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101-2007&2008-3-F10-August2008

101-2007&2008-3-F10-August2008 - Kuwait University...

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Unformatted text preview: Kuwait University Math-101 August 02, 2008 Math. 8: Comp. Sci. Dept. Final Exam Time: 2 hours Calculators and Mobile Phones are NOT allowed. Each question counts 4 points. 1) Use the definition of the limit to show that lim (4:: — 5) = *3. 35—»1 2 % + 3 if I g 2 , _ 2) Let fix) = $2 + Agingzw22 if a: > 2 . Find the constant A such that f($) is a —4 - continuous at 5': = 2. 3) Use the definition of the derivative to find f’(1), where f(.1:) = .7: 4— \/E w 1. 4) Find the sit-coordinate of the point where the tangent line to the curve 3; = (21—1)1/3 +592+7 is vertical. n+3 5) Let HI) = / t(5—t) alt, a: E R. Show that fix) has a maximum value at 3 : 1. - I 6) The slope of a curve y : fix) is given by 771(55) z 2:): +sina‘ + 1. Find f(a:) knowing that this curve passes through the point 13(0, 1). 7) Evaluate the following integrals: tan6+7 2 a) /< cost? ) d6, b 2--—‘T‘2 d ) jg) v2m3+9 $— 8) Set up an integral that can be used to find the area of the region bounded by the J's-axis and the curve 1; z 51:3 — 9:5. 9) Set up an integral for the volume of the solid obtained when the region bounded by y = :52 + 3 and y = 4a: is revolved about: a) y-axis, b) y 2 —1. 10) Find the average value of fins) = 3\/9_: on [0.4], and determine the number c satisfying the Mean Value Theorem for Integrals. . i E \ .9 v Yuma—LCM / Fined. 5mm Caz/08mg W \. \eJc E>obacg>wenJ \wa_'-:-3+—5\z_ (=2? u\x_‘m\¢_e => \x_\'.1_\ /_ e/q. \eJr a; age/q, V-e.>.o ‘3 S>o€.t. UL. \ 7c--‘1-,_\1_<5 :2) \uir‘5+‘$\£_€ :tp um una— ‘5 - x—Ir-‘IL 2-“ Kim QCX1=UM EE+3=CA+3 veal“ 72—71- 1- \Ln’\ ‘E—C KW \Ju-n 7. (A =3KACXJL‘1 D. "—- 7C + ___—..____ .: LE-Jr --— ' 7‘ “'51-“ 7: +1? ("x-13 c 7: +13, H “US, (AA-'5 :‘ L»+% :2) Q:‘+I%- b. ‘€‘cn = Um ‘Ech—Em = UM m-H‘i—z _ Kim Lx-Ll *fi_c-,(_1‘)-fi X-Jw K—l ‘ x—wn 7-H. “14-“ 7<—\ bun—m fiznsxa-u : um 70—— ux—eu —-X -: him ————_.__... x—ap-x (GK—n Qx-l—S’ffi x._-.-\ \X-flk‘x-‘L—fi] - (x-an—L+\ _1_, ': \A. —...__ — _“___, _. 3 flW—lxcxankx._'2.—G~\ "' '1 ‘ ,1 mil-5 L». Fun: ”ii—£13441 +17: V?” Cm): Ukra‘lq. Um 63mm :00. ‘fi-‘a-‘h. 7?: Curve. Was r; uefl-LCmQ \anczemk Qme x:‘f1... 5- ‘%‘gm = Qc+33\’_vs-cx+n] .. xvi—9m , - 457:; 6. ?\cx\= 0 © x =\ mack ?\Cx1=-640.W5,9—L~7M uH-GX. FR“ : be+fimx+ \de: x12. (33K +><+c. E’cm:\=) \-:.—\_+C. 2.9 ill-:7. Qua QCXI: “7&1— CDSX+X+1. 1-- OJ SLM\IO\:9: SKK'fimS-{kfi-wfl'é C19 " C9361 — #LKQACQ—Fl‘} +6. u fine '3.- cl::5ec1;o\9 1'5 2‘5 ‘93 — :1 -‘—- KL\ 2: E-- 351:3d +0) U: ‘1:3+5 ‘6 55%: 7 a 3 oLLA- 6):.“ch <3 '5. a. “345% =0 2—; X: 43,033. (A: Sk-x'tcbxmfidx +S(Qfl'><-7’+%X\d> '3- a s. «I’m-acux =3 21—HKA-"5co :> 7C:\)_'33 cm Ziagdfixtbtx— Lye—E'sN'SdX- m v = ‘35 w [\wx+\? .. Lxh—m 101x. 9% 311 LP \O-%au:iq§35kdX2-§L—k-33—7K\O:L+ 3E:k+=>C=-L£3- CO ...
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