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ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
1.
f ( x)  (2 x  1)( x  3)( x  3) ;
g ( x)  2 x3  5x 2  21x  36 ;
Sickles Invitational
h( x)  ( x  2)4
A: for the roots of g , r1 , r2 and r3 , and r1  r2  r3 the value of A is r3  r1 .
B: for function f , when in standard form ( f ( x)  a1 x n  a2 x n 1  ...) , B is the coefficient
of the quadratic term.
C: when the function h is translated so that the minimum point of y  h( x  h)  k is (-1, 3)
C is the value of h  k .
D= the number of distinct real roots for the function y  h  x  .
-----------------------------------------------------------------------------------------------------------------------------2. A student used Cramer's Rule to solve a system of linear equations, and
4 1
correctly used x 
94 4
to solve for x .
2 1
3 4
A = the value of the x-coordinate of the solution of the system.
B = the value of the y-coordinate of the solution of the system.
a x  b1 y  c1
C: when the system is written  1
, for a1 and a2 positive prime integers,
a2 x  b2 y  c2
and both equations in standard form, C = the least slope of the two graphs of the
two linear equations.
a x  b1 y  c1
D: when the system is written  1
, for a1 and a2 positive prime integers,
a
x

b
y

c
2
2
 2
and both equations in standard form, D = the (positive) distance between the two
points on the graphs of the two linear equations when x  5 .
------------------------------------------------------------------------------------------------------------------------------3. x  8k and y  4k . i  1 .
A = the simplified complex number in a  bi form, for the product x y when k  2 .
B = the simplified rationalized complex number in a  bi form, for the quotient
x2
y
when k  2 .
C = 0 if x 2 y 3 is a complex number, and C=1 if x 2 y 3 is not a complex number.
D = an expression in terms of k for x y 2 , with NO absolute value or radical signs,
if k  0 . The expression should be simplified.
---------------------------------------------------------------------------------------------------------------------------
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
4. A = the number of integer solutions to the inequality 3x  1  20 .
B = the number of integer solutions to the inequality 4  3x  x 2  0 . If there is an
infinite number of integer solutions, let B = 1 .
1
C = the least integer in the solution set to the conjunction x  5 and 4  x 2  0 .
2
If there is no least integer in the solution set, let C = 100 .
D: One point with coordinates ( a , b ) exists so that it satisfies the inequalities
y  2 x 2  8x  6.5 and y  0 , where a and b are integers. D = a  b .
------------------------------------------------------------------------------------------------------------------------------x 1
5. f ( x)  2 x3  1 , g ( x)  2  4 x  8 , h( x) 
.
3 x
 3
A = the real value of f 1    , for f 1 ( x) denoting the inverse of f .
 4
B = p  q , where p is the x-intercept of the graph of g , and q is the x-intercept
of the graph of h .
C = the range of g , for domain x  2 , written in interval notation.
D = g (1  f ( f ( f (1)))) .
------------------------------------------------------------------------------------------------------------------------------6. A = the distance from the point (2, 5) to the graph of ( x  5)2  ( y  9)2  2 .
B = the length of the major axis of the graph of 25x 2  20 y 2  50 x  40 y  7955  0 .
C = the value of k if the asymptotes of the graph of the graph of 4 x2  9 y 2  36 y  72  0
have equations 2 x  3 y  k and 2 x  3 y  k .
D = the maximum y-value of the graph of y  9  8x  x2 .
----------------------------------------------------------------------------------------------------------------------------1
7. A = log 3    log 2 (32)  log 4 8 .
 81 
B = the sum of the real solutions to the equation 32 x 1  28  3x   9  0 .
C = the square root of the product of the real solutions to the
equation (log x)2  log  x 4   5 =0.
1 
D = the real solution to the equation log  x   2 log 2  1 .
2 
----------------------------------------------------------------------------------------------------------------------------
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
8.
A = the remainder (written as A, instead of
Sickles Invitational
A
) when  x 4  5 x 3  14 x  2  is divided by
x5
the binomial ( x  5) .
B = the largest value of k so that the discriminant of (3x 2  kx  x  1) is 28.
C = the value of p  q so that f ( x)  ( x  3)4 ( x  p  2)2  q  2 has exactly one distinct
real root.
D = the sum of the non-real roots of y  x3  1 .
------------------------------------------------------------------------------------------------------------------------x
1
   , written in simplified fraction form.
9. A = the real solution to 2 4
8
4
1
B = the positive real solution to 2 x 3  written in simplified fraction form.
8
C = the real solution x so that x  1 to 2 x  x  1  2 , written in simplified fraction form.
2 x1
D = the real solution to
5( x  2)  5( x  2)  5( x  2)  ...  x  1 .
----------------------------------------------------------------------------------------------------------------------------3
x 3
x 2  3x  4
10. f ( x) 
. g ( x) 
. h( x ) 
2
x2
x5
x 1
A = 2 p  q where x  p and y  q are asymptotes of the graph of f .
B = m  2n where the range of the graph of f is (, m)  (m, n)  (n, ) .
C = j  k  t where the domain of y  h( f ( x)) is "All real numbers not equal
to x  j , and not equal to x  k and not equal to x  t ."
3 x  15
D = r  s for g (h( x)) 
.
rx  s
-----------------------------------------------------------------------------------------------------------------------------3
11. A = the value of a for the equation  5  4i   (10  8i) 2 (a  bi) .
B = the value of  i 2016 
2016
.
C = r  s , where the least positive integer value of k so that 12k is both a perfect squared
integer and a perfect cubed integer.
3x
D = the value of x so that
42
8
x
3
 1 
 
 32 
x 1
. Write D in reduced fraction form.
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
x  2 y  4

12. For parts A and B, consider the system 4 x  y  2 .
y  2

A = the area on the xy-plane of the graphed
solution to the system of inequalities above.
y




x
.
B = the least possible value of x  y for ( x, y ) a solution
to the system.










C and D: The values for the perpendicular bisector
of the segment with endpoints (3, 10) and (-1, 2)
have equation 2x  Cy  D .
---------------------------------------------------------------------------------------------------------------------------13. Abel can fill an empty tub in two hours, as he is using a drinking glass.
Barnaby can fill the same empty tub in 1.5 hours, as he is using a "Big Gulp" ® glass.
When the drain is open, the full tub drains completely in 3 hours.
Use these constant rates, where appropriate, for each question below.

A = the number of minutes it would take to completely fill the tub, initially empty, when
both Abel and Barnaby are putting water into the tub at their respective rates.
Give your answer as a reduced fraction if the answer is not an integer.
B = the number of minutes it would take to completely fill the tub, initially empty, when
Abel is putting water into the tub, while the drain is open. Give your answer as a
reduced fraction if the answer is not an integer.
C = the number of minutes it would take to completely fill the tub, initially empty, when
Abel is working alone to fill the tub for 10 minutes, and then Barnaby begins to help
him. They complete the task together. Give your answer as a reduced fraction if the
answer is not an integer.
D = the number of minutes it would take to completely fill the tub, initially empty, when
Abel and Barnaby work to fill the tub for 30 minutes, and then Abel opens the drain
and walks away. Barnaby finishes the task alone. Give your answer as a reduced
fraction if the answer is not an integer.
__________________________________________________________________________
14. Consider the equation 3 4k  3 8m  10 so that k and m are both positive and integers.
A = the value of k  m if
B = the value of k  m if
C = the value of k  m if
4k  2 .
3
4k  4 .
3
4k  6 .
3


ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
D = the value of k  m if 3 4k  8 .
15. A = the sum of the distinct y-coordinates for the intersection points of the graphs of
x2  y 2  68 and y  2 x 2 when graphed on the xy-coordinate plane.
B = the sum of the distinct x-coordinates for the intersection points of the graphs of
x2  y 2  68 and y  2 x 2 when graphed on the xy-coordinate plane.
C = the coordinates of the point in quadrant II which is an intersection point of
x 2  y 2  3 and x 2  2 y 2  6 when graphed on the xy-coordinate plane.
D: the asymptote with positive slope to x 2  y 2  3 intersects the graph of
x 2  2 y 2  6 on the xy-coordinate plane at point P in quadrant I.
D = the sum of the coordinates of P.
-------------------------------------------------------------------------------------------------------------------ANSWERS:
Part A
7
10
Part B
1
16
3. 8 2
4. 13
1
5.
2
6. 5  2 or
 2 5
21
7. 
or 10.5
2
8. 72
1
9.
7
10. 3
1.
2.
5
or -1.25
4
18
360
7
66
8
11. 
12.
13.
14.
15.
4i 2
4
0
Part C
6
3
 or -0.75
4
0
2
(, 2]
Part D
2
55
3
or 13
or 13.75
4
4
8k
1
2
40
6
25
1
100
80
5
1
8
6
7
5
4
1
4
3
or 1.5
2
3888
14
1
0
360
43
0
4
330
7
62
1
5
7
26
75

129
2 2
notes
part C must be in
interval notation
A, B, C must be
in fraction form
D must be in
fraction form
A,C must be
fraction form
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
SOLUTIONS:
3
and r3  4 . A= r3  r1 =7.
2
B: f ( x)  (2 x  1)( x  3)( x  3) = 2 x3  x 2  18 x  9 . B= -1.
C: h has a minimum point at (2, 0). We want to translate up 3 and left 3.
h( x  3)  3  ( x  1)4  3 . h  k =3+3=6. C=6.
D: The original h function is tangent to the x-axis once at x=2. The absolute value of x
in the argument keeps the right side the same ( x  0 ) and reflects that over the y-axis.
D=2 distinct real roots at -2 and 2.
y= f(x)
y=f(|x|)
1. A: The roots of g are r1  3, r2 

y









































y
x











x





2x  y  4
. Multiply the first equation by 4 and add to get
3 x  4 y  94
11x=110 so x=10. Substitute to get 20-y=4 and y=16. A=10. B=16.
C: the slopes of the lines whose equations are written in part A are 2 and -3/4.
C= the least slope = -3/4.
D: At x=5, the lines have values 6 and 79/4. Distance is a vertical distance which is
79/4-6= 79/4-24/4= 55/4 or 13 3/4 or 13.75.
2
8i
x2
3. A = 16 8  4i 2i 2  8 2 ; B =
= 16 / 8  4i 4i /(2i 2) 
 4i 2 .
y
2
C = 0 since imaginary numbers are a subset of complex numbers.
D: Since k  0 , x  8k  i 8 ( k ) . y  4k  2i  k . x y 2 = i 8 ( k ) 2i k 2 =
2.
A and B: The system is


k k  k 2  k since –k is positive. (Verify for k= -2:
16 8 2 = part A times 2 = 16 .) 8k = -8(-2)= -16.
D = 8k
4. A: 20  3x 1  20 . 19  3x  21.  6.3...  x  7 . Integer solutions are -6, -5, …, 0, ….5, 6.
which are 13 in number. A=13.
B: x 2  3x  4  0 . ( x  4)( x  1)  0 . The graph of y  ( x  4)( x  1) is below the x-axis for
x=-3, -2, -1, 0 which are 4 integer solutions.
B=4.
1 2
C: 5  x  5 and  x  4. x 2  8 .  8  x  8 . The intersection of these sets has
2
integers from -2 through 2. The least integer solution is -2.
D: y  2 x 2  8x  6.5 = 2( x2  4 x)  6.5  2( x 2  4 x  4)  6.5  8 = 2( x  2) 2 1.5 . So y must be
greater than -1.5 and less than 0. y= -1. x= 2 from the vertex. A= a  b =2-1=1.
8i 2 k  8k . Note that
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
3
1
1
1
5. A: 2 x 3  1   . 2 x 3  . x3  . x  A  .
4
4
8
2
B: h has x-intercept 1, when the numerator is 0. For g, we solve 2  4 x  8  0 . 4 x  8  2
4x+8=4. x= -1. The sum is 1 + -1 = 0.
C: The graph of the absolute function given by g is downward, starting from 4x+8=0 or x= -2
and y=2. The range is y  2 , or (, 2] .
D: f (1)  1 so f ( f ( f (1)))  1 . g (1  f ( f ( f (1)))) = g (1  1)  g (2)  2  16  2  4  2 . D= -2.
6. A: The center of the circle is (5, 9) and distance from that to (2, 5) is 32  42  5 . Subtract
the radius 2 to get A= 5  2 .
B: 25( x 2  2 x  1)  20( y 2  2 y  1)  7955  25  20 . 25( x  1)2  20( y  1)2  8000 .
( x  1)2 ( y  1) 2

 1 . Major axis has length 2 400  40 . B=40.
320
400
2
x 2 ( y  2)2
C: 4( x2 )  9( y 2  4 y  4)  72  36 .

 1 . Asymptotes are ( y  2)   ( x) and
3
9
4
both go through (0, -2). So in standard form: 2 x  3 y  6 . k = 6. C= 6.


D: Vertex of y  9  8x  x2  use x= -b/(2a) = 8/(-2)= -4. y= 9  32 16  25 . D=25.
3
21
1
7. A = log 3    log 2 (32)  log 4 8 = 4  5   9  1.5  10.5   .
2
2
 81 
1
B : 32 x 1  28  3x   9  0 . Let b  3x . 3b 2  28b  9  0 . (3b  1)(b  9)  0 . 3x  or 3 x  9
3
x= -1 or 2. Sum of the solutions is 1.
1
C: Let (log x)2  4log  x   5 = (log x  5)(log x  1)  0 . Solutions 105 and
. Product 10 4
10
C  104  100 .
1
x
1
1 
D: log  x   2 log 2  1 . log 2  1 . x  10 . x=D=80.
8
4
2 
8. A =  54  5(5)3  14(5)  2  = 70+2 = 72. (Remainder Theorem)
B: (3x 2  kx  x  1) has discriminant (1  k )2  4(3)  (1  k )2  12  28 . 1  k  4 . k=5 or -3.
B = the largest value of k which is 5.
C: y= ( x  3)4 ( x  p  2)2 appears to have 2 real roots, on the x-axis. But if x-3 = x+p+2
then p= -5. Then there is one distinct real root. But to keep this on the x-axis, q+2=0
and q= -2. Sum is -7.
D: x= -1 is one root, and we know the sum of the roots of the graph are –b/(2a)=0. So the
sum of the non-real roots must be 1. D=1.
ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
Sickles Invitational
3
9. A: 2 2
2(2 x 1)
2
3 x
4
1
1
 1 4 1
. 4x  2  1  3x . 7 x  1 . x  A  . B: x 3  . x  B    =
7
16
 16  8
C: 2 x  2  x  1 . 4 x2  8x  4  x  1. 4 x 2  9 x  5  0 . (x-1)(4x-5)=0. Since x  1 in the
question, x=5/4. Therefore C=5/4.
D: 5( x  2)  ( x  1)  x 2  2 x  1 . x 2  2 x  8  0 . ( x  4)( x  2)  0 . x= -2 an extraneous solution
but x=4 gives a valid answer. D=4.
x 2  3x  4 ( x  1)( x  4)
10. A: f ( x) 
=
has a removable discontinuity at x= -1, and asymptotes
x2 1
( x  1)( x  1)
at x=1 and y=1. 2p+q = A = 3.
B: the removable discontinuity is at (-1, 5/2) so 5/2 is not part of the range. The asymptote
at y=1 means 1 is also not in the range. Range = (,1)  (1, 2.5)  (2.5, ) . You have to
sketch the graph to make sure we have all other range values covered.
(, m)  (m, n)  ( n, ) gives m=1, n=2.5. m+2n=6.
x4
x 2  3x  4
 5 .
 5 . Reduce:
2
x 1
x 1
5x  5  x  4 . 6 x  9 . x  1.5 . The sum of these three values is 1.5.
3 x  15 3 x  15
3
3x  15
D: g (h( x)) 
=
=
=
. r= -1, s= -13.
 x 3
x  3  2( x  5)  x  13 rx  s

2
 x5
r+s= -14.
5
3
i .
11. A:  5  4i   (2(5  4i)) 2 (a  bi) = 4(5  4i) 2 ( a  bi) . (5  4i )  4(a  bi ) . a  bi 
4
a= A = -5/4 or -1.25
C: f omits the domain values 1 and -1. Then
B =  i 2016 
2016
= 1 since 2016 is divisible by 4 and i 4  1 .
C: 12k  22 3k , we need powers to be 6 to get a perfect cube and square.
k  24 35 will give us 26 36 . So k=16(243)=3888.
D:
2
3x
1
x
1
x  2 y  4

4 x  y  2
y  2

y
  2
5( x 1)

5
. 2x  5x  5 . 7 x  5 . x   .
7
2
1
1
12. A = bh   9  4  since x  2(2)  4. x  8 for the
2
2
right vertex. 4x  2  2 , x= -1. A=Area= 18 .
B = 0 since the least value of k is 0 and the origin
is in the solution, so x  y  0 .



x






C and D: slope is 2 for the segment so the perpendicular
2 1

is -1/2. So for 2x  Cy  D , slope
and C= 4.
C
2
The midpoint is (1, 6) so D=2(1)+4(6)= 26. D= 26.







ALGEBRA TWO TEAM QUESTIONS
Sponsors' Copy : February 27, 2016
13.
A: t= hours.
B:
Sickles Invitational
1
2
6
360
t  t  1 . Multiply by 6: 3t+4t=6. Minutes= (60) 
.
2
3
7
7
1 1
t  t  1 . 6 hours = 360 minutes.
2 3
1 1 1

of the tub in ten minutes.
2 6 12
1
2
11
11
330
t  t  . Times 12: 6t  8t  11 . Minutes =
(60) 
.
2
3
12
14
7
1 1 2 1 7
2 1
5
5


D: Filled in a half hour is
of the tub. t  t  . Minutes = 60  75 .
2 2 3 2 12
3 3 12
4
3
14. A: 4k  8. k  2 . 8m  8 . 8m  8(8)(8) . m=64. k+m=66.
C: Abel can fill half the tub in 1 hour so he will fill
B: 4k  64. k  16 .
3
8m  6 . 8m  23 33 . m=27. k+m=43.
C: 4k  6(6)(6). k  3(3)(6)  54 .
3
8m  4 . 8m  4(4)(4) . m=8. k+m=62.
D: 4k  8 8 8. k  64(2)  128 . 8m  2 . m=1. k+m=129.
15. A: We sketch an ellipse with center at the origin, and a upward opening parabola with
vertex at the origin. Due to symmetry, we are pretty sure that there is only one distinct
1
y-coordinate to the two intersection points. y  y 2  68. 2 y 2  y  136  0 .
2
(2 y  17)( y  8)  0 . y= -17/2 is extraneous since the parabola has range y  0 .
y=8 is the only value. So A=8.
B: Because of symmetry, there are two points of intersection. 8  2x 2 . x  2 . B=0.
C: Subtract the equations to get 3 y 2  3. y  1 . In quadrant II, y= 1. x2  2  6. x  2
C = 1 + (-2) = -1.
D: The asymptotes of x 2  y 2  3 have slopes 1 and -1. So the equation of the positive
3
sloped asymptote is x=y. x 2  2 y 2  6 becomes 3 y 2  6. y   2 . P is in quadrant I
so P is at


2, 2 and so D= 2 2