2e AP Exam Prep Solutions Ch 08 WEB

AP? Exam Practice Questions for Chapter 8 1

AP? Exam Practice Questions for Chapter 8

1. Evaluate each series.

I:

3 n=1 3 n

=

3

n =1

1 n1

3

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

diverges.

II:

e n

n =1

sin

2

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

diverges.

III:

5

( n=1 n + 1)!

lim

n

(n

5 +

2)!

(n

+ 5

1)!

=

lim

n

1 n+2

= 0 1

p

>

2.

lim

n

5(n + 3) (n - 1)p

n p -1 1

=

lim

n

5n p + 15n p-1

(n - 1)p

= 5

Because

lim

n

is finite and positive,

5(n + 3) n=2 (n - 1) p

converges when p > 2.

So, the answer is C.

5.

cos x

=

n=0

(-1)n

x2n

(2n)!

( ) cos x2

=

(-1)n x2 n=0 (2n)!

2n

( ) ( ) ( ) ( ) x2 2

x2 4

x2 6

x2 8

= 1 - 2! + 4! - 6! + 8! -

= 1-

x4 2!

+

x8 4!

-

x12 6!

+

x16 8!

-

x2 cos x2

=

x2

-

x2 x4 2!

+

x2 x8 4!

-

x2 x12 6!

+

x2 x16 8!

-

=

x2

-

x6 2!

+

x10 4!

-

x14 6!

+

x18 8!

-

So, the answer is C.

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

2 AP? Exam Practice Questions for Chapter 8

6.

n =1

(-1)n +1

3n3 + 1

an

=

1 3n3 + 1

an +1 < 0.001

1

1

3(n + 1)3 + 1 < 1000

1000 < 3(n + 1)3 + 1

999 < 3(n + 1)3

333 < (n + 1)3

n > -1 + 3 333 n > 5.9313 n=6

7.

1 1+ x

= 1-

x+

x2

-

x3

+ + (-1)n xn

+

Graph

f (x)

=

1 1+ x

and

y

=

x2.

y y = x2

5

4

3

1

f(x) =

1 x + 1

-1 -1

x 12345

The graphs intersect at x = 0.755.

So, the answer is C.

So, the answer is B.

8. (a) P1(x) = g(3) + g(3)(x - 3) = 50 + 1630(x - 3)

So,

g(3.1)

50

+

160 3

(3.1

-

3)

55.333.

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

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

will lie below the graph of g. So, the approximation is less

than the actual value of g(3.1).

(b)

P3(x) = g(3) + g(3)(x - 3) + g2(!3)(x - 3)2 +

g(3)

3!

(

x

-

3)3

=

50

+

160 3

(

x

-

3)

+

414 12!(x

-

3)2

+

231! (x

-

3)3

=

50

+

160 3

(

x

-

3)

+

1481( x

-

3)2

+

7 2

(

x

-

3)3

So, g(3.1) 50 + 1630(3.1 - 3) + 1841(3.1 - 3)2 + 72(3.1 - 3)3

55.513.

(c) g(3.1) - P3(3.1) = R3(3.1)

=

g

(4) (

4!

z)

(

x

-

3)4

811243!(3.1 - 3)4

= 0.0005849 < 0.0006

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.

? 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 3

9. (a)

ex

=

1+

x

+

x2 2!

+

x3 3!

++

xn n!

+

e-x

= 1 + (- x)

+

(- x)2

2!

++

(- x)n

n!

+

=

1-

x

+

x2 2!

-

x3 3!

+

+

(-1)n x

n!

+

xe- x

=

x-

x

x

+

x x2 2!

-

x x3 3!

++

(-1)n x xn

n!

+

=

x

-

x2

+

x3 2!

-

x4 3!

++

(-1)n xn+1

n!

+

So, the first four nonzero terms are

x

-

x2

+

x3 2!

-

x4 . 3!

(b)

lim

x0

f (x)

-x x3

+

x2

=

lim

x0

xe- x

-x x3

+

x2

=

1 2

(c)

g(x) =

x te-t dt

0

=

x

t

0

-

t2

+

t3 2!

-

t4 3!

++

(-1)n tn+1

n!

+

dt

=

x

t

0

- t2

+

t3 2

-

t4 6

++

(-1)n tn+1

n!

+ dt

=

1 2

t

2

-

1 3

t

3

+

1 8

t

4

-

1 30

t

5

+

+

(-1)n (n + 2)n!

t

n

+

2

+

x

0

=

1 x2 2

-

1 x3 3

+

1 x4 8

-

1 x5 30

++

(-1)n xn+2 (n + 2)n!

+

So,

g

1 5

1 12 2 5

-

1 1 3 3 5

+

1 8

1 5

4

.

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.

(d)

an +1

<

1 90,000

( )-1 n+1 xn+3 (n + 3)(n + 1)!

<

1 90,000

(1 5)6 (6)(4!)

<

1 90,000

1 2,250,000

<

1 90,000

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

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

4 AP? Exam Practice Questions for Chapter 8

10. (a)

cos x

=

1-

x2 2!

+

x4 4!

-

+

(-1)n x2n (2n)!

+

( ) ( ) ( ) ( ) f (x) = cos x2 = 1 -

x2 2 2! +

x2 4 4! -

x2 6 6!

++

(-1)n x2 (2n)!

2n

+

= 1-

x4 2

+

x8 24

-

x12 720

++

(-1)n x4n (2n)!

+

(b)

lim

n

( )-1 n+1 x4(n+1) 2(n + 1)!

(2n)! (-1)n x4n

= lim x4 n

1

(2n + 2)(2n + 1)

=0

Because the series converges for all x, R = .

(c)

cos x2

1-

x4 2

+

x8 24

cos(1)

1-

(1)4

2

+

(1)8

24

=

13 24

an +1

=

x12 6!

=

(1)12

6!

=

1 720

<

1 500

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.

? 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 5

11. (a)

f (x)

= 1-

x2

+

x4 2!

-

x6 3!

+

x8 4!

-

x10 5!

+

f (x)

=

-2x

+

4x3 2

-

6x5 6

+

8x7 24

-

10x9 120

+

=

-2x

+

2x3

-

x5

+

1 x7 3

-

1 x9 12

+

f (x)

=

-2

+

6x2

-

5x4

+

7 x6 3

-

3 x8 4

+

So, f (0) = 0 and f (0) = -2.

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

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

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

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

(b)

f (x)

=

1-

x2

+

x4 2!

-

x6 3!

f (1)

= 1 - (1) +

(1)4

2!

-

(1)6

3!

=

1 2!

-

1 3!

Use the Alternating Series Test to find the error.

an +1

=

x8 4!

=

(1)8

4!

=

1 24

<

1 10

So,

f (1)

=

1 2!

-

1 3!

with an error

=

1 24

<

1. 10

(c)

y

=

f (x) = 1 - x2 +

x4 2!

-

x6 3!

+

x8 4!

-

x10 5!

+

y

=

-2x

+

2x3

-

x5

+

1 x7 3

-

1 x9 12

+

y

+

2xy

=

-

2x

+

2x3

-

x5

+

1 x7 3

-

1 x9 12

+

+ 2x1 -

x2

+

x4 2!

-

x6 3!

+

x8 4!

-

x10 5!

+

=

-

2x

+

2x3

-

x5

+

1 x7 3

-

1 x9 12

+

+ 2 x

-

2x3

+

x5

-

1 x7 3

+

1 x9 12

-

1 60

x11

=0

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

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download