1
Final Exam Review SOLUTIONS
1.
A(t ) = A0 e −0.00012t
The bone contains 22% of the amount A0 , which means A(t ) = 0.22 A0 .
0.22 A0 = A0 e −0.00012t
0.22 A0 A0 e −0.00012t
=
A0
A0
0.22 = e −0.00012t
ln 0.22 = −0.00012t
t=
ln 0.22
 12, 618 years
−0.00012
2. There are two available models for this problem.
Use a base of 2:
Divide by A0
Apply ln on both sides
Solve for t
Use a base of e:
n(t ) = n0  2
A(t ) = A0 e rt
10000 = n0  214/4
2 A0 = A0 e r (4)
10000 n0  214/4
= 14/4
214/4
2
10000
n0 = 14/4 = 884
2
2 A0 A0 e 4 r
=
A0
A0
t /4
2 = e4r
ln 2 = 4r
ln 2
r=
4
A(t ) = A0 e
10000 = A0 e
10000
e
ln 2
14
4
=
ln 2
t
4
ln 2
(14)
4
A0 e
ln 2
14
4
ln 2
14
4
e
A0 = 884
1− x
3.
3
 
5
= 23 x
1− x
3
ln   = ln 23 x
5
3
(1 − x) ln   = 3 x ln 2
5
Apply log or ln on both sides
Distribute ln(3/5)
3
3
ln   − x ln   = 3x ln 2
5
5

3
 3 
ln   = x  3ln 2 + ln   
5
 5 

3
ln  
5
x=
3
3ln 2 + ln  
5
Group x terms on same side and factor out x
Solve for x
2
4. log 2 x + log 2 ( x − 6) = 4
log 2 ( x( x − 6) ) = 4
Write as a single log
log 2 ( x 2 − 6 x ) = 4
x 2 − 6 x = 24
Write in exponential form
x − 6 x = 16
2
x 2 − 6 x − 16 = 0
( x − 8)( x + 2) = 0
x = 8, −2
x =8
5.
Factor
Check these two possible answers in the
original equation: -2 does not work, 8 works.
83 x −2 = 164 x
There are two available methods to solve this equation.
Apply a logarithm on both sides:
3 x−2
ln 8
= ln16
Convert to same base:
(23 )3 x − 2 = (24 ) 4 x
4x
(3 x − 2) ln 8 = 4 x ln16
29 x −6 = 216 x
3 x ln 8 − 2 ln 8 = 4 x ln16
9 x − 6 = 16 x
3 x ln 8 − 4 x ln16 = 2 ln 8
−7 x = 6
x ( 3ln 8 − 4 ln16 ) = 2 ln 8
x=−
2 ln 8
ln 82
ln(23 ) 2
x=
=
=
3ln 8 − 4 ln16 ln 83 − ln164 ln(23 )3 − ln(2 4 ) 4
x=
6
7
ln 26
6 ln 2
6 ln 2
6
=
=
=
−
ln 29 − ln 216 9 ln 2 − 16 ln 2 −7 ln 2
7
1


6.  log a x − log a y  + 3log a z
2


( log
log a
a
x − log a y1/2 ) + log a z 3
x
+ log a z 3
1/2
y
xz 3
xz 3
log a 1/2 = log a
y
y
Write the coefficients as exponents
Apply the division law
Apply the multiplication law
3
 y4 x 

7. log8 
2 
3
z


3
 y4 x 
3log8 
2 

 3z 
(
Write the exponent in front
)
3log8 y 4 x − 3log8 ( 3 z 2 )
Apply the division law
3log8 ( y 4  x1/2 ) − 3log8 ( 3 z 2 )
3log8 ( y 4 ) + 3log8 ( x1/2 ) − 3log8 (3) − 3log8 ( z 2 )
Apply the multiplication law
3
12 log8 ( y ) + log8 ( x ) − 3log8 ( 3) − 6 log 8 ( z )
2
Drop down all exponents
8. 3x + y − 24 x + 6 y + 45 = 0
2
2
( 3x
3( x
3( x
3( x
2
− 24 x ) + ( y 2 + 6 y ) = −45
2
− 8 x ) + ( y 2 + 6 y ) = −45
Group terms according to variable
2
− 8x +
)+(y
− 8 x + 16 ) + ( y
Complete the square
2
2
2
) = −45
+ 6 y + 9 ) = −45 + 48 + 9
+ 6y +
Note: 16 x 3 = 48, so add 48 to other side
3 ( x − 4 ) + ( y + 3) = 12
2
3( x − 4)
12
( x − 4)
4
2
2
( y + 3)
+
12
2
( y + 3)
+
=1
Center: ( 4, −3)
(
(
Foci: 4, −3  2 2
)
=
2
12
Vertices: 4, −3  2 3
2
)
12
12
Divide by 12
4
9.
x2 y 2
−
=1
9 27
Center: ( 0, 0 )
Vertices: ( 3,0 )
Foci: ( 6,0 )
Asymptotes: y =  3 x
10. ( y − 2) = 4 x + 20
2
( y − 2) 2 = 4( x + 5)
( y − 2) 2 = 4 1( x + 5)
Factor out 4
Note: p = 1
Vertex: ( −5, 2 )
Focus: ( −4, 2 )
Directrix: x = −6
11. Given vectors u and v in the figure, find w = 2u − v and do the following:
a. Express w in component form
For the vector v: tail is (6, 1) and head is (3, 7) so v = −3, 6
For the vector u: tail is (2, 1) and head is (4, 3) so u = 2, 2
w = 2 2, 2 − −3,6 = 7, −2
b. Express w in terms of i and j
w = 7i − 2 j
c. Find the magnitude and direction of w
w = 72 + (−2)2 = 53
 2
 = tan −1  −  = −15.95o
 7
5
12. an = (−1) n 3n
2
a1 = (−1)1
a1 = −
3
21
3
2
a2 = (−1) 2
a2 =
3
4
3
22
a3 = (−1)3
a3 = −
3
23
3
8
a4 = (−1) 4
a4 =
3
16
3
24
a5 = (−1)5
a5 = −
3
25
3
32
13. 1 + 1 + 1 + 1 + ...
8
64
512
This is an infinite sum of terms of geometric sequence with a1 = 1 and r = 1 .
8
8
The infinite series converges because r  1 and it converges to S =
= .
1 7
1−
8
1
14. 7 + 7 + 7 + 7 + L
3
9
27
i

7 7 7
 1 1 1

1
7+ + +
+ L = 7 1 + + +
+ L  = 7   .
3 9 27
 3 9 27

i =0  3 
1
3
This is an infinite sum of terms of geometric sequence with a1 = 7 and r = .
The infinite series converges because r  1 and it converges to S = 7 = 21 .
1−
1
3
2
15. The third term of an arithmetic sequence is 13 and the sixth term is 31. Find the fifth term, a5 .
a3 = 13
a6 = 31
a6 = d + a5 = d + d + a4 = d + d + d + a3
a6 = 3d + a3
31 = 3d + 13
3d = 18
d =6
a6 = d + a5
31 = 6 + a5
a5 = 25
Replace a6 with 31 and a3 with 13
Solve for the common difference
6
61o
16.
Angle A:
A = 180o − 90o − 61o
Side a:
Side c:
111.1
c
111.1
c=
sin 61o
c = 127.03 ft
111.1
a
111.1
a=
tan 61o
a = 61.58 ft
sin 61o =
tan 61o =
A = 29o
17. There are two approaches to solve this problem.
Law of Sines on the big triangle:
a
32
Tangents using the two smaller triangles:
72o
o
60 − y
60 ft
y
76o
sin 32o sin 76o
=
60
a
o
a sin 32 = 60sin 76o
60sin 76o
sin 32o
a  109.86 ft
60 − y
x
60 − y
x=
tan18o
tan18o =
a=
x
109.86
x = 109.86 cos18o  104.5 ft
cos18o =
tan14o =
x=
y
60 − y
=
o
tan14
tan18o
y tan18o = (60 − y ) tan14o
y tan18o = 60 tan14o − y tan14o
y ( tan18o + tan14o ) = 60 tan14o
y=
60 tan14o
 26.05 ft
tan18o + tan14o
x=
y
26.05
=
= 104.5 ft
o
tan14
tan14o
y
x
y
tan14o
7
18. If the point ( −1, −3) lies on the terminal side of  , find the following: sin  , cos , tan  , csc ,
sec , cot 
c 2 = (−1)2 + (−3) 2
c 2 = 10
c = 10
−1
−3
10
sin  =
sin  =
−3
10
−3 10
10
cos  =
−1
10
cos  =
− 10
10
10
−3 csc  =
tan  =
−3
−1
10
tan  = 3
csc  = −
3
10
sec  =
−1
sec  = − 10
−1
−3
1
cot  =
3
cot  =
19. Find the exact value of each the following:
7
1
a. sin
=−
6
2
3
 7 
b. cos  −
=−
2
 6 
c. tan5 = 0
4
d. tan
= 3
3
e. sin
f.
csc
g. cos

6
− cos

3
=
1 1
− =0
2 2
2 2 3
=
3
3
7
 3 4
= cos 
+
12
 12 12




2 1
2 3
2− 6

  
 −

=
 = cos  +  = cos cos − sin sin =
4
3
4
3
2 2 2 2
4

4 3
8
20. Find the exact value of each the following:

−1  1 
a. sin  −  = −
6
 2

−1
b.
( 3) = 3
tan
2
−1  1 
c. cos  −  =
 2 3

1 
1
6
d. cot  cos −1    =
=
24 12
 5 

5
24

1
(
)
−1
e. sin tan (1) =
f.
1
cos −1   = 
5

1
6
 1 
cot  cos −1    = cot ( ) =
=
24 12
 5 

2
2

 7  7
cos  cos −1    =
 9  9


 8   289
g. sec  2sin −1  −   =
 17   161

8
 8
sin −1  −  =   sin  = −
17
 17 

1
1
 8 
sec  2sin −1  −   = sec ( 2 ) =
=
=
cos ( 2 ) 1 − 2sin 2 
 17  

1
 8 
1− 2 − 
 17 
2
=
289
161
21. sin   0 in quadrants III and IV. sec  0 in quadrant I and IV. Therefore,  must be in quadrant
IV.
cos  =
15
4
−1
csc  =
15
4
4
= −4
−1
9
22. Find the amplitude, period, any roots, and the equation for all asymptotes of the following
functions, and sketch one period of the graph.
a.


y = −3sin   x − 
2

Amplitude: 3
Period: 2
Roots: 1 + n where n is any integer
2
b. y = − tan 3 x
Period:
Roots:

3
n
3
Asymptotes:
where n is any integer

6
+
n
3
where n is any integer
10
23. Prove the identity:
1−
a.
1−
b.
sin 2 
= cos  (1 + cos   0)
1 + cos 
sin 2 
1 − cos 2 
(1 + cos  )(1 − cos  )
= 1−
= 1−
= 1 − (1 − cos  ) = cos 
1 + cos 
1 + cos 
1 + cos 
sin( +  )
= 1 + cot  tan 
sin  cos 
sin( +  ) sin  cos  + sin  cos  sin  cos  sin  cos 
cos  sin 
=
=
+
= 1+
= 1 + cot  tan 
sin  cos 
sin  cos 
sin  cos  sin  cos 
sin  cos 
c. 1 = tan  + cot
sec csc
sin  cos 
sin 2 
cos 2 
sin 2  + cos 2 
1
+
+
tan  + cot  cos  sin  sin  cos  sin  cos 
=
=
= sin  cos  = sin  cos  = 1
1
1
1
1
sec  csc 
cos  sin 
cos  sin 
cos  sin 
cos  sin 
24. sin(2 ) + 3cos  = 0
2sin  cos  + 3cos  = 0
cos  (2sin  + 3) = 0
cos  = 0
=

+ k
2
where k is any integer
Use double angle formula
Factor out cosine
2sin  + 3 = 0
3
sin  = −
2
no solution
Set both factors equal to zero
25. Solve the equation on the interval [0, 2 ) :
a. 8sin 2  + 10sin  = 3
8sin 2  + 10sin  − 3 = 0
(4sin  − 1)(2sin  + 3) = 0
Factor
4sin  − 1 = 0 2sin  + 3 = 0
Set both factors equal to zero
1
sin  =
4
3
sin  = −
2
 1  no solution
 = sin −1  
4
 = 0.2527, 2.8889 radians
Use inverse sine to find angle in radians.
11
b. 1 − sin x = cos 2x
1 − sin x = 1 − 2sin 2 x
Use double angle formula
2sin 2 x − sin x = 0
sin x(2sin x − 1) = 0
Factor out sine
sin x = 0 2sin x − 1 = 0
x = 0, 
1
sin x =
2
 5
x= ,
6 6
 5
x = 0, , , 
6 6
Set both factors equal to zero
c. tan 3x + 3 = 0
tan 3 x = − 3
2
5
+ 2 k
3x =
+ 2 k
3
3
1
1 2
1
1 5
( 3x ) =  + 2 k  ( 3x ) =  + 2 k 
3
3 3
3 3
 3

2 2 k
5 2 k
x=
+
x=
+
9
3
9
3
2 6 k
5 6 k
x=
+
x=
+
9
9
9
9
2 8 14
5 11 17
x=
,
,
x=
,
,
9 9 9
9 9
9
3x =
When does tangent equal − 3 ?
The equation will be true at every
multiple of 2π.
Divide by 3
Let k = 0, 1, 2
These are all the solutions on the
interval [0, 2 )
26.
a.
Angle B:
sin 50o sin B
=
15
10
15sin B = 10sin 50o
10sin 50o
15
 10sin 50o
B = sin −1 
15

Angle C:
C = 180o − 50o − 30.71o
C  99.29o
Side c:
sin 50o sin 99.29o
=
15
c
o
c sin 50 = 15sin 99.29o
15sin 99.29o
c=
 19.32
sin 50o
sin B =

o
  30.71

12
b.
Side a:
a 2 = 182 + 10 2 − 2(18)(10) cos 40 o
a  12.17 cm
Angle B:
182 = 12.17 2 + 102 − 2(12.17)(10) cos B
cos B =
182 − 12.17 2 − 102
−2(12.17)(10)
 182 − 12.17 2 − 102 
B = cos −1 

 −2(12.17)(10) 
B  108.14o
27.
Angle A:
102 = 142 + 152 − 2(14)(15) cos A
cos A =
102 − 142 − 152
−2(14)(15)
 102 − 142 − 152 
A = cos 

 −2(14)(15) 
−1
A  40.15o
1
28. x = 2t + 1, y =  t + 
 2
a.
2
Side x:
sin 40o sin 40.15o
=
14
x
o
x sin 40 = 14sin 40.15o
14sin 40.15o
x=
 14.05
sin 40o
Angle C:
C = 180o − 40o − 108.14o
C  31.86o
13
b.
Solve for t in first equation:
Substitute into other equation:
2
 1
y = t + 
 2
x = 2t + 1
t=
x −1
2
 x −1 1 
y=
+ 
2
 2
x
y= 
2
x2
y=
4
2
2



29. Given v = −4 2 + 4 2 i and w = 2  cos + i sin  ,
6
6

a. Write v in polar form.
Modulus: v =
( −4 2 ) + ( 4 2 )
2
2
=8


Direction:  ' = tan −1  4 2  = tan −1 (−1) = − 


 −4 2 
3
=
4
4
v is in quadrant II
3
3 

v = 8  cos
+ i sin

4
4 

b. Write the exact value of
vw in
polar form.

 3  
 3   
v  w = 8  2  cos 
+  + i sin 
+ 
6
6 
 4
 4


 11 
 11  
v  w = 16  cos 
 + i sin 

12


 12  

c. Write the exact value of
v
in polar form.
w
v 8
 3  
 3   
=  cos 
−  + i sin 
− 
w 2
6
6 
 4
 4
v

 7
= 4  cos 
w
 12

Get a common denominator to
add the angles

 7
 + i sin 

 12



Get a common denominator to
subtract the angles
d. Find the exact values of the cube roots of v in polar form.
14
3
, n = 3 (cube roots)
4
r = 8,  =
wk =
n
r ( cos  k + i sin k ) = 3 8 ( cos  k + i sin k ) , k = 0,1, 2
3
+ 2 k
135o + 360o k 135o 360o k
k = 4
=
=
+
= 45o + 120o k
3
3
3
3
0 = 45o
1 = 165o
2 = 285o



w0 = 2 ( cos 45o + i sin 45o ) = 2  cos + i sin 
4
4

11
11 

w1 = 2 ( cos165o + i sin165o ) = 2  cos
+ i sin

12
12 

19
19 

w2 = 2 ( cos 285o + i sin 285o ) = 2  cos
+ i sin

12
12 

e. Find the exact value of w4 in rectangular form.
 1
  
2
2 
3
  

w4 = 24  cos  4   + i sin  4    = 16  cos + i sin  = 16  − + i  = −8 + 8i 3
3
3 
2 
 6 

  6
 2
30. Write x − y = 4 in polar form.
2
2
( r cos  ) − ( r sin  )
2
2
=4
r 2 cos 2  − r 2 sin 2  = 4
r 2 ( cos 2  − sin 2  ) = 4
r 2 ( cos 2 ) = 4
r 2 = 4sec 2
31. Write r =
Square
Factor out r2
Use double angle formula
Solve for r2
2
in rectangular form.
1 − cos 
2
(1 − cos  )
1 − cos 
r (1 − cos  ) = 2
r (1 − cos  ) =
r − r cos  = 2
r−x=2
Multiply by the denominator on both sides
Distribute r
Solve for r
r = 2+ x
r 2 = (2 + x) 2
x2 + y 2 = 4 + 4x + x2
y2 = 4 + 4x
32. Sketch the graph of each polar equation:
Square both sides
Solve for the y term
Solve for r2
15
a)
b)
n
33. Prove that 2 + 4 + L + 2n = n(n + 1) for all positive integers n. Note: 2 + 4 + L + 2n =  2k .
k =1
Step 1: Check that the statement is true for n = 1
2 = 1(1 + 1)
2=2
Step 2: Assume that the statement is true for n = k
Assume that 2 + 4 + L + 2k = k (k + 1)
Step 3: Show that the statement is true for n = k +1
Show that 2 + 4 + L + 2k + 2(k + 1) = (k + 1) ( (k + 1) + 1) is true
Start with the left-hand side:
2 + 4 + L + 2k + 2(k + 1)
Distribute the 2
2 + 4 + L + 2k + (2k + 2)
21444442
+ 4 + L44444
+ 2k3 + (2k + 2)
k (k + 1)
+ (2k + 2)
k + k + 2k + 2
The first k terms are equal to k(k + 1) by our
assumption
Replace 2 + 4 + … + 2k with k(k + 1)
2
k + 3k + 2
Distribute
2
(k + 1)(k + 2)
The right-hand side:
(k + 1) ( (k + 1) + 1)
(k + 1)(k + 2)
34.
Factor
Both sides are equal.
So if the kth statement is true, then the k+1th
statement is also true by induction
16
Amplitude: 10 inches
Period: 5 seconds
2
B
2
5=
B
2
B=
5
Solve for the value of B using the period
Period =
 2
d (t ) = 10sin 
 5

t

It is a sine model because at t = 0 it is at rest
position (d = 0)
35. Find the partial decomposition of the rational expression:
3
3
=
x − 5 x + 6 ( x − 2)( x − 3)
3
A
B
=
+
( x − 2)( x − 3) x − 2 x − 3
A x −3
B x−2
=

+

x −2 x −3 x −3 x −2
A( x − 3) + B( x − 2)
=
( x − 2)( x − 3)
3 = A( x − 3) + B ( x − 2)
2
let x = 3 : let x = 2 :
3 = B(1)
3 = A(−1)
B=3
A = −3
3
3
3
=
−
x − 5x + 6 x − 3 x − 2
2
3
x − 5x + 6
2
Factor the denominator
Linear factors have constant numerators
Get common denominator
Set the numerators equal
Select x values that will make one factor
equal to zero.
Solve for A and B