Download Math 11 Final Fall 2010

Columbia College
Mathematics 11
Final Examination
December 2010
(3 hours)
1. Find the slope-intercept equation of the line that contains the points (3,2) and (5,18)
and prove that the point (10,28) is also on this line.
(4)
2. Find the equation of the perpendicular bisector of the line segment that has the
endpoints (1,3) and (7,1).
(4)
3. Use elimination to solve the system of equations:
1. 2 x  3y  30
2. 4x  5y  16
(4)
4. Solve the system of equations:
1. 2x  y  z  18
2. 3x  y  2z  18
3. 4x  y  z  0
(6)
5.
Using set notation, describe the shaded region shown at
the left.
(6)
6. By completing the square, find the vertex of the parabola y  2x 2  12x  16 and then
accurately draw the graph indicating the vertex and two points on the parabola that are to
the right of the vertex and two points on the parabola that are to the left of the vertex.
(6)
7. In the same coordinate plane draw the graphs of y  x 2  6x  6 and y   x  4 and
find their points of intersection by solving:  x 2  6x  6   x  4 .
(6)
8. Solve the equation 9 x 2  54 x  56  0 by completing the square.
(4)
9. Solve the equation 15x 2  32x  7  0 by factoring.
(4)
10. What is a function?
(2)
11. Find the domain of f ( x ) 
2x  10
x 2  8x  7
(4)
12. If g ( x )  3x  2 and f ( x )  2x 2  5x  2 , express the composite function f(g(x)) in
standard form.
(4)
13. If g ( x ) 
x2
3x  2
and g (f ( x )) 
, find f(x).
2x  1
6x  9
(4)
14. Find all the solutions of
9  3x  x  15 .
(4)
15. Find all the solutions of
2x  1 x  3

x 4 x 5
(4)
16. Let P( x )  3x 4  5x3  2x 2  3x  4 and D( x )  x 2  2x  3 . Divide P(x) by D(x)
and then check.
(4)
17. Given that 1 and 2 are the two real zeros of P( x)  x 4  2x3  x 2  4x  4 , find the
other two zeros. Express these zeros in standard form.
(4)
18. Use the Rational Zero theorem to solve 2x 3  3x 2  7 x  3  0 .
(4)
2
19. Let f ( x )   x  3 be restricted to the domain [6,3]
3
a) In the same coordinate plane, draw the graphs of f and f 1 .
(4)
b) State the domain and the range of f and the domain and the range of f
1
.
(2)
c) Find the formula for f
1
.
(2)
20. Let f ( x )  3x 2  36x  46 be restricted to the domain (,6] . Find the formula for
f 1 .
(4)
21. Solve the following:
a) 23x  15  77
(2)
b) x  4  2x  11  17
(4)
c) 5  3x  7
(2)
22.
Given that O is the centre of the circle and
BDA = 40o , find BAC
(4)
23.
The smaller circle with centre O1 has
radius = 1 and the bigger circle with
centre O2 has radius = 2. O1 is on the
bigger circle and the two circles
intersect at points A and C. The line
AC intersects the line O1O2 at point
B.
a) Find the length of the line segment
(2)
O1O2 .
b) Prove that ABO 2  CBO2 (5)
24.
Given that AB and CD are chords of
equal length in a circle with centre O
and that the lines AC and BD intersect
at E, prove that BEA  CED .
(4)
Total = 113
Answers.
18  2
 2 . Therefore the equation of the line is of the form
5  ( 3)
y = 2x+C for some constant C. Since (5,18) is a solution: 18  2  5  C  C  8 .
Therefore the equation of the line is y = 2x + 8. Since 28  2 10  8 , (10,28) is a
solution of the equation and therefore a point on the line. (4)
1. The slope of the line is
2. The slope of the given line segment is
1 3
1
  . Therefore the slope of its
7 1
3
1
perpendicular bisector is 3 (because   3  1 ) The midpoint of the given line
3
1  7 3  1
,
segment is 
  (4,2) . Therefore the equation of the perpendicular bisector is
2 
 2
of the form y = 3x + C where (4,2) is a solution: 2  3  4  C  C  10 . Therefore the
equation of the perpendicular bisector is y  3x  10 . (4)
3.
Therefore the system has the unique solution (9,-4). (4)
4. Adding equation 1. to equation 3. yields 6x = 18 and therefore x = 3.
Adding equation 1. to equation 3. yields 5x + 3z = 36 and therefore 15 + 3z = 36 and z
= 7. Substituting x = 3 and z = 7 into all three equations yields y = 5. Therefore the
system has the unique solution (3,5,7). (6)
5. The line containing (-2,1) and (2,1) is the horizontal line y = 1. The line containing
(-2,1) and (1,-2) is y   x  1 and the line containing (2,1) and (1,-2) is y  3x  5 .
Therefore the shaded area is the set {( x, y) : y  1 and y  x  1 and y  3x  5} . (6)


6. y  2x 2  12x  16  2( x 2  6x)  16  2 ( x  3)2  9  16  2( x  3)2  2 . Therefore
the vertex of the parabola is (3,-2). The graph is on the next page. (6)


7. a) y  ( x 2  6x)  6   (x  3)2  9  6  ( x  3)2  3 is a parabola with vertex
(3,3). The line y   x  4 has intercepts (0,4) and (4,0). The graph is on the next page.
(3)
b)  x  4  x 2  6x  6  x 2  7x  10  0  (x  2)( x  5)  0 which means that the
two intersections are (2,2) and (5,-1). (3)


8. 9x 2  54x  56  0  9(x 2  6x)  56  0  9 (x  3)2  9  56  0
 9( x  3)2  25  (x  3)2 
25
5
14
4
 x 3    x 
or . (4)
9
3
3
3
9. 15x 2  32x  7  0  (5x  1)(3x  7)  0  5x  1  0 or 3x  7  0
 5x  1 or 3x  7  x 
1
7
or x   . (4)
5
3
10. Definition: A function is a rule or a set of rules that pairs each number in a set called
the domain with exactly one number in a set called the range. (2)
11. Domain of f = {x : 2x  10  0 and x 2  8x  7  ( x  7)( x  1)  0}
= {x : x  5 and x  7} . (4)
12. f (g(x ))  f (3x  2)  2(3x  2)2  5(3x  2)  2  2(9x 2  12x  4)  15x  10  2
= 18x 2  39x  20 . (4)
13. g(f ( x )) 
14.
f ( x )  2 3x  2

 f ( x )  3x  4 . (4)
2f ( x )  1 6x  9
9  3x  x  15  9  3x  x 2  30x  225  x 2  33x  216  0
 ( x  9)( x  24)  0  x  9 or x  24 but only x  9 is a solution of the
given equation. (  24 is extraneous). (4)
15. After multiplying both sides of the equation by the denominators:
(2x  1)( x  5)  (x  3)( x  4)  2x 2  9x  5  x 2  x  12  x 2  8x  7  0
 ( x  1)( x  7)  0  x  1 or x  7 . (4)
16.
(4)
17. Since 1 and 2 are zeros of P(x), x  1 and x  2 are factors of P(x) by the Factor
theorem and therefore ( x  1)9x  2)  x 2  3x  2 is also a factor of P(x):
Therefore P( x )  ( x  1)( x  2)( x 2  x  2) so
that the other zeros of P(x) are the zeros of
1  8
1
7
 
i and
x2  x  2 :
2
2
2
1
7
 
i . (4)
2 2
18. Since the coefficients of the polynomial P( x )  2x3  3x 2  7x  3 are integers, we
have by the Rational zero theorem that if r is a rational zero of P then it must be a number
of the form p/q where p is a factor of the constant term, 3, and q is a factor of the leading
coefficient, 2. Since p =  1 or  3 and q =  1 or  2 , r must be in the set:
1
3
1
1
{1, ,3, } . P(1)  3 , P( )  0 . Therefore x  is a factor of P(x) and by
2
2
2
2
1
dividing P(x) by this factor we have: P( x )  ( x  )( 2 x 2  2 x  6) = (2x  1)( x 2  x  3)
2
 1   11
1
11
i and
Therefore the other zeros of P are the zeros of x 2  x  3 :
=  
2
2
2
1
11
 
i . (2)
2
2
19.
b) Domain of f = [-6,3], Range of f = [1,7], Domain of f 1( x ) = [1,7], Range of f 1( x ) =
[-6,3]. (2)
2
2
c) y   x  3 (where y = f(x)) Interchange x and y to get x   y  3 (where
3
3
2
3
9
3
9
y  f 1( x) ), x  3   y  y   x  . Therefore f 1( x )   x  . (2)
3
2
2
2
2


20. f (x )  3( x 2  12x)  46  3 (x  6)2  36  46  3( x  6)2  62 . Therefore f is one-toone on the interval (,6] (the graph of f is the left arm of a parabola with its vertex at
(6,61) ). Setting y  3( x  6)2  61 where y = f(x) and then interchanging x and y to
x  61
x  61
but
 ( y  6)2  y  6  
3
3
since the values of y must be  6 (the domain of f) it follows that
x  61
. (4)
f 1( x)  6 
3
get x  3( y  6)2  61 where y  f 1( x) ):
21. a) 23x  15  77  23x  15  77 or 23x  15  77  23x  92 or 23x  62
 x  4 or 
62
. (2)
23
b) There are three cases to consider:
i) When x  4 then x  4  x  4 and 2x  11  2x  11 and the equation becomes:
 3x  10  x  
10
which is not less than -4. Therefore there are no solutions from
3
case 1.
11
then x  4  x  4 and 2x  11  2x  11 and the equation
2
becomes  x  2 or x  2 which is in the interval and therefore is a solution.
11
iii) When x 
then both expressions in the absolute values are positive and the
2
equation becomes: 3x = 24 or x = 8 which is a solution. Therefore the equation has two
solutions: x  2 and x = 8. (4)
ii) When  4  x 
c) 5  3x  7  5  3x  7 or 5  3x  7  3x  2 or  3x  12  x  
or x  4 . (2)
2
3
22.
BDA  40o (Given)
 BCA  40o (Both angles in the same arc
AB)
CBA  90o (Angle subtended by a
diameter)
 BAC  180o  CBA  BCA  50o
(Sum of the angles in a triangle = 180 o . (4)
23.
a) O1O2  2 (the radius of the big circle) (2)
b) First prove: O1AO2  O1CO2 :
(radii of the small circle)
O1A  O1C
O1O2  O1O2 (common side)
AO2  CO2 A (radii of large circle)
Therefore O1AO2  O1CO2 by the Side-side-side theorem and therefore:
(corresponding angles of congruent triangles)
BO2A  BO2C
To prove ABO 2  CBO2
(common side)
BO 2  BO 2
BO2A  BO2C (proved)
(radii of large circle)
O2A  O2C
Therefore ABO 2  CBO2 by the Side-angle-side theorem. (5)
24. BEA  CED
Proof: ABE  DCE (both angles subtended by the chord AD)
AB = BC
(given)
BEA  CED (both angles subtended by the chord BC)
Therefore BEA  CED by the Angle-side-angle theorem. (4)