101-2006&2007-1-M10-November2006

101-2006&2007-1-M10-November2006 - Kuwait...

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Unformatted text preview: Kuwait University Math 101 Date: November 9, 2006 Dept. of Math. 8.: Comp. Sci. First Exam Duration: 90 minutes Calculators, mobile phones, pagers and all other mobile communication equipment are not allowed Answer the following questions: 1. Evaluate the following limits, if they exist: (a) lim[tan2($ % 1)]2 31—41 $2 “— 1 (b) lim (V2932 + % V23)? — 5) (2 pts.) IE—“OO 2. Find the vertical and horizontal asymptotes, if any: for the graph of f($):lJf—1l(l—2CE) 4 t. 2524—55—2 (pg) 3. Find the :c—coordinates of the points at which the function f is discontinuous, where 3:3 + 8 Classify the types of discontinuity of f as removable, jump, or infinite. (4 pts.) 4, (a) State The Intermediate Value Theorem. (1 pt.) (b) Show that the equation .1" — cos (7m) 2 0 has a real solution. (3 pts.) 5. Use the definition of the derivative to find f’ , where f = 35 + (3 pts.) , 2 , dy 7r 6. (a) Let y = x/3u2 + on + 3 and u 2 2 — sec :5 a COtLE.F1Hd E— at 3: = Z (3 pts.) I (b) Find 3—3:, where y : [cos (:1:2 — UP. (3 pts.) .9 99 Math 1 0 1 First Exam Kuwait University Dept. of Math. 8c Comp. Sci. Date: November 9, 2006 Answers Key 1. 2. . [ten2(3:—1)]"2 _ tan2(a:—1) 2 , tan2(3:—1) 2 ——-w--— 2 41 ————«— : ——+_ = (“$13 (z—1)2 Mas—1) 45311 2(1— 1) V2352 + 3 + V2552 — 5 (2:02 + 3) — (23:2 — 5) b 11m \/2$2+3—\/21"2—5 X—-w—— = 11mm_-— ()w*w( ) V2562 + 3 + v2$2 — 5 remix/23:2 + 3 + V2322 — 5 :E . _ . 1w—11(1—2$)_ _ _ mhrjnmflm) — 1113100 (I _1)(I + 2) — :> y * —2 & y — 2 are HA for the graph of f. lim f(:c) = iim ICC—“M =z> is VA for the raph off We we: (a: — 1) (13+ 2} ' g ' lim f 2 => f has an infinite discontinuity at m = 3. zfiai I131121” (x) — : f has a removable discontinuity at x 2. (b) f (:5) : :5 ~ eos(7r.1:] is continuous on [0,1], f (O) ——1 < O and f (1) I 2 > O. From The Intermediate Value Theorem, there is at least one c 6 (0,1) such that f (c) : O.Thue, c is a solution of the equation f (m) = D. , . [(x+h)+ x+h~[az+\/§] , h+\/a:+h—\/:E 5. f 2 11m = lim—_ 1H0 h hAO h ha h i 1 :hmEHm—ME X____\/$++\/E=1+hm I+h I 11+ h—>0h mo h \/:c+h+\/E ’HOh (\/m+h+\/E) 2%; dy 6u+5 dy 1 6. 1 =—1,—— =—-——n~— _ =__ d E = —25ee2:ntanz — csczccgc —u = —6 dz: d1: 3:: dy i dig du __ 1 fl dz 1::1 fi- du uz—l X d3 132-? u 2 x ( —- 4 (b) y’ = 5 [cos (32 — 1)]4 [—2z sin (x2 — 1)] . O. 99 ...
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101-2006&amp;amp;2007-1-M10-November2006 - Kuwait...

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