AP® Exam Practice Questions for Chapter 8 1

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AP® Exam Practice Questions for Chapter 8

1. Evaluate each series.

I: 1 33

1 1

3 13

n n nn

∞ ∞

= ==

Because 1 3p = and 0 1,p< < the series

diverges.

II: 1 sin 2

n

n

e∞

=

Because sin 2 2.99 1,r e= ≈ > the series

diverges.

III: ( )1

5

1 !n n

∞

= +

( )

( )1 !5 1lim lim

2 ! 5 2

0 1

n n

nn n→∞ →∞

+⋅ =

+ +

= <

By the Ratio Test, the series converges. So, the answer is C.

2. ( )

( )

3

1

3 3

1 1

3

1

4

1

4 4

lim 04

n

nn

n n n

nn

n

n na

n

∞

=

+ +

→∞

−

+= <

=

By the Alternating Series Test, the series converges.

( ) ( )( )

133 3

1

1lim lim lim

4 44

11

4

nn nn nnn n nn

nn n→∞ →∞ →∞

−= =

= <

By the Root Test, the series converges absolutely.

So, the answer is B.

3. ( ) ( )2ln 3 ln 3

1 ln 32! !

n

n+ + + + +

Let ln 3.x =

2

ln 3

12! !

3

n

x

x xxn

e

e

= + + + + +

=

==

So, the answer is C.

4. ( )

( )2

5 3

1p

n

nn

∞

=

+

−

Use the Limit Comparison test to compare with

12

1,p

n n

∞

−= which converges for 1 1 2.p p− > >

( )

( ) ( )1 15 3 5 15

lim lim11 1

5

p p p

p pn n

n n n nn n

− −

→∞ →∞

+ +⋅ =− −

=

Because limn→∞

is finite and positive, ( )

( )2

5 3

1p

n

nn

∞

=

+

−

converges when 2.p >

So, the answer is C.

5. ( ) ( )

( ) ( )( )

( ) ( ) ( ) ( )

2

0

222

0

2 4 6 82 2 2 2

4 8 12 16

cos 12 !

1cos

2 !

12! 4! 6! 8!

12! 4! 6! 8!

nn

n

nn

n

xxn

xx

n

x x x x

x x x x

∞

=

∞

=

= −

−=

= − + − + −

= − + − + −

2 4 2 8 2 12 2 16

2 2 2

6 10 14 182

cos2! 4! 6! 8!

2! 4! 6! 8!

x x x x x x x xx x x

x x x xx

⋅ ⋅ ⋅ ⋅= − + − + −

= − + − + −

So, the answer is C.

2 AP® Exam Practice Questions for Chapter 8

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6. ( ) 1

3 31

1 1

3 1 3 1

n

nn

an n

+∞

=

− =

+ +

( )( )( )

( )

1

3

3

3

3

3

0.001

1 1

10003 1 1

1000 3 1 1

999 3 1

333 1

1 333

5.9313

6

na

n

n

n

n

nnn

+ <

<+ +

< + +

< +

< +

> − +>

=

So, the answer is B.

7. ( )2 311 1

1n nx x x x

x= − + − + + − +

+

Graph ( ) 1

1f x

x=

+ and 2.y x=

The graphs intersect at 0.755.x =

So, the answer is C.

8. (a) ( ) ( ) ( )( ) ( )1160

3 3 3 50 33

P x g g x x′= + − = + −

So, ( ) ( )1603.1 50 3.1 3 55.3.

3g ≈ + − ≈

Because ( )g x′ is increasing on [ ]2, 4 , ( )3 0.g′′ > So, ( )g x

is concave upward at 3,x = and the tangent line at 3x =

will lie below the graph of g. So, the approximation is less than the actual value of ( )3.1 .g

(b) ( ) ( ) ( )( ) ( )( ) ( )( )

( ) ( ) ( )

( ) ( ) ( )

2 33

2 3

2 3

3 33 3 3 3 3

2! 3!160 141 21

50 3 3 33 4 2! 3!

160 141 750 3 3 3

3 8 2

g gP x g g x x x

x x x

x x x

′′ ′′′′= + − + − + −

= + − + − + −⋅

= + − + − + −

So, ( ) ( ) ( ) ( )2 3160 141 73.1 50 3.1 3 3.1 3 3.1 3

3 8 255.51308.

g ≈ + − + − + −

≈

(c) ( ) ( ) ( )( ) ( )

( ) ( )

( ) ( )( ) ( )

11

3

44 4

3.1 3.1 3.11 !

11233 3.1 3 0.0005849 0.0006

4! 8 4!

nn

ng z

g P R x cn

g zx

++− = = −

+

= − = − = <⋅

−1 1 2 3 4 5−1

1

3

4

5

y = x2

1x + 1

f(x) =

x

y

Reminders: Round each answer to at least three decimal places to receive credit for the approximation.

You do not need to simplify the coefficients in these Taylor polynomials.

In these approximations, be sure to write

rather than Because this is an

approximation, a point may be deducted if an equal sign is used.

AP® Exam Practice Questions for Chapter 8 3

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9. (a) 2 3

12! 3! !

nx x x xe x

n= + + + + + +

( ) ( ) ( )

( )

2

32

12! !

11

2! 3! !

nx

n

x xe x

n

xx xxn

− − −= + − + + + +

−= − + − + + +

( )

( )

2 3

13 42

1

2! 3! !

1

2! 3! !

n nx

n n

x xx x x xxe x x xn

xx xx xn

−

+

− ⋅ ⋅⋅ ⋅= − ⋅ + − + + +

−= − + − + + +

So, the first four nonzero terms are 3 4

2 .2! 3!

x xx x− + −

(b) ( ) 2 2

3 30 0

1lim lim

2

x

x x

f x x x xe x xx x

−

→ →

− + − += =

(c) ( )

( )

( )

( )( )

( )( )

0

13 42

0

13 42

0

2 3 4 5 2

0

2 3 4 5 2

1

2! 3! !

1

2 6 !

11 1 1 1

2 3 8 30 2 !

11 1 1 1

2 3 8 30 2 !

x t

n nx

n nx

xnn

nn

g x te dt

tt tt t dtn

tt tt t dtn

t t t t tn n

x x x x xn n

−

+

+

+

+

=

− = − + − + + +

− = − + − + + +

−= − + − + + +

+

−= − + − + + +

+

So, 2 3 4

1 1 1 1 1 1 1.

5 2 5 3 5 8 5g ≈ − +

(d)

( )( )( )

( )( )( )

1

1 3

6

1

90,000

1 1

3 1 ! 90,000

1 5 1

6 4! 90,000

1 1

2,250,000 90,000

n

n n

a

xn n

+

+ +

<

−<

+ +

<

<

1 pt: answer with justification [using results from part (a)]

Reminders: This approximation does not need to be simplified.

Write rather than

Because this is an approximation, a point may be deducted if an equal sign is used.

Reminder: This error bound does not need to be simplified.

4 AP® Exam Practice Questions for Chapter 8

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10. (a) ( )

( )22 4 1

cos 12! 4! 2 !

n nxx xxn

−= − + − + +

( ) ( ) ( ) ( ) ( ) ( )( )

( )( )

2 4 6 22 2 2 22

44 8 12

1cos 1

2! 4! 6! 2 !

11

2 24 720 2 !

nn

n n

x x x xf x x

n

xx x xn

−= = − + − + + +

−= − + − + + +

(b) ( ) ( )

( )( )

( ) ( )( )

1 4 14

4

1 2 ! 1lim lim 0

2 2 2 12 1 ! 1

n n

n nn n

x nx

n nn x

+ +

→∞ →∞

−⋅ = =

+ + + −

Because the series converges for all x, .R = ∞

(c)

( ) ( ) ( )

4 82

4 8

cos 12 24

1 1 13cos 1 1

2 24 24

x xx ≈ − +

≈ − + =

( )

12

1

12

6!

1

6!

1 1

720 500

nxa + =

=

= <

Reminder: You do not need to simplify the coefficients in these Taylor polynomials.

Reminders: Write rather

than Because this is an

approximation, a point may be deducted if an equal sign is used.

This error bound does not need to be simplified.

AP® Exam Practice Questions for Chapter 8 5

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

11. (a) ( )4 6 8 10

212! 3! 4! 5!

x x x xf x x= − + − + − +

( )3 5 7 9

3 5 7 9

4 6 8 102

2 6 24 1201 1

2 23 12

x x x xf x x

x x x x x

′ = − + − + − +

= − + − + − +

( ) 2 4 6 87 32 6 5

3 4f x x x x x′′ = − + − + − +

So, ( ) ( )0 0 and 0 2.f f′ ′′= = −

Because ( )0 0,f ′ = f has a critical value at 0.x =

Because ( )0 0,f ′′ < f is concave downward at 0.x =

So, by the Second Derivative Test, f has a relative maximum at 0.x =

(b) ( )

( ) ( ) ( ) ( )

4 62

4 6

12! 3!

1 1 1 11 1 1

2! 3! 2! 3!

x xf x x

f

= − + −

= − + − = −

Use the Alternating Series Test to find the error.

( )88

1

1 1 1

4! 4! 24 10n

xa + = = = <

So, ( ) 1 11

2! 3!f = − with an error

1 1.

24 10= <

(c)

2

2 2

2

2 0

2

2

12

12

ln

when 1

x C

x x

y xyy xy

dy xydx

dy x dxy

dy x dxy

y x C

y e

y Ce e C

− +

− −

′ + =′ = −

= −

= −

= −

= − +

=

= = =

( ) ( ) ( )

( )

2

2 3

2 32 22

4 62

12! 3!

12! 3!

12! 3!

x

x

x xe x

x xe x

x xx f x

−

= + + + +

− −= + − + + +

= − + − + =

So, ( ) 2xf x e−= is a solution of the differential

equation 2 0.y xy′ + =

Reminder: Explicitly identify each function by name. Referring to “it,” “the function,” or “the graph” will not receive credit on the exam.

6 AP® Exam Practice Questions for Chapter 8

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12. (a) ( ) ( ) ( )1

1

1 1

n n

n

xf x

n

−∞

=

− −=

( )

( )( )

11 1

lim lim1 1

1lim

11

nn

nn nn

n

xa na n x

n xn

x

++

→∞ →∞

→∞

− = ⋅

+ − −

=+

= −

1 1

1 1 1

0 2

x

xx

− <

− < − << <

When 0:x = ( ) ( ) ( )1 2 1

1 1 1 1

n n n

n n n

+ +− − −= = −

The series diverges (divergent p-series).

When 2:x = ( ) ( ) ( )1 1

1 1 1

n n n

n n

+ +− −=

The series converges by the Alternating Series Test.

So, the interval of convergence is ( ]0, 2 .

(b) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

2 3 4

2 3

2 3

1 1 11

2 3 4

1 0 1 1 1

1 1 1 1 1 1n n

x x xf x x

g x f x x x x

x x x x

− − −= − − + − +

′= = − − − + − − − +

= − − + − − − + + − −

(c) ( ) ( ) ( ) ( ) ( ) ( )2 31 1 1 1 1 1

1

n ng x x x x x

x

= − − + − − − + + − −

=

(d) ( ) ( )( ) ( ) ( )

( )( )

3

3 2

3 2

23

2

3

1

1 3

1 3

13

1

3

1

h x f x

h x f x x

g x x

xx

xx

= +

′ ′= + ⋅

= +

= ⋅+

=+

1 pt: answer

2 pts: finds

[using g from part (c)]

AP® Exam Practice Questions for Chapter 8 7

© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

13. (a) Because ( )2 4,g′ = the equation of the tangent line

at ( )2, 3 is

( )3 4 2

4 5.

y x

y x

− = −

= −

So, ( ) ( )2.4 4 2.4 5 4.6.g ≈ − =

Because ( ) ( )2 0, g g x′′ > is concave upward at 2,x =

and the tangent line at 2x = will lie below the graph

of g. So, the approximation is less than ( )2.4 .g

(b) ( ) ( ) ( )

( )

2.4

20.2 2.1 2.3

0.2 5 8

2.6

g x dx g g′ ′ ′= +

= +

=

( ) ( ) ( )2.4

22.4 2

3 2.6

5.6

g g g x dx′≈ +

= +=

(c) ( ) ( ) ( )( )

( ) ( ) ( )( )

2.2 2 0.2 2

3 0.2 4 3.8

2.4 2.2 0.2 2.2

3.8 0.2 7 5.2

g g g

g g g

′≈ + = + =

′≈ + = + =

(d) ( ) ( ) ( )( ) ( )( )

( ) ( )

( ) ( )

2

2

2

2

2 22 2 2

2!20

3 4 2 22

3 4 2 10 2

g xP x g g x

x x

x x

′′ −′= + − +

= + − + −

= + − + −

So, ( ) ( ) ( )22.4 3 4 2.4 2 10 2.4 2

6.2.

g ≈ + − + −

=

Error: ( )( )

332 6 2 8 1

2.4 23! 6 5 125 10

g′′′ − = = <

Reminders: This approximation does not need to be simplified.

Write rather than Because

this is an approximation, a point may be deducted if an equal sign is used.

Reminder: Write rather than

Because this is an approximation,

a point may be deducted if an equal sign is used.

Reminders: This approximation does not need to be simplified.

Write rather than Because

this is an approximation, a point may be deducted if an equal sign is used.

Reminders: This approximation does not need to be simplified.

Write rather than Because

this is an approximation, a point may be deducted if an equal sign is used.