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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—xT) .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 +2W. 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 13.1,000310
.— PORK 5' 3:3. "Doc :31\ [loop = gulf. “0.0 C. 1,000370
_ 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($)2
A. —12
A“ M = 1:2 2 :2: B. o
W90 ha): 0* 0* C. 00
:  0° ® _00 E. impossible to determine (Nu. an is afequAj an ,
’11.: deummew ‘ts aﬂvmdamﬂ O Mmék Posmw wéuu. mmw‘s 6%? i4 Wm M WM I‘vu'fhmi'db %‘
W 7% 8‘}. 5,, aaW4n'a‘kﬁ ~00. 11 0
12. lim 6352—“: C : e '2 1 . A. x—>1 MA 161/161E EXAM 1 Spring 2001 February 1, 2001 13. If £13318“) 2 —2, and lrn;[3r(t) + 2s(t)] = —1, then £13: r(t) = @1 We (3rM+2sze))=—l 3—1 {791+ C. ~2
,_ D. 0 .—> 3 ‘ g; r16] + 7";e’;?f 5&1 ’ —‘ E. cannot be determined.
,7 g 533W”) 3» 2+1) = '1
——7 3 EM r62) = 3 L—W
‘4 /€‘M rlt’l : 1 b9? 14. The graph of y = f (x) is reﬂected about the y—axis, then translated down 4 units
and to the right 3 units, and ﬁnally compressed horizontally by a factor of 2. The
resulting graph has equation 1
Y‘C‘HLO" aloo‘d' ‘1'“;5 ___> V: {(‘X) A. y=f<— (Em—3))—4
. = — ilzI:——3 4
{wstah dwn‘fwih —> Vﬂ'xl’” B y f( (2 )>+
C. y=f(—(2:L'—~3))+4
WW +0 ri5kl’ 3 um’rs @ y=f((2w—3))—4
~42, \j = €Cd(~&_3)) —I+ 13 y=f(—(2$+3))—4 Congress Mﬁkﬁ‘ﬂg (9'7 M *2,
whit/his same. nu guaée WM 19.] ’F‘Ufyﬁti
"' t9: :50 43—333'4
= {(——('Lx3)) ”Ll. 8 ...
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