Uploaded by Viveka Mehrotra

AP Calc BC 2019

advertisement
20
Please note: Some of the questions in this former practice exam may
no longer perfectly align with the AP exam. Even though these
questions do not fully represent the 2020 exam, teachers indicate
that imperfectly aligned questions still provide instructional value.
Teachers can consult the Question Bank to determine the degree to
which these questions align to the 2020 Exam.
7 L
F
PP
E
F
E L
LF
L LE
7 L
P L S R LG G E
ROO
R G R
PS S
LR 7 F
S PL G R
GR OR G
P
L O
GP
FRSL
R
L
L
G
L
FO
RRP
L
R O 7R P L
L
F L R
L
P
F
R OG FROO F OO P
L O
L GPL L
LR
G
S
PL
F
ORF LR
L
P
L
L
F
L
L
F
FL F
T
L
F
P
L
GGL LR O GL LE LR L L
LRO LR R
ROO
R G FRS L
SROLFL
GP
O L
PL
LR R
F LF
P FF
R R
F RRO
OO
PR O R FF
RR
R OL
LF
F
7 F
RPP L
G 2 OL
FR
SR
k
20 R
R
LE
L
VW R
R
R W
:
R
R
R
AP Central is the official online home for the AP Program: apcentral.collegeboard.org
URP W H
OF O
p
U FWLFH
F
PL L WU WLR
F
P
F
F
L HOH H
PL
SUR L H E W H ROOH H R U RU
P SUHS U WLR
H F HU UH
SHUPLWWH WR R OR W H P WHUL O
P H FRSLH WR H LW W HLU
W H W L
FO
URRP HWWL R O
R P L W L W H HF ULW R W L H P
WH F HU
R O FROOHFW OO P WHUL O
WHU W HLU PL L WU WLR
HHS
W HP L
HF UH ORF WLR
k
V
P
V
9V
F F
F
LWLR O L WULE WLR L L
LRO WLR R W H ROOH H R U
FRS UL W SROLFLH
P UH OW
L W H WHUPL WLR R U FWLFH
P
FFH
RU R U F RRO
HOO
W H
UHPR O R FFH WR RW HU R OL H
HU LFH
F
W H
H F HU
RPP LW
2 OL H FRUH HSRUW
T
F
P
AP Central is the o cial online home for the AP Program: apcentral.collegeboard.org.
DP ,Q W FW RQ
W G QW Q
K
FW RQ , 0 W S
KR F
FW RQ ,,
0 WS
DP
SRQ
Q
KR F
FW RQ
W RQ
SRQ
KR F
R
W IR WK 0 W S
W RQ
.
RN
QP QW DQG DW RQD
FR Q
G
Q
FR Q :R N K W
W RQ
F SWR
DQG
IR PDQF
DWD
1RW
K S E FDW RQ KR
WK SD Q PE
WKDW DSS D G Q
WK
−
P
ERRN DQG Q WK DFW D
DP
K S E FDW RQ D QRW SD QDW G WR E Q WK SD
k
K R
RD G R
RD G G DQF G DF P QW R DP
DQG WK DFR Q R R D
W G W DG PD N RI WK R
RD G
RWK S RG FW DQG
F PD E W DG PD N RI WK
S FW
R Q
P
RQ WR
FRS
KW G R
RD G PDW D PD E
T
W G RQ Q DW
FR
ERD G R
T
W IR P
7XH GD PRU L
)ULGD PRU L
7RWDO 7LP
RXU D G
1XPE U RI 4X
G S
G
PE
WLRQ
IT
I P I
PL XWH
PD
D
DP
3 UF QW RI 7RWDO 6FRU
:ULWLQJ ,Q WUXP QW 3H FLO UHTXLUHG
7RWDO 7LP
RXU D G
1XPE U RI 4X
PL XWH
WLRQ
3 UF QW RI 7RWDO 6FRU
:ULWLQJ ,Q WUXP QW (LW HU SH FLO RU SH
LW EODFN RU GDUN EOXH L N
1RW For Section II, if students nish
3DUW EHIRUH W H H G RI W H WLPHG
PL XWH
IRU 3DUW
W H FD RW EH L
RUNL
R 3DUW % 6WXGH W PX W DLW X WLO W H
EH L L
RI W H WLPHG
RXU IRU 3DUW %
R H HU GXUL
W H WLPHG SRUWLR IRU
3DUW % WXGH W PD
RUN R W H TXH WLR
L 3DUW
LW RXW W H X H RI D FDOFXODWRU
0D
0D
3DUW
7LP
RXU
1XPE U RI
4X WLRQ
1
3DUW
7LP
PL XWH
3DUW
7LP
PL XWH
1XPE U RI
4X WLRQ
D
D
DS
T
G
1XPE U RI
4X WLRQ
D
D
G
D
3 UF QW RI 6 FWLRQ ,,
6FRU
DS
T
G
3DUW
7LP
D
D
RXU
3 UF QW RI 6 FWLRQ ,,
6FRU
1 PE U RI
VW RQV
1
D
D
D
G
IRU L WULEXWLQJ DP
HFN W DW W H WLWOH R DOO H DP FR HU L
RU
%H XUH WR
distribute the correct exam—AB or BC—to the students. If there are any exam booklets with a di erent title, contact
W H 3 FRRUGL DWRU LPPHGLDWHO
( DP SDFNHW
HU
R WDL HU IRU WXGH W
HHW
( WUD 1R
3 6WXGH W 3DFN
( WUD SH
G D
7 L ERRN
3H FLO
RPH F RRO 6HOI 6WXG
FDOFXODWRU
DUSH HU
LI HHGHG
LW HUD HU
LW EODFN RU GDUN EOXH L N
( WUD SDSHU
6WDSOHU
DUW WHPSODWH
RGH D G
( WUD UDS L
D
( DP
3 ( DP 6HDWL
6F RRO
0D
SH FLO
HOHFWUR LF GH LFH
:DWF
RGH
6L
IRU W H GRRU WR W H WH WL
URRP
( DP L 3UR UH
3 R H RI D NL G DUH SUR LELWHG GXUL
WH W DGPL L WUDWLR L FOXGL EUHDN
3
O
O V
PV
W H
P
6
7,1
32 ,
25 3
VW Q : Q R
W
8 86
1
PV
P Q VW U
W
RROV Q W
QW
W W V
Q
3 UWR 5 R Q
U Q ,VO Q V
5H XODUO
6F HGXOHG ( DP
6WXGH W PX W EH HDWHG
R OH W D
IHHW DSDUW
/DWH 7H WL
6WXGH W PX W EH HDWHG
R OH W D
IHHW DSDUW
( DP
8 86
; 06
PV
P Q VW U
W
RROV 2 WV
W
QW
W W V
Q
3 UWR 5 R
Q W
U Q ,VO Q V
6WXGH W PX W EH HDWHG
R OH W D
IHHW DSDUW
UDS L FDOFXODWRU DUH UHTXLUHG WR D
HU RPH RI W H TXH WLR R W H 3 DOFXOX
( DP %HIRUH WDUWL W H H DP DGPL L WUDWLR PDNH XUH HDF WXGH W D D UDS L
FDOFXODWRU IURP W H DSSUR HG OL W R SD H
RI W H
G D
0D D ,I D
WXGH W GRH RW D H D UDS L FDOFXODWRU IURP W H DSSUR HG OL W RX PD SUR LGH R H
IURP RXU XSSO ,I W H WXGH W GRH RW D W WR X H W H FDOFXODWRU RX SUR LGH RU GRH
RW D W WR X H D FDOFXODWRU DW DOO H RU H PX W D G FRS GDWH D G L W H UHOHD H
WDWHPH W R SD H
RI W H
G D
0D D
XUL W H DGPL L WUDWLR RI 6HFWLR , 3DUW % D G 6HFWLR ,, 3DUW
WXGH W PD D H
R PRUH W D W R UDS L FDOFXODWRU R W HLU GH N
DOFXODWRU PD RW EH DUHG
O O WRU P PRU V R QRW Q
WR E O U E IRU RU IW U W
P 6WXGH W
LW
H OHWW 3DFNDUG
6HULH D G D LR );
UDS L FDOFXODWRU PD X H
FDUG GH L HG IRU X H LW W H H FDOFXODWRU 3URFWRU
RXOG PDNH XUH L IUDUHG SRUW
H OHWW 3DFNDUG DUH RW IDFL HDF RW HU Q
U S Q
O O WRUV Q E V
WR VWRU
W Q O
Q W W SUR WRUV V R O PRQ WRU W W VW
QWV U V Q W U
O O WRUV SSURSU W O
WW PSWV E VW
QWV WR V W
O O WRU WR U PR
P T VW RQV Q RU QV UV IURP W URRP P U V OW Q W
Q OO W RQ RI
3
P V RU V
7 H 3 DOFXOX % ( DP D G W H 3 DOFXOX % ( DP RXOG EH DGPL L WHUHG
LPXOWD HRX O 7 H PD EH DGPL L WHUHG L HSDUDWH URRP RU L W H DPH URRP LI LW L
PRUH FR H LH W
0XOWLSO
RLF
D
D
D
D
D
G
I
7 H H H DP L FOXGH XU H TXH WLR
DGGLWLR WR W H DFWXDO WH W WDNL WLPH
7 H WLPH DOOR HG IRU W H XU H TXH WLR
L L
0DNH XUH RX EH L W H H DP DW W H GH L DWHG WLPH 5HPHPEHU RX PX W FRPSOHWH D
HDWL F DUW IRU W L H DP 6HH SD H
IRU D HDWL F DUW WHPSODWH D G L WUXFWLR
6HH W H
G D
0D D IRU H DP HDWL UHTXLUHPH W SD H
,I R U
Q W U
O UO V
O
P V
,W L 7X GD PRUQLQJ 0D
DQG RX LOO E WDNLQJ LW
DP RU W
3 DOFXOX
DP
UW
,I R U
Q W
OW UQ W
P IRU O W W VW Q V
,W L ULGD PRUQLQJ 0D
DQG RX LOO E WDNLQJ LW U W
DP RU W
3 DOFXOX
DP
3
O
O V
PV
3 DOFXOX
3 DOFXOX
P
,I R U
Q W
P V
Look at your exam packet and con rm that the exam title is “AP Calculus AB.”
Raise your hand if your exam packet contains any title other than “AP Calculus AB,”
DQG , LOO OS RX
,I R U
Q W
P V
Look at your exam packet and con rm that the exam title is “AP Calculus BC.”
Raise your hand if your exam packet contains any title other than “AP Calculus BC,”
DQG , LOO OS RX
,I R U
Q ERW W
P Q
P Q
W V P URRP V
Look at your exam packet and con rm that the exam title is “AP Calculus AB” or
“AP Calculus BC,” depending upon which exam you are taking today. Raise your
DQG LI RXU DP SDFN W FRQWDLQ DQ RW U WLWO DQG , LOO OS RX
Once you con rm that all students have the correct exam, say:
,Q D PRP QW RX
DJU WR DOO RI W
LOO RS Q W
DP SDFN W
RS QLQJ W L SDFN W RX
3 3URJUDPo SROLFL DQG SURF GXU RXWOLQ G LQ W
RX PD QR U PRY W
ULQN UDS IURP W RXW LG RQO RI RXU DP
SDFN W R QRW RS Q W 6 FWLRQ , ERRNO W GR QRW U PRY W
ULQN UDS IURP
the Section II materials. Put the white seals and the shrinkwrapped Section II
ERRNO W D LG
DU IXOO U PRY W
3 DP ODE O IRXQG Q DU W
ERRNO W FRY U 3ODF LW RQ SDJ
RI RXU DQ
U
the top right corner that reads “AP Exam Label.”
WRS O IW RI RXU DP
W RQ W OLJ W EOX ER Q DU
,I WXGH W DFFLGH WDOO SODFH W H H DP ODEHO L W H SDFH IRU W H XPEHU ODEHO RU LFH
HU D DG L H W HP WR OHD H W H ODEHO L SODFH 7 H
RXOG RW WU WR UHPR H W H ODEHO
W HLU H DP FD WLOO EH SURFH HG FRUUHFWO
L W Q FDU IXOO WR DOO P LQ WUXFWLRQ , LOO JLY RX WLP WR FRPSO W DF
W S 3O D ORRN XS DIW U FRPSO WLQJ DF W S 5DL
RXU DQG LI RX DY
DQ TX WLRQ
L H WXGH W H RX
5 DG W
WLPH WR FRPSOHWH HDF
WDW P QW RQ W
WHS
IURQW FRY U RI W
6LJQ RXU QDP DQG ULW WRGD o GDW
1R SULQW RXU IXOO O JDO QDP
7XUQ WR W
EDFN FRY U RI RXU
L H WXGH W D IH
U W
R W PR H R X WLO DOO WXGH W DUH UHDG
6 FWLRQ , ERRNO W
U LQGLFDW G
DP ERRNO W DQG U DG LW FRPSO W O
PL XWH WR UHDG W H H WLUH FR HU
U DQ TX
WLRQ
RX LOO QR WDN W PXOWLSO F RLF SRUWLRQ RI W
DP RX RXOG DY LQ
IURQW RI RX W PXOWLSO F RLF ERRNO W DQG RXU DQ
U
W RX PD Q Y U
GL FX W PXOWLSO F RLF
DP FRQW QW DW DQ WLP LQ DQ IRUP LW DQ RQ
LQFOXGLQJ RXU W DF U DQG RW U WXG QW ,I RX GL FOR W PXOWLSO F RLF
DP FRQW QW W URXJ DQ P DQ RXU 3 DP FRU
LOO E FDQF O G
2S Q RXU DQ
U
W WR SDJ
RX PX W FRPSO W W DQ
U
W X LQJ
D 1R S QFLO RQO 0DUN DOO RI RXU U SRQ
E JLQQLQJ RQ SDJ
RI RXU
answer sheet, one response per question. Completely ll in the circles. If
RX Q G WR UD
GR R FDU IXOO DQG FRPSO W O 1R FU GLW LOO E JLY Q IRU
3
O
O V
PV
P
DQ W LQJ ULWW Q LQ W
DP ERRNO W 6FUDWF SDS U L QRW DOOR G EXW RX PD
X W PDUJLQ RU DQ EODQN SDF LQ W
DP ERRNO W IRU FUDWF
RUN
6 FWLRQ , L GLYLG G LQWR W R SDUW
DF SDUW L WLP G SDUDW O DQG RX
PD
RUN RQ DF SDUW RQO GXULQJ W WLP DOORWW G IRU LW DOFXODWRU DU
QRW DOOR G LQ 3DUW 3O D SXW RXU FDOFXODWRU XQG U RXU F DLU U W U
DQ TX WLRQ
RX DY
RXU IRU 3DUW 3DUW TX WLRQ DU QXPE U G W URXJ
0DUN
RXU U SRQ
IRU W
TX WLRQ RQ SDJ
RI RXU DQ
U
W 2S Q RXU
6 FWLRQ , ERRNO W DQG E JLQ
HFN W DW WXGH W DUH PDUNL W HLU D
D G W DW W H DUH RW ORRNL EH R G 3DUW
D L W RX L PR LWRUL
WXGH W
RUN
IW U
HU L SH FLO R SD H RI W HLU D
HU
7 H OL H RI
DW W H WRS RI HDF SD H
HHW
LOO
PQ W V V
7
U DU
IW U
PLQXW
U PDLQLQJ
PQ W V V
6WRS RUNLQJ RQ 3DUW
DQG WXUQ WR SDJ
LQ RXU 6 FWLRQ , ERRNO W
On that page, you should see an area marked “PLACE SEAL HERE.” Making sure
DOO RI RXU RW U DP PDW ULDO LQFOXGLQJ RXU DQ
U
W DU RXW RI W
D WDN RQ RI RXU DO DQG SU
LW RQ W DW DU D DQG W Q IROG W
DO RY U
W RS Q GJ WR W IURQW FRY U
XU RX GRQoW DO W 3DUW
FWLRQ RI W
ERRNO W RU O W W
DO WRXF DQ W LQJ F SW W PDUN G DU D
IW U OO VW
QWV
V
O
3 UW
V
UDS LQJ FDOFXODWRU DU U TXLU G IRU 3DUW
RX PD J W RXU FDOFXODWRU IURP
XQG U RXU F DLU DQG SODF W P RQ RXU G N 3DUW TX WLRQ DU QXPE U G
W URXJ
ROG RXU DQ
U
W R RQO SDJ
L
R LQJ DQG PDUN RXU
U SRQ
IRU W
TX WLRQ RQ W DW SDJ
RX DY
PLQXW IRU 3DUW
RX
PD E JLQ
HFN W DW WXGH W D H HDOHG W HLU ERRNOHW SURSHUO D G DUH R
RUNL R 3DUW %
7 H ODU H % L D DOWHU DWL
DGHG SDWWHU DW W H WRS RI HDF SD H LOO D L W RX L
PR LWRUL W HLU RUN 3URFWRU
RXOG PDNH XUH W DW WXGH W DUH X L W HLU FDOFXODWRU
DSSURSULDWHO 3URFWRU
RXOG DO R PDNH XUH H OHWW 3DFNDUG FDOFXODWRU L IUDUHG SRUW
DUH RW IDFL HDF RW HU
IW U
PQ W V V
7
U DU
IW U
PLQXW
U PDLQLQJ
PQ W V V
6WRS RUNLQJ DQG WXUQ WR SDJ
RX DY
PLQXW WR DQ
U
Questions 91–94. These are survey questions and will not a ect your score.
Note that each survey question has ve answer options. You may not go back to
RUN RQ DQ RI W
DP TX WLRQ
L H WXGH W DSSUR LPDWHO
3
O
O V
PV
PL XWH WR D
HU W H XU H TXH WLR
P
QV
OR
RXU ERRNO W DQG SXW RXU DQ
U
W RQ RXU G N IDF XS 0DN XU
RX DY RXU 3 QXPE U ODE O DQG DQ 3 DP ODE O RQ SDJ
RI RXU DQ
U
W 6LW TXL WO
LO , FROO FW RXU DQ
U
W
ROOHFW D D
HU HHW IURP HDF WXGH W
3 XPEHU ODEHO D G D
3 ( DP ODEHO
IW U OO QV
UV
WV
E
Q ROO
HFN W DW HDF D
W
HU
HHW D D
V
1R
RX PX W DO RXU 6 FWLRQ , ERRNO W 5 PRY W U PDLQLQJ
LW
DO
IURP W EDFNLQJ DQG SU
RQ RQ DF DU D RI RXU DP ERRNO W FRY U
marked “PLACE SEAL HERE.” Fold each seal over the back cover. When you
have nished, place the booklet on your desk, faceup. I will now collect your
6 FWLRQ , ERRNO W
ROOHFW D 6HFWLR , ERRNOHW IURP HDF
FR HU RI W H HDOHG 6HFWLR , ERRNOHW
WXGH W
7 HUH L D
PL XWH EUHDN EHW HH 6HFWLR
:
U
W RQ , P W U OV
EU N V
Q OO
IRU W
E
HFN W DW HDF
WXGH W D
L
HG W H IUR W
, D G ,,
Q ROO
W
Q
R QW
IRU Q
R
U
3O D OL W Q FDU IXOO WR W
LQ WUXFWLRQ E IRU
WDN D
PLQXW EU DN
OO LW P RX SODF G XQG U RXU F DLU DW W E JLQQLQJ RI W L
DP LQFOXGLQJ
RXU 6WXG QW 3DFN PX W WD W U DQG RX DU QRW S UPLWW G WR RS Q RU
DFF
W P LQ DQ
D
DY RXU ULQN UDSS G 6 FWLRQ ,, SDFN W RQ WRS
RI RXU G N GXULQJ W EU DN RX DU QRW DOOR G WR FRQ XOW W DF U RW U
WXG QW QRW
W WERRN RU DQ RW U U RXUF GXULQJ W EU DN RX PD
QRW PDN S RQ FDOO
QG W W P
DJ
X
RXU FDOFXODWRU F FN PDLO X
D RFLDO Q W RUNLQJ LW RU DFF
DQ O FWURQLF RU FRPPXQLFDWLRQ G YLF
RX
PD QRW O DY W G LJQDW G EU DN DU D 5 P PE U RX PD Q Y U GL FX
W PXOWLSO F RLF
DP FRQW QW LW DQ RQ DQG LI RX GL FOR W FRQW QW
W URXJ DQ P DQ
RXU 3 DP FRU
LOO E FDQF O G U W U
DQ TX WLRQ
RX PD E JLQ RXU EU DN 7
U
IW U W
EU
0D , DY
5
WLQJ LOO U
XP DW
SRQ
N V
Y U RQ o DWW QWLRQ 3ODF
RXU 6WXG QW 3DFN RQ RXU G
N
RX PD QR U PRY W
ULQN UDS IURP W 6 FWLRQ ,, SDFN W EXW GR QRW RS Q
W 6 FWLRQ ,, DP ERRNO W XQWLO RX DU WROG WR GR R
5 DG W EXOO W G WDW P QW RQ W
when you have nished. . . .
IURQW FRY U RI W
DP ERRNO W
RRN XS
1R WDN DQ 3 QXPE U ODE O IURP RXU 6WXG QW 3DFN DQG SODF LW RQ W
DG G ER ,I RX GRQoW DY DQ 3 QXPE U ODE O
ULW RXU 3 QXPE U LQ W
box. Look up when you have nished. . . .
5 DG W
OD W WDW P QW
Using your pen, print the rst, middle, and last initials of your legal name in the
ER
DQG SULQW WRGD o GDW
U LQGLFDW G 7 L FRQ WLWXW
RXU LJQDWXU
DQG RXU DJU P QW WR W
WDW P QW RQ W IURQW FRY U
3
O
O V
PV
P
1R WXUQ WR W EDFN FRY U 8 LQJ RXU S Q FRPSO W ,W P
“Important Identi cation Information.”. . .
5 DG ,W P
U W
U DQ TX
W URXJ
XQG U
WLRQ
,I W L L RXU OD W 3 DP RX PD N S RXU 6WXG QW 3DFN 3ODF LW XQG U RXU
F DLU IRU QR 2W U L LI RX DU WDNLQJ DQ RW U 3 DP W L
DU O DY
RXU 6WXG QW 3DFN RQ RXU G N DQG , LOO FROO FW LW QR
5 DG W U PDLQLQJ LQIRUPDWLRQ RQ W EDFN FRY U RI W
DP ERRNO W SD LQJ
FDU IXO DWW QWLRQ WR W EXOO W G WDW P QW LQ W LQ WUXFWLRQ
R QRW RS Q W
DP ERRNO W RU EU DN W
DO LQ W
DP ERRNO W XQWLO RX DU WROG WR GR R
Look up when you have nished. . . .
ROOHFW W H 6WXGH W 3DFN IURP WXGH W
R DUH WDNL
D
RW HU
3 ( DP W L
HDU
QV
U W
U DQ TX
WLRQ
6 FWLRQ ,, DO R D W R SDUW W DW DU WLP G SDUDW O RX DU U SRQ LEO IRU
SDFLQJ RXU OI DQG PD SURF G IU O IURP RQ TX WLRQ WR W Q W LW LQ
DF SDUW UDS LQJ FDOFXODWRU DU U TXLU G IRU 3DUW
R RX PD N S RXU
FDOFXODWRU RQ RXU G N RX PX W ULW RXU DQ
U LQ W DSSURSULDW SDF
LQ W
DP ERRNO W X LQJ D 1R S QFLO RU D S Q LW EODFN RU GDUN EOX LQN
R QRW EU DN W
DO IRU 3DUW DW W L WLP
U W U DQ TX WLRQ
RX DY
PLQXW WR DQ
WR FRPSO W RXU U SRQ
SDS U RX X
ULW RQO
ƒ
ƒ W
UW
UDL
TX WLRQ LQ 3DUW ,I RX Q
RXU DQG W W WRS RI DF
G PRU SDS U
WUD
W RI
RXU 3 QXPE U
ƒ W
DP WLWO
TX
R QRW ULW
DQG
WLRQ QXPE U RX DU
RXU QDP
2S Q RXU
RUNLQJ RQ
DP ERRNO W DQG E JLQ
HFN W DW WXGH W DUH RUNL R 3DUW R O D G ULWL W HLU D
HU L W HLU H DP
ERRNOHW X L SH FLO RU SH
LW EODFN RU GDUN EOXH L N 7 H SD H IRU W H 3DUW
TXH WLR DUH PDUNHG LW ODU H
RU
DW W H WRS RI HDF SD H WR D L W RX L
PR LWRUL W HLU RUN
IW U
PQ W V V
7
U DU
IW U
PLQXW
U PDLQLQJ LQ 3DUW
PQ W V V
6WRS RUNLQJ RQ 3DUW
DOFXODWRU DU QRW DOOR
RXU FDOFXODWRU XQG U RXU F DLU
G IRU 3DUW
3O D
SXW DOO RI
7XUQ WR SDJ
RX DY
RXU IRU 3DUW
XULQJ W L WLP RX PD JR EDFN WR
3DUW EXW RX PD QRW X
RXU FDOFXODWRU 5 P PE U WR R
RXU RUN DQG
ULW RXU DQ
U WR DF SDUW RI DF SUREO P LQ W DSSURSULDW SDF LQ W
DP ERRNO W U W U DQ TX WLRQ
Using your nger, break open the seals on Part B. Do not peel the seals away
IURP W ERRNO W RX PD JR RQ WR W Q W SDJ DQG E JLQ 3DUW
3
O
O V
PV
P
IW U
7
PQ W V V
U DU
IW U
PLQXW
U PDLQLQJ LQ 3DUW
PQ W V V
6WRS RUNLQJ DQG FOR
RXU
DP ERRNO W 3ODF LW RQ RXU G
N IDF XS
,I D
WXGH W X HG H WUD SDSHU IRU D TXH WLR L W H IUHH UH SR H HFWLR
D HW R H
students staple the extra sheet(s) to the rst page corresponding to that question in their
IUHH UH SR H H DP ERRNOHW
RPSOHWH D , FLGH W 5HSRUW DIWHU W H H DP D G UHWXU
W H H IUHH UH SR H ERRNOHW
LW W H H WUD HHW DWWDF HG L W H , FLGH W 5HSRUW UHWXU
H HORSH HH SD H
RI W H
G D
0D D IRU FRPSOHWH GHWDLO
QV
5 PDLQ LQ RXU
DW
LW RXW WDONLQJ
ROOHFW D 6HFWLR ,, H DP ERRNOHW IURP HDF
LO W
DP PDW ULDO DU FROO FW G
WXGH W
HFN IRU W H IROOR L
ƒ ( DP ERRNOHW IUR W FR HU 7 H WXGH W SODFHG D
D G SUL WHG W HLU L LWLDO D G WRGD
3 XPEHU ODEHO R W H
DGHG ER
GDWH
ƒ Exam booklet back cover: The student completed the “Important Identi cation
, IRUPDWLR
DUHD
: H DOO H DP PDWHULDO D H EHH FROOHFWHG D G DFFRX WHG IRU UHWXU WR WXGH W D
HOHFWUR LF GH LFH RX PD D H FROOHFWHG EHIRUH W H WDUW RI W H H DP
,I R U
Q W U
O UO V
O
P V
RX PD QRW GL FX RU DU W IU
U SRQ
DP FRQW QW LW DQ RQ
XQO
LW L U O D G RQ W
ROO J RDUG
E LW LQ DERXW W R GD
RXU 3
DP FRU U XOW
LOO E DYDLODEO RQOLQ LQ -XO
,I R U
Q W
OW UQ W
P IRU O W W VW Q V
1RQ RI W FRQW QW LQ W L
DP PD Y U E GL FX
G RU DU G LQ DQ
DQ WLP
RXU 3 DP FRU U XOW
LOO E DYDLODEO RQOLQ LQ -XO
,I Q VW
QWV RPSO W
W
3 Q PE U
U
WW
E
QQ Q RI W V
D DW
P V
3O D U P PE U WR WDN RXU 3 QXPE U FDUG LW RX RX LOO Q G W
LQIRUPDWLRQ RQ W L FDUG WR YL
RXU FRU DQG RUG U 3 FRU U SRUWLQJ
UYLF RQOLQ
QV
RX DU QR GL PL
IW U
G
DP 7D N
%H XUH WR L H W H FRPSOHWHG HDWL F DUW WR W H 3 FRRUGL DWRU 6F RRO PX W UHWDL
HDWL F DUW IRU DW OHD W L PR W
X OH W H WDWH RU GL WULFW UHTXLUH W DW W H EH
UHWDL HG IRU D OR HU SHULRG RI WLPH 6F RRO
RXOG RW UHWXU D
HDWL F DUW L W HLU
H DP LSPH W X OH W H DUH UHTXLUHG D SDUW RI D , FLGH W 5HSRUW
,I RX DGPL L WHUHG H DP WR WXGH W
LW DFFRPPRGDWLR
UH LH W H
G D
0D D D G W H
G
IRU L IRUPDWLR DERXW FRPSOHWL
W H 1R WD GDUG GPL L WUDWLR 5HSRUW 1 5 IRUP D G UHWXU L W H H H DP
3
O
O V
PV
P
7 H H DP SURFWRU RXOG FRPSOHWH W H IROOR L WD N LI D NHG WR GR R E W H
3 FRRUGL DWRU 2W HU L H W H 3 FRRUGL DWRU PX W FRPSOHWH W H H WD N
ƒ RPSOHWH D , FLGH W 5HSRUW IRU D
WXGH W
R X HG H WUD SDSHU IRU W H IUHH UH SR
HFWLR , FLGH W 5HSRUW IRUP DUH SUR LGHG L W H FRRUGL DWRU SDFNHW H W LW
W H H DP LSPH W
V IRUPV P VW E RPSO W
W
1R S Q O ,W L
EH W WR FRPSOHWH D L OH , FLGH W 5HSRUW IRU PXOWLSOH WXGH W SHU H DP XEMHFW SHU
DGPL L WUDWLR UH XODU RU ODWH WH WL
D OR D DOO UHTXLUHG L IRUPDWLR L SUR LGHG
, FOXGH DOO H DP ERRNOHW
LW H WUD HHW RI SDSHU L D , FLGH W 5HSRUW UHWXU
H HORSH HH SD H
RI W H
G D
0D D IRU FRPSOHWH GHWDLO
ƒ 5HWXU DOO H DP PDWHULDO WR HFXUH WRUD H X WLO W H DUH
3 3UR UDP 6HH SD H
RI W H
DERXW HFXUH WRUD H %HIRUH WRUL
SD H RI W H D
HU HHW D G
PDWHULDO
LSSHG EDFN WR W H
G D
0D D IRU PRUH L IRUPDWLR
F HFN W H 6F RRO 8 H 2 O
HFWLR R
Š )LOO L W H DSSURSULDWH HFWLR
XPEHU FLUFOH L RUGHU WR DFFH D HSDUDWH
3 , WUXFWLR DO 3OD L 5HSRUW IRU UH XODUO F HGXOHG H DP R O RU XEMHFW
FRUH UR WHU DW W H FOD
HFWLR RU WHDF HU OH HO 6HH 3R W ( DP FWL LWLH L W H
G D
0D D
Š Check your list of students who are eligible for fee reductions and ll in the
DSSURSULDWH FLUFOH R W HLU UH L WUDWLR D
3
O
O V
PV
HU
HHW
H
1
I
P
1
,
1
AP Calculus BC Exam
®
2019
SECTION I: Multiple Choice
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
1 hour and 45 minutes
Number of Questions
45
Percent of Total Score
50%
Writing Instrument
Pencil required
Part A
Number of Questions
30
Instructions
Section I of this exam contains 45 multiple-choice questions and 4 survey questions. For
Part A, fill in only the circles for numbers 1 through 30 on the answer sheet. For Part B,
fill in only the circles for numbers 76 through 90 on the answer sheet. Because Part A and
Part B offer only four answer options for each question, do not mark the (E) answer circle
for any question. The survey questions are numbers 91 through 94.
Indicate all of your answers to the multiple-choice questions on the answer sheet. No
credit will be given for anything written in this exam booklet, but you may use the booklet
for notes or scratch work. After you have decided which of the suggested answers is best,
completely fill in the corresponding circle on the answer sheet. Give only one answer to
each question. If you change an answer, be sure that the previous mark is erased
completely. Here is a sample question and answer.
Time
1 hour
Electronic Device
None allowed
Part B
Number of Questions
15
Time
45 minutes
Electronic Device
Graphing calculator
required
Use your time effectively, working as quickly as you can without losing accuracy. Do not
spend too much time on any one question. Go on to other questions and come back to
the ones you have not answered if you have time. It is not expected that everyone will
know the answers to all of the multiple-choice questions.
Your total score on the multiple-choice section is based only on the number of questions
answered correctly. Points are not deducted for incorrect answers or unanswered
questions.
Form I
Form Code 4PBP4-S
68
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
CALCULUS BC
SECTION I, Part A
Time—1 hour
Number of questions—30
NO CALCULATOR IS ALLOWED FOR THIS PART OF THE EXAM.
Directions: Solve each of the following problems, using the available space for scratch work. After examining the
form of the choices, decide which is the best of the choices given and fill in the corresponding circle on the answer
sheet. No credit will be given for anything written in this exam booklet. Do not spend too much time on any one
problem.
In this exam:
(1) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which
f (x ) is a real number.
(2) The inverse of a trigonometric function f may be indicated using the inverse function notation f − 1 or with the
prefix “arc” (e.g., sin−1 x = arcsin x).
-3-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
∫1 (4x
2
1.
(A)
27
2
3
)
− x dx =
(B) 27
(C) 36
(D) 57
2. Let f be the function defined by f (x ) = x 3 − 3x 2 − 9x + 11. At which of the following values of x does f
attain a local minimum?
(A) 3
(B) 1
(C) −1
(D) −3
-4-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
d
2(sin x )2 =
dx
(
3.
 1
(A) 4 cos 
 2 x
)



(B) 4 sin x cos x
(C)
2 sin x
x
(D)
(
2 sin x cos x
x
)
4. The position of a particle is given by the parametric equations x(t ) = ln t 2 + 1 and y(t ) = e3−t . What is the
velocity vector at time t = 1 ?
(A)
1, e2
(B)
1, −e2
(C)
1 2
,e
2
-5-
(D)
1
, −e2
2
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
∞
en
∑ n is
n=1 p
5.
(A)
p
p−e
(B)
e
p−e
(C)
x
f (x )
2
4
e
p ln
(D) divergent
()
p
e
f ¢(x ) f ≤(x )
g(x )
−3
−2
3
g ¢(x ) g ≤(x )
5
1
6. The table above gives values of the twice-differentiable functions f and g and their derivatives at x = 2. If h is
f ¢(x )
the function defined by h(x ) =
, what is the value of h¢(2) ?
g(x )
(A)
9
4
(B)
3
5
(C) −
3
2
(D) −
21
4
-6-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
( )
7. Which of the following is the Maclaurin series for x cos x 2 ?
8.
(A) x −
x5 x9
x13
+
−
+L
2! 4!
6!
(B) x −
x3 x5
x7
+
−
+L
2!
4!
6!
(C) x 3 −
x 7 x11
x15
+
−
+L
3!
5!
7!
(D) x 3 −
x5 x7
x9
+
−
+L
3! 5!
7!
lim
x→∞
(A) −2
10 − 6x 2
is
5 + 3e x
(B) 0
(C) 2
(D) nonexistent
-7-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
9. The function f is not differentiable at x = 5. Which of the following statements must be true?
(A) f is not continuous at x = 5.
(B) lim f (x ) does not exist.
x→5
(C) lim
x→5
(D)
f (x ) − f (5)
does not exist.
x−5
5
∫0 f (x) dx does not exist.
10. The second derivative of a function f is given by f ≤(x ) = x(x − 3)5 (x − 10)2. At which of the following values
of x does the graph of f have a point of inflection?
(A) 3 only
(B) 0 and 3 only
(C) 3 and 10 only
(D) 0, 3, and 10
-8-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
p
11.
⌠ ex − 1
dx =
 x
⌡0 e − x
(A) e p − p − 1
(B) ln(e p − p) − 1
(D) ln(e p − p)
(C) p − ln p
x
0
4
8
12
16
f (x )
8
0
2
10
1
12. The table above gives selected values for the differentiable function f. In which of the following intervals must
there be a number c such that f ¢(c) = 2 ?
(A) (0, 4)
(B) (4, 8)
(C) (8, 12)
(D) (12, 16)
-9-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
dy
= x + 2y with initial condition f (0) = 2. What is
dx
the approximation for f (−0.4) obtained by using Euler’s method with two steps of equal length starting
13. Let y = f (x ) be the solution to the differential equation
at x = 0 ?
(A) 0.76
(B) 1.20
(C) 1.29
(D) 3.96
14. What is the slope of the line tangent to the curve
(A) −3
(B) −
1
3
(C) 1
(D)
x +
9 1 
y = 2 at the point  ,  ?
 4 4 
4
3
-10-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
∞
⌠
6

dx is

3 2
⌡1 (x + 3) /
15.
(A)
16. If
3
4
(B) 3
(C) 6
(D) divergent
dy
= 2 − y , and if y = 1 when x = 1, then y =
dx
(A) 2 − e x−1
(B) 2 − e1−x
(C) 2 − e−x
-11-
(D) 2 + e−x
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
17. Which of the following series converges?
(A)
∞
n=1
(B)
∞
(D)
 n + 1

2n 
∑ (−1)n 
n=1
(C)
1 − n 

n 
∑ (−1)n 
∞




∞
2 n 

n 
 2
n n
(−1)

∑
 3 n
n=1
∑ (−1)n 
n=1
-12-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
18. Let R be the region in the first and second quadrants between the graphs of the polar curves f (q ) = 3 + 3 cos q
and g(q ) = 4 + 2 cos q , as shaded in the figure above. Which of the following integral expressions gives the
area of R ?
6
(A)
∫− 2 ( g(q ) − f (q)) dq
(B)
∫0 ( g(q ) − f (q)) dq
(C)
1
2
∫0 ( g(q ) − f (q))
(D)
1
2
∫0 (( g(q))
p
p
p
2
2
dq
)
− ( f (q ))2 dq
-13-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
19. Which of the following statements about the series
∞
(− 1)n n
n=1
n2 + 1
∑
(A) The series can be shown to diverge by comparison with
is true?
∞
1
∑ n.
n=1
(B) The series can be shown to diverge by limit comparison with
∞
1
∑ n.
n=1
(C) The series can be shown to converge by comparison with
∞
∑
1
2
n=1 n
.
(D) The series can be shown to converge by the alternating series test.
-14-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
20. If
4
∫1
f (x ) dx = 8 and
4
∫1 g(x ) dx = −2, which of the following cannot be determined from the information
given?
(A)
(B)
(C)
(D)
1
∫4 g(x ) dx
4
∫1 3f (x) dx
4
∫1 3 f (x) g(x) dx
4
∫1 (3f (x) + g(x)) dx
-15-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
∞
21. Which of the following is the interval of convergence for the series
(A) [−9, 1)
(B) [−5, 5)
(C) [1, 9)
∑
(x + 4)n
n=1 n
◊ 5n+1
?
(D) (−∞, ∞)
22. What is the slope of the line tangent to the polar curve r = 3q at the point where q =
(A) −
p
2
(B) −
2
p
(C) 0
p
?
2
(D) 3
-16-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
23. The definite integral
4
∫0
x dx is approximated by a left Riemann sum, a right Riemann sum, and a trapezoidal
sum, each with 4 subintervals of equal width. If L is the value of the left Riemann sum, R is the value of the
right Riemann sum, and T is the value of the trapezoidal sum, which of the following inequalities is true?
(A) L <
4
∫0
(B) L < T <
(C) R <
4
∫0
(D) R < T <
x dx < T < R
4
∫0
x dx < R
x dx < T < L
4
∫0
x dx < L
-17-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
24. The graph of the piecewise linear function f is shown above. What is the value of
(A) −8
(B) −6
(C) 0
12
∫0
f ¢(x ) dx ?
(D) 22
-18-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
25. The function f has a continuous derivative. If f (0) = 1, f (2) = 5, and
(A) 3
(B) 6
(C) 10
2
∫0
f (x ) dx = 7, what is
2
∫0 x ◊ f ¢(x) dx ?
(D) 17
26. Which of the following series are conditionally convergent?
∞
I.
(− 1)n
∑
n=1 n!
II.
(− 1)n
n
n=1
III.
(− 1)n
n=1 n + 2
∞
∑
∞
∑
(A) I only
(B) II only
(C) II and III only
-19-
(D) I, II, and III
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
27. Let R be the region in the first quadrant bounded by the graph of y =
x − 1 , the x-axis, and the vertical line
x = 10. Which of the following integrals gives the volume of the solid generated by revolving R about the
y-axis?
10
(A) p
∫1
(B) p
∫1
10
3
(x − 1) dx
(100 − (x − 1)) dx
(
2
))
⌠
(C) p  10 − y 2 + 1
⌡0
3
⌠
(D) p 
⌡0
(

100 − y 2 + 1

(
dy
2
)  dy
-20-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
28. If
dy
d 2y
dx
= 5 and
= sin t 2 , then 2 is
dt
dt
dx
()
()
(A) 2 t cos t
2
(B)
()
2 t cos t 2
5
(C)
()
2t cos t 2
25
-21-
(D) undefined
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
29. Which of the following expressions is equal to lim
n →∞
1 1
1
1
1 

+
+
+L+
n  ?
1
2
3
n 
2 + 
2
+
2
+
2
+

n 
n
n
n
2
⌠ 1
dx
(A) 
⌡1 x
1
⌠ 1
dx
(B) 
⌡0 2 + x
2
1
⌠
dx
(C) 
⌡0 2 + x
3
1
⌠
dx
(D) 
⌡2 2 + x
-22-
GO ON TO THE NEXT PAGE.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
1 3
x + L, and the Maclaurin series converges to
3
f (x ) for all real numbers x. If g is the function defined by g(x ) = e f (x), what is the coefficient of x 2 in the
30. A function f has a Maclaurin series given by 2 + 3x + x 2 +
Maclaurin series for g ?
(A)
1 2
e
2
(B) e2
(C)
5 2
e
2
(D)
11 2
e
2
END OF PART A
IF YOU FINISH BEFORE TIME IS CALLED,
YOU MAY CHECK YOUR WORK ON PART A ONLY.
DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.
-23-
B
B
B
B
B
B
B
B
B
CALCULUS BC
SECTION I, Part B
Time—45 minutes
Number of questions—15
A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUESTIONS ON THIS PART OF THE EXAM.
Directions: Solve each of the following problems, using the available space for scratch work. After examining the
form of the choices, decide which is the best of the choices given and fill in the corresponding circle on the answer
sheet. No credit will be given for anything written in this exam booklet. Do not spend too much time on any one
problem.
BE SURE YOU FILL IN THE CIRCLES ON THE ANSWER SHEET THAT CORRESPOND TO
QUESTIONS NUMBERED 76–90.
YOU MAY NOT RETURN TO QUESTIONS NUMBERED 1–30.
In this exam:
(1) The exact numerical value of the correct answer does not always appear among the choices given. When this
happens, select from among the choices the number that best approximates the exact numerical value.
(2) Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which
f (x ) is a real number.
(3) The inverse of a trigonometric function f may be indicated using the inverse function notation f − 1 or with the
prefix “arc” (e.g., sin−1 x = arcsin x).
-26-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
76. The function f is continuous on the closed interval [0, 5]. The graph of f ¢, the derivative of f, is shown above.
On which of the following intervals is f increasing?
(A) [0, 1] and [2, 4]
(B) [0, 1] and [3, 5]
(C) [0, 1] and [4, 5] only
(D) [0, 2] and [4, 5]
77. If
2
dy
= 6e−0.08 (t − 5) , by how much does y change as t changes from t = 1 to t = 6 ?
dt
(A) 3.870
(B) 8.341
(C) 18.017
(D) 22.583
-27-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
78. The function f has the property that lim− f (x ) = +∞ , lim+ f (x ) = −∞ , lim f (x ) = 2, and lim f (x ) = 2 .
x→1
x→1
x→−∞
x→+∞
Of the following, which could be the graph of f ?
(A)
(B)
(C)
(D)
-28-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
79. Tara’s heart rate during a workout is modeled by the differentiable function h, where h(t ) is measured in beats
per minute and t is measured in minutes from the start of the workout. Which of the following expressions gives
Tara’s average heart rate from t = 30 to t = 60 ?
60
(A)
∫30 h(t ) dt
(B)
1
30
∫30 h(t) dt
(C)
1
30
∫30 h¢(t) dt
(D)
h¢(30) + h¢(60)
2
60
60
-29-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
80. The function f has derivatives of all orders for all real numbers with f (2) = −1, f ¢(2) = 4 , f ≤(2) = 6,
and f ¢¢¢(2) = 12 . Using the third-degree Taylor polynomial for f about x = 2, what is the approximation
of f (2.1) ?
(A) −0.570
(B) −0.568
(C) −0.566
81. Let f be a function with derivative given by f ¢(x ) =
(D) −0.528
x 3 + 1 . What is the length of the graph of y = f (x) from
x = 0 to x = 1.5 ?
(A) 4.266
(B) 2.497
(C) 2.278
(D) 1.976
-30-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
82. Let h be a continuous function of x. Which of the following could be a slope field for a differential equation of
dy
= h(x ) ?
the form
dx
(A)
(B)
(C)
(D)
-31-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B


k 3 + x for x < 3


f (x ) = 
 16

for x ≥ 3

2

k − x

83. Let f be the function defined above, where k is a positive constant. For what value of k, if any, is f continuous?
(A) 2.081
(B) 2.646
(C) 8.550
(D) There is no such value of k.
84. Let f be a differentiable function. The figure above shows the graph of the line tangent to the graph of f
at x = 0 . Of the following, which must be true?
(A) f ¢(0) = − f (0)
(B) f ¢(0) < f (0)
(C) f ¢(0) = f (0)
(D) f ¢(0) > f (0)
-32-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
85. A referee moves along a straight path on the side of an athletic field. The velocity of the referee is given by
v(t ) = 4(t − 6) cos(2t + 5), where t is measured in minutes and v(t) is measured in meters per minute. What is
the total distance traveled by the referee, in meters, from time t = 2 to time t = 6 ?
(A) 3.933
(B) 14.578
(C) 21.667
(D) 29.156
-33-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
86. The graph of the function f shown above consists of three line segments. Let h be the function defined by
h(x ) =
x
∫0 f (t) dt . At what value of x does h attain its absolute maximum on the interval [−4, 3] ?
(A) −4
(B) −2
(C) 0
(D) 3
-34-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
87. The position of a particle moving in the xy-plane is given by the parametric equations x(t ) = e− t and
y(t ) = sin(4t) for time t ≥ 0. What is the speed of the particle at time t = 1.2 ?
(A) 1.162
(B) 1.041
(C) 0.462
(D) 0.221
1 4 2 3 1 2 1
x − x + x − x . For how many values of x in the open
4
3
2
2
interval (0, 1.565) is the instantaneous rate of change of f equal to the average rate of change of f on the closed
88. Let f be the function defined by f (x ) =
interval [0, 1.565] ?
(A) Zero
(B) One
(C) Three
(D) Four
-35-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
B
89. The population P of rabbits on a small island grows at a rate that is jointly proportional to the size of the rabbit
population and the difference between the rabbit population and the carrying capacity of the population. If the
carrying capacity of the population is 2400 rabbits, which of the following differential equations best models
the growth rate of the rabbit population with respect to time t, where k is a constant?
(A)
dP
= 2400 − kP
dt
(B)
dP
= k(2400 − P)
dt
(C)
dP
1
= k (2400 − P)
dt
P
(D)
dP
= kP(2400 − P)
dt
90. A region is bounded by two concentric circles, as shown by the shaded region in the figure above. The radius of
the outer circle, R, is increasing at a constant rate of 2 inches per second. The radius of the inner circle, r, is
decreasing at a constant rate of 1 inch per second. What is the rate of change, in square inches per second, of
the area of the region at the instant when R is 4 inches and r is 3 inches?
(A) 3p
(B) 6p
(C) 10p
(D) 22p
-36-
GO ON TO THE NEXT PAGE.
B
B
B
B
B
B
B
B
END OF SECTION I
IF YOU FINISH BEFORE TIME IS CALLED,
YOU MAY CHECK YOUR WORK ON PART B ONLY.
DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO.
_______________________________________________________
MAKE SURE YOU HAVE DONE THE FOLLOWING.
• PLACED YOUR AP NUMBER LABEL ON YOUR ANSWER SHEET
• WRITTEN AND GRIDDED YOUR AP NUMBER CORRECTLY ON YOUR
ANSWER SHEET
• TAKEN THE AP EXAM LABEL FROM THE FRONT OF THIS BOOKLET
AND PLACED IT ON YOUR ANSWER SHEET
-37-
B
AP Calculus BC Exam
®
2019
SECTION II: Free Response
DO NOT OPEN THIS BOOKLET OR BREAK THE SEALS ON PART B UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
1 hour and 30 minutes
Number of Questions
6
Percent of Total Score
50%
Writing Instrument
Either pencil or pen with
black or dark blue ink
Weight
The questions are
weighted equally, but
the parts of a question
are not necessarily
given equal weight.
Part A
Instructions
Number of Questions
The questions for Section II are printed in this booklet. Do not break the seals on Part B
until you are told to do so. Write your solution to each part of each question in the space
provided. Write clearly and legibly. Cross out any errors you make; erased or crossed-out
work will not be scored.
2
Time
30 minutes
Electronic Device
Graphing calculator
required
Percent of Section II Score
33.33%
Part B
Number of Questions
4
Time
1 hour
Electronic Device
None allowed
Percent of Section II Score
66.67%
Manage your time carefully. During Part A, work only on the questions in Part A. You
are permitted to use your calculator to solve an equation, find the derivative of a function
at a point, or calculate the value of a definite integral. However, you must clearly indicate
the setup of your question, namely the equation, function, or integral you are using. If you
use other built-in features or programs, you must show the mathematical steps necessary
to produce your results. During Part B, you may continue to work on the questions in
Part A without the use of a calculator.
As you begin each part, you may wish to look over the questions before starting to work
on them. It is not expected that everyone will be able to complete all parts of all questions.
• Show all of your work, even though a question may not explicitly remind you to do so.
Clearly label any functions, graphs, tables, or other objects that you use. Justifications
require that you give mathematical reasons, and that you verify the needed conditions
under which relevant theorems, properties, definitions, or tests are applied. Your work
will be scored on the correctness and completeness of your methods as well as your
answers. Answers without supporting work will usually not receive credit.
• Your work must be expressed in standard mathematical notation rather than calculator
syntax. For example,
5
1
x2 dx may not be written as fnInt(X2, X, 1, 5).
• Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If you
use decimal approximations in calculations, your work will be scored on accuracy.
Unless otherwise specified, your final answers should be accurate to three places after
the decimal point.
• Unless otherwise specified, the domain of a function f is assumed to be the set of all
real numbers x for which f x is a real number.
Form I
Form Code 4PBP4-S
68
CALCULUS BC
SECTION II, Part A
Time—30 minutes
Number of questions—2
A GRAPHING CALCULATOR IS REQUIRED FOR THESE QUESTIONS.
-3-
GO ON TO THE NEXT PAGE.
1
1
1
1
1
1
1
1
1
1
1. The rate at which cars enter a parking lot is modeled by E (t ) = 30 + 5(t − 2)(t − 5)e− 0.2t . The rate at
which cars leave the parking lot is modeled by the differentiable function L. Selected values of L(t ) are given
in the table above. Both E (t ) and L(t ) are measured in cars per hour, and time t is measured in hours after
5
A.M.
(t = 0). Both functions are defined for 0 £ t £ 12.
(a) What is the rate of change of E (t ) at time t = 7 ? Indicate units of measure.
Do not write beyond this border.
Do not write beyond this border.
(b) How many cars enter the parking lot from time t = 0 to time t = 12 ? Give your answer to the nearest
whole number.
-4-
Continue question 1 on page 5.
1
1
1
1
1
1
1
1
1
1
(c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate
12
L(t ) dt . Using correct units, explain the meaning of
12
∫2
L(t ) dt in the context of this problem.
(d) For 0 £ t < 6 , 5 dollars are collected from each car entering the parking lot. For 6 £ t £ 12, 8 dollars are
collected from each car entering the parking lot. How many dollars are collected from the cars entering the
parking lot from time t = 0 to time t = 12 ? Give your answer to the nearest whole dollar.
-5-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
∫2
2
2
2
2
2
2
2
2
2
2
2. A laser is a device that produces a beam of light. A design, shown above, is etched onto a flat piece of metal
Both x and y are measured in centimeters, and t is measured in seconds. The laser starts at position (0, 0) at
time t = 0, and the design takes 3.1 seconds to complete. For 0 £ t £ 3.1,
dy
= 4 cos(2.5t).
dt
dx
= 3 cos t 2 and
dt
( )
(a) Find the speed of the laser at time t = 3 seconds.
(b) Find the total distance traveled by the laser from time t = 1 to time t = 3 seconds.
-6-
Continue question 2 on page 7.
Do not write beyond this border.
Do not write beyond this border.
using a moving laser. The position of the laser at time t seconds is represented by (x(t), y(t )) in the xy-plane.
2
2
2
2
2
2
2
2
2
2
(d) What is the difference between the y-coordinates of the laser’s highest position and lowest position
for 0 £ t £ 3.1 ? Justify your answer.
-7-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
(c) The laser is farthest to the right at time t = 1.253 seconds. Find the x-coordinate of the laser’s rightmost
position.
END OF PART A
IF YOU FINISH BEFORE TIME IS CALLED,
YOU MAY CHECK YOUR WORK ON PART A ONLY.
DO NOT GO ON TO PART B UNTIL YOU ARE TOLD TO DO SO.
-8-
CALCULUS BC
SECTION II, Part B
Time—1 hour
Number of questions—4
NO CALCULATOR IS ALLOWED FOR THESE QUESTIONS.
DO NOT BREAK THE SEALS UNTIL YOU ARE TOLD TO DO SO.
-13-
3
3
3
3
3
3
3
3
3
3
NO CALCULATOR ALLOWED



9 − x2


f (x ) = 
px

−x + 3 cos


2


3. Let f be the function defined above.
( )
for −3 £ x £ 0
for 0 < x £ 4
Do not write beyond this border.
Do not write beyond this border.
(a) Find the average rate of change of f on the interval − 3 £ x £ 4.
(b) Write an equation for the line tangent to the graph of f at x = 3.
-14-
Continue question 3 on page 15.
3
3
3
3
3
3
3
3
3
3
NO CALCULATOR ALLOWED
(d) Must there be a value of x at which f (x ) attains an absolute maximum on the closed interval
− 3 £ x £ 4 ? Justify your answer.
-15-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
(c) Find the average value of f on the interval − 3 £ x £ 4.
4
4
4
4
4
4
4
4
4
4
4. The continuous function f is defined for −4 £ x £ 4. The graph of f, shown above, consists of two line
1
1
5
segments and portions of three parabolas. The graph has horizontal tangents at x = − , x = , and x = .
2
2
2
It is known that f (x ) = −x 2 + 5x − 4 for 1 £ x £ 4 . The areas of regions A and B bounded by the graph
of f and the x-axis are 3 and 5, respectively. Let g be the function defined by g(x ) =
x
∫− 4 f (t) dt .
(a) Find g(0) and g(4).
-16-
Continue question 4 on page 17.
Do not write beyond this border.
Do not write beyond this border.
NO CALCULATOR ALLOWED
4
4
4
4
4
4
4
4
4
4
NO CALCULATOR ALLOWED
(c) Find all intervals on which the graph of g is concave down. Give a reason for your answer.
-17-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
(b) Find the absolute minimum value of g on the closed interval [−4, 4]. Justify your answer.
5
5
5
5
5
5
5
5
5
5
NO CALCULATOR ALLOWED
(a) Show that
dy
−4x
=
.
dx
3(y + 1)
(b) Using the information from part (a), find
Do not write beyond this border.
Do not write beyond this border.
5. The graph of the curve C, given by 4x 2 + 3y 2 + 6y = 9, is shown in the figure above.
d 2y
dx 2
in terms of x and y.
-18-
Continue question 5 on page 19.
5
5
5
5
5
5
5
5
5
5
NO CALCULATOR ALLOWED
(c) In polar coordinates, the curve C is given by r =
3
dr
for 0 £ q £ 2p . Find
.
2 + sin q
dq
As q increases, on what intervals is the distance between the origin and the point (r, q) increasing?
(d) Let S be the region inside curve C, as defined in part (c), but outside the curve r = 2 . Write, but do not
evaluate, an integral expression for the area of S.
-19-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
Give a reason for your answer.
6
6
6
6
6
6
6
6
6
6
NO CALCULATOR ALLOWED
∞
(−1)n+1(x − 3)n
6. Consider the series ∑
, where p is a constant and p > 0.
5n n p
n=1
◊
(b) For p = 1 and x = 8, does the series converge absolutely, converge conditionally, or diverge? Explain
your reasoning.
-20-
Continue question 6 on page 21.
Do not write beyond this border.
Do not write beyond this border.
(a) For p = 3 and x = 8, does the series converge absolutely, converge conditionally, or diverge? Explain
your reasoning.
6
6
6
6
6
6
6
6
6
6
NO CALCULATOR ALLOWED
(d) When p = 1 and x = 3.1 , the series converges to a value S. Use the first two terms of the series to
approximate S. Use the alternating series error bound to show that this approximation differs from S by
less than
1
.
300,000
-21-
GO ON TO THE NEXT PAGE.
Do not write beyond this border.
Do not write beyond this border.
(c) When x = −2, for what values of p does the series converge? Explain your reasoning.
STOP
END OF EXAM
____________________________________
THE FOLLOWING INSTRUCTIONS APPLY TO THE COVERS OF THE
SECTION II BOOKLET.
• MAKE SURE YOU HAVE COMPLETED THE IDENTIFICATION
INFORMATION AS REQUESTED ON THE FRONT AND BACK
COVERS OF THE SECTION II BOOKLET.
• CHECK TO SEE THAT YOUR AP NUMBER LABEL APPEARS IN
THE BOX ON THE FRONT COVER.
• MAKE SURE YOU HAVE USED THE SAME SET OF AP
NUMBER LABELS ON ALL AP EXAMS YOU HAVE TAKEN
THIS YEAR.
-22-
I
P
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
,
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
0
S
RF
F R
R
4
−
Topic
FUN-6.B
The Fundamental
Theorem of Calculus
and Definite Integrals
)
=
−
=(
− )−
( − )=
−
=
Incorrect. This response would result if the antidifferentiation was
not done and the endpoints were substituted directly into the
integrand, as follows.
−
=(
− )−( − )=
−
=
Incorrect. This response would result if the integrand was
differentiated rather than antidifferentiated, as follows.
−
(D)
Learning Objective
Correct. This question involves using the basic power rule for
antidifferentiation and correctly substituting the endpoints and
evaluating, as follows.
∫(
(C)
R
Q
1.E
(B)
F
R
Skill
(A)
R
=(
− )−(
− )=
−
=
Incorrect. This response would result if the powers of were not
divided by the new exponent when the power rule for antiderivatives
was applied, as follows.
−
=(
− )−( − ) =
−
=
4
Q
Skill
3.D
(A)
Learning Objective
Topic
FUN-4.A
Using the First
Derivative Test to Find
Relative (Local) Extrema
Correct. A local minimum for
will occur where the derivative
changes sign from negative to positive.
′( ) =
−
−
=
(
−
−
)=
′
( + )( − )
The zeros of ′ are at = − and = A sign chart shows that
′ changes from negative to positive at = Another way to see
this is to observe that the graph of ′ is a parabola opening up so
that the graph crosses the -axis from negative to positive at the
larger zero, =
(B)
Incorrect. This is a zero of the second derivative
′′( ) =
− = ( − ) where ′′ changes sign from negative to
positive. Therefore, = is a point of inflection of the graph of
not the location of a local minimum. This response might also have
been chosen if the derivative had been incorrectly factored as
( − )( + ) Then = is a zero of this expression where the
sign changes from negative to positive.
(C)
Incorrect. This is a zero of the derivative ′ where ′ changes sign
from positive to negative. Therefore, = − is the location of a local
maximum of
not a local minimum.
(D)
Incorrect. This response might have been chosen if the derivative
had been incorrectly factored as ( − )( + ) and the zero where
the derivative changes sign from positive to negative was selected.
4
Q
Skill
Learning Objective
Topic
1.E
FUN-3.C
The Chain Rule
(A)
Incorrect. This response might come from incorrectly applying the chain rule
twice as
(
(B)
(
(
) = ⋅(
⋅
(
)
)=
(
⋅ ⋅
( )
)= ( )
)
)=
⋅
(
)⋅
(
)=
⋅
(
)⋅
Incorrect. This response might come from using the chain rule only once,
with the innermost “inside” function,
as follows.
(
(D)
)
Incorrect. This response might come from correctly applying the chain rule
the first time but not the second, as follows.
(
(C)
( ( ( ) ) ) = ′( ′( ) ) as follows.
(
)
)=
⋅
(
)⋅
( )=
⋅
(
)⋅
Correct. The chain rule must be used twice for this composition of three
functions.
(
(
)
)=
=
⋅
(
)⋅
=
⋅
(
)⋅
=
⋅
(
(
)⋅
⋅
⋅
(
(
))
(
)
)
4
Q
Skill
1.E
(A)
(B)
Learning Objective
Topic
FUN-8.B
(BC ONLY) Solving Motion
Problems Using Parametric
and Vector-Valued Functions
Incorrect. This response would result if the chain rule was not used
during the differentiation of the ( ) component of the position vector
resulting in the loss of the negative sign, as follows.
()=
(
+
)
−
=
()=
(
+
)
−
=
()=
(
+
)
−
=
()=
(
+
)
−
=
−
⇒ ( )=
+
Correct. The components of the velocity vector are the derivatives of the
components of the position vector.
−
(C)
⇒ ( )=
−
+
Incorrect. This response would result if the chain rule was not used
during the differentiation of both components of the position vector, as
follows.
(D)
−
⇒ ( )=
+
Incorrect. This response would result if the chain rule was not used
during the differentiation of the ( ) component of the position vector,
as follows.
−
+
−
−
⇒ ( )=
−
4
Q
Skill
Learning Objective
Topic
3.D
LIM-7.A
(BC ONLY) Working
with Geometric Series
(A)
Incorrect. This response might be chosen if the series is correctly
identified as a convergent geometric series with common ratio
=
in
(B)
π
but the first term was taken to be
−
=
−
=
π
π
π −
=
converges to
π
=
−
Since this ratio is less than
π
−
=
π
=
π
and
the series
π −
Incorrect. The sum of the series might have been taken to be the
value of the definite integral used in performing the integral test for
( )=
convergence. Let
∫
∞
( )
=⌠

⌡
∞
(π )
(π )
=
( )
⌠

→∞ ⌡ π
( )
( ) ( )
(D)
resulting
π
Correct. The series is a geometric series with first term
common ratio
(C)
rather than
=
→∞
(π )
(π )


 π

π
=
− π = − π
=
=

→∞ 
π

−
π


π
π
π
π


Incorrect. This response might be chosen if the series is correctly
( )
( )
identified as a geometric series with common ratio
ratio is thought to be greater than
( )
=
π
but the
4
Q
Skill
Learning Objective
Topic
1.E
FUN-3.B
The Quotient Rule
(A)
Correct. The derivative of is found by using the quotient rule.
′′( ) ( ) − ′( ) ′( )
′( ) =
( ( ))
The values for the functions and derivatives at = are obtained from the table.
′′( ) ( ) − ′( ) ′( ) ( )( − ) − ( − )( ) − +
′( ) =
=
=
=
(− )
( ( ))
(B)
Incorrect. This response would result if the derivative of a quotient was taken to be
the quotient of the derivatives, as follows.
′′( )
′′( )
′( ) =
⇒ ′( ) =
=
′( )
′( )
(C)
Incorrect. This response comes from only differentiating the numerator in the
quotient, as follows.
′′( )
′′( )
′( ) =
⇒ ′( ) =
=
=−
−
( )
( )
(D)
Incorrect. This response would result if the terms in the numerator were added
rather than subtracted in the quotient rule, as follows.
′′( ) ( ) + ′( ) ′( )
( )( − ) + ( − )( ) − −
′( ) =
⇒ ′( ) =
=
=−
(− )
( ( ))
4
Q
Skill
Learning Objective
Topic
2.C
LIM-8.G
(BC ONLY) Representing
Functions as Power Series
(A)
Correct. Beginning with the Maclaurin series for
( )
obtain the Maclaurin series for
is substituted to
This series is then multiplied term-
by-term by
= −
( )=
( )=
(B)
+
−
−
( )

 −

+
+
−
( )
+L = −

+ L =

−
−

=  −

−
+
+
−
+L
−
+L
−
+
+
( ) =( )−

+ L =

−
( )=



−
( )
−
−
+
−
+L
was
+L
+
( )
+
−
−
( )
+L =

+ L =

−
−
+
+
−
−
−
=
+
−
+L

 −

+
−

+ L =

−
+
−
+L
+L
Incorrect. This response would result if the Maclaurin series for
used, and the series was multiplied by
=
was
+L
Incorrect. This response would result if the Maclaurin series for
used instead of the series for
=
(D)
( )
+
Incorrect. This response would result if the Maclaurin series for
used, but there was no substitution of
before multiplication by
= −
(C)
+L
+L
was
4
Q
Skill
2.B
(A)
Learning Objective
Topic
LIM-2.D
Connecting Limits at
Infinity and Horizontal
Asymptotes
Incorrect. This response might come from treating the problem like
the limit of a rational function as goes to infinity when the
numerator and denominator are polynomials of the same degree. If
only the coefficients of the
term and the
term are considered,
it might be thought that the limit would be
(B)
Correct. The numerator of
−
+
−
=−
is a translated power function
and the denominator is a translated exponential function. Since the
exponential function
grows faster than the power function
the relative magnitude of the denominator compared to the
numerator will result in this expression converging to as goes to
infinity.
(C)
Incorrect. This response might come from treating the problem like
the limit of a rational function as goes to
If only the constant
terms are considered, it might be thought that the limit would be
=
(D)
Incorrect. It might be thought that the limit is nonexistent since the
numerator goes to −∞ and the denominator goes to +∞ as goes
to infinity, but this does not take into account the relative magnitude
of the exponential function in the denominator compared to the
quadratic term in the numerator as gets larger.
4
Q
Skill
3.B
Learning Objective
Topic
CHA-2.B
Defining Average and
Instantaneous Rates of
Change at a Point
(A)
Incorrect. This statement could be false. A function can be
continuous at a point without it being differentiable at that point. For
example, ( ) = − is continuous at = but not
differentiable there.
(B)
Incorrect. This statement could be false. For example, if
then
is not differentiable at = but
( )= −
( )=
→
(C)
(D)
Correct. The expression
( )− ( )
is one form of the
−
definition of the derivative of
at = Since
is not
differentiable at = however, this limit does not exist.
→
Incorrect. This statement could be false. The definite integral can be
found if
is continuous even if
is not differentiable at one of
the endpoints. For example, if ( ) = − , then
is not
differentiable at
=
but
∫
( )
=
( )( ) =
4
Q
Skill
2.B
Learning Objective
Topic
FUN-4.A
Determining Concavity
of Functions over Their
Domains
(A)
Incorrect. This response would result if the change of sign of the
second derivative ′′ at = is detected, but the change in sign at
= is overlooked.
(B)
Correct. A point of inflection occurs where the second derivative ′′
changes sign. The zeros of ′′( ) occur at =
= and
=
Constructing a sign chart with the three factors
( − )
and ( − ) shows that ′′( ) is positive for <
negative for < < positive for < <
and positive for
>
Therefore, the graph of
has a point of inflection only at
= and = where ′′( ) changes from positive to negative
and then from negative to positive, respectively.
(C)
Incorrect. The second derivative ′′ is positive both to the left and to
the right of =
so there is no sign change in ′′( ) at =
Therefore, the graph of
does not have a point of inflection at
=
There is a sign change in ′′( ) from positive to negative at
= so the graph of
does have a point of inflection at = in
addition to the one at =
(D)
Incorrect. These are the three zeros of the second derivative
′′( ) only changes sign at = and =
′′ but
4
Q
Skill
Learning Objective
Topic
1.E
FUN-6.D
Integrating Using Substitution
(A)
Incorrect. This response would result if the substitution
an attempt to evaluate the definite integral.
=
−
⇒
=
− ⇒
When = π
= π −π
Only the substitutions for
−
however, and not for
⌠

⌡
(B)
π
−
−
=
∫
π −π
−
⇒
=
=π
When
=
=
−
=
and for the limits of integration were made,
as follows.
π −π
=
⌠

⌡
π
−
−
− ⇒
π
=
−
Substituting for
=
( π − π) −
When
−
=
=
−
=
−π
for
π −π
=⌠
=
and for the limits of integration gives
π −π
=
⌡
( π − π) −
=
(π
=
)
−π −
Incorrect. This response would result if the integrand was incorrectly simplified
−
as
or
⌠

⌡
(D)
When
−
was used in
−
Incorrect. This response would result if the substitution =
− was correctly
used in an attempt to evaluate the definite integral, but the evaluation used
= rather than
=
=
(C)
=
=
and then either the antiderivative was only evaluated at the upper limit
was taken to equal
π
−
−

=⌠
π
⌡
as follows.
(− )
π
=( −
) =π −
π
Correct. This integral can be evaluated by using substitution of variables with
=
−
=
−
⇒
=
=π
When
⌠

⌡
=
−
−
( π − π)
π
=
−
Substituting for
π
− ⇒
=⌠
⌡
π −π
=
When
−
=
=
−
=
−π
for
and for the limits of integration gives
=
π −π
=
( π −π) −
=
(π
)
−π −
4
Q
Skill
Learning Objective
Topic
3.D
FUN-1.B
Using the Mean Value
Theorem
(A)
Incorrect. This interval might be chosen because of an error in
computing the average rate of change over the interval as
( )− ( )
−
=
= rather than
( )− ( )
−
=
−
−
=−
(B)
Incorrect. This interval might be chosen because of an error in
computing the average rate of change over the interval as
( )− ( )
−
−
=
=
=
= rather than
−
−
( )− ( )
(C)
Correct. The function
is continuous on the closed interval [
]
and differentiable on the open interval (
) By the Mean Value
Theorem, there is a number in the interval (
) such that
( )− ( )
−
′( ) =
=
=
−
Incorrect. This response would result if the Intermediate Value
Theorem was used instead of the Mean Value Theorem to select the
open interval (
) where ( ) = for some number in the
interval.
(D)
4
Q
Skill
Learning Objective
Topic
1.E
FUN-7.C
(BC ONLY) Approximating Solutions
Using Euler's Method
(A)
Correct. Euler’s method for the differential equation
written as
+
=
(
)= +
can be
is an
) Δ , where Δ is the step size. Then
−
−
= − 0.2, since there
) Here the step size is Δ =
(
+
approximation for
=
(
are two steps of equal length.
=
= −
=−
=
+
(B)
=
=
(
+
) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.2 ) = 2 + ( 4 )( −0.2 ) = 1.2
) Δ = 1.2 + ( −0.2 + 2 (1.2 ) ) ( −0.2 ) = 1.2 + ( 2.2 )( −0.2 ) = 0.76
(
Incorrect. This response would result if only the first step was done using Euler’s
method for the differential equation
=
= −
(C)
(
)= +
=
=
(
+
) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.2 ) = 2 + ( 4 )( −0.2 ) = 1.2
Incorrect. This response would result if the step size was taken to be Δ = −0.1
rather than Δ = −0.2. As a result, this response was the approximation for
( − ) rather than for ( − )
=
= −
=−
=
+
(D)
=
=
=
(
(
+
) Δ = 2 + ( 0 + 2 ( 2 ) ) ( −0.1) = 2 + ( 4 )( − 0.1) = 1.6
) Δ = 1.6 + ( −0.1 + 2 (1.6 ) ) ( − 0.1) = 1.6 + ( 3.1)( − 0.1) = 1.29
Incorrect. This response would result if the step size was taken to be Δ = 0.2 rather
than Δ = −0.2. As a result, this response was the approximation for ( ) rather
than for ( − )
=
=
=
=
+
=
=
+
(
) Δ = 2.8 + ( 0.2 + 2 ( 2.8 ) ) ( 0.2 ) = 2.8 + ( 5.8 )( 0.2 ) = 3.96
(
) Δ = 2 + ( 0 + 2 ( 2 ) ) ( 0.2 ) = 2 + ( 4 )( 0.2 ) = 2.8
4
Q
Skill
Learning Objective
Topic
1.E
FUN-3.D
Implicit Differentiation
(A)
Incorrect. This response would result if the equation obtained by
implicit differentiation was solved incorrectly for
+
( )
=
=−
⇒
=−
=−
=−
as follows.
It would also be obtained if the expression for
had been correctly
found but the - and -values were reversed when substituting into
the first derivative, as follows.
+
( )
(B)
=
=−
⇒
=−
=−
=−
Correct. The slope of the tangent line is the value of
at the point
( ) During the implicit differentiation to find
both the
power rule and the chain rule are needed.
+
The point
=
( ) is on the curve, since
equation
+
+
() ()
=
=
=
and
=
satisfy the
At this point,
⇒
+
=
⇒
=−
(C)
Incorrect. This response might come from observing the symmetry
of the expressions for and in the equation of the curve and
concluding that the derivative expressions will be symmetric, leading
to a slope of without considering that the - and -coordinates of
the given point are not the same.
(D)
Incorrect. This response would result if the method of implicit
differentiation was incorrectly applied by taking the derivative of the
expression
without consideration of the chain rule and
+
setting the result equal to
=
+
⇒
as follows.
( )
=
+
=
+ =
4
Q
Skill
Learning Objective
Topic
1.E
LIM-6.A
(BC ONLY) Evaluating
Improper Integrals
(A)
Incorrect. This response would result if the integrand was just
evaluated at the lower limit, = rather than antidifferentiated, as
follows.
=
( + )
(B)
(C)
(D)
=
=
=
Incorrect. This response would result if in applying the power rule
for antiderivatives, the power of ( + ) was not divided by the new
exponent, as follows.


∞
⌠
⌠


=
=
−


→∞ 
→∞ 
⌡ ( + )
⌡ ( + )
 ( + )





−
= + =
=
+
→∞ 

 ( + )

Correct. This is an improper integral, since the region over which
the integrand is being integrated is unbounded. The
antidifferentiation is an application of the power rule.


∞
⌠
⌠


=
=
−


→∞ 
→∞ 
⌡ ( + )
⌡ ( + )
 ( + )





−
= +
=
+
=
→∞ 

+
)
 (

Incorrect. This response might come from assuming that the definite
integral of a function integrated over an unbounded region will
always diverge.
4
Q
Skill
1.E
(A)
Topic
FUN-7.D
Finding General
Solutions Using
Separation of Variables
Incorrect. This response would result if a chain rule error was made
during the antidifferentiation of the
term.
=
−
⇒
⌠

⌡ −
+
−
Since
−
⇒
⇒
=
−
−
=
−
at the initial value
>
−
−
=
or
⇒
⌠

⌡ −
=
−
=
−
−
Since
=
the solution would be
−
−
∫
+
⇒−
⇒
=
−
=
−
−
− ⇒
>
=
+
= −
−
=− + ⇒
−
=
−
at the initial value
=
the solution to the
−
or
=
−
differential equation is
=
−
−
Incorrect. This response would result if an arbitrary constant was not
included during the antidifferentiation.
=
−
⇒
⌠

⌡ −
=
Since
−
(D)
+
Correct. The differential equation can be solved using separation of
variables and the initial condition to determine the appropriate value
for the arbitrary constant.
=
(C)
=
= −
− ⇒
−
=
=
−
∫
=
=
(B)
Learning Objective
−
∫
⇒−
−
=
at the initial value
>
−
=
=
−
or
⇒
=
−
=
−
the solution would be
−
= −
Incorrect. This response would result if an arbitrary constant was not
included during the antidifferentiation and the incorrect sign was
taken for the absolute value when solving for
=
−
⌠

⌡ −
−
⇒
=
=−
−
=
−
∫
⇒−
⇒
=
−
+
−
=
⇒
−
=
−
4
Q
Skill
3.D
(A)
Learning Objective
Topic
LIM-7.A
(BC ONLY) Alternating
Series Test for
Convergence
Incorrect. This series diverges by the
−
= ≠
(− ) − =
term test since
Incorrect. This series diverges by the
+
= ≠
(− ) + =
term test since
Incorrect. This series diverges by the
term test since
→∞
(B)
→∞
→∞
(C)
→∞
(D)
→∞
(− )
=
Correct. The series
→∞
∞
∑ ( − ) 
=
does not exist.
=


∞
∑ (− )
=
three conditions: (1) the series is alternating, (2)
≥
→∞
=
and
<
for all
+
Therefore, the series converges by the alternating series test.
(3) the terms
=
( ) satisfies the
are decreasing since
4
Q
Skill
1.C
(A)
π
( ) − ( ))
( ( )−
=
∫− (
( ) − ( ))
( ))
=
∫
π
( ( ) − ( ))
Incorrect. The square of the difference between the two polar curves
was integrated rather than the difference of the squares.
∫
(D)
CHA-5.D
(BC ONLY) Finding the
Area of the
Region Bounded by Two
Polar Curves
Incorrect. This response comes from using a rectangular form of the
area instead of a polar form for area, as if the curves were treated as
functions of in terms of
but expressing the limits in terms of the
polar angle θ
∫
(C)
Topic
Incorrect. This response comes from using a rectangular form for area
instead of a polar form for area, as if the curves were treated as
functions of in terms of
The region appears to lie above the
interval [ −
] on the -axis.
∫− (
(B)
Learning Objective
π
( (θ ) − (θ ) )
∫
θ =
π
( (θ ) − (θ ) )
θ
Correct. The area bounded by the two polar curves can be found with a
definite integral. Let be the smaller radius, and let
be the larger
radius. The graphs of the two polar curves bound the region
the domain ≤ θ ≤ π Then the area of
is given by
∫
=
π
( (θ ) )
∫ ((
π
θ −
∫
π
( (θ ) )
(θ ) ) − ( (θ ) )
)
θ
θ =
∫ ((
π
(θ ) ) − ( (θ ) )
over
)
θ
4
Q
Skill
3.B
(A)
Learning Objective
Topic
LIM-7.A
(BC ONLY) Alternating
Series Test for
Convergence
Incorrect. The series
+
∞
∑=
is the divergent harmonic series. However,
so this inequality goes the wrong way to use the
<
comparison test to show that the series
∞
∑
addition, the divergence of the positive series
that the series
(B)
∞
∑=
+
=
that the series
∞
∑=
→∞
=
+
∞
∑=
+
the limit comparison test shows
also diverges. However, it cannot be
divergence of the positive series
(C)
∞
∑
=
does not imply
=
is the divergent harmonic series. Since
concluded from that the alternating series
series
∞
∑
diverges.
Incorrect. The series
→∞
diverges. In
+
=
∞
∑
=
∞ −
(
)
∑=
+
diverges. The
does not imply that the
diverges.
Incorrect. The series
∞
∑
is a convergent
=
-series with
for all > so this inequality
+
goes the wrong way to use the comparison test to show that the series
=
>
∞
∑
+
convergent.
=
However,
<
converges and that therefore
∞
∑
=
(− )
+
is absolutely
4
(D)
Q
Q Q
Correct. The series
∞
∑=
(− )
series is alternating, (2)
+
→∞
satisfies the three conditions: (1) the
and (3) the terms
=
+
=
are decreasing. Therefore, the series converges by the
+
alternating series test.
=
To verify that the terms
following.
+
<
⇔
⇔
⇔
+
<
+
<
( + ) +
+
+
<
+
+
+
+
+
≥
The last inequality is true for all
Alternatively, if
( )=
is decreasing because
for
>
are decreasing, consider the
+
then
′( ) =
(
=
)−
( + )
+
( ) and the function
=
(
−
+
)
<
4
Q
Skill
Learning Objective
Topic
1.C
FUN-6.A
Applying Properties of
Definite Integrals
(A)
Incorrect. The value of this integral can be determined using the
properties of the definite integral, as follows.
∫
(B)
= ⋅∫
( )
= − (− ) =
( )
= ⋅ =
Correct. It is not true in general that
∫
( ) ( )
∫
values of
∫
( )=−
∫
and
∫
∫
∫
and
∫
then
⋅∫
( )
∫
( ) ( )
( ) ( )
( )=
=
( )
the value of
(D)
( )
Incorrect. The value of this integral can be determined using the
properties of the definite integral, as follows.
∫
(C)
= −∫
( )
( )
( )
( )
so the individual
cannot be used to determine
For example, if
∫
=
=⌠ −
( )
=−
( − ) and ( ) = − ( − ) then
( )
( ) ( )
and
and
However, if
=−
⌡
( )=
∫
( )
=
= − as before, but now
=⌠ −
( − )
=−
⌡
Incorrect. The value of this integral can be determined using the
properties of the definite integral, as follows.
∫
=
(
( ) + ( ))
⋅∫
( )
+
=
∫
∫
( )
( )
=
+
⋅
∫
( )
+ (− ) =
4
Q
Skill
Learning Objective
Topic
3.D
LIM-8.D
(BC ONLY) Radius and Interval of
Convergence of Power Series
(A)
Correct. The ratio test can be used to determine the interval of convergence.
+
→∞
⇒
+
=
<
+
( + )
→∞ ( + ) ⋅
⇒− <
⋅
+
⋅ +
( + )
=−
At
=
the series is
∞
(− )
∑=
∞
∑ (−
=
⋅ +
=
multiple of the alternating harmonic series.
the series is
∞
∑=
+
⋅
=
∞
∑=
+
=
+
<
)
which converges since it is a
, which diverges since it is a multiple of
the harmonic series.
Therefore, the interval of convergence is − ≤
(B)
⋅
+
→∞
<
Now check the endpoints of the interval.
At
=
<
Incorrect. The ratio test might have been used to correctly determine that the radius
of convergence
is
but the interval was taken to be centered at
+
→∞
At
=−
At
=
=
→∞
+
( + )⋅
the series is
the series is
∞
∑=
(− )
+
⋅
∞
∑=
⋅
⋅
+
⋅
+
=
→∞
+
⋅
=
⇒
=
which converges by the alternating series test.
which diverges by the
term test.
(C)
Incorrect. This response would result if the ratio test was not used to determine the
radius of convergence. Looking at the form of the general term, the center was
thought to be at = and the radius of convergence was taken to be
It was
assumed that the series must converge at one endpoint and diverge at the other.
(D)
Incorrect. Errors in applying the ratio test might have led to the conclusion that the
radius of convergence was infinite.
4
Q
Skill
1.E
(A)
Learning Objective
Topic
FUN-3.G
(BC ONLY) Defining Polar
Coordinates and
Differentiating in Polar Form
Incorrect. This response would result if the slope was taken to be
θ
=
θ
=
θ = θ
θ ⇒
=
θ = θ
θ ⇒
θ=
π
=
θ
=
θ
θ=
−
θ
θ
θ=
θ=
π
π
π
=−
θ− θ
=
θ θ =π = −
θ+ θ
=
θ θ =π =
π
π
(B)
Correct. The slope of the polar curve is
θ
=
θ
=
θ = θ
θ ⇒
=
θ = θ
θ ⇒
θ=
(C)
=
θ
=
θ
θ=
π
−
θ
θ
π
θ=
θ=
π
π
=−
θ− θ
=
=
θ θ =π = −
θ+ θ
π
θ θ =π =
π
Incorrect. This response would result if the product rule was applied
incorrectly by taking the product of the derivatives.
=
θ = θ
θ ⇒
=
θ = θ
θ ⇒
θ=
(D)
π
π
π
=
θ
=
θ
θ=
π
−
θ
θ
=−
=
θ⇒
θ ⇒
θ
θ
θ=
θ=
π
π
=−
=
θ θ =π = −
θ θ =π =
=
Incorrect. This response would result if the slope was taken to be
rather than
θ
θ=
π
=
θ=
π
=
θ
4
Q
Skill
Learning Objective
Topic
1.F
LIM-5.A
Approximating Areas
with Riemann Sums
(A)
Incorrect. It was correctly determined that the left Riemann sum is
an underestimate and the right Riemann sum is an overestimate of
the definite integral because the function ( ) =
is increasing on
the interval [
is concave down,
] Because the graph of
however, the trapezoidal sum is an underestimate, not an
overestimate.
(B)
Correct. Because the function ( ) =
is increasing on the
interval [ ] the left Riemann sum is an underestimate and the
right Riemann sum is an overestimate of the definite integral, so
<
∫
<
Since the graph of
is concave down, the
trapezoidal sum is also an underestimate, but it is a closer
approximation to the definite integral than the left Riemann sum.
Therefore,
<
<
∫
<
(C)
Incorrect. This would be the correct response for the graph of a
function that is decreasing and concave up. The graph of
however, is increasing and concave down, so all the
( ) =
inequalities are going in the wrong direction.
(D)
Incorrect. It was correctly determined that the trapezoidal sum is an
underestimate of the definite integral since the graph of ( ) =
is concave down on the interval [
is
] Because the function
increasing, however, the left Riemann sum is an underestimate and
the right Riemann sum is an overestimate of the definite integral.
This response has reversed those two inequalities.
4
Q
Skill
3.D
(A)
′( )
=
at
= ( ) − ( ) = (− ) −
and
=
=−
where the values of
are obtained from the graph.
( )
=
( )( ) − ( )( ) − ( )( ) = −
Incorrect. This response would result if the Fundamental Theorem of
Calculus was incorrectly applied, as follows.
∫
(D)
FUN-6.B
The Fundamental
Theorem of Calculus
and Definite Integrals
Incorrect. This response would result if the function
was
integrated over the interval [
rather
than
as
follows.
′
]
∫
(C)
Topic
Correct. By the Fundamental Theorem of Calculus,
∫
(B)
Learning Objective
′( )
= ′( ) − ′( ) = ( − ) − ( − ) =
Incorrect. This response is the total area bounded by the graph of
and the -axis over the interval [
]
∫
( )
=
( )( ) + ( )( ) + ( )( ) = + + =
4
Q
Skill
Learning Objective
Topic
1.E
FUN-6.E
(BC ONLY) Using Integration by
Parts
(A)
Correct. This definite integral can be found using integration by parts, where
∫
=
=
∫
(B)
=
=
=
′( )
( ) −∫
⇒
( )
=
( )
=
( )− =
−
=
Incorrect. This response would result if the technique of integration by parts was
applied incorrectly as follows.
= ⇒
=
= ′( )
⇒ = ( )
⋅ ′( )
=
( ) − ( ) =(
( ) − ) − ( ( ) − ( )) =
−
Incorrect. This response would result if each factor in the integrand was
antidifferentiated separately, as follows.
∫
(D)
⇒
∫
⋅ ′( )
∫
(C)
−
⋅ ′( )
=
( )
( ) =
( )− =
Incorrect. This response would result if an error was made in the integration by
parts when addition was used instead of subtraction, resulting in
∫
∫
=
=
⇒
⋅ ′( )
+
∫
=
=
=
′( )
( ) +∫
⇒
( )
=
=
( )
( )+ =
+
=
=
4
Q
Skill
3.D
(A)
Learning Objective
Topic
LIM-7.A
(BC ONLY)
Determining Absolute
or Conditional
Convergence
Incorrect. Series I is absolutely convergent, not conditionally
convergent, since
(B)
∞
∑=
converges by the ratio test.
Incorrect. Series II was correctly identified as being conditionally
convergent. Series III is also conditionally convergent, since
∞ −
(
∑=
)
converges by the alternating series test but
+
∞
∑=
+
diverges by the limit comparison test with the harmonic series
(C)
Correct. A series
∑
∞
∑=
is conditionally convergent if the series
converges but the series of absolute terms
∑
diverges. Each of
the three series in this problem converges by the alternating series
test.
The series
∞
∑=
(− )
is not conditionally convergent, since
∞
∑=
converges by the ratio test (so this series is absolutely convergent).
The series
∞
∑
The series
∞
=
series
(D)
∑
(− )
=
is conditionally convergent because the series
diverges, since it is a
=
∑
∞
∞ −
(
∑
=
+
)
=
-series with
<
is conditionally convergent because the series
diverges by the limit comparison test with the harmonic
+
∞
∑
=
Incorrect. This response might be chosen because all three series
converge by the alternating series test. To determine whether they
are conditionally convergent, however, it is also necessary to
consider the series of absolute terms
∑
Series I is absolutely
convergent, not conditionally convergent, since
the ratio test.
∞
∑
=
converges by
4
Q
Skill
1.C
(A)
Learning Objective
Topic
CHA-5.C
Volume with Washer
Method - Revolving
Around - or -axis
Incorrect. This response is the volume of the solid generated by
revolving
about the -axis rather than the -axis. A typical slice
rotated about the -axis would form a disk of radius = =
−
( )=π
with cross-sectional area
solid would be
(B)
∫
= π ( − ) The volume of the
π( − )
Incorrect. This response is the volume obtained by rotating about the
-axis the region bounded by the graph of =
the
−
horizontal line =
and the vertical lines = and =
A
typical slice rotated about the -axis would form a washer with
−π
cross-sectional area ( ) = π
=
=
− and the outer radius is
(C)
Incorrect. For this response, the cross-sectional area of the washer
obtained by rotating a typical slice about the -axis was taken to be
π ( − ) rather than π
=
(D)
where the inner radius is
−π
where the inner radius is
+ and the outer radius is
=
Correct. Rotating a typical slice about the -axis will form a washer
with inner radius and outer radius
where each radius must be
expressed in terms of
Since =
the inner radius is
−
=
=
+
Since the region is bounded on the right by the
vertical line =
the outer radius is =
The cross-sectional
area of a typical slice rotated about the -axis is therefore
( )=π
−π
=π
(
−
(
+
)
) The volume of the solid
is found by using the definite integral of the cross-sectional area for
between and
as follows.
− =
⌠ π

⌡
(
−
(
+
)
)
⌠
= π
⌡
(
−
(
+
)
)
4
Q
Skill
1.E
(A)
Learning Objective
Topic
CHA-3.G
(BC ONLY) Second
Derivatives of
Parametric Equations
Incorrect. This response is the value of
=



=


( ( ))
not the value of
The chain rule must be used to find
=






and then used again to find
(B)
=
Incorrect. The chain rule was correctly used to find
( )
 however. Another application of the



 by
chain rule would be needed to find
by dividing




This response is just



(C)
Correct. By the chain rule,
=
=
( )
The chain rule is
needed again to find the second derivative, as follows.
=



=







 =
( )
=
( )
(D)
Incorrect. This response comes from thinking that
Since
=
the quotient would be undefined.
=
4
Q
Skill
2.C
(A)
Learning Objective
Topic
LIM-5.C
Riemann Sums, Summation
Notation, and Definite
Integral Notation
Incorrect. The sum inside the limit can be interpreted as a right Riemann




=
sum in the form ∑ 
 = ∑ ( + Δ ) Δ , where ( )
=  +
=



and Δ =
. The value of Δ corresponds to an interval of length
Since Δ = 2 +
for the initial value
=
the interval starts at
Therefore, the limit of the Riemann sum would be equal to
∫
(B)
( )
=⌠
not ⌠
⌡
⌡
Correct. The sum inside the limit can be interpreted as a right Riemann




=
sum in the form ∑ 
 = ∑ ( Δ ) Δ , where ( )
+
=  +
=



and Δ =
. The value of Δ corresponds to an interval of length
Since Δ =
for the initial value
=
the interval starts at
Therefore, the limit of the Riemann sum is equal to the definite integral
∫
(C)
( )
=⌠
⌡
+
Incorrect. The sum inside the limit can be interpreted as a right Riemann




=
sum in the form ∑ 
 = ∑ ( Δ ) Δ , where ( )
+
=  +
=



and Δ =
. The value of Δ corresponds to an interval of length
Since Δ =
for the initial value
=
the interval starts at
therefore the limits on the definite integral should go from
to
(D)
to
and
not
Incorrect. The sum inside the limit was correctly interpreted as a right
Riemann sum on an interval of length starting at = With that
choice of interval, however, the Riemann sum would be written as




=
The right Riemann
∑
 = ∑ ( + Δ ) Δ , where ( )
=  +
=



sum using
=
not
( )=
=
+
would correspond to an interval starting at
4
Q
Skill
2.B
(A)
Learning Objective
Topic
LIM-8.A
(BC ONLY) Finding Taylor
Polynomial Approximations
of Functions
Incorrect. This response would result if the chain rule was never used in
finding the first and second derivatives of
and ′′(
( )
)=
series for
(B)
in the Maclaurin
( )
′′( ) =
was computed as
=
Incorrect. This response would result if the coefficient of
in the
Maclaurin series for
was taken to be ′′( ) without dividing by
In
addition, the chain rule was never used in finding the first and second
derivatives of
(C)
As a result, the coefficient of
( )
′( ) =
leading to
( )
′′( ) =
leading to
=
Incorrect. This response would result if the chain rule was correctly used
in finding ′( ) but was not used again when the product rule was used
during the computation of the second derivative, as follows.
′( ) = ′( ) ( )
′′( ) = ′′( ) ( ) + ′( )
( )
′′( ) = ′′( ) ( ) + ′( ) ( ) =
The coefficient of
+
=
in the Maclaurin series for
was found to be
′′( ) =
(D)
Correct. The coefficient of
in the Maclaurin series for
( )
Since ( ) =
the chain rule gives ′(
product rule and the chain rule again gives
′′( ) =
′′( )
( )
+
′( )
(
The values of ( )
′( ) and
Maclaurin series for
=
( )=
′( ) =
=
′′( ) =
Therefore,
⋅
(
′′( ) =
the coefficient of
′( )
)=
(
( )
( )
) = ′( )
)=(
′′( )
is
Using the
′′( ) + ( ′( ) )
)
( )
′′( ) can be determined from the
⋅ =
′′( ) + ( ′( ) )
)
( )
=( + )
in the Maclaurin series for
is
and so
=
(
)=
4
Skill
2.E
Q
Learning Objective
Topic
FUN-4.A
Determining Intervals
on Which a Function Is
Increasing or
Decreasing
(A)
Incorrect. The graph of
is concave up where ′ is increasing.
This response might come from switching the roles of the function
and its derivative and thinking that
is increasing where the graph
of ′ is concave up. The graph of ′ is concave up on the intervals
(
) and (
)
(B)
Incorrect. This response might come from treating the given graph
as the graph of
rather than the graph of ′ These are the two
intervals where ′ is increasing.
(C)
Incorrect. These are the intervals where both
increasing.
(D)
Correct. The function
is increasing on closed intervals where ′
is positive on the corresponding open intervals. The graph indicates
that ′( ) > on the intervals (
is
) and ( ) so
increasing on the intervals [
] and [ ]
and
′ are
4
Q
Skill
1.E
(A)
Learning Objective
Topic
CHA-4.D
Using Accumulation
Functions and Definite
Integrals in Applied
Contexts
changes from = to
Incorrect. This response is how much
The amount by which
= that is, ′( ) − ′( ) =
changes from = to = is ( ) − ( ) which can be computed
by using the Fundamental Theorem of Calculus.
(B)
Incorrect. This response is the approximation to the change in
along the line tangent to the graph of at = It can also be
interpreted as the approximation to the integral
∫
′( )
= ( ) − ( ) by using a left Riemann sum with one
interval of length Δ = 5.
Δ ≈ ′(1) Δ = ′(1) ⋅ 5 = 8.341
(C)
Incorrect. This response is the approximation to the integral
∫
′( )
= ( ) − ( ) by using the trapezoidal sum
approximation over one interval of length Δ = 5.
 ′( ) + ′( )  ⋅ Δ =  ′( ) + ′( )  ⋅ 5 = 18.017








(D)
Correct. The change in from = to
Fundamental Theorem of Calculus,
( )− ( )=
∫
′( )
=
∫
−
=
(− )
is ( ) − ( ) By the
=
numerical integration is done with the calculator.
where the
4
Q
Skill
2.D
(A)
Learning Objective
Topic
LIM-2.D
Connecting Infinite
Limits and Vertical
Asymptotes
Incorrect. This graph displays the appropriate behavior as it
approaches the horizontal asymptote at = but in this graph
( ) = − ∞ and
( ) = +∞ which is the opposite
→ −
→ +
behavior for what the graph of
the vertical asymptote at =
should be doing as it approaches
(B)
Incorrect. This graph displays the appropriate behavior as it
approaches the vertical asymptote at = but it has = − as a
horizontal asymptote rather than =
(C)
Correct. Since
→ −
( ) = +∞ and
→ +
approaches the vertical asymptote at
( ) = −∞ the graph of
=
in the upward
direction as approaches from the left and approaches the vertical
asymptote in the downward direction as approaches from the
right. Since
( ) = and
( ) = the graph of
→− ∞
→+ ∞
approaches the horizontal asymptote at = in both horizontal
directions. This graph exhibits these properties and therefore could
be the graph of
(D)
Incorrect. This graph has a vertical asymptote at = and a
horizontal asymptote at = rather than a vertical asymptote at
= and a horizontal asymptote at =
4
Q
Skill
3.F
Learning Objective
Topic
CHA-4.B
Finding the Average
Value of a Function on
an Interval
(A)
Incorrect. The definite integral was not divided by the length of the
interval [
] over which the averaging is done.
(B)
Correct. The average value of a function
is
−
∫
( )
Tara’s average heart rate from =
is the average value of the function
(D)
−
∫
]
to =
over the interval [
would therefore be given by the expression
(C)
over an interval [
] and
()
Incorrect. This response is the average rate of change of Tara’s heart
rate from =
to =
not the average of her heart rate over
that interval. By the Fundamental Theorem of Calculus, this
( )− ( )
expression is equal to
−
Incorrect. This response is the average of the rate of change of Tara’s
heart rate at the two times =
and =
not the average of her
heart rate over the interval from =
to =
4
Q
Skill
1.E
(A)
Topic
LIM-8.B
(BC ONLY) Finding Taylor
Polynomial Approximations
of Functions
Incorrect. This response comes from only using three terms of the Taylor
polynomial rather than the Taylor polynomial of degree
(− +
(B)
Learning Objective
( − )+ ( − )
)
=
=−
Correct. The third-degree Taylor polynomial for
following.
( ) = ( ) + ′( )( − ) +
=− + ( − )+
( − ) +
about
′′( )( − ) +
=
is the
′′′( )( − )
( − )
=− + ( − )+ ( − ) + ( − )
(
(C)
)≈ (
)=− + (
) + (
) =−
Incorrect. This response would result if the coefficient of the degree
term was taken to be
(− +
(D)
)+ (
( )
( − )+ ( − ) +
( − )
Incorrect. The coefficient of the degree
rather than
(−
+
( )
( )
( ) rather than
)
=
( )
=−
term was taken to be
( )
( − )+ ( − ) +
( − )
)
=
=−
( )
( )
4
Q
Skill
1.E
(A)
Topic
CHA-6.A
(BC ONLY) The Arc Length of a
Smooth, Planar Curve and
Distance Traveled
Incorrect. This response comes from not taking the square root in the
integrand of the definite integral for the length of a curve, as follows.
∫ (
(B)
Learning Objective
+ ( ′( ) )
)

=⌠
  +
⌡ 
(
+
Correct. The length of the graph of
by the definite integral
∫

=⌠



=
∫ (
=
( ) from
=
(
)
)
+
(C)
+
=
+
is given
=
Incorrect. This response comes from not squaring the derivative in the
integrand of the definite integral for the length of a curve, as follows.
∫
(D)
+
=
=
to
∫
⌡
where the numerical integration is done with the calculator.
+ ( ′( ) )
)
+ ′( )
=
∫
+
(
+
)
Incorrect. This response is the change in
(
)− ( )=
∫
′( )
=
∫
=
=
from
+
=
to
=
4
Q
Skill
Learning Objective
Topic
2.D
FUN-7.C
Sketching Slope Fields
(A)
Correct. In the slope field for a differential equation of the form
= ( ) the slope at a point (
) depends only on the value of
The line segments in the slope field at each point on a vertical
line perpendicular to the -axis should therefore all have the same
slope. The line segments in this slope field show that behavior and
therefore this could be a slope field for a differential equation of the
form
(B)
= ( )
Incorrect. In the slope field for a differential equation of the form
= ( ) the slope at a point (
) depends only on the value of
The line segments in the slope field at each point on a vertical
line perpendicular to the -axis should therefore all have the same
slope. The line segments in this slope field do not exhibit that
behavior. The slopes of the line segments depend on both the - and
the -values.
(C)
Incorrect. In the slope field for a differential equation of the form
= ( ) the slope at a point (
) depends only on the value of
The line segments in the slope field at each point on a vertical
line perpendicular to the -axis should therefore all have the same
slope. The line segments in this slope field do not exhibit that
behavior. The slope of the line segment at a point depends only on
the value of
This could be a slope field for a differential equation
of the form
(D)
=
( )
Incorrect. In the slope field for a differential equation of the form
= ( ) the slope at a point (
) depends only on the value of
The line segments in the slope field at each point on a vertical
line perpendicular to the -axis should therefore all have the same
slope. The line segments in this slope field do not exhibit that
behavior. The slopes of the line segments depend on both the - and
the -values.
4
Q
Skill
Learning Objective
Topic
3.D
LIM-2.C
Removing
Discontinuities
(A)
Correct. The limit at
limits are equal.
=
( )=
( )⇒
exists if the left-hand and right-hand
+
=
−
The solution to this equation for > is =
With this
value of
is
( ) exists and is equal to ( ) Therefore,
→ −
→ +
→
=
continuous at
(B)
Incorrect. This response comes from trying to make the left-hand
and right-hand limits of the derivative equal at = as follows.
<


′( ) = 
>
( − )

→ −
′( ) =
→ +
′( ) ⇒
=
(
>
The solution to this equation for
(C)
−
)
Incorrect. In trying to set the left-hand and right-hand limits of
equal at = the might have been substituted for the parameter
rather than the variable
as follows.
+
=
−
=
The positive solution to this equation is
(D)
=
is
Incorrect. This response might come from errors that lead to an
equation that has no positive solution. For example, it might come
from trying to make the left-hand and right-hand limits of the
derivative equal at = but also making a chain rule error in the
derivative of the piece for > as follows.


′( ) = 


→
−
<
(
′( ) =
−
−
→
>
)
+
′( ) ⇒
=
This equation has no solution for
(
−
−
)
4
Q
Skill
2.D
Learning Objective
Topic
CHA-2.C
Defining the Derivative
of a Function and Using
Derivative Notation
(A)
Incorrect. The slope of the line tangent to the graph of
at =
is ′( ) Therefore, ′( ) is negative. Since the tangent line goes
through the point (
( ))
( ) is also negative. Therefore, it
cannot be true that ′( ) = − ( ) since both are negative.
(B)
Incorrect. The slope of the line tangent to the graph of
at =
is ′( ) and the tangent line goes through the point (
( ))
Therefore, ′( ) = − and ( ) = − so it is not true that
′( ) < ( )
(C)
Correct. The slope of the line tangent to the graph of
at =
′( ) and the tangent line goes through the point (
( ))
Therefore, ′( ) = − and ( ) = − so ′( ) = ( )
(D)
Incorrect. The slope of the line tangent to the graph of
at =
is ′( ) and the tangent line goes through the point (
( ))
Therefore, ′( ) = − and ( ) = − so it is not true that
′( ) > ( )
4
Q
Skill
1.E
(A)
is
Learning Objective
Topic
CHA-4.C
Connecting Position,
Velocity, and
Acceleration Functions
Using Integrals
Incorrect. This response is the referee’s displacement over the time
interval ≤ ≤ rather than the total distance the referee traveled.
∫
()
=∫
( − )
( + )
=
(B)
Incorrect. This response is the absolute value of the referee’s change
in velocity from time = to time =
( )− ( ) =
(C)
Correct. The referee’s total distance traveled on the time interval
≤ ≤
is
∫
()
=
∫
( − )
( + )
=
where the numerical integration is done with the calculator.
(D)
Incorrect. This response comes from averaging the referee’s initial
and final velocities on the time interval ≤ ≤
then multiplying
by the length of the time interval.
( )+ ( )
( )+ ( )
⋅Δ =
⋅ 4 = 29.156
4
Q
Skill
2.B
(A)
Learning Objective
Topic
FUN-5.A
Interpreting the Behavior of
Accumulation Functions Involving
Area
Correct. Because is continuous, the Extreme Value Theorem guarantees the
existence of an absolute maximum on the closed interval [ −
] and that the
maximum will occur at a critical value or at one of the endpoints. Since is an
= − and = Therefore, the
antiderivative of
′( ) = ( ) = at
candidates are = −
=−
= and = Evaluate at each candidate
and select the largest.
(− ) =
(− ) =
∫
∫
−
−
()
= −∫
()
= −∫
−
−
( )=
∫
()
=
( )=
∫
( )
=−
(
()
()
(
= − − ( )( ) +
=−
(
)
( )( ) = − ( − + ) =
)
( )( ) = −
)
( )( ) + ( + ) = − ( +
The maximum occurs at
)=−
=−
(B)
Incorrect. Both critical values might have been found, and the relative minimum
was picked while not accounting for the endpoints.
(C)
Incorrect. Both critical values might have been found, and the relative maximum
was picked while not accounting for the endpoints.
(D)
Incorrect. The four candidates might have been identified, but the computation
of ( ) found the area of the region from = to = and did not account
for the region being below the horizontal axis, as follows.
( )=
∫
()
=
(
)
( )( ) + ( + ) = ( +
)=
4
Q
Skill
1.E
(A)
Learning Objective
Topic
FUN-8.B
(BC ONLY) Solving
Motion Problems Using
Parametric and VectorValued Functions
Incorrect. This response would result if the problem was treated like
rectilinear motion. The velocity of the particle was taken to be the
value of the derivative
=
=
=
=
at time
( )
−
(
=
−
−
=
)
−
=−
=
Then the speed was taken to be the absolute value of velocity.
(B)
Incorrect. This response is the magnitude of the position vector at
time =
not the magnitude of the velocity vector at that time.
( ( )) + ( ( ))
(C)
)
−
+(
( ⋅
Correct. The speed of the particle at time
′( )
velocity vector
( ′( ) ) + ( ′( ) )
(D)
(
=
=
′( ) = −
=
=
(−
−
−
)) =
is the magnitude of the
( ) at time
)
+(
( ⋅
)) =
Incorrect. This response would result if the components of the vector
′( )
′( ) = −
−
( ) were not squared when finding the
=
magnitude of the vector at
′( ) + ′( )
=
=
(−
−
)+(
( ⋅
)) =
4
Q
Skill
1.E
Learning Objective
Topic
CHA-3.A
Interpreting the
Meaning of the
Derivative in Context
(A)
Incorrect. This response might be chosen if the calculation of the
average rate of change resulted in a value that was greater than or
less than −
It would also be chosen if the average rate of change
but the instantaneous rate of
was correctly found to be −
change was taken to be the second derivative of
not the first
derivative. In either case, the resulting equation would have no
solution in the interval [
]
(B)
Incorrect. This response would be chosen if the average rate of
change was correctly found to be −
but the graph of
not
′ was drawn to determine the number of intersection points with
the horizontal line = −
It would also be chosen if the
instantaneous rate of change was correctly identified as the derivative
of
but the average rate of change over the interval [
] was
thought to be the average at the endpoints,
( )+ (
)
=−
or the average value of the function
over the interval,
∫
( )
=−
In all these
cases, the resulting equation would have only one solution in the
interval [
]
(C)
Correct. The average rate of change of
(
)− ( )
=−
[
] is
−
of change of
is the derivative,
′( ) =
on the closed interval
The instantaneous rate
−
+
−
The
graph of ′ produced using the calculator, intersects the horizontal
line = −
three times in the open interval (
)
(D)
Incorrect. This response might be chosen because the function
a polynomial of degree
is
4
Q
Skill
3.F
Learning Objective
Topic
FUN-7.H
(BC ONLY) Logistic
Models with Differential
Equations
(A)
Incorrect. This differential equation is for a rate of change that is
equal to the difference between the carrying capacity and a term that
is proportional to the size of the population. There is no joint
proportionality with the size of the population and the difference
between the size of the population and the carrying capacity.
(B)
Incorrect. This differential equation would be for a model where the
population grows at a rate that is only proportional to the difference
between the carrying capacity and the size of the population. There is
no joint proportionality with the size of the population.
(C)
Incorrect. This differential equation would be for a model where the
population grows at a rate that is jointly proportional to the
reciprocal of the size of the population, not the population itself, and
the difference between the carrying capacity and the size of the
population.
(D)
Correct. The model for logistic growth that arises from the statement
“The rate of change of a quantity is jointly proportional to the size of
the quantity and the difference between the quantity and the carrying
( − ) where represents the quantity that is
=
capacity” is
changing and is the carrying capacity. Here the population
is
changing according to the logistic growth model with carrying
capacity
Therefore, the differential equation for the model
would be
=
(
− )
4
Q
Skill
Learning Objective
Topic
1.E
CHA-3.E
Solving Related Rates
Problems
(A)
Incorrect. This response would result if the area of the region was
taken to be = π ( − ) instead of π
In addition, an
−π
error was made in the power rule when doing the differentiation, as
follows.
=π( − )
(B)
(
− ) ( − ( − )) = π
Incorrect. This response would result if the area of the region was
taken to be
=
= π ( − ) instead of
π
−π
= − (since the inner radius
and
the instant when
= π( − )
(C)
) =π(
−
(
=
and
)=
−
=
It is given that
is decreasing). At
this gave
π ( − ) ( − ( − )) = π
Incorrect. This response would result if the area of the region was
correctly taken as = π
Using the chain rule gave the
−π
rate of change of the area with respect to time as
= π
− π
It is given that
=
but
was taken
to be rather than − by not taking into account that the inner
radius is decreasing. At the instant when = and = this
= π( )
gave
(D)
− π( )
= π ( ( ) − ( )) =
π
Correct. The area of the region is = π
Using the chain
−π
rule gives the rate of change of the area with respect to time as
= π
− π
(since the inner radius
and =
= π( )
It is given that
=
=−
and
is decreasing). At the instant when
− π( )
= π ( ( ) − ( − )) =
π
=
3
6 25 1
4
6 BC
1 6
LRQ
′( ) =
7
V
RI F
R
E
∫
F
∫
()
P
V
=
F V
R
R
()
7R
P
RI
V
=
()
{
=
V
RO
PE
R P =
≈ ( − )⋅
+(
= ⋅
( )+ ( )
− )⋅
+
F V
+
+ ⋅
OR I RP
( )+ ( )
+ ( − )⋅
( )+ ( )
V
+(
+ ⋅
−
)⋅
+
( )+ ( )
+ ⋅




R
OV P
R P R
O
R
+
=
∫
()
V
PE
R VE
∫
()
7R
P
=
RI F V
=
0
+
V
∫
ROO
()
O
OR
30
=
=
ROO V
9V
FROO F
I RP P
=
R




OP V
V
FR V
V
3
6 25 1
6
1 6
4
LRQ
( ′( ) ) + ( ′( ) ) =
E
V
7
F
V
P
V U
UV F
P
∫
( ′( ) ) + ( ′( ) )
=
7
=
UV
V F
F U
V V
F
E
)= +⌠
(
F
V
F
=
V F
U
V U U P P
UV
P
V UVU
⇒
=
V V
U
=
P V
U
=
=
V
( )=
(
7
)=
+⌠
=−
⌡
+⌠
=
⌡
U F E
V
F
V
V
V
9V
U
V
( )− ( )=
{
V







=
=
⌡
V
U
U
U
U
V
( )=
( )= +⌠
{
P
=
⌡
7
=
U
V UV
V
V V
FU F
U
V U
=
V
M V
F
3
6 25 1
4
RI F
1 6
LRQ
( ) − (− ) − −
=
− (− )
=
6 BC
V
=−
( )
( ) = − + FRV π = −
E
)R
<
(
(
( π )) ⋅ π = − +
′( ) = − + − V
T
R IR
F 7
O
O
− ( − ) ∫−
∫−
( )
RI
∫−
O − ≤
∫
=
∫
( )
 − + FRV π
=⌠
→ −
⌡
−
(
( )
( )= ( )=
=
F V
V FR
V
≤
)(
π
=
V
π
( ))
=
OV RI
− ≤ ≤
− )
( )
( )
OP
∫−
+
∫−
− ( − ) ∫−
(
=− + − +
V
R
( )











π
( )
=
′( )
T
R
( π )) ⋅ π
′( ) = − + − V
<



= −

+
V
( π )
V FR
R V
π
O
RI
R
≤
∫−
− + FRV
−
RI
( )
π
V
=−
(π− )
OP
→ +
( )=
VR
R V R [−
(
P 9 O
]
V
EVRO P P P R [ −
9V
7 R P
]
{
FR
V
=
M V IF R
≤
3
6 25 1
4
∫−
()
( )=
∫−
( )
=
∫−
()
( )=
E
=
+
−
∫
+
+
=
+
+ −

=
+
+ − ⋅

=
+
+
1 6
LRQ
=





( )
∫ (−
=
6 BC
+
+
(
+
O RI
−


⋅
− ⋅
( − (− )) =
) − (−
O
⋅
+
( )
R
RI R
)
− ⋅ 

⋅
[−





]F
(− ) =
′( ) = ( )
I V =−
=
VF
V
V
M V IF R
( )=
EVRO
F 7
P
RI
<
V
<
P P
O
V FR F
<
<
R [−
RI
R
R
E F V
OV
9V
≤ ≤
()
′( ) = ( ) V
IR − < <
OV
7 V
EVRO P P P
R O RFF
=− R
=
7
R
)
−
+
∫
( )
] V (− ) =
OV − <
′( ) =
( ) V
<−
F
V
R
{
OV
VR
3
6 25 1
4
(
+
⇒
+
+
⇒ (
+ )
⇒
=
E
=
1 6
LRQ
{
( )
=
−
+
( + )
−
)⋅
=
( ( + ))
U
>
+V θ
( + V θ) >
π
7
π
≤θ ≤
=
U
F E
π
( + )−





+
( ( + ))
≤θ ≤ π
⌠

⌡
π
π



UP



θ




P V
UP
U
V
T
U
U
U V
θ
F Vθ <
V
θ
>
( θ ) V FU V
U
⇒
+V θ
=
U
π
<θ <
V
U
−
=
−
=
⋅F Vθ
θ ( + V θ)
) U
P F
U F
=−
− ⋅ ( + ) − (−
=
F
)=
+
6
+V θ =
 −
+ V θ 
9V
⇒ V θ =−

 θ

⇒ θ =
π
U
Uθ =
π
F
V
U
3
6 25 1
6
1 6
4
7
V U V
=
S=
∞ −
(
∑
V
∑=
+
(− )
7 U
∞
∑=
=
U
V
S V U V
∞ −
(
V U V
∞ −
(
+
∑=
)
∑
(− )
∞
=
S=
F
⋅
+
∞
∑
=
=






=
F F
>
U V
+
)
=
⋅
∞ −
(
∑=
)
+






V U V V
UP
U
∞ −
(
FV U V
+
)
∑
V U V
=−
V U V
F
⋅
=
V
∞ −
(
+
)
∑
F V
UV
∞
∑
=
F
U V EV
F
F
V
UV
∞ −
(
∑
=
F
U VF
U V
∞
∑=
V
S V U V
U
V U V
S
7 U
F F
∞ −
(
∑
∞ −
(
∑=
)
+
(
⋅
8V
∞ −
(
∑=
− ) =
UPV
)
+
(
)
⋅
∞ −
(
∑=
≈
∞
∑ −S
=
∞
= −∑





S
=
U V
V
+
)
(
(
)
⋅
U





)
=
V U V
UPV
FU V
P
≈
P
U
EV
7 U
U
V U V UU U E
UU U E
E
−
<
U
P
⋅
9V
U
⋅
=
F
V
V
UV
U
∞
∑=
−
U S>
⋅
−
V
=
S>
F
S
=
S=
S
(− )
⋅
=
V U V
(− )
U V
+
)
U VF
⋅
=
7
∑
+
)
=
7 U
F 7
∑=
U
V
∞ −
(
U V EV
∞ −
(
V
U VE
=
S=
F
⋅
=
E 7
+
)
∑
V U V
+
)
=
∞
LRQ
<
UP
V
⋅
U P
V V U UP
V U V
UU U E
S
)
+
F
FW R
0 WS
FR
:R
RF
×
1 P
R
R W RI
W
FW R
4
W
6 W RQ 6 R
R QRW R Q
SR
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
R QRW R Q
6 P
W
6 W RQ
6 R
R QRW R Q
RPSR W
FR
W
6 W RQ 6 R
6 W RQ
W
6 R
RPSR W 6 R
5R Q WR Q D W
R Q P
36 R
RQ
RQ
D W
D
RPSR W
6 R 5DQ
36 R
W
F
4
FW R
0 WS
W RQ
FR
RF
×
1 P
R
R W RI
W
FW R
4
W
6 W RQ , 6 R
R QRW R Q
SR
×
W RQ
R W RI
4
R QRW R Q
×
W RQ
R W RI
4
R QRW R Q
×
W RQ 4
R W RI
R QRW R Q
6 P
W
6 W RQ ,,
6 R
R QRW R Q
RPSR W
FR
W
6 W RQ , 6 R
W
6 W RQ ,, 6 R
RPSR W 6 R
5R Q WR Q D W
R Q P
36 R
RQ
RQ
D W
D
6
R
RPSR W
6 R 5DQ
36 R
FR
:R
W
RQ
F
0
Question
Skill
Learning Objective
R
DF
DQ
RF
R PDQF
DD
RQ
Topic
The Fundamental Theorem of Calculus
and Definite Integrals
Using the First Derivative Test to Find Relative (Local)
Extrema
Key
% Correct
A
95
A
82
1
1.E
FUN-6.B
2
3.D
FUN-4.A
3
1.E
FUN-3.C
The Chain Rule
D
85
FUN-8.B
(BC ONLY) Solving Motion Problems Using
Parametric and Vector-Valued Functions
B
80
4
1.E
5
3.D
LIM-7.A
(BC ONLY) Working with Geometric Series
B
66
6
1.E
FUN-3.B
The Quotient Rule
A
88
7
2.C
LIM-8.G
(BC ONLY) Representing Functions as Power Series
A
69
B
81
C
68
B
67
Integrating Using Substitution
D
70
Connecting Limits at Infinity
and Horizontal Asymptotes
Defining Average and Instantaneous Rates
of Change at a Point
Determining Concavity of Functions over
Their Domains
8
2.B
LIM-2.D
9
3.B
CHA-2.B
10
2.B
FUN-4.A
11
1.E
FUN-6.D
12
3.D
FUN-1.B
Using the Mean Value Theorem
C
80
(BC ONLY) Approximating Solutions
Using Euler's Method
A
74
13
1.E
FUN-7.C
14
1.E
FUN-3.D
Implicit Differentiation
B
74
15
1.E
LIM-6.A
(BC ONLY) Evaluating Improper Integrals
C
61
Finding General Solutions Using Separation
of Variables
B
46
16
1.E
FUN-7.D
17
3.D
LIM-7.A
(BC ONLY) Alternating Series Test for Convergence
D
57
D
65
18
1.C
CHA-5.D
(BC ONLY) Finding the Area of the Region Bounded
by Two Polar Curves
19
3.B
LIM-7.A
(BC ONLY) Alternating Series Test for Convergence
D
69
20
1.C
FUN-6.A
Applying Properties of Definite Integrals
C
78
A
71
B
59
(BC ONLY) Radius and Interval of Convergence
of Power Series
(BC ONLY) Defining Polar Coordinates
and Differentiating in Polar Form
21
3.D
LIM-8.D
22
1.E
FUN-3.G
23
1.F
LIM-5.A
Approximating Areas with Riemann Sums
B
67
24
3.D
FUN-6.B
The Fundamental Theorem of Calculus
and Definite Integrals
A
65
25
1.E
FUN-6.E
(BC ONLY) Using Integration by Parts
A
33
C
53
D
58
C
40
B
37
D
22
26
3.D
LIM-7.A
27
1.C
CHA-5.C
28
1.E
CHA-3.G
29
2.C
LIM-5.C
30
2.B
LIM-8.A
(BC ONLY) Determining Absolute or Conditional
Convergence
Volume with Washer Method - Revolving Around
x- or y-axis
(BC ONLY) Second Derivatives of Parametric
Equations
Riemann Sums, Summation Notation,
and Definite Integral Notation
(BC ONLY) Finding Taylor Polynomial Approximations
of Functions
RQ
Question
Skill
F
R
DF
DQ
Learning Objective
76
2.E
FUN-4.A
77
1.E
CHA-4.D
78
2.D
LIM-2.D
79
3.F
CHA-4.B
80
1.E
LIM-8.B
81
1.E
CHA-6.A
82
2.D
FUN-7.C
83
3.D
LIM-2.C
R PDQF
DD
Topic
Key
% Correct
D
92
D
82
C
95
B
70
B
80
B
68
Sketching Slope Fields
A
71
Removing Discontinuities
A
72
C
63
C
62
A
49
C
72
Determining Intervals on Which a Function
Is Increasing or Decreasing
Using Accumulation Functions and Definite Integrals
in Applied Contexts
Connecting Infinite Limits and Vertical Asymptotes
Finding the Average Value of a Function
on an Interval
(BC ONLY) Finding Taylor Polynomial Approximations
of Functions
(BC ONLY) The Arc Length of a Smooth, Planar Curve
and Distance Traveled
Defining the Derivative of a Function
and Using Derivative Notation
Connecting Position, Velocity, and Acceleration
Functions Using Integrals
Interpreting the Behavior of Accumulation Functions
Involving Area
(BC ONLY) Solving Motion Problems Using
Parametric and Vector-Valued Functions
84
2.D
CHA-2.C
85
1.E
CHA-4.C
86
2.B
FUN-5.A
87
1.E
FUN-8.B
88
1.E
CHA-3.A
Interpreting the Meaning of the Derivative in Context
C
42
D
66
D
60
89
3.F
FUN-7.H
(BC ONLY) Logistic Models
with Differential Equations
90
1.E
CHA-3.E
Solving Related Rates Problems
RQ
RQ
Question
Skill
Learning Objective
Topic
Mean Score
1
1.E|3.D|3.F|4.D|4.A|4.B|4.C|4.E
CHA-2.D|CHA-4.E|LIM-5.A
2.3|8.3|6.2|8.3
7.33
2
1.D|1.E|3.B|3.D|4.D|4.E
FUN-8.B
9.6
5.05
3
1.D|1.E|3.C|3.D|4.A
CHA-2.A|CHA-2.C|CHA-4.B|FUN-1.C
2.1|2.2|8.1|5.2
3.03
4
1.D|1.E|2.B|2.E|4.A
FUN-6.A|FUN-4.A
6.6|5.5|5.6
4.89
5
1.D|1.E |3.E|3.G|4.C
FUN-3.D|FUN-4.E|FUN-3.G|CHA-5.D
3.2|5.12|9.7|9.9
5.04
6
1.E|3.B|3.D|3.E|4.A
LIM-7.A|LIM-8.C
10.9|10.5|10.12
3.64
Download