Mat 243 review exam2 sols(2)

Mat 243 review exam2 sols(2) - MAT 243 REVIEW EXAM 2(1 Fill...

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Unformatted text preview: MAT 243 REVIEW EXAM 2 (1) Fill in the blank in the statements below: (a) A function f : A —> B is one-twone if and only if V5.11, e A g chl ) = 9’04») "—2)" = Kb) (13) A function f : A —} B is not one-to—one if and only if 3 3%: 6A ( E: liki) A XI :f Xe) (c) A function f : A —> B is onto if and only if Vfie C) EKG/H 2-55 J =13.) (d) A function f : A —> B is not onto if and only if 2%6 biXéfli 2&23 ‘3) (m) A sequence is a function whose domain f§ ill 541' CL Viola-N (144': 'Ue “Juana flame“. 2 (n) An arithmetic sequence is a CAM whoee domain {'5 H: 5!} 09— Vick - W 6. rm. CHE (3145. (o) A geometric sequence is a in LX & 2‘11? "0 hffiabwhose domain {3 gfl SJ‘OE Mom —M$344.‘oe (2) Prove that f : Z —) 2; f (n) = 3n — 5 is one-to—one but not onto. Ch“ 9%- . aha-lo ”om ' Assm 3n, —S= 3% "5* Wm n, “if. m ##- QILQA'PIAMa. 84903115. 77”» 3": :— 3“; ”I in; ' 'i N0; Ohb' 0G)" [a ‘5‘ TM ('5' 140 (“pail “1&9“, Pb Sud: +4.2! . I." (t £Ck):3n—S=S. TL: 043 §aMxxA rs. 1% win» 5 he; 5“" ‘chl‘g'l, l (3) Prove that g : R —> HP U {0}; 9(1) = (33 + 3)2 is not one—to-one but it is onto. Mo+m-fom. gun-Jo Md are. TM» .- 1: : "' GP X7131 (gay—('5).— Cale) of G) a“ 015%; 2 m 243 REVIEW ms 2 {4) Letf : IR —> R; f(:r;) = 12. (a) Find m—s, 2)): E0: ‘11 (b) Find f-luo, 1)) = H 0) W"! ') (5) Letf : R —> R; fix) = sins: (21.) Find m—gm); 5%,?) (b) Find r1({0.5}):§ JE+ 192"} £41— + @217] W“ lg M a ma {Kleqmi (6) Let} = 1R: a Ha; Ha“) = li—EJ- (a) Find m—Lm): 2-0 0,5 3% (b) Find f—l{{15}) .2 [80' 953 [Q 4 3!. a. :1 g as x 4 95 (7) Evaluate the sum and simplify as much as possible Show all your work. Calculator an— swers will not be accepted. 30 = Zii2(k+3)2- 2 (2+3) — 2 CM)" (2.0%) ‘G l2: I '2" 2*: (:296+62k+2300_ IQ- 3g3|£+y323l+210l6 k-l 2" 2-1 (““155 + 2.qu +210 #16 - ‘l (8) Evaluate the sum and simplify as much as possible. Show all your work. Calculator an— ‘21—”? swers will not be accepted? ’2' Z. 2{:+1) Z s. i L': = 2" ( ) k ‘15' '2’; I: has '2 ‘ .9— __ 21(2 G'g’l'jé)):2[($)w l—l—g: . i=0 25¢ _?_ -l (9) Find the sum. Show all your work. Calculator answers will not be accepted. I! C (—6)+(—2)+2+6+10+...+442 "’ £3: 2 (‘3') I __‘l 2—6442"! = ”1‘7“" 5‘ '3 I2=O In. :05) + ‘12 (ans 4— filfi’lf “3 (—eDII'erz :12, us =2‘t03‘r MAT 1243 REVIEW EXAM 2 3 (10) Find a recursive definition for the following seuences defined 1) the closed formulas: (a) a" = —3 — 5n. 5 a. nun l (4. so 6...} c‘k-I [ EEK-1M (b) an = (—5)-3“ -‘ J «r a «Au—flew h BEE! (11) Fill in the blank statement below. (a) A function f3") isfi’i'g-O {if 9(1) if and orily if L141 (Xi-54 (0145th C. and [’2 suck“. it) .40 late" at x». (b) A function fix) is big-Q of 9(1‘) if and only if 1124!. “(3" CMMS C “ilk SucL M 10—69??— claim) vP—XHa. (c) A function f(:.-:) is big-6 of g(:c) if and onl_ if {.0065 0038:) we! (109:2. 0:). (c) Mien f(;r) is big-9 of 9(1') then we say that fix) and 9(3) are of “PL! Saw Odd—n (12) Use the definition of "f(:r) is O(g(r))" to Show that flat) = 4534 + 2513 — 3(log(x))4 — 821:2 — 781' + 10023 is C(34). Find witnesses C and is so that |45.1:" + 251?2 — 3(log(:1:))4 — 82:11".2 — 782: + 10023| S C - |:r“] whenx>k. 4 x 4 K \ H§x*t1g{1+éxb8x)?) +6ng1) +C-‘1L80i- [0023‘ .. fix)! LISVHIJJS'VE’I 4- 3\’("I+811KLI “WM HX‘? é: Ioozsex‘ aim/1mm Luz“! m M 449le +lx“( '-= ”L x” 23¢: 1/] 6:2“ ,2 =11 w ct»: Choose a function 9(3) as simple as possible no u e 5m. est order such 1'. at 0(9(I)) x 'f __ . _ 4— m + '3‘ “( lq x! '3 j O( 3 ' p to. --‘t 3 3) ‘4 00“) =- ._ ,( 92.1»2 (2) V" “(2'1" MAT 243 REVLEW' EXAM 2 3 (10) Find a recursive definition for the following sequences defined by the closed formulas: (a)an=-—3—5n ell-g.) 0.292;:77—3/ m .51 (11) Fill in the blank statement below. (a) A function f (1:) is big-0 of g(;r:) if and only if (b) A function f(z) is big-Q of g(:c) if and only if (c) A function f(:i:) is big-9 of 9(35) if and only if (c) When f (1:) is big-8 of 9(3) then we say that f(:::) and g(.L') are of (12) Use the definition of ” f (I) is O(g(:c))” to show that fix) = 453:" + 25352 — 3(log(z))‘i — 32172 — 781? + 10023 is 0(1“). Find witnesses C and I: so that |45$4 + 251-2 — 3(log(:c))" — 82st:2 — 78:: + 10023| 5 C" II‘I when 3: > k. (13) Give a big 0 mtimate of fix) = (21+91nx+5-3$)-(m2+91nx—9x3)+(—9x+5-2I+2x?)-(2+3—1.0001=41m). Choose a function 9(3) as simple as possible with the smallest order such that f(:c) = 081(1)) . 3 fl ' x ‘1- _ ' olx_ EMA" (Zx+qfln¥+s-s‘)- (WWW! “W (W l 1’ 3 N x ~ c. 1 Loan 3. Moszlhx13J= O( wool") 10({5 @W W 0( 2810001”) 3 00-0092: 4 MAT 243 REVIEW' EXAM 2 (14) Give a big-O estimate of the number of operations, where an operation is an addition or a multiplication used in the segment of the algorithm below ‘ . at! 16°F tz=0 dimer-Io) 003951,”; We. on ms: 9‘ __ L fol-i;=10ton2+3né’ (h) Tine-5} l°°l° has 0(2 4-51.. 2. )" for} := 2" to 2"+5n p 0 ‘ I 'MaJ'I‘G“: Each r'l-I’LJ-ws 00'») E 13==t+ij <5— ladd-‘m‘ 'M at) 0(h)‘ k on) oP‘chmHs. ““5 endfor CM ‘6“): GUM-F‘Q‘} end fer ‘HJ- S'LOOP WP oymdwwa 1 40 1’” 6 loop has 004 +3“. ) 2:003) 9 e.‘ FmJath. Eac‘a. (15) Also book pg 216 3, 4 and pg 229 problem 1,2,3,4 (16) Use induction for prove that. 2221““ — 1) = 371.2 + 271 for all positive integers n. cal-L} PUG-”(i (Gh—IFBn‘i-Zw) - I basessi‘l-PI :0.) ;= (jCGEHFB-I‘H-I) M lea! 55:13: £01 “A an.“ 1-4 W“ c' L l u. .C lhdouc} be S “09 43-? M ff;:3(h‘_011 205+”) (*5 +4.4, _________________ P4912. Poe-H)“- ( ‘2 (GE '2.- P700 : 2‘ (62.4), CCnH)—l -C(k+l)*l = 3y; 4-9994—5- k" um: 50m; HM): and. (5‘3, PM: P0») ism ' 1.0..” p51,.“ In @3313: 1. w W)" (”’4”) 4“” le‘f- 5W7 roe W M w, “.2 ii”) . h . ”’lmowflizifiqupch) 4's 1w»:- flat '44 We" M“? {Blow- yochn):= CM+l)-'73 ‘ w w mot.- (mm)!eh'.CM+l) >3“‘-(a+a).>.. 3 ' 3 :3 --————- e \E'ejH/ “H73 61 ”CE-7' ...
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  • Spring '06
  • Callahan
  • Math, Arithmetic progression, Continuous function, Geometric progression, Review Exam, Mohacsy, blank statement

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