Answer Key
International Practice Exam MCQ Answer Key
-
University:
Lynn University -
Course:
1202320 | Calculus BC Academic year:
2023
-
Views:
219
Pages:
7
Author:
Gudrunemend
Name:
АРе Calculuв ВС
Anвwer Sheet
for :Мultiple-Choice Section
No.
1
,д
2
с
76
4
'('\
. .д..
6
1Л
81
В2
83
з
б
7
8
9
10
11
.-
No.
Answer
12
13
14
15
16
17
18
19
20
21
22
23
;-;:,,
с..
А3
,1..1.
84
85
86
87
88
89
90
1)
с
[)
r.z)
Г)
R
7
л
(__
с
с
28
f?�
30
-
?,
)
)
'f)
29
во
А-
24
25
26
27
77
78
LL
с
i2,
.,
-
Answer
с__
i1
С'
п,
�
[)
а
(.,1...
'<
Е
с..
!?.
f,.
�
·р, АР ® CALCULUS ВС
2017 SCORING GUIDELINES
Question 1
(а) ,/(х'(2))2 + (у'(2))2 = 3.272461
2
:
{ 1 : expression for specd
1 : answer with units
The speed ofthe particle at tiine t = 2 seconds is 3.272 meters per
seco11d.
2
2
2
2
(Ь) s(t) = ,/(х'(1)) + (у'(1)) = ,/(2cos(2!)) + (2t- 1)
2
·{ 1 : conside1-s s'(4)
• 1 : answer with reaso11
2
_ { 1 : iпtegraJ
• 1 : ans�1er
s'( 4} = 2.16265
Since s'( 4) > О, the speed ofthe particle is incгeasiпg at time 1 = 4.
(с)
s:
2
2
,/(x'(r)) + (y'(t)) dt = 22.381767
Tl1e total distat1ce ihe pa11icle travels ove,· the tiшe interval
О ;;; t ;;; 5 seconds is 22.3 82 (or 22.381) inete1·s.
(d) x(IO) =х(8)+х'(8)·2 =sin 16+х'(8)-2 = -4.118541
2
у(\О) = у(8)+ у'(8)-2 = (8 -8)+ у'(8}·2 = 86
The position of the particle at time t = \ О seconds is (-4.119, 86)
(01· (-4.118, 86)).
3:
{ 1 : uses position at 1 = 8
1 : uses velocity at t = 8
J : position at 1 = 10 АР Ф CALCULUS вс
2017 SCORING GUIDELINES
Ouestion 2
(а) Jr4.s a(t) dt = 66.532128
o
2:
j
1 : integral
1 : ans\ver
At time t = 4.5, tank А contains 66.532 liters of,vater.
(Ь) a(k) = 20.5 � k = 0.892040
J: (20.5 - a(r)) dt = 10.599191
3 :
At tinie 1 = k, the difference in the amounts ofwate,· in the tanks is
10.599 liters.
[2.4\б
(с) Jo
s:
12.416
b(I) dt = Jfk
o Ь(!) dt + k
b(I) dt
2:
{ 1 : sets a(k) = 20.5
1 : integral
1 : answer
{
1 :
J
2.416
k
a(t) dr
1 : a11swer
b(t) dt = 20.5 • k = 18.286826
•
2.416
On k < t < 2.416, tank А receives 1k
a(t) dt = 44.497051 lite1,; of
waler, wblch is 14.470 mo,·e lite1·s ofwater than tank В.
J2-416
Тherefore, k
2.416
b(t)dt+ k
Jr;
o
J
2.4l6
b(t) dt = ;
J
a(r) dt -14.470 = 30.027051.
b(1)d1=48.3l3876
Attime 1 = 2.416, tank В contains48.3l4(or48.ЗJЗ) litersof,vaieг.
(d) и/(3.5)- а'(З.5) = -1.14298 < О
The diffe1,ance w(r) - a(t) is decreasing at r = 3.5.
2·
•
{ 1 : w'(З.5) - а'(З.5) < О
1 : conclt,sion AP CALCULUS ВС
2017 SCORING GUIDELINES
Question З
(а) g(5) - g(-5) = 12 - (,r + 7) = 5 - ,..
5 - (-5)
10
10
3
(Ь) g'(x) = f(x)
l : answer
: { 1 : difference quotient
2: answer
g'(З) = 1(3) = 4
The instan!aneous i-ate of change of g at х = 3 is 4.
(с) The graph of g is concave up оп -5 < х < -2 and О < х < 3,
because g'(x) = f(x) is increasing оп these intervals.
2 : intervals with justification
(d) g'(x) = /(х) is defined at all х with -5 < х < 5.
g'(x) = f(x) = О at х = -2 and х = \.
Therefore, g has critical points at х = -2 and х = \.
g has neither а local inaxiinum по,· а local шiniшum at х = -2
because g' does not change sign there.
g has а Iooal minimшn at х = l because g' changes from negative to
positive there.
3
:
1
1 : considers /(х) = О
1 : critica\ poin!s at
х = -2 and х = 1
1 : answers with justifications АР ® CALCULUS ВС
2017 SCORING GUIDELINES
Question 4
(а)
f f'(x)dx"" 2·3.5+2•0.8+2·5.8 = 20.2
/(6) - /(О)= J: f'(x)
6
f'(x) dx"' 20 + 20.2 = 40.2
j l : шidpoint sшn
1 : Fundamental Theorem
3
ofCalcL1lus
1 : answer
J: f'(x) dx S: б • 7 = 42.
2
0
/(6) = /(О)+
s:
(Ь) Since f'(x) s 7,
dx
1(6) - f(0) s 42 � /(6) s: 20 + 42 = 62
:
. ( 1 : integral bound
• 1 : answer with reasoning
Thei·efore, the actual value of f( 6) could not Ье 70.
(с)
J: f''(x) dx = f'(4)- f'(2) = 1.7 - 2 = -0.3
(d) lim (f(x) - 20е') = О
.х➔О
lim (0.5/(х) - 10) = О
,➔О
Using L'Hospiia]'s Rule,
х
lim f(x) - 2Ое·' = lim f'(x) - 20е = 4 - 20 = _8
0.5(4)
х ➔оО.5/(х)-10 х ➔ о 0.5f'(x)
2:
{ 1 : Ft1ndamentaJ Theore111
of Calculus
1 : answer
2·
•
\ l : L'Hospital's Rule
1 : aнswer AP @ CALCULUSBC
2017 SCORING GUIDELINES
Question 5
2
= 2/ - у -2.,.у
2
х
(Ь)
dy 1
= -1 + _± = О
4
dx (,,у)=(4, 2)
2:
2:
d1y
= 2, 8 - 4 - 16 = _ _!_ < 0
16
4
dx2 1 (х,у)=(4, 2)
Ву the Second De,·ivative Test, g has а гelative 111axi111u111 at х = 4.
(с) :Z:lr_,,.,.)=(1,2) = -1 +
Т
=3
3:
{ I : quotient rule
d1
J. ---2.
• dx2
{ 1 : considers dy \
dx (х,у)=(4, 2)
1 : ans1ver withjustification
!
1 : uses dyl
dx (х. y)=(I. 2)
1
and d yl
dx1 (х, y)=(I. 2)
2 : Taylor polynomial
11,е second-deg1·ee Taylo1· polynomiaJ for h about х = l is
'
'
8
Т2 {х) = 2 + З(х -1) + (х - 1)- = 2 + З(х -1) + 4(х -1)-.
21
(d)lh(l.l)-Als
max
lMx �1. 1
"'
lh (x)I 11 . 1 - 11
З!
3
< 60 __
1_=_!0,__=__l_
- 6 1000 1000 100
2·
•
{ J : form ofthe eiтor boнnd
J : analysis j
АР ® CALCULUS ВС
2017 SCORING GUIDELINES
Question 6
(а)
r00
[-
ь
_ _l_ dx = lim r _,_I_ dx = lim 1 tan- 1
Ь➔оо з
Ь➔00Jз ,.- + 9
Jз х-, + 9
ь
(-зх)lз )
= lim (l1an- 1 (/J_) _ _l_tan- 1 (1)) = tr__ - -"- = -"---ь➔оо З
3
3
6 12 12
(Ь) The function f is continuo11s, positive, and deci-easing оп [3, оо).
Ву the integ1--al test_, since
1 : antide,;vative
3 : { 1 : limit expression
1 : answer
2 : conclusion ,vith conditions
00
f3О()f(x) dx converges, Lf(n)
converges.
11,=]
-OR1
1 r1or п >
о
_ :,.
0 < -2 п +9 <2
п
L-
00 1 converges, the series "'
00
1
Since the series L
L f( 11) =
2
2-11=З n
11=Э
11=3 11 + 9
converges Ьу the compш·ison test.
оо
оО 2 9
1
(с) Considcr the series L
= L 11 � .
,,
)
(
,,:1 е , f(n)
»:J е
lim
11➔СО
е"
<Х)
1
Ву the ratio test, L
converges.
n:I ( е,, • J(n) )
00 ( 1)"
The1·efo1·e, L
converges absolutely.
n•I (е,, • f(n) )
4: !
1
1
1
1
: sets up ratio
: computes limit ofratio
: conc]usion ofratю test
: coпverges absolutely
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