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

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪


5 − 2𝑥, 𝑥 > 3





(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

⎧

⎨⎪⎪

⎩⎪⎪


5 − 2𝑥, 𝑥 > 3





(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

⎧

⎨

⎪⎪

⎩

⎪⎪


5 − 2𝑥, 𝑥 > 3





(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

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪


5 − 2𝑥, 𝑥 > 3





(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


5 − 2𝑥, 𝑥 > 3





(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

2020 AB mock exam AB1 and AB2 Solutions › uploads › 8 › 7 › 5 › 9 › 8759495 ›...

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Transcript of 2020 AB mock exam AB1 and AB2 Solutions › uploads › 8 › 7 › 5 › 9 › 8759495 ›...