PreCalculus Individual Solutions– Vero Beach Invitational 2013
For all questions, R  All Real Numbers and i  1 .
xy  3  x 2  y 2
1. D
xy  1  x  y
3  x 2  y 2  1  x  y
4  x2  y 2  x  y
A circle has infinitely many distinct points.
2. D
i.
f  {(0,1), (1, 2), (2,3)} Yes- the x’s do not repeat.
ii.
g  {(1, 0), (0,1), (1, 2), (0,3)} No – the x’s repeat.
iii. h  {( x, x) x  R}
iv.
j  {( x, 2) x  R}
Yes, it is the line y = x.
Yes, it is a horizontal line y = 2.
v. k  {( x, y ) x  y  2, x, y  R}
2
2
No, fails the vertical line test.
3. B C (6, 2)(3x3 )2 (2 x 2 )4  2160 x 2 .
4. C; x(1+20)-5(-1-0)+2(-5-0)=0 x = 5/21
5. B y 
x2  2 x  1
( x  1)2

3x 2  4 x  1 (3x  1)( x  1)
At x = 1, there is a hole. The VA is from setting the remaining denominator factor equal to zero. The
HA comes from the ratio of the leading coefficients.
6. A Given log10 2  0.3010 and log10 3  0.4771
1
1
1
1
log10 4 48  log 48  log  24 3  log  24   log 3  log 2  log 3
4
4
4
4
 0.3010  .4771/ 4
 0.420275  0.42
7. A
(n !)(n  1)(n  2)!
(n!)
1


(n  1)!(n  1)!
(n  1)! (n  1)
PreCalculus Individual Solutions– Vero Beach Invitational 2013
10 x  3
x2  x 1
4 x 5  10 x 2  1
4 x3  4 x 2  8 x  2 
x2  x 1
10 x  3
x2  x 1
10(1)  3
r (1) 
 13
(1) 2  (1)  1
8. C r ( x) 
9. B Given the following four sets, A  {1, 2} B  {1, 2,3} C  {2,3} D  {3}
A  D   No values in common, true.
ii. A  A  B A x B is a Cartesian product, so no, A cannot be a subset.
i.
iii.
A  D  A  B A x D = {(1,3),(2,3)} A x B = {(1,1),(2,1),(1,2),(2,2),(1,3),(2,3)}, so yes.
iv.
A  B  B  A Yes, the order of the union does not matter.
v.
A  C  B Yes.
0.202202202...  n
202.202202202...  1000 n
10. B 202  999n
202
n
999
202  999  1201
U  (T  S )  ( S  V )
11. E {blair , erin, francis, ira, alex, hunter , casey, drew, jade, glen}  { jade}  {drew}
{blair , erin, francis, ira, alex, hunter , casey, glen}
U  (T  S  V )
a. {blair , erin, francis, ira, alex, hunter , casey , drew, jade, glen}  {alex, hunter , casey, drew, jade, glen}
{blair , erin, francis, ira}
U  (S  V )
b. {blair , erin, francis, ira, alex, hunter , casey, drew, jade, glen}  {drew}
{blair , erin, francis, ira, alex, hunter , casey, jade, glen}
c.
(T  S )  ( S  V )
{ jade}  {drew}  { jade, drew}
U  (T  S )  ( S  V )
d. {blair , erin, francis, ira, alex, hunter , casey, drew, jade, glen}  { jade}  {alex, hunter, casey, jade, drew, g
{blair , erin, francis, ira}
e. None of the Above
PreCalculus Individual Solutions– Vero Beach Invitational 2013
12. D
cos(60 )  cos(30 )  cos(90 )
i.
1
3

0
2 2
sec
ii.

cos

 tan
6
6
1
(2)( )  1
2

4
2 tan 30  tan 60
iii.
2
3
 3
3
csc
iv.
1
2
2

4
sin

4
 cot

4
2
1
2


sin(  )  sin
3
3
2

 sin
v. sin
3
3
3
3

2
2
sin(90  30 )  cos(60 )
vi.
sin(60 )  cos(60 )
3 1

2
2
2 cos x  sin(2 x)
sec x
3
 

cos  
 sin cos 2  sin
2
2

13. E 
2 cos x  2sin x cos x 1
cos x
 (2)
[1  sin x]  1  sin x
(1)(1)  (1)(1)
2
cos x
PreCalculus Individual Solutions– Vero Beach Invitational 2013

 8 
 40  
cos  Arc sin    Arc tan   
14. D
 17 
 9 

(15 /17)(9 / 41)  (8 /17)(40 / 41)  455 / 697
 sin x  sin 2 x  0
sin x(1  2 cos x)  0
15. C sin x  0;cos x  1/ 2
x  0,  , 2 , 
 x  5


, 5
3
3
16. C
y  3sin( x  2)  2
a. It has a phase shift of
2

 c / b
b. It has a vertical shift of -2.
c. It has an amplitude of -3. (Amplitude is
a positive value)
i
17. A 1
0
j
k
0
1  i(0  1)  j (3  0)  k (1  0)  i  3 j  k
d. It has a frequency of
1
1
1
.


2 period 2

e. None of the Above
1 3
18. C A point on 15 x  8 y  68  0 is (0,-68/8). The distance between 15 x  8 y  51  0 and that point is found
68
(15)(0)  (8)(
)  51
Ax  By  C
0  68  51
8


7
by:
2
2
2
2
A B
15  8
152  82
19. B
x 2  3xy  y 2  4 x  5 y  3  0
i. The graph is a hyperbola. B 2  4 AC  5  0; hyperbola (T)
ii. The graph has been rotated 45 degrees from the x-axis.
tan 2 
  45
B
3

AC 0
PreCalculus Individual Solutions– Vero Beach Invitational 2013
20. A
x2  6 x  9  y 2  4 y  4  3  9  4
( x  3)2  ( y  2)2  16
8x  1
A
C


x  13 x  42 x  B x  D
8x  1
A
C


2
x  13 x  42 x  6 x  7
21. B Ax  7 A  Cx  6C  8 x  1
2
AC  8
8  6  7  5
22. A 2, 3, 1
1.
k 1
 k 
z

z

z


i
  i  k 1  zi
i 1
i 1
 i 1 
2.
z
k 1
1
i 1
i
1
 z1   zi
i 1
3. Assume true for n = k:
k
k
i 1
i 1
 zi  z1  z2  ...  zk  z1  z2  ...  zk   zk
48 14
23. C r  ( )  2
3
so the means would be: 6,12,24. Their product is 1728.
24. C
4(13C 3) 22

52C 3
425
( x  2i )( x  2i )  x 2  4
x3
25. B x  4 x  3 x  bx  c
2
3
2
r ( x)  bx  4 x  c  12  0
b  4; c  12
26. C
27. E They are all coterminal.
310
 1240
a.
45
b. 1640  1240
3760  1240
62
 1240
d. 
9
c.
PreCalculus Individual Solutions– Vero Beach Invitational 2013
3 

6
28. A (i  1)   2cis
  2 cis9  64cis  64
4


12
12
a. -64
b. 64
c. 128i
29. D e
i

4
 cis
30. D
d. 128-128i
e. None of the Above

4

2
2

i
2
2