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Sol-161E1-S01 - MA 161/161E EXAM 1 Spring 2001 1 The...

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Unformatted text preview: MA 161/161E EXAM 1 Spring 2001 1. The distance between the points (1,1) and (3,0) is 1 2. The domain of the function f (:13) = —$+———— is ‘/|2x+3|—1 {n+3} 'l 7/0 /> 22w; 7/1 or n+3 6—! #3 2x 7/ '2 0" 1X 5 “’7‘- ...3 X 7, ., \ or ’ x é -'L I", “.5 February 1, 2001 .«5 C.\/§ .(—oo, —2)U(—1,oo) B. (—1,oo) C. (—oo,0) U (1,00) 0D. (~00, 00) E. There is no solution MA 161 /161E EXAM 1 Spring 2001 February 1, 2001 3. Let L be a straight line through (0, 1) and parallel to the line 23: + 3y + 1 = 0. Then an equation for L is _ A. 3x+2y+3=0 flew. “+33 H —o ——> 35 =:1)<-| ———> 3'--X 'L 8 2. ' 2 —° W ”‘l M 3 D 5m—y+1=0 m L L» Mk ’% m Wupwo.) E “511:” a“ L w m. efi'uw‘fith g—H 'éYH) ,..2 ">9 ’ ‘3'” 1 4. tan(sin'1m) A. cos‘ x -1 __ X B. v1+tan§x W93$MX Msme—x-T) .m M X 1 MA )4! 2'1 , D 1:362 9 :1: I—x" @W .-| - __§.__. m than“ x , me»- L MA 161/161E EXAM 1 Spring 2001 February 1, 2001 5. The graph of f is given. Then f(f(0)) : A. —2 «3(0) = - . @ s ,F(,F(03) : {1—2) :: 3 E. undefined 6. A rectangle has area 25 (square inches) and one of its sides has length L (inches). Express the perimeter P (in inches) as a function of L. . L- 74m. 2; : LW A. P=2L+% W W @P=2L+% L C. P=L+§L9 D. P=L+50L2 P =: 2L +2-W. E. P:2L—25L2 N 2§=LW -—» W: T 9 ‘11.“. p:21,+2(2§),= (n+7: MA 161/161E EXAM 1 Spring 2001 February 1, 2001 7. The graph of y— — (1 — Incl) )2 looks like (i=(l’W3L {—0 —x) F’l’XZO {(|+X)L13LX‘0. gu. boflml‘f [”521 19‘5wa 8. In a certain colony of bacteria population triples every 7 hours. Suppose initially there are 1,000 bacteria. After 10 hours the population is 0/7 PR1) : [.000 : 3°. I‘ooo : 3/ - l‘ooo A~ 1:000'37 P(7) : 3 . (‘ooo = 3I - I‘M) ‘— 31—, - l‘ooo 13.1,000-310 .— PORK 5' 3:3. "Doc :31\ [loop = gulf. “0.0 C. 1,000-370 _ 3 _ 'Jl‘l _ 7 mm: 2‘34",“ ~3 ~ "0” w 31’0”“ 3“” - ,n -- pm 37" two A H—‘_~——n—Q—“———\~—-——i_.’—u—~__ MA 161/161E EXAM 1 Spring 2001 February 1, 2001 9. The solution(s) of the equation 1n(1nx) = 0 is (are) A. a; = 1 fiéfl X) :0 o .3626 .9 (fix) 5 e C.a;=1ande ——-) AX f; I D. 33:62 ' E. The equation has no solution 10. Given the following graph of f (x), which statement is true? 3! A. ' = B. $139+ f($) = 0 C. lim f (at) does not exist m—-)2‘ D. lim f(m) = 1 and 0 :1:—>2 @ lim f(x) = 1 (I: 32—)2’ MA 161/161E EXAM 1 Spring 2001 February 1, 2001 11. If lim f(:c) = 4, lim 9(1)) 2 —3, and lim 11(35): 0, it follows that lim “gym is :1:—>a w—ML w—ra z—m, h($)2 A. —12 A“ M = 1:2 2 :2: B. o W90 ha):- 0* 0* C. 00 : - 0° ® _00 E. impossible to determine (Nu. an is afequAj an , ’11.: deummew ‘ts aflvmdamfl O Mmék Posmw w-éuu. mmw‘s 6%? i4 Wm M WM I‘vu'fhmi'db %‘ W 7% 8‘}. 5,, aaW4n'a‘kfi ~00. 1-1 0 12. lim 6352—“: C : e '2 1 . A. x—>1 MA 161/161E EXAM 1 Spring 2001 February 1, 2001 13. If £13318“) 2 —2, and lrn;[3r(t) + 2s(t)] = —1, then £13: r(t) = @1 We (3rM+2sze))-=—l 3—1 {791+ C. ~2 ,_ D. 0 .—> 3 ‘ g; r16] + 7";e’;?f 5&1 ’ —‘ E. cannot be determined. ,7 g 533W”) 3» 2+1) = '1 -——7 3 EM r62) = 3 L—W ‘4 /€‘M rlt’l : 1 b9? 14. The graph of y = f (x) is reflected about the y—axis, then translated down 4 units and to the right 3 units, and finally compressed horizontally by a factor of 2. The resulting graph has equation 1 Y‘C‘HLO" aloo‘d' ‘1'“;5 ___> V: {(‘X) A. y=f<— (Em—3))—4 . = — il-zI:——3 4 {wstah dwn‘fwih —-> Vfl'xl’” B y f( (2 )>+ C. y=f(—(2:L'—~3))+4 WW +0 ri5kl’ 3 um’rs @ y=f(-(2w—3))—4 ~42, \j = €Cd(~&_3)) -—-I+ 13- y=f(—(2$+3))—4 Congress Mfikfi‘flg (9'7 M *2, whit/his same. nu guaée WM 19.] ’F‘Ufyfiti -"' t9: :50 43—333'4 = {(——('Lx-3)) ”Ll. 8 ...
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