 1 2 m1 i. 0139 Lin ["9 “710.13” oak _ Tkil OI eacn SL Dull—GIVE! 13 i _L_J3fl n n aux A n i' flux 3'] ,A n .— A_.3-1 A /n m fi\‘[/4*\/ ,, \ fi\\l~/é-‘€\Ié fi\‘inn llé'k‘:\iié fi\‘ir\r7 “K‘é Lifm.ll: ) Iii/LiillZA—JIIZA—1i: ) I lfllfi: ) ILi—I—‘I IH: ) IAI_ A W ‘ “d/ Lgd‘m" " W " L4” \ an Inn Lgi \m I Im LAI and m V ‘ V ‘ \ Il/ / lb V ‘ I \ ID / I II/ V ‘ I ID I ID [€21 K721 ‘ ’ K721 L ‘ ’ J [(221 "' T 5. [10 Marks] Find the area A between the curve 3; = 3:3 — 23:2 —I— a: + 3 and the line y = 93: —I— 3. NOTE: THE PROBLEM ON THE EXAM WAS STATED INCORRECTLY. THEREFORE IT IS BEING MARKED VERY LENIENTLY. Solution This curve and this line intersection Where 3:3 — 23:2 + a: + 3 = y = 937 + 3, i.e. Where 0—333 23:2 837—513(372 2x 8)—m(at 4)(wi2). Thus they intersect at a: = —2, at a: : 0 and at .T : 4. Then I" i i I' iI/3 n2n n\ In n\1i Il/fj n2n n\ /n.n\1 A : ill {1” ~41" +3I+j)—t3§r+j)al‘]+]ll KT _1T +JI+j}—|\UT+3}HQZ |J72 I IJV ._I'\ 1 1 _/l I" I I 17 l 3 n 2 n 1 I ll .3 n 2 n 1 : III 9." —A§U _DIQIT]+]I 93 _1T _D£E[lril" I‘d/7’) | I“! .4 I I V n | i .Ai . 1 r) i" | i1 5) |“‘| | / 1fl \| | 100 i A A Q r oi I ll /1 A Q r all | I iU ,\I IA LAO .A — _'1‘7.__rr”_{rr“i ill—m’__’r"_{m“il— |._l|_i_ ._|‘}||_L|h|._ ._|>< AW n‘” “WI i ' i AW n‘“ “W i | | l*' n *“ll ' I"JL n ‘V 4 .3 i ,‘I i4 .3 Inl | \ .3 II | .3 i—A] II IU] I \ /| I lRi i Rni R 5252 on U] _ I u I u UV uu n : i I—i—i | : —+— : : .32 ’)i ' i ‘)| Q Q ’) U] I U] U U U 332/3 6. [20 Marks] Consider the function : 1—2 — 27 (r\ 11:“JL r 3. \dl [flu but: 11 I A Solution f = 0 Where the numerator of f is zero. Thus f has one xiintercept at a: = 0. The y intercept of f occurs at y = f(0) = 0. (L\ 1—5:.AA1L-1ka “Nimrnj rannulrnlfln ff .L' \U) I‘ 11 U LLLU V51 blLd/l dbyl lprbUb U1 vi Solution The vertical asymptotes of f occur Where the denominator of f is zero7 i.e. at at : #1 and at m = 1. {0‘1 Finri f‘hn lnn‘r-ivnn 9| acxrmhfnfoc rd? 1? \c, l “N we mummy i WJMWWW w J Solution Note that ‘ O . 172/3 . $43 hm —2 : 0 and hm —9 : 0. wflool—x xfiiool_lr- Hence f has the miaxis as a horizontal asymptote on both the left and the right. 2 2+42: > 0 3x1/3(l 73:2)2 — Note 2 + 43:2 Z 0 and (l — 3:2)2 2 0 for all :6. Thus f is decreasing for :c g 0 and increasing for m 2 0. I \ _ . In] I‘nr‘qfo and \V, uuuwuu w u Solution Note fl is never zero. Thus the only critical point of f occurs at a: = 0 Where f’ does not exist. By the First Derivative Test, f has a local minimum at a: = 0. _.1,A~ A .i H _ “TWA L, kn» in 1U V61 blbdul lid) 1356 Mb. Solution Note ’_(0) : —00 and = +00. Therefore has a vertical cusp at 17 = O. Ila“ q‘ nfnl’] ‘hn (TY'Qh‘ nt' 1? ‘0, WWW“ we DWI,“ w J . Solution ...
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This note was uploaded on 12/22/2010 for the course MATHEMATIC Math 1300 taught by Professor Kochman during the Fall '07 term at York University.

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SFE - SOLUTIONS TO FINAL EXAM PART I (5 marks each) 1. Let...

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