ap® exam practice questions for chapter 8 · 2020. 11. 25. · ap® exam practice questions for...
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AP® Exam Practice Questions for Chapter 8 1
© 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.
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.