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PRE-CALCULUS TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
1. Arc-functions have the traditional restricted range.
  7  

 5  
A = Arc sin  sin 
   Arc cos  cos 
  . Write A in fraction form.
 6 
  6 

3

3
B = sin(2 x) for sin x  and  x 
. Write B in fraction form.
5
2
2
7

 
C = sin   for sin  
and 0    . Write C in fraction form.
2
4
2
h k
1

3 
when sin   , 0    and sin   ,     .
2
2
5 2
10
-----------------------------------------------------------------------------------------------------------------------------
D = h  k for sin(   ) =
2. f ( x)  a cos(bx)  2 for a  0 .
g ( x)  sin(2 x  c )  d for c  0 .
Ferris Wheel, for
parts C and D
A = a  b if the minimum value of f is 14 , and the period of the graph of f is

.
4

to the right.
8
C = 2m  n if d (t )  m cos(4t )  n gives the distance above ground in feet that a rider of a
Ferris wheel is, if the rider's seat ranges from 2 feet to 200 feet above ground. m  0 .
d (t ) is measured in feet above ground.
D = p  q if D(t )  p cos(qt )  3 gives the distance above ground in feet that a rider of
a Ferris wheel is, if the wheel makes a complete rotation in 30 seconds, t is
measured in minutes, and the rider has a maximum height of 20 feet above ground.
D(t ) is measured in feet above ground. p  0 .
B = c  d if the minimum value of g is 12, and the phase shift of g is
PRE-CALCULUS TEAM QUESTIONS
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---------------------------------------------------------------------------------------------------------------------------R
U
3.
12
10
42
30o
o
S
10
T
V
12
W
Acute RST and obtuse UVW have SR=VW=12, ST=UV=10, mS  42o , and mW  30o .
mT  mR  mS . Triangles may not be drawn to scale.
A = the value of p  q for ( RT )2  p  q , in RST , using
for sin  42o  .
B = the length of the shortest altitude of RST , using
2
as an approximation
3
2
as an approximation for sin  42o  .
3
Write B as a reduced fraction.
C = the value of h  k , given that UW= h  k . V is the obtuse angle of UVW .
D = the y-coordinate for vertex U, if V is at the origin on the xy-coordinate plane,
and W is at the point (12, 0) on the x-axis.
4.
E1 : 9 x 2  y 2  18 x  6 y  9  0
E2 : x 2  4 y 2  40 y  104  0 .
A = the larger y-coordinate of the two foci of the conic determined by E1 above.
B = the coordinates of the endpoint of the transverse axis of the conic determined
by E2 above, written as (a, b) with a  0 . Put your answer as a coordinate pair.
C = the sum of the focal radii of the conic determined by E1 above.
D = the product of the eccentricities of the two conics defined by E1 and E2 above.
Write D as a reduced fraction.
---------------------------------------------------------------------------------------------------------------------------5.
E1 : y 2  4 x  6 y  1  0 .
E2 : ( x  2) 2  8 y  40 .
E3 : x 2  y 2  2 x  2 y  9  0
A = the sum of the x- and y-coordinates of the focus of the parabola defined by E1 .
B = the sum of the x- and y-coordinates of the focus of the parabola defined by E2 .
C = the distance between the vertex and focus of the graph of the parabola with
3
directrix equation y  x  5 and focus (10, 5).
4
D = the area of the circle with equation E3 .
PRE-CALCULUS TEAM QUESTIONS
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---------------------------------------------------------------------------------------------------------------------x
6.
x 1
f ( x)  3
.
1
1
g ( x )    . h( x ) 
.
log 3 x
9
A = the real solution to the equation 81 f ( x)  g ( x) .
B = the real solution to the equation h( f ( x))  g (1) , written in reduced fraction form.
C = f 1 (3 3)  g 1 ( 3) , for f 1 and g 1 the inverses of the functions
f and g respectively.
D: h1 ( x)  3m( x ) for h 1 the inverse of h . Give an expression for m( x) in terms of x .
--------------------------------------------------------------------------------------------------------------------7. f ( x)  4 x3  14 x 2  6 x . g ( x)  x3  13x 2  112 x  100 . h( x)  a4 x 4  a3 x 3  a2 x 2  a1 x  a0
A = the coefficient of the quadratic term, a2 , of the function h above, given that
a4  1 and the roots of h are 3  2i, 3  2i, i and  i .
B = the value of r2 given that the roots of f are r1 , r2 and r3 , and r1  r2  r3 .
C = the sum of absolute values of the roots of g .
D = sum of the roots of the functions y  f ( x  1) , y  g ( x  1) and y  h( x  1) .
---------------------------------------------------------------------------------------------------------------------
8. Consider the function y  f ( x) defined by parametric equations
x  t 1 and y  t 2  2t  2 , for t  0 .
A = the minimum y-value of the graph of y  f ( x) for the function defined above.
B = the change in x-coordinates as the y-coordinate changes from 2 (at t  0 )
to 10.
C = the value of k if the directrix of the graph of y  f ( x) has equation y  k .
(Note: consider all real values of t for the graph of f for this part.)
D: the graph of y  f ( x) for t  0 is reflected over the y-axis , and the corresponding
parametric equations for the reflected function are x2  a1t  a2 and y2  a3t 2  a4t  a5
and t  0 . D = a1  a2  a3  a4  a5 .
---------------------------------------------------------------------------------------------------------------------5
ab
9. A = the value of
for 1  i 3  a  bi and i  1 .
16
B = the sum of the distinct y-intercepts when the polar graph r  12  4sin  is graphed
on the xy-plane.
C = the area of the graph of r  8cos .
D = the distance between points P and Q on the graph of r 2  9sin(2 ) , given that


PRE-CALCULUS TEAM QUESTIONS
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the maximum radius of the graph occurs at points P and Q for 0    2 .
------------------------------------------------------------------------------------------------------------------------10. A = the sum of p and q for the unit vector p, q which has the same direction as 6,8 .
B = the angle between the vectors a  2.5 3, 2.5 and b  3, 3 3 .
C = m  n so that vector 3, 4 is perpendicular to vector m, n and the length of m, n
is twice that of 3, 4 .
7
and magnitude 4.
12
-------------------------------------------------------------------------------------------------------------------------ax 2  bx  40
11. f ( x) 
for a and c relatively prime integers.
cx 2  d
2
The graph of y  f ( x) has asymptotes at x  4 , y  , and
3
a removable discontinuity at x  4 .
D = the sum r  s , of the vector r , s with direction
A = a  b for the function f described above.
B = c  d for the function f described above.
C = the x-intercept of the graph of f described above.
D = the y-intercept of the graph of f described above. Write D in reduced fraction form.
-------------------------------------------------------------------------------------------------------------------------12. A is the sum of the values of a for 0  a  2 so that 2 cos(a )  1  0 .
B is the sum of the values of b for 0  b  2 so that 1  cos 2 b  3 cos(b) .
C is sum of the values of c for 0  b  2 so that sin(2c)  tan c .
D is the product of all values of cos x that correspond to each value of x in
16
16
. That is, if each solution to sin 2 x 
is  xn ,cos( xn ) 
0  x  2 for which sin 2 x 
25
25
for 0  xn  2 , give the product of all values of cos( xn ) .
----------------------------------------------------------------------------------------------------------------------------- 
 
13. A = sin    cos   .
 12 
 12 
B = the angle of inclination (with the positive x-axis: 0  B   ), in radians, of the line
with equation 3 x  y  3 .

1
C = the value of cos 2   sin 2  for  so that 0   
and sin   cos   .
2
3
pq 2
14

4
D = p  q for  sin   cos   
if tan  2  
, and 0    .
9
4
7
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14.
Diagram 1 shows a
semi-elliptical arch with
base width 600 feet.
The center of the arch
is 100 feet high.
Diagram 2 shows a
parabolic arch with
base width 600 feet.
The center of the arch
is 100 feet high.
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100 ft
100 ft
600
600ftft
Diagram 1
600 ft
Diagram 2
Both diagrams have vertical symmetry in the center of the diagram.
A = the height in feet of the arch in diagram 1, ten feet to the right of the center.
"Ground level" is the major axis of the ellipse.
B = the height in feet of the arch in diagram 2, ten feet to the right of the center,
(All heights are of outer arch shown.)
C = the distance x in feet horizontally from the center of diagram 1, if the height at that
distance x is 50 feet above ground.
D = the distance x in feet horizontally from the center of diagram 2, if the height at that
distance x is 50 feet above ground.
15.
E1 : x 2  4 y 2  2 x  4 y  1  0 .
E2 : x 2  y 2  2 x  4 y  5  0
E3 : y 2  x  2 y  4  0
E4 : x 2  y 2  10 x  2 y  30  0
Statement S: The equation produces a graph on the xy-plane = 2 points.
Statement T: The equation produces no graph on the xy-plane = 3 points.
Statement U: The equation represents a parabola with a vertical axis of
symmetry = 4 points.
Statement V: The equation produces a parabola which opens to the right = 5 points.
A = the number of points that would be assigned to E1 based on the four statements
S, T, U and V. More than 1 statement may apply to E1 .
B = the number of points that would be assigned to E2 based on the four statements
S, T, U and V. More than 1 statement may apply to E2 .
C = the number of points that would be assigned to E3 based on the four statements
S, T, U and V. More than 1 statement may apply to E3 .
D = the number of points that would be assigned to E4 based on the four statements
S, T, U and V. More than 1 statement may apply to E4 .
PRE-CALCULUS TEAM QUESTIONS
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ANSWERS: (fraction) means answer must be in reduced fraction form.
2. A= 24
2
(fraction)
3
24
53
B= 
(fraction)
B=
or 13.25
25
4
C= 299
2
C=
(fraction)
4
D = 23
D = 21
Thrown out
3. A = 32244
------------------------5. A = 2
B = 9
------------------------7. A = 14
1
B=
2
C = 21
1. A =
C=
3
D = 11
------------------------9. A= 1  3 (no i )
B = 8
C = 16
D=6
------------------------2
13. A= 
2
2
B=
3
17
C=
9
D = 17
20
(fraction)
3
C = 116
B=
D = 43 3
------------------------6. A = 1
10
B=
(fraction)
9
9
C=
or 2.25
4
1
or x 1
D=
x
------------------------1
10. A = 0.2 or
5

B=
6
C=2
D= 2 2
------------------------10 899
14. A =
3
899
8
B=
or 99
9
9
C = 150 3

 1
1. A: Arc sin      .
6
 2
D = 150 2
Arc cos( 3 / 2) 
25
or 12.5
2
------------------------11. A = 0
D=
B = 45
C=5
5
D=
(fraction)
6
------------------------15. A=2
B=2
C=2
D=3
5
2
. Sum =
6
3
4. A = 3  2 2
B = (2, 5 )
C=6
10
3
------------------------8. A = 1
B=4
D=
3
or 0.75
4
D = 1
C=
------------------------12. A = 2
B = 2
C = 5
81
D=
625
-------------------------
PRE-CALCULUS TEAM QUESTIONS
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B = 2sin x cos x  2
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3 4 24

for the angle x with terminal side in Quadrant II.
5 5
25
3
4  4  3  2 . Note: Use cos x  1  2sin 2  x  . Find cosx= 3 .
 
4
2
8
4
2
1
C=
1 4
3 3 4 3 3
h k



=
. 4  27  h  k . h+k= 23.
2 5
2 5 10 10
10
D thrown out: not specified as integers
2 
 . b  8 . a  b = 24 .
2. A: the graph will go from a  2 to a  2 . a  2  14 . a  16 .
b
4
c 
1
53
 so c  . B = 13.25 or
B: The function has range 1  d , so 1  d  12 . d=13.
.
2
8
4
4
C: m  n =2 and m  n  200 . Solve to get n=101. m=99. 2m+n=299.
2 1
D: We want a period of 1/2 minute so
 . q  4 . p+3=20. p=17. p+q=21
q 2
D = sin(   ) 
3. A: Use the law of cosines to get ( RT )2  144  100  2(120) cos 42o . Using 2/3 for sin  42o 
5
. ( RT )2  244  80 5  p  q  p=244 and q=6400(5)=32000
3
and p+q=32244.
1
2
B: Area = (12)(10)  40 . Since mT  mR  mS , T is the largest angle so 12 is the
2
3
1
20
largest side, and the shortest altitude h gives (12)h  40 . h 
2
3
C: Use the two right triangles shown to the
U
right to get UW=8+ 6 3 . h  k = 8  36(3)
10
6
h+k = 8+108 = 116.
D: Drop the altitude on the diagram to the right
30o
and use the results from part C to get the height
V
W
from U is half of 8+ 6 3 or 4  3 3 .
( x  1)2 ( y  3) 2
4. E1 : 9 x 2  y 2  18 x  6 y  9  0 . 9( x 2  2 x  1)  ( y 2  6 y  9)  9  9  9 .

 1.
1
9
x 2 ( y  5)2
E2 : x 2  4 y 2  40 y  104  0 . x 2  4( y 2  10 y  25)  104  100 .

1.
4
1
A: Foci will be 9  1  2 2 from the vertex, vertically. Largest y-value is 3  2 2 .
B: The transverse axis is horizontal with endpoints (2, -5) and (-2, -5). Answer = (2, -5).
C: The sum of the focal radii of an ellipse is 2a which is 2(3)=6.
2 2
5
10
D: eccentricity is c/a so for E1 this gives
. For E2 this gives
. D=
.
3
2
3
gives cos 42o 
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5. E1 : y 2  4 x  6 y  1  0 . ( y  3)2  4 x  1  9 . ( y  3)2  4( x  2) . V( -2, 3), opens right.
E2 : ( x  2)2  8 y  40 . E2 : ( x  2)2  8( y  5) . V(-2, -5). Opens down.
E3 : x 2  y 2  2 x  2 y  9  0 . ( x  1)2  ( y  1)2  11 . A circle with C(-1, 1), r= 11
A: focus is 1 unit to the right of V(-2, 3) which gives (-1, 3). Sum = 2.
B: focus is 2 units down from (-2, -5) which gives (-2, -7). Sum = -9.
C: Distance from a point to a line using the formula with line 3x-4y= -20 and point
3(10)  4(5)  20
(10, 5).
 6  4  4  6 . Half of this is vertex to focus, which is 3.
9  16
D=
 11
2
 11
1
. Base 3 gives powers 4  x 1  2x gives x= -1.
9x
1
1
1
10
 9 . 9x  9  1 . x= .
 9.
B: h( f ( x)) 
. g(-1)=9.
x 1
x1
x 1
9
log 3 3
log 3 3
6. A: 81 3x 1 
x
3
5 1
1
C: 3  3 3 . Powers of 3: x  1  . f 1 (3 3) = .    3 . Powers of 3: 2 x 
2
2 9
2
1
9
so g 1 ( 3) =  . Sum = .
4
4
1
1
1
D: x 
. x log3 y  1 . log3 y  . y  31/ x . D= .
x
x
log 3 y
x1
7. A: Sum of 3  2i is 6= -b/a. Product is 13=c/a. So a quadratic factor is x 2  6 x  13 .
Similarly the other two roots give x 2  1 . Multiply to get a quadratic factor of 1+13=14.
1
1
B 2 x(2 x 2  7 x  3)  2 x( x  3)(2 x  1) so r1  r2  r3 gives r1  0, r2  , r3 =3. B= r2 = .
2
2
2
C: Since g(1)=0, divide synthetically by x-1 to get x  12 x  100 . Use the Quadratic Formula
12  144  400
to get
and divide by 2 to get 6  36  100  6  8i . Absolute values are
2
1, 62  82  10 and 62  82  10 for a sum of 21.
D: Since the (x+1) makes each root 1 less (corresponding to moving the graph to the
left 1), making each root 1 less, we just add -1 to each root. (Note: you can check that
the non-real roots also are made less by 1 by comparing x 2  1 to ( x  1)2  1 and
comparing the roots  i to 1  i . Back to the problem: Sum of the roots is –b/a.
f sum is 7/2. g sum is 13. h sum is 6. Subtract 10 for the number of roots, to get
7/2 + 13 + 6 -10 = 12.5 or 25/2.
8. A: y  ( x  1)2  2( x  1)  2  x 2  4 x  5 . y is min at x= -b/(2a)=2. y= 4  8  5  1. Check that
time is in the given interval: (2, 1) gives time 1. So A=1.
B: time changes from 0 to 4, since t 2  2t  2  10 , (t  4)(t  2) gives t=4. x changes from
1 to 5. B=4.
C: y  x 2  4 x  5  ( x  2)2  1 when you complete the square. Directrix is 1/4 unit down from
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vertex (2, 1) to get y=0.75.
D: replace x with (-x) to reflect over the y-axis. So we get  x  t  1 and the y-values remain
the same at y  t 2  2t  2 . D= -1+-1 + (1-2+2)= -1.
5

 5 
  
9. A: Using DeMoivre's Theorem, 1  i 3 =  2cis    = 32cis  

 3 
 3 

1
ab
3
32  1
3
 
32  cis   32   i
, this is
 . Since we want
 
  1 3
16
3
2 
16  2 2 

2

3
B: y-intercept is where   or
or coterminal angles with them. r= 8 or 16. Since at
2
2
3
the graph is at y= -16, the sum is 8-16= -8. To verify we can change the equation
2


5
to rectangular x 2  y 2  12 x 2  y 2  4 y . The y-intercept occurs at x=0 so y 2  12 y  4 y .
Either y 2  8 y  0 or y 2  16 y  0 . y  0 so y = 8 or -16, for a sum of -8.
C: This is a circle with diameter 8, so area is 16 .

D: This is a lemniscate and a max radius is at   where sine is at its maximum.
4
Points P and Q must have a radius of 9  3 so distance is 6.
1
6 8
,
10. A: magnitude is 10 so the vector is
for a sum of .
5
10 10
7.5 3  7.5 3
3


= 3 . B=
56
2
6
C: Reverse the components and negate one of them to get a perpendicular. Since the
magnitude of the original vector is 5, we want magnitude 10. So 8, 6 or 8,6 both
B: cos  
give a sum of components with absolute value 2.
7

D: For cosine and sine of
, you can use the half angle identities or addition of
and
12
4
7  6  2
7
6 2



, or memorization. cos
. sin
. 4 times the sum is 2 2
12
4
12
4
3
11. a=2 and c=3 since the horizontal asymptote is y=2/3. The denominator must be
3( x  4)( x  4) since there is a vertical asymptote x=4 and a removable discontinuity at
x= -4. So d= -48. The numerator is 2( x  4)( x  ___) due to the removable discontinuity
and since we need to multiply to get constant -40 so the gives 2( x  4)( x  5) to give an
x-intercept of 5. That makes the numerator 2 x 2  2 x  40 and b= -2.
40 40 5

 .
A = a  b  2  2  0 . B = c  d  3  48  45 . C = 5. D =
d
48 6
PRE-CALCULUS TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
1 2 4

 2 .
12. A: cos a   .
2 3
3
sin b
B: sin b  3 cos b .
 3 . cosb must be positive but sinb can be either sign. So in
cos b

5
QI or QIV, tan b  3 . b  or
for a sum of 2 .
3
3
2
sin c
 3 5 7
C: 2sin c cos c 
. sin c  0 or 2 cos 2 c  1 . cos c  
. 0,  ; , , ,
. Sum of 5 .
cos c
4 4 4 4
2
16
4
3
D: sin 2 x 
 sin x  which occurs four times in 0  x  2 and cos x is twice and
25
5
5
3
81
twice  for a product of
.
5
625
2
3 2 1 2
6 2
6 2
  
  
13. A = sin    
. and cos    
; the difference 
.


2
4
4
3 4 2 2 2 2
3 4
 2
B: slope is  3 = tan  . The angle of inclination for a negative is   
.
3
3
1
1
8
C: Square both sides of sin   cos   to get 1  sin(2 )  . sin(2 )  .
3
9
9
1
Since cos 2   sin 2  = (cos   sin  )(cos   sin  ) = (cos   sin  ) and since  is in
3
17
8
Use sin(2 )  to get cos(2 ) 
. Due to the absolute value, we don't have to worry
9
9
about the sign of the radical.
2
14
7
in QI, sin(2 ) 
(using the Pythagorean Th.) and cos(2 ) 
3
7
3
2
2 2 2 11 2 2 p  q 2
(sin   cos  ) 2  1  sin(2) 1 
 
. Square: 1  
=
. p  q  17 .
3
9
3
9
3
9
x2
y2
14. A: Set center ground at (0, 0).

 1 is the entire ellipse. x=10 gives
3002 1002
10 899
100
y2
899
y2
1

.
. y

3
300 300 100 100 900 100 100
2
B: Set center ground at (0, 0). y  ax  100 . Use (300, 0) to get 300(300)a  100
1
899
(100)  100 . 99 8/9 feet or
a= -1/900. x=10 gives y 
.
900
9
300 3
x2
50 50
x2
3
 150 3


1
C: See part A.
.
 .x=
2
2
2
300
100
300(300) 4
D: tan  2  
PRE-CALCULUS TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
1 2
x2
x  100 . y=50. 50  
 100 . y  50 900  150 2
900
900
E1 : x 2  4 y 2  2 x  4 y  1  0 . ( x  1)2  4( y  0.5)2  1.75 . hyperbola with a graph.
D: See part B. y  
15.
E2 : x 2  y 2  2 x  4 y  5  0 . ( x  1)2  ( y  2)2  0 . Graph of a single point.
E3 : y 2  x  2 y  4  0 . ( y  1)2  1( x  3) . Parabola, open to the left.
E4 : x 2  y 2  10 x  2 y  30  0 . ( x  5)2  ( y  1)2  4 . No graph.
A : E1 has 2 for statement S. A=2.
B : E2 has 2 points for statement S. B=2.
C : E3 has 2 points for statement S. C=2.
D : E4 .has 3 points for statement T. D=3.