5 Steps to a 5 AP Calculus BC 2019

AP Calculus BC Practice Exam 2 427

88. The correct answer is (A).

∑∞

n= 1

n!+ 2 n

2 n·n!

=

∑∞

n= 1

n!

2 n·n!

+

∑∞

n= 1

2 n

2 n·n!

=

∑∞

n= 1

1

2 n

+

∑∞

n= 1

1

n!

The series

∑∞

n= 1

1

2 n

is a geometric series, so

∑∞

n= 1

1

2 n

=

1 / 2

1 − 1 / 2

=1.

Compare the series

∑∞

n= 1

1

n!

to the known

MacLaurin series

∑∞

n= 0

xn

n!

= 1 +x+

x^2

2!

+

x^3

3!

···=ex, and

atx=1,

∑∞

n= 0

xn

n!

=

∑∞

n= 0

1

n!

= 1 + 1 +

1

2!

+

1

3!

···=e^1. Thus,

∑∞

n= 1

1

n!

=e^1 −1.

(Note that the summation index changes from

n=0ton=1.) Therefore,

∑∞

n= 1

1

2 n

+

∑∞

n= 1

1

n!

= 1 +e− 1 =e. (Also, you

could have evaluated

∑∞

n= 1

1

n!

using your T1-89

calculator.)

89. The correct answer is (B).

y=

1

x^2

−

1

x^3

=x−^2 −x−^3

⇒y′=− 2 x−^3 + 3 x−^4 ⇒y′′= 6 x−^4 − 12 x−^5. Set

the second derivative equal to zero and solve.

6

x^4

−

12

x^5

=

6 x− 12

x^5

= 0

⇒ 6 x− 12 = 0 ⇒x=2. Alsoy′′< 0

forx<2 andy′′>0 forx>2.

90. The correct answer is (D).
    The logistic growth model to the logistic
    differential equation
    dp
    dt
    =kp

(

1 −

p

M

)

with

Mequal to the maximum value is

p=

M

1 +Ae−kt

. In this case,p=

200

1 +Ae−.^25 t

and

t=0, you havep=20. Thus,

20 =

200

1 +Ae(−.25)(0)

or 20=

200

1 +A

orA=9,

Thus, atp=100, 100=

200

1 + 9 e−.^25 t

and

entering the equation into your calculator, you

havet≈ 8. 789 ≈9 years.