2020 AB mock exam AB1 and AB2 Solutions › uploads › 8 › 7 › 5 › 9 › 8759495 ›...
Transcript of 2020 AB mock exam AB1 and AB2 Solutions › uploads › 8 › 7 › 5 › 9 › 8759495 ›...
2020 AB Mock Exam AB1 and AB2 SG
Exam created by Bryan Passwater [email protected]
Solutions by Ted Gott [email protected]
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(a)Showthat𝑔iscontinuousat𝑥 = 3.
g 3( ) = f 3( )2−5( ) = f 4( )
limx→3−g x( ) = f 4( ) lim
x→3+g x( ) = 5− 2 3( ) = −1
f 6( )− f 4( ) = ′f x( )dx4
6
⌠⌡
−4( )− f 4( ) = −3⇒ f 4( ) = −4( )− −3( ) = −1
limx→3g x( ) = g 3( )⇒ g x( ) is continuous at x = 3
3:
1: considers left and right sided limits
1: f 4( ) using FTC
1: conclusion with reasoning
⎧
⎨
⎪⎪⎪
⎩
⎪⎪⎪
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(b)Findtheabsolutemaximumvalueof𝑓ontheinterval[−2,10].Justifyyouranswer.
candidates are where ′f x( ) changes from positive to negative
and end points x = −2,x = 10
x = 4⇒ ′f x( ) changes from positive to negative
f 6( )− f −2( ) = ′f x( )dx−2
6
⌠⌡
f −2( ) = f 6( )− ′f x( )dx−2
6
⌠⌡ = −4− −5+11− 3⎡⎣ ⎤⎦ = −7
f 6( )− f 4( ) = ′f x( )dx4
6
⌠⌡
f 4( ) = f 6( )− ′f x( )dx4
6
⌠⌡ = −4− −3⎡⎣ ⎤⎦ = −1
f 10( )− f 6( ) = ′f x( )dx6
10
⌠⌡
f 10( ) = f 6( )+ ′f x( )dx6
10
⌠⌡ = −4+ 7 + 4⎡⎣ ⎤⎦ = 7
absolute maximum is f 10( ) = 7
3:
1: sets ′f x( ) = 0
1: identifies x = 4 as candidate1: conclusion with reasoning
⎧
⎨⎪⎪
⎩⎪⎪
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(c)For𝑥 ≠ 3, thefunction𝑘isdefinedby𝑘(𝑥) =3∫ 𝑓"(𝑡)𝑑𝑡#$
% − 4𝑥3𝑒!&($))* − 𝑥
.Itisknownthat lim$→#
𝑘(𝑥)
canbeevaluatedusingL"Hospital"sRule.Find𝑓(3)andevaluate lim$→#
𝑘(𝑥).
L'Hospital's Rule ⇒ limx→3
3e2 f x( )+5 − x( ) = 0
3e2 f 3( )+5 − 3= 0⇒ e2 f 3( )+5 = 1⇒ 2 f 3( )+5= 0
f 3( ) = − 52
limx→3
3 ′f t( )dt4
3x
⌠⌡ − 4x
3e2 f x( )+5 − x= lim
x→3
3 ′f 3x( ) 3( )− 4
3e2 f x( )+5 2 ′f x( )( )−1
=3 ′f 9( ) 3( )− 4
3e0 2 ′f 3( )( )−1=
3 0( ) 3( )− 4
3e0 2 3.5( )( )−1= −4
3 7( )−1= − 1
5
3:
1: f 3( )1: chain rule with 3e2 f x( )+5
1: answer with L'Hospital's Rule
⎧
⎨
⎪⎪
⎩
⎪⎪
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(d)Let𝑦 = ℎ(𝑥)beafunctionsuchthat𝑑𝑦𝑑𝑥
= 𝑓"(2𝑥) ∙ (3𝑦 − 1)where𝑦 >13.Find
𝑑!𝑦𝑑𝑥!
and
usethisexpressiontodetermineifthegraphofℎ(𝑥)isconcaveupordownatℎ(1) = 2.
dydx
= ′f 2x( ) ⋅ 3y −1( )d 2 ydx2 = ′′f 2x( ) 2( ) 3y −1( )+ ′f 2x( ) ⋅ 3
dydx
⎛⎝⎜
⎞⎠⎟
dydx 1,2( )
= ′f 2( ) ⋅ 3 2( )−1( ) = 5( ) 5( ) = 25
d 2 ydx2
1,2( )= ′′f 2( ) 2( ) 3 2( )−1( )+ ′f 2( ) ⋅ 3 25( )( )
= 0( ) 2( ) 3 2( )−1( )+ 5( ) ⋅ 3 25( )( ) = 375
d 2 ydx2
1,2( )= 375> 0⇒ h x( ) concave up at h 1( ) = 2
3:
1: product rule
1: expression for d 2 ydx2
1:d 2 ydx2 at 1,2( ) with
conclusion
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⎨
⎪⎪⎪
⎩
⎪⎪⎪
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(e)Considerthefunction𝑦 = ℎ(𝑥)frompart(d).For − 1 < 𝑥 < 5, onwhatinterval(s)isthe
functionℎ(𝑥)increasing? Giveareasonforyouranswer.
1: correct interval with reasoning
dydx
= ′f 2x( ) ⋅ 3y −1( ) y > 13⇒ 3y −1( ) > 0
′f x( ) > 0⇒ 0 < x < 4⇒ ′f 2x( ) > 0⇒ 0 < 2x < 4
′f 2x( ) > 0⇒ 0 < x < 2
h x( ) is increasing on 0 < x < 2
𝐀𝐁𝟏:Let𝑔bethefunctiondefinedby𝑔(𝑥) = 8𝑓(𝑥! − 5), 𝑥 ≤ 3
5 − 2𝑥, 𝑥 > 3
Thefunction𝑓istwicedifferentiableontheclosedinterval[−2, 10]andsatisfies𝑓(6) = −4.
Thegraphof𝑓", thederivativeof𝑓, isshowninthefigureabove.Thegraphof𝑓"hashorizontal
tangentlinesat𝑥 = 2, 𝑥 = 5, 𝑥 = 8, and𝑥 = 9.Theareasoftheregionsbetweenthegraph
of𝑓"andthe𝑥axisarelabeledinthefigure.
(f)Considerthedifferentialequationfrompart(d).Useseparationofvariablestofindageneral
solutionfor𝑦 = ℎ(𝑥)oftheform𝑙𝑛|3𝑦 − 1| = 𝑎𝑓(𝑏𝑥) + 𝐶.
dydx
= ′f 2x( ) ⋅ 3y −1( )⇒ 13y −1
dy = ′f 2x( )dx
13y −1
dy⌠⌡⎮
= ′f 2x( )dx⌠⌡ ⇒ 1
3ln 3y −1 = 1
2f 2x( )+C
ln 3y −1 = 32f 2x( )+C
2 :1: separation of variables1: answer with
supporting work
⎧
⎨⎪
⎩⎪
𝐀𝐁𝟐.Alargepizzachaindeliverspizzasnightlyfrom6PMto2AM.OnaFridaynight, theratethatthepizza chainreceivescallsfordeliveryismodeledbythedifferentiablefunction𝐶"(𝑡),measuredinhundreds ofcallsperhour, and𝑡ismeasuredinhourssince6PM(𝑡 = 0).Selectedvaluesof𝐶"(𝑡)are giveninthetableabove. (a)Usethedatainthetabletoapproximate𝐶""(7).Usingcorrectunits, interpretthemeaningof𝐶""(7) incontextoftheproblem.
The rate that the pizza chain receives delivery calls is decreasing at a rate of 0.4 hundred or 40 calls per hour per hour.
(b)Dothedatainthetablesupporttheconclusionthatthereisatime𝑡, 0 ≤ 𝑡 ≤ 8, atwhichthe pizzachainreceives300deliverycallsperhour? Giveareasonforyouranswer.
Yes. is differentiable and therefore continuous on
and 300 is between and so the IVT guarantees there is a time t = c between 0 and 8 when
(c)Usingcorrectunits, explainthemeaningofthedefiniteintegral z 𝐶"(𝑡)𝑑𝑡,
-incontextoftheproblem.
UsearightRiemannsumwiththefoursubintervalsindicatedbythedatainthetabletoapproximate
thevalueof z 𝐶"(𝑡)𝑑𝑡,
-.
is the hundreds of delivery calls received from t = 0 and
t = 8 hours or 6 PM to 2 AM.
or 1440
calls
′′C 7( ) ≈ ′C 8( )− ′C 6( )8− 2
=0.2( )− 1( )2
= −0.82
= −0.42 :
1: approximation1: interpretation with units⎧⎨⎩
′C t( )0 ≤ t ≤ 8 ′C 0( ) = 500 ′C 0( ) = 20
′C c( ) = 300
2 :1: approximation1: interpretation with units⎧⎨⎩
′C t( )dt0
8
⌠⌡
′C t( )dt0
8
⌠⌡ ≈ 4.5( ) 1( )+ 2.5( ) 3( )+ 1( ) 2( )+ 0.2( ) 2( ) = 14.4
3:1: interpretation with units1: right Riemann sum1: approximation
⎧
⎨⎪
⎩⎪
𝑡 (hours) 0 1 4 6 8
𝐶"(𝑡) (100!scalls hour⁄ ) 5 4.5 2.5 1 0.2
(d)ThetotalnumberofcallsthatthepizzachainreceivesfordeliveryonFridaynight, in100"𝑠ofcalls,
canbemodeledbythefunction𝐹, givenby𝐹(𝑥) =𝑥!
200 − 2𝑥where𝑥isthenumberofdelivery
drivers.Whenthereare60drivers, thenumberofdriversisdecreasingatarateof724driversper
hour.Accordingtothismodel, whatistherateofchangeoftotaldeliverycallswithrespecttotime,
in100"𝑠pizzasperhour, atthetimewhenthereare60driversdeliveringpizzas.
F x( ) = x2
200− 2x⇒ calls drivers( ) x = 60⇒ dx
dt= − 724
dFdt
= dFdx
⋅ dxdt
=200− 2x( ) 2x( )− −2( )x2
200− 2x( )2⎛
⎝⎜⎜
⎞
⎠⎟⎟dxdt
dFdt x=60
=200− 2 60( )( ) 2 60( )( )− −2( ) 60( )2
200− 2 60( )( )2⎛
⎝⎜⎜
⎞
⎠⎟⎟
− 724
⎛⎝⎜
⎞⎠⎟
= 168006400
⎛⎝⎜
⎞⎠⎟
− 724
⎛⎝⎜
⎞⎠⎟= 21
8⎛⎝⎜
⎞⎠⎟
− 724
⎛⎝⎜
⎞⎠⎟= − 4964
3:
1: quotient rule
1:ddtF x( )( )
1: answer
⎧
⎨⎪⎪
⎩⎪⎪
𝐏𝐎𝐈𝐍𝐓𝐕𝐀𝐋𝐔𝐄𝐒
𝑨𝑩𝟏
(a)Showthat𝑔iscontinuousat𝑥 = 3.
3: �1: considersleftandrightsidedlimits1: 𝑓(4)usingFTC1: conclusionwithreasoning
(b)Findtheabsolutemaximumvalueof𝑓ontheinterval[−2,10].Justifyyouranswer.
3: �1: sets𝑓"(𝑥) = 01: identifies𝑥 = 4and𝑥 = 9ascandidates1: answerwithjustification
(c)For𝑥 ≠ 3, thefunction𝑘isdefinedby𝑘(𝑥) =3∫ 𝑓"(𝑡)𝑑𝑡#$
% − 4𝑥3𝑒!&($))* − 𝑥
.Itisknownthat lim$→#
𝑘(𝑥)
canbeevaluatedusingL"Hospital"sRule.Find𝑓(3)andevaluate lim$→#
𝑘(𝑥).
3: �1: 𝑓(3)1: chainrulewith3𝑒!&($))*1: answerwithL"HRule
(d)Let𝑦 = ℎ(𝑥)beafunctionsuchthat𝑑𝑦𝑑𝑥
= 𝑓"(2𝑥) ∙ (3𝑦 − 1)where𝑦 >13.Find
𝑑!𝑦𝑑𝑥!
and
usethisexpressiontodetermineifthegraphofℎ(𝑥)isconcaveupordownatℎ(1) = 2.
3:
⎩⎪⎨
⎪⎧1: productrule
1: expressionfor𝑑!𝑦𝑑𝑥!
1:𝑑!𝑦𝑑𝑥!
at(1,2)withconclusion
(e)Considerthefunction𝑦 = ℎ(𝑥)frompart(d).For − 1 < 𝑥 < 5, onwhatinterval(s)isthe
functionℎ(𝑥)increasing? Giveareasonforyouranswer.
1:correctintervalswithreasoning
(f)Considerthedifferentialequationfrompart(d).Useseparationofvariablestofindageneral
solutionfor𝑦 = ℎ(𝑥)oftheform𝑙𝑛|3𝑦 − 1| = 𝑎𝑓(𝑏𝑥) + 𝐶.
2: 81: separationofvariables1: answerwithsupportingwork
𝐏𝐎𝐈𝐍𝐓𝐕𝐀𝐋𝐔𝐄𝐒
𝑨𝑩𝟐
(a)Usethedatainthetabletoapproximate𝐶""(3).Usingcorrectunits, interpretthemeaningof𝐶""(3) incontextoftheproblem.
2: 81: approximation1: interpretationwithunits
(b)Dothedatainthetablesupporttheconclusionthatthereisatime𝑡, 0 ≤ 𝑡 ≤ 8, atwhichthe pizzachainreceives300deliverycallsperhour? Giveareasonforyouranswer.
2: 81: C"(8) < 3 < C′(0)1: conclusionusingIVT
(c)Usingcorrectunits, explainthemeaningofthedefiniteintegral z 𝐶"(𝑡)𝑑𝑡,
-incontextoftheproblem.
UsearightRiemannsumwiththefoursubintervalsindicatedbythedatainthetabletoapproximate
thevalueof z 𝐶"(𝑡)𝑑𝑡,
-.
3: �1: interpretationwithunits1: rightRiemannsum1: approximation
(d)ThetotalnumberofcallsthatthepizzachainreceivesfordeliveryonFridaynight, in100"𝑠ofcalls,
canbemodeledbythefunction𝐹, givenby𝐹(𝑥) =𝑥!
200 − 2𝑥where𝑥isthenumberofdelivery
drivers.Whenthereare60drivers, thenumberofdriversisdecreasingatarateof724driversper
hour.Accordingtothismodel, whatistherateofchangeoftotaldeliverycallswithrespecttotime,
in100"𝑠pizzasperhour, atthetimewhenthereare60driversdeliveringpizzas.
3: �
1: quotientrule
1:𝑑𝑑𝑡(𝐹(𝑥))
1: answer