CALCULUS BC PRACTICE EXAM #2 SOLUTIONS

Revised: 3/26/2015

1 ( ) ( ) ( ) ( ) ( )

5 5 5 55

11 1 1 1

2 2 1 2 1 2 2 4 5 1 0 A f x f x f x x= ⇒ − ⇒ − = − = − − = ∴∫ ∫ ∫ ∫

2 3/ 6/ 62 3

00

1 1 1 1 1 12 13cos sin cos sin sin A3 3 2 2 24 24 24

x x x x xπ

π ⎛ ⎞⎤+ ⇒ + ⇒ + = + = ∴⎜ ⎟⎦ ⎝ ⎠∫

3 33 3 3sin cos AND 3 B

cos

tt tdx dy dy ex t t y e e

dt dt dx t= ⇒ = = ⇒ = ⇒ = ∴

4 ( )2

20 0

sin 3 sin3lim limx x

x xx→ →

⇒x

sin3x⎛ ⎞⎜ ⎟⎝ ⎠ x

33

⎛ ⎞⎜ ⎟⎝ ⎠

33

⎛ ⎞⎜ ⎟⎝ ⎠

9 D⎛ ⎞ = ∴⎜ ⎟⎝ ⎠

5 ' 4 2cos 2 '' 4 4sin 2 0 sin 2 1 2 A

2 4y x x y x x x xπ π= + ⇒ = − = ⇒ = ⇒ = ⇒ = ∴

6 ( ) ( )

2 2 '

2 2

4 4 0 2, 2 lim lim 2 2 4 E2 2 0 1

L H

x x

x x xf x xx x→ →

− −= ≠ ⇒ = ⇒ = = ∴− −

7 ( ) ( ) ( )

443/ 2 1/ 2 5/ 2 3/ 2

11

5 3 2 2 2 32 8 1 1 2 38 76 Ex x x x ⎤+ = + ⇒ + − + ⇒ = ∴⎡ ⎤⎣ ⎦⎦∫

8 4 2 3

2 2

2

33 13 3 1/ 3 1 0 3lim lim Ccos2 cos 1/ 2 0 22x x

x x x xxx x x

x→∞ →∞

−⎡ ⎤− −⇒ = = ∴⎢ ⎥+ +⎣ ⎦ +

9 ( ) 2

2 2

1, 2,0 sin cos C 1 2 C C 3sin 2

So, cos 3 cos 3 B2 2

dy x ydy xdx y xdx y

x xy y

= ⇒ = ⇒ − = + ⇒ − = + ⇒ = −

− = − ⇒ = − + ∴

∫ ∫

10 ( )

( ) ( ) ( ) ( ) ( )

2 2

/ 2/ 2

0 0

1/ 2 1/ 2 1/ 2 1/ 2

1 100

00

sinlim ln 1 cos lim ln 1 0 ln 1 cos 0 Yes1 cos

lim 1 2 1 lim 2 1 1 2 0 1 2 No

1 1 1lim lim2 2

aa aab b

b b

bbx x

b b

x x ax

x x b

xe e

ππ

+ +

− −

→ →

−

→ →

− −

→∞ →∞

= − ⎤ ⇒ ⎡ − − − ⎤ = − −∞ = ∞ ∴⎦ ⎣ ⎦−

⎤ ⎡ ⎤− = − − ⇒ − − − = − − = ∴⎦ ⎣ ⎦

⎤ ⎛ ⎞= − ⇒ −⎜ ⎟⎥⎦ ⎝ ⎠

∫

∫

∫ ( )2

1 11 0 1 No A2 2be

⎡ ⎤ ⎛ ⎞− ⇒ − − = ∴ ∴⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦

11 ( ) ( )1/ 2 1 ' 2 1

2f x x x ⎛ ⎞= − + ⎜ ⎟⎝ ⎠

( ) ( )1/ 22 1 2x −− ( ) ( )1/ 2 1 14 ' 5 9 5 D3 3

f ⎛ ⎞⇒ = + = ∴⎜ ⎟⎝ ⎠

12

( ) ( )

( )

0 0

0

sin 1 sin sin2 2 2lim lim sin ' cos ' cos 0

2 2

sin 12 lim ' where sin E

2

h h

h

h hf x x f x x f

h h

hf f x x

h

π π ππ π

ππ

→ →

→

⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟ ⎛ ⎞⎝ ⎠ ⎝ ⎠⇒ ⇒ = ⇒ = ⇒ = =⎜ ⎟⎝ ⎠⎛ ⎞+ −⎜ ⎟ ⎛ ⎞⎝ ⎠∴ = = ∴⎜ ⎟⎝ ⎠

CALCULUS BC PRACTICE EXAM #2 SOLUTIONS

Revised: 3/26/2015

13 ( )2 2

2 24

00

1 1Area 1 A2 2

x xxe dx e e⎤= = = − ∴⎦∫

14 ( ) ( )

( ) ( )

( ) ( )

2 2

2

I. lim 0 only if is alternating

1 1II. Converges and So by Comparison Test converges

1III. No. If then may diverge B

xf x f x

f x f xx x

f x f xx

→∞=

<

> ∴

∑ ∑

∑

15 ( ) ( ) ( )( )

( ) ( ) ( )

( ) ( )

2 2 2 22 2

2 2 22 2 2

0 04 4 2

2 16 2 2 2 32 4 32 2 ' 0 32 2 0 16 4, 4 16 16 16

2 4 1' Absolute Max at 4 D4 16 4

x x x x x xf x x x xx x x

f x x y− + −−

+ − + − −= = = = ⇒ − = ⇒ = ⇒ = −+ + +

←⎯⎯⎯⎯⎯⎯→ ∴ = ⇒ = = ∴+

16 ( )( )( )

I. Yes. By IVT, ' 0 somewhere where 1 2

II. No. ' might equal zero, but we can't guarantee it.

III. No. ' might equal zero, but we can't guarantee it. A

f x x

f x

f x

= < <

∴

17 ( ) ( )1 9 8MVT: ' slope 2 somewhere where 1 3 B

3 1 4f c x− −= = = = − − ≤ ≤ ∴

− −

18 ( )

( )

( )

I. "Height" 0

II. "Taller" Increasing ' 0

II. "Grown Lesser Amount" Increasing at a DECREASING rate '' 0 B

h t

h t

h t

⇒ >

⇒ ⇒ >

⇒ ⇒ < ∴

19 ( ) ( )( )

( ) ( ) [ ]

22 2 2 2

22

0 03 3

6 9 6 2' 0 6 54 12 0 6 54 0 9 3, 3

9

' 3,3 E Remember to plug into "original" '

x x xy x x x x x

x

f x y− + −−

+ −= = ⇒ + − = ⇒ − + = ⇒ = ⇒ = −

+

←⎯⎯⎯⎯⎯⎯→ ⇒ ∴ −

20 ( ) ( ) ( ) ( )

( ) ( )( )

( ) ( )( )

( )

( )( )

und2 2 1

3 und3 1

1 1 ' ' 1 1 1

2 '' 2 1 '' 1

I. ' 0 No maxima.

II. '' 0 2 No inflection point at 0. Note: There IS an

x xxf x f x f xx x x

f x x f xx

f x

f x

+ +−

− + −−

+ −= ⇒ = = ⇒ ←⎯⎯⎯→

+ + +−= − + = ⇒ ←⎯⎯⎯→+

≠ ∴

= − ∴ = [ ]( )

inflection point at 1

III. 1 and YES there is a Vertical Asymptote at 1 B

x

f x

= −

− =∅ = − ∴

CALCULUS BC PRACTICE EXAM #2 SOLUTIONS

Revised: 3/26/2015

21 ( ) [ ] ( )( ) ( )

12 2 ! 1 Ratio Test lim • lim 2 0 1 for all ! 1 ! 12

So ROC E

n n

nn n

x x n x xn n nx

+

→∞ →∞

− − ⎛ ⎞⇒ ⇒ ⇒ − = <⎜ ⎟+ +⎝ ⎠−

= ∞ ∴

∑

22 ( )( )

( )

2 2

222

1

6Intersections: 4 6 4 4 3 2 0 2 1 0 2,11

6Washer Volume 4 C1

x x x x x x x x xx

x dxx

π

= − ⇒ = − + − ⇒ − + = ⇒ − − = ⇒ =+

⎛ ⎞⇒ = − − ∴⎜ ⎟−⎝ ⎠∫

23 ( )

( ) ( ) ( ) ( ) ( )1,1 1,1 1,1 1,1 1,1

From looking at graph of slope field, at 1,1 0

A) 1 B) 1 C) 1 D) 1 E) 1

From looking at graph of slope field, at 0, 0

dydx

dy dy dy dy dydx dx dx dx dx

dyydx

− − − − −

− >

= − = − = − = =

= = E∴

24 ( ) ( ) ( )( ) ( )( ) ( )( ) ( ) ( )( )

6

0

1 1 1 1 5 36 1 6 2 6 2 1 2 3 1 3 1 3 3 8 3 14 4 18 D2 2 2 2 2 2

g f x= = + + + + + + = + + + + = + =∫

25 ( ) ( ) ( )' 1 AND ' 2 0 AND ' 4.5 0 Bf f f=∅ < = ∴ 26

( )( ) ( ) ( )9 10 A B 9 10 A 2 B 2 32 3 2 2 3 2

3 7 72 28 7B B 4 AND A A 1 2 2 2

A B 1 4 1 ln 2 3 4ln 2 C A2 3 2 2 3 2 2

x x x xx x x x

x x

x xx x x x

+ ⇒ + ⇒ + = − + ++ − + −

− − −= = ⇒ = = = ⇒ =

+ = + ⇒ + + − + ∴+ − + −

∫ ∫

∫ ∫

27 ( ) ( ) ( ) ( )2 2 2 2 2 0 2 1 Bt t tx t e v t e a t e a e e− − − −= ⇒ = − ⇒ = ⇒ = = = ∴ 28

( ) ( ) ( )

( )

( ) ( ) ( )

2

1 2

2

1 1 2 2

2 2

t cos,sin C , C 2

cos 00 1C 1 C 1 AND C 2 C 2 2

t cos 1 2 cos 2 11, 2 2 1, 2 3,2 E2 2

tv t t t s t

ts t s

πππ

ππ π

π ππ π π π

⎛ ⎞−= ⇒ = + + ⇒⎜ ⎟⎝ ⎠

−+ = ⇒ = + = ⇒ = + ⇒

⎛ ⎞ ⎛ ⎞− −= + + + ⇒ = + + + = ∴⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

End of part A

CALCULUS BC PRACTICE EXAM #2 SOLUTIONS

Revised: 3/26/2015

29 Graph velocity (derivative of

position) and see how many times velocity is zero from 0 to 2 seconds (window).

Answer is 3 times C∴

30 ( )

( ) ( )

( ) ( )( )

( )( )

32

1

n

1A lim 0, Alternating Series convergesln( 1)

1B Compare with converges 1n2n !3 1C lim lim 0 1 converges Answer is D

2 2 ! 3 2 2 2 1

n

n

n n

n

p

n n n

→∞

+

→∞ →∞

= ⇒+

⇒ >

× ⇒ = < ∴+ + +

∑

31 ( )( )26 12 12 "10" 5 600 (E) ds dx dsSA x xdt dt dt

⎛ ⎞= ⇒ = ⇒ = = ∴⎜ ⎟⎝ ⎠

32 ( ) ( ) ( )1 cos 1 sin sin 1 sin cos C Ax x dx x x xdx x x x+ = + − = + + +∫ ∫ 1u x= + sinv x= du dx= cosdv xdx=

33 ( )

( )3

2

0

Graph 4sin 3 on right

1 4sin 3 4.189 2

C

r

d

π

θ

θ θ

=

≈

∴

∫

34 2 2 2 2 2

2

49 49 49 1 1Area Area 492 2

Simply Graph Area and CALCULATE MAX.

Max Area 12.25 B

x y y x y x

xy x x

+ = ⇒ = − ⇒ = −

= ⇒ = −

∴ = ∴

35

( ) ( )

( )

23 2 2

2 2

2

3 33 0 3 3 3 2 0 2 3

3 3 0 3 3 12 12A) 0,0 0 B) 2,4 02 3 0 2 3 8 63 3C) 2,2 2 3

dy dy dy y xx xy y x x y ydx dx dx y x

dy y x dy y xdx y x dx y xdy y xdx y

−⎛ ⎞ ⎛ ⎞− + = ⇒ − − + = ⇒ =⎜ ⎟ ⎜ ⎟ −⎝ ⎠ ⎝ ⎠

− − −⇒ = = ≠ ⇒ = = =− − −−⇒ =−

6 12 0 B4 6x−= ≠ ∴−

36 2 223 Geometric Series 1 1 5 5 5 5 A

5 5

nr r r r

⎛ ⎞⇒ − < < ⇒ − < < ⇒ − < < ∴⎜ ⎟

⎝ ⎠∑

CALCULUS BC PRACTICE EXAM #2 SOLUTIONS

Revised: 3/26/2015

37

( ) ( )2 23 3

2 2

0 0

L L cos 1 sin 1.04 Adx dy t tdt dt

π π

⎛ ⎞ ⎛ ⎞= + ⇒ = + − ≈ ∴⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∫ ∫

38

3

ln 23

"proportional to its size" "exponential growth"ln 2 2 1 ln 2 3

3ln 2 3ln 33 1 ln 3 4.755 C

3 ln 2

kt k

t

y ne e k k

e t t

=

= ⇒ = ⇒ = ⇒ =

= ⇒ = ⇒ = = ∴

39 ( ) ( )

12 1 1 1 1 1 1 1ln ln ln ln ' ln ln A

2 2 2 2 2 2f x x x x x x x x x f x x x x

x⎛ ⎞ ⎛ ⎞⎛ ⎞= = = = ⇒ = + = + ∴⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

40 ( )

( )2 2 2

2t=3

1cos sin AND ln 1 1

1Speed Speed sin 3 0.287 A4

dx dyx t t y tdt dt t

dx dydt dt

= ⇒ = − = + ⇒ =+

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + ⇒ = − + = ∴⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

41 ( ) ( ) ( ) ( ) ( )

( )( )

1 2 23 3 3

23

1 2 1 12 1 ' 2 1 2 3 2 1 0 B3 0 23 2 1

f x x f x x x xx

−= + ⇒ = + = = ⇒ + = ⇒ = − ∴+

42 ( ) ( ) ( )( ) ( ) ( )3.9 4 ' 4 3.9 4 3.9 5 2 0.1 4.8 Cf f f f≈ + − ⇒ ≈ + − = ∴ 43

( )2 3

1 1 Geometric Series 1 and 2 1 2 1 2

1 1 2 4 8 A1 2

a r xx x

x x xx

⇒ ⇒ = = − ⇒+ − −

≈ − + − ∴+

∑ ∑

∑

44 ( ) ( )

( ) ( )

3 3

3 6 6

1 1 21 C 1 C C 3 3 3

1 2 1 2 1 4 2 2 C3 3 3 3 3 3

t tx

t

V t e x t e t

x t e t x e e

= − ⇒ = − + ⇒ = + ⇒ = ⇒

= − + ⇒ = − + = − ∴

45 Graph tan and CALCULATE DERIVATIVE @

4

' 2.57 C4

y x x x

f

π

π

= =

⎛ ⎞ ≈ ∴⎜ ⎟⎝ ⎠

JJJJJ End of Exam JJJJJ