1. Grade 11 Pre-Cal
June 2001
Answer Key
Part 1:
1. b
2. a
3. c
4. c
5. c
6. a
7. b
8. a
9. d
10. c
Part II: Short Answer
x!2
11. f !1 ( x) =
3
12. therefore 2 real roots
b 2 ! 4ac > 0
13. ( 3, -1 )
14. ! 4x + y ! 6 = 0
15. !BCE = 41o
17. f ( g (2)) = 5
18. Not possible, the absolute value can not be negative
19. k= –3
25
20. k=
12

2. Part III: Short Answer
21. f(-2) = 0 k=8
22. Sum of roots = 4 + 3 + 4 ! 3
=8
Product of roots = ( + 3 )4 ! 3 )
(
4
= 13
x 2 ! 8 x + 13 = 0
2 cos quot; = !1
23. Quad I ! = 120 o
! r = 60 o
!1
cos quot; =
2
Quad III ! = 240 o
24. Graph
25. Critical values are at – 5, and 9. Restricted value is x = – 5
because denominator would be 0.

3. Solution: (quot; !,quot;5)U [9, ! )
26. Draw the perpendicular bisector for each chord. The perpendicular
bisector is also the diameter. The place where the two bisectors
intersect is the center of the circle.
27. Factor the polynomial to get (2 x ! 3)( x + 5) = 0
3
Solution: x = and x = !5
2
28. or quot; (x quot; 3) ! 5
xquot;3! 5
U
Solution: (quot; !,quot;2] [8, ! )
x +1
(2 x ) = x + 1
29. f !1 ( x) = !1
f
2 2
Part IV: Long Answer
30. x 2 + 9 = 4 x 2 ! 12 x + 9
0 = x(3x – 12)
x = 0, 4
31. Use factors of constant term (-6) and the remainder theorem to find
the first factor. P(a) = 0. Use synthetic division to find the other
factor. Repeat steps to determine if you can continue factoring
until all factors are “x” terms only, if possible.

4. 32. BC = CD therefore !BAE = !DAC
arc BC = arc CD because chords are the same length
angle BCA = 30 inscribed in arc AB
angle BEC = 30 isosceles triangle
33. sin x tan x ! sin x = 0
sinx = 0 at 0, and 180 degrees
tanx = 1 at 45, and 225 degrees
'2 1 !2
a. & ,'4 #
34. b. x =
$ !
%3 3quot; 3
1#
&
c. { x quot; !} d. % y y ( '4 quot;
x
3!
$
35. A = (20 + 4 x )(30 ! 2 x )
w = 20 + 4 x
2
A = !8(x ! 5) + 800
l = 30 ! 2 x
Solution: x = 5 therefore w = 40, l = 20
sin C1 sin 30 o
36. C1 = 53.1o C 2 = 126.9 o
=
8 5
A1 = 96.9 o A2 = 23.1o
a1 = 9.9units a 2 = 3.9units
37. 2 x + 12 y ! 4 z = 10
Answer: x = 1, y = 0, z = – 2
2 x + y ! 4 z = 10
38. Omit question

5. 39. P(1) = a (1) 2 + b(1) ! 3 P(!1) = a (!1) 2 + b(!1) ! 3
Solve system of equations to find: a = 2, b = 1
5(10)
40. !ABC =
2
41. 3 tan 2 quot; + 5 tan quot; ! 2 = 0
Factor to solve
1 tan quot; = !2
tan ! =
3
quot; r = 63.4 o
! = 18.4 o
quot; = 116.6 o
! = 198.4 o
quot; = 296.6 o