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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