Core 1 Revision Day_Master_v3x

Report
Further Mathematics Support Programme
www.furthermaths.org.uk
Core 1 Revision Day
Let Maths take you Further…
Outline of the Day
10:00 11:00
11:00 11:15
11:15 12:15
Algebra
Break
Co-ordinate Geometry
12:15 1:00pm
Lunch
1:00
2:00
Curve Sketching and Indices
Calculus
2:00
3:00
3:00pm
Home time!
2
ALGEBRA
3
4
For AS-core you should know:
• How to solve quadratic equations by factorising,
completing the square and “the formula”.
• The significance of the discriminant of a quadratic
equation.
• How to solve simultaneous equations (including one
linear one quadratic).
• How to solve linear and quadratic inequalities.
QUICK QUIZ
5
6
7
8
Question 1
The expression (2x-5)(x+3) is equivalent to:
A) 2x2 + x - 15
B) 2x2 - x - 15
(2x-5)(x+3)
2
C) 2x + 11x - 15
2 +6x – 5x -15
=
2x
D) 2x2 - 2x - 15
= 2x2 +x – 15
E) Don’t know
Question 2
The discriminant of the quadratic equation
2x2 +5x-1=0
is:
A) 17
b2 – 4ac
B) 33
= 52 – 4 x 2 x (-1)
C) 27
= 25 + 8
D) -3
E) I don’t know
=33
Question 3
Consider the simultaneous equations:
x + 3y = 5
3x – y =5
The correct value of x for the solution is:
3 x (2) 9x –3 y =15
A) x=1
B) x= -1
+ x + 3y = 5
C) x=2
10x
= 20
D) x= -2
x=2
E) I don’t know
(3)
(1)
12
Worked Example
a) Write the expression x2 -8x – 29 in the form (x+a)2 + b, where a and b are
constants.
b) Hence find the roots of the equation x2 -8x – 29 = 0. Express the roots in the form
c±d√5 where c and d are constants to be determined.
a) x2  8x  29  ( x  4)2 16  29
Take away 16 since the -4
in the bracket will give us
an extra 16.
 ( x  4)2  45
So a=-4 and b=-45.
b) x2  8x  29  0  ( x  4)2  45  0
 ( x  4)2  45
We must
complete the
question
Don’t forget the plus and
minus. A very common
error through out Alevel!
 x  4   45
 x  4  45
 x  4  5 9  4  3 5
 x  4  3 5 so c=4 and d=3
13
Worked Example
Solve the simultaneous equations
1) x – 2y = 1,
2) x2 + y2 = 29.
Label the equations 1 and 2
Equation 1 does not have any
squared terms, so it is easier
to expression x in terms of y
Equation 1  x  1  2 y
Using this in Equation 2 gives
(1+2y)2  y 2  29
 1  4 y  4 y 2  y 2  29  0
 5 y 2  4 y  28  0
You must know how to solve
quadratic equations with ease!
You could use the formula if
you wanted!
 (5 y  14)( y  2)  0
 y  2, y  14
5


 When y  14 , x=1+2  14 = 23
14 5
5
5
Worked Example
a) Find the set of values of x for which 6x+3>5-2x.
b) Find the set of values of x for which 2x2 -7x > >-3.
c) Hence, or otherwise, find the set of values of x for which
6x+3>5-2x and 2x2 -7x > >-3.
a)6 x  3  5  2 x  8 x  2  x  1
4
b)2x 2  7 x  3  2x 2  7 x  3  0  (2 x  1)(x  3)  0
Draw a quick sketch to help you.
The function changes signs at x=3, x= 1 .
2
½
So we need x< 1 or x>3
2
3
c) Draw a number line
¼
½
Since both inequalities must be true
3
we look for where there is red and blue
15
1 x 1
4
2
x3
16
Question for you to try
Question 1
17
Question for you to try
Question 2
18
Question for you to try
Question 3
19
20
Solutions
Question 1
For part (i) your values of a and b are:
A)
B)
C)
D)
E)
a =30, b = 2;
a = 120, b = 2;
a = 30, b = 5;
a = 120, b = 5;
None of these.
21
Worked Solution
Question 1
i)5 8  4 50  5 4  2  4 25 2
 5 2 2  4 5 2
 10 2  20 2
 30 2
a=30 and b=2
(6  3) 6 3  3 6 3  3 2 3 1




36  3
33
11 11
6  3 6  3 (6  3)
1
2
 p  ,q 
11
11
(6  3)  (6  3)  36  6 3  6 3  ( 3)2  36  3
ii)
3

3

22
Solutions
Question 2
The formula for r is given by:
A) r 
V
3h
B)
r
3V
h
C)
r
3V
h
D)
r
Vh
3
E) None of these.
23
Solutions
Question 3
The set of values for x is:
A)
-3<x<1
B)
-3>x>1
C)
-3<x or x>1
D)
-3>x or x>1
E)
None of these.
24
25
Question 1
C1(AQA) Jan 2006
26
Question 3
C1(AQA) Jan 2007
27
Question 7
C1(AQA) Jan 2007
28
Question 1
C1(Edexcel) Jan 2006
29
Question 5
C1(Edexcel) Jan 2006
30
Question 2
C1(Edexcel) Jun 2006
31
Question 6
C1(Edexcel) Jun 2006
32
Question 8
C1(Edexcel) Jun 2006
33
Question 2
C1(Edexcel) Jan 2007
34
Question 5
C1(Edexcel) Jan 2007
35
Question 1
C1(Edexcel) Jun 2007
36
Question 6
C1(Edexcel) Jun 2007
37
Question 7
C1(Edexcel) Jun 2007
38
Question 2
C1(Edexcel) Jan 2008
39
Question 3
C1(Edexcel) Jan 2008
40
Question 8
C1(Edexcel) Jan 2008
41
COORDINATE GEOMETRY
42
43
For AS-core you should know:
• How to calculate and interpret the equation of a straight line.
• How to calculate the distance between two points, the midpoint of two
points and the gradient of the straight line joining two points.
• Relationships between the gradients of parallel and perpendicular lines.
• How to calculate the point of intersection of two lines.
• Calculating equations of circles and how to interpret them.
• Circle Properties.
QUICK QUIZ
44
Gradient = change in y = y2 – y1
change in x
x2 – x1
y = mx + c
m = gradient
c = y intercept
Distance between two points
Equation of a circle:
Centre: (a, b)
Radius: r
47
Question 1
A straight line has equation 10y = 3x + 15.
Which of the following is true?
A) The gradient is 0.3 and the y-intercept is 1.5
B) The gradient is 3 and the y-intercept is 15
C) The gradient is 15 and the y-intercept is 3
y =and
3/10
+ 15/10 is 0.3
D) The gradient is 1.5
thexy-intercept
E) Don’t know
y = 0.31 x + 1.5
Question 2
A is the point (1, 5), B is the point (4, 7) and C is
the point (5, 2). Triangle ABC is
A) right-angled
B) scalene with no right angle
C) equilateral
(4  1)  (7  5)  9  4  13
D) isosceles
(4  5)  (7  2)  1  25  26
E) Don’t know
(5  1)  (2  5)  16  9  25  5
2
2
2
2
2
2
The sides are all different lengths
Question 3
• A circle has the equation (x + 3)² + (y − 1)² = 4.
Which of the following statements is false?
A) The y coordinate of the centre is −1
B) The radius of the circle is 2
C) The x coordinate of the centre is −3
The equation represents a
D) The point (−3,−1)
lies on the circle
circle with centre
E) Don’t know
(-3, 1) and radius 2.
So the statement is incorrect
51
Worked Example
y
NOT TO SCALE
B(3,4)
O
A
C
x
The line AB has equation y=5x-11 and passes through the point B(3,4) as shown above.
The line BC is perpendicular to AB and cuts the x-axis at C. Find the equation of the line BC
and the x-coodinate of C.
Gradient of line AB is 5
So gradient of line BC is -1/5.
Gradient of perpendicular
= -1 / (gradient of original)
Equation of BC is: y = -1/5 x + c
Using B(3,4) we get: 4 = -1/5 * 3 + c.
So c = 4 + 3/5 = 23/5
So Equation of BC is y = -1/5 x + 23/5
X coordinate of C is given when y = 0. So 0 =52-1/5 x + 23/5. So x = 23.
Worked Example
A circle has equation (x-2)2 + y2 = 45.
a) State the centre and radius of this circle.
b) The circle intersects the line with equation x + y = 5 at two points, A and B.
Find algebraically the coordinates of A and B.
c) Compute the distance between A and B to 2 decimal places.
a) Centre of circle is (2,0) and radius is √45
b) Equation of line implies: x = 5-y.
Using this in the equation of the circle
gives:
(5-y-2) 2 + y2 = 45
(3-y) 2 + y2 = 45
9-6y+y2 +y2 =45
2 y2 -6y + 9 =45
2 y2 -6y -36 =0
y2 -3y -18 =0
(y-6)(y+3)=0
So y = 6 or y = -3.
When y=6, x = 5 – 6 =1.
When y=-3, x = 5-(-3) = 8.
So coordinates are (1,6) and (8,-3)
“State” means you should be able to
write down the answer.
Equation of circle with centre (a,b)
and radius r is (x-a)2 + (y-b)2 = r2
c) Draw a diagram:
(1,6)
Watch the
minus signs
(8,-3)
Distance =
(8  1) 2  (6  (3)) 2
 72  92
53
 49  81  130  11.40
54
Question for you to try
Question 1
55
Question for you to try
Question 2
56
Question for you to try (part 1)
Question 3 (Part One)
57
Question for you to try (part 2)
Question 3 (Part Two)
58
59
Solution to Question 1
Question 1
The equation of the line is:
A)
B)
C)
D)
E)
3x + 2y = 26
3x + 2y = 13
-3x + 2y = 26
-3x + 2y = 13.
Don’t know.
60
Solution to Question 1
Question 1
Any line parallel to 3x + 2y = 6 must have the same gradient, but a different intercept.
(Short method): So any line parallel to given line has the form 3x + 2y = c (constant)
If the line goes through (2,10) then 3 * 2 + 2 * 10 = c, so c =26.
Hence equation is 3x + 2y = 26.
(Long method): Rearrange equation to get y = 3 – (3/2) x.
Gradient is –(3/2).
So new line must have the equation y = -(3/2) x + c
Use the point (2,10) to get
10 = -(3/2) * 2 + c. So c = 10 + 3 = 13.
Thus y = (-3/2)x + 13.
(This is the same as the last line since we have 2y = -3x + 26, so 3x + 2y =26.
61
Question 2
Solution to Question 2
The radius of the circle in (ii) is:
A)
B)
C)
D)
E)
√45
½ √45
√85
½ √85
Don’t know.
62
Solution to Question 2
i)
ii)
Gradient of AB =(8-0)/(9-5) = 8/4 =2
Gradient of BC = (1-0)/(3-5) = 1/(-2) = -½
Gradient = change in y/change in x
Product of gradients = 2 x (-½) = -1,
so perpendicular.
If AC is diameter then midpoint of AC is centre of the circle.
Midpoint of AC =  9  3 , 8  1   6,4.5
 2
iii)
2 
AC = √ ((9-3)2 + (8-1)2 ) = √ (62 + 72 ) = √85
So diameter is √85 and hence radius is ½√85
So equation of the circle is (x-6)2 + (y-4.5)2 = (½√85)2 =85/4
So equation is (x-6)2 + (y-4.5)2 =85/4
Coordinates of B(5,0) give (5-6)2 + (0-4.5)2 = 1 + 81/4 = 85/4. So B lies on the circle.
Let (x,y) be coordinates of D.
D(x,y)
Midpoint of BD is centre of circle (6,4.5).
So  5  x 0  y 
,

  6,4.5  5  x  12, y  9,  x  7, y  9.
 2
(6,4.5)
2 
So coordinates of D are (7,9).
63
B(5,0)
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Question 2
C1(AQA) Jan 2006
65
Question 5
C1(AQA) Jan 2006
66
Question 7
C1(AQA) Jun 2006
67
Question 4
C1(AQA) Jan 2007
68
Question 2
C1(AQA) Jan 2007
69
Question 1
C1(AQA) Jun 2007
70
Question 5
C1(AQA) Jun 2007
71
Question 1
C1(AQA) Jan 2008
72
Question 4
C1(AQA) Jan 2008
73
Question 3
C1(Edexcel) Jan 2006
74
Question 10
C1(Edexcel) Jun 2006
75
Question 11
C1(Edexcel) Jun 2006
76
Question 10
C1(Edexcel) Jun 2007
77
Question 11
C1(Edexcel) Jun 2007
78
Question 4
C1(Edexcel) Jan 2008
79
CURVE SKETCHING
(AND INDICES)
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81
For AS-core you should know:
• How to sketch the graph of a quadratic given in completed square form.
• The effect of a translation of a curve.
• The effect of a stretch of a curve.
QUICK QUIZ
82
85
Question 1
The vertex of the quadratic graph y=(x-2)2 - 3
is :
A) Minimum (2,-3)
B) Minimum (-2,3)
The
graph
has
a
minimum
C) Maximum (2,-3)
point, since the coefficient of
D) Maximum
(-2,-3)
x² is positive.
E) Don’t know
The smallest possible value of
(x-2)2 is 0, when x = 2.
[When x = 2 y = -3]
Question 2
The quadratic expression x2 -2x-3 can be written
in the form:
A) (x+1)2 - 4
B) (x-1)2 - 4
C) (x-1)2 - 3
2 -2x-3
2 -1 -3 =(x-1)2 - 4
2
x
=(x-1)
D) (x-1) - 2
E) Don’t know
The -1 is present to correct
for the +1 we get when
multiplying out (x-1)2
Question 3
The graph of y=x2 -2x-1 has a minimum point at:
y = x2 -2x-1
2 -1 -1
=
(x-1)
A) (1,-1)
= (x-1)2 - 2
B) (-1,-1)
So minimum point is (1,-2)
C) (-1,-2)
D) (1,-2)
E) Don’t know
89
Worked Exam Question
i) Write x2 -2x - 2 in the form (x-p)2 + q.
ii) State the coordinates of the minimum point on the graph y= x2 -2x - 2 .
iii) Find the coordinates of the points where the graph of y= x2 -2x - 2 crosses the
axes and sketch the graph.
iv) Show that the graphs of y= x2 -2x - 2 and y= x2 +4x - 5 intersect only once. Find the
x coordinate of the point of intersection.
x2 -2x – 2 = (x-1)2 -1 -2 = (x-1)2 -3.
So p=1 and q =-3.
ii) (x-1)2 has its smallest value when x=1, y value at this point is -3.
y
So minimum point is (1,-3).
iii) Graph crosses x-axis when y=0.
x2 -2x – 2 =0 implies (x-1)2 -3 =0.
So (x-1) = ±√3.
So x= 1 ±√3.
(1-√3,0)
Coordinates are (1 +√3,0) and (1 -√3,0)
Graph crosses y-axis when x=0.
(0,-2)
So coordinates are (0,-2)
90
i)
x
(1+√3,0)
(1,-3)
Worked Exam Question
i) Write x2 -2x - 2 in the form (x-p)2 + q.
ii) State the coordinates of the minimum point on the graph y= x2 -2x - 2 .
iii) Find the coordinates of the points where the graph of y= x2 -2x - 2 crosses the
axes and sketch the graph.
iv) Show that the graphs of y= x2 -2x - 2 and y= x2 +4x - 5 intersect only once. Find the
x coordinate of the point of intersection.
iv) Intersection when x2 -2x – 2 = x2 +4x – 5.
So -2x-2 = 4x -5, giving 6x – 3 =0. So x= ½.
Intersection when x=½.
91
92
Question for you to try
Question 1
93
Question for you to try
Question 2
94
Question for you to try
Question 3
95
Question for you to try
Question 4
96
97
Solution to Question 1(i)
Question 1
Solution to (i) is:
A)
B)
C)
D)
E)
4/27
8/81
4/3
4/81
Don’t know
98
Solution to Question 1 (ii)
Question 1
Solution to (ii) is:
B)
3a10 b 8
c2
3
C)
3a10b8c 2
A)
8
3
b
D)
c2
E)99 Don’t know
Question 3
Solution to Question 3
(-3,9)
The graph of y=4x2 -24x + 27 is:
A)
C)
(0,27)
(0,27)
(-4.5,0)
x
(1.5,0)
x
(1.5,0)
(4.5,0)
(3,-9)
B)
(3,9)
(0,27)
D)
x
(-4.5,0)
(-1.5,0)
(-3,-9)
(1.5,0)
(0,-27)
100
E)
Don’t know
(4.5,0)
x
101
Question 3
C1(AQA) Jan 2006
102
Question 1
C1(AQA) Jan 2007
103
C1(AQA) Jun 2007
Question 3
104
C1(AQA) Jan 2008
Question 5
105
C1(Edexcel) Jun 2006
Question 3
106
C1(Edexcel) Jun 2006
Question 9
107
C1(Edexcel) Jan 2007
Question 10
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CALCULUS (NON-MEI)
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110
For AS-core you should know:
• How the derivative of a function is used to find the gradient of its curve at
a given point.
• What is meant by a chord and how to calculate the gradient of a chord.
How the gradient of a chord can be used to approximate the gradient of a
tangent.
• How to differentiate integer powers of x and rational powers of x.
• What is meant by a stationary point of a function and how differentiation
is used to find them.
• Using differentiation to find lines which are tangential to and normal to a
curve.
QUICK QUIZ
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114
Question 1
The gradient of the curve y=3x2 – 4 at the point
(2,8) is :
A) 12
B) 6x
2–4
If
y=3x
C) 48
then
dy/dx
=
6x
D) 8
x=2
E) Don’t know dy/dx = 6x2 = 12
So gradient of curve at (2,8) is
12.
Question 2
If
3t 3  23t t2 3  2t 2 3
x  x 2
 t 1
2
2t
2t
2
dx 3
  1 .5
dt 2
then
A)
B)
C)
D)
E)
dy/dx =1.5
dy/dx = 3/2 - t
dx/dt = 1.5
dt/dx = 1.5
Don’t know
Question 3
Solution to Question 3
The correct answer is (B)
POINT A: the gradient is positive (sloping upwards from left to right) when x = 0.
Hence, the graph of the derivative crosses the y-axis at a positive value of y.
POINT’S B: the gradient is zero, this means that the graph of the derivative must
cross the x-axis at the points labelled B’.
The original curve looks like the graph of a cubic, so we would expect the graph of
its derivative to be a quadratic graph (a parabola), passing through the points
labelled A’ and B’.
119
Worked Example
A curve has equation y = x² – 3x + 1.
i) Find the equation of the tangent to the curve at the point where x = 1.
ii) Find the equation of the normal to the curve at the point where x = 3.
i) If y = x² – 3x + 1 then
dy/dx = 2x -3.
When x=1, gradient is given by dy/dx evaluated at x=1: gradient = 2x1 -3 = -1.
Equation of a straight line y=mx +c
So y = -x + c
When x = 1, y = 1² – 3x1 + 1 = -1.
So line passes through (1,-1).
So -1 = -1 + c, so c= 0.
Equation of tangent is y=-x.
ii) When x=3, gradient is given by dy/dx: gradient = 2 x 3 – 3 = 3.
Gradient of normal = -1/gradient of tangent.
So gradient of normal is -1/3.
When x=3, y = 3² – 3x3 + 1 = 1.
So line passes though (3,1).
Equation of line is y = -1/3 x + c
Using the point (3,1) gives: 1 = (-1/3) x 3 + c. So x = -2.
Equation of normal is y = -1/3 x -2.
Worked Example
A curve has equation y = 2x3 -3x2 – 8x + 9.
i)
Find the equation of the tangent to the curve at the point P (2, -3).
ii)
Find the coordinates of the point Q at which the tangent is parallel to the tangent at P.
i) If y = 2x3 -3x2 – 8x + 9, then dy/dx = 6x2 – 6x -8.
When x = 2, dy/dx = 6x22 – 6x2 -8. = 24 – 12 - 8 =4.
Tangent has gradient 4 and passes through (2,-3).
Using y = mx + c we have
y = 4x +c.
Using the point (2,-3) we have
-3 = 4 x 2 + c, so c = -11.
Hence equation of tangent is y = 4x -11.
ii) Tangent at P has gradient 4.
Tangent is parallel when gradient is the same.
So dy/dx = 6x2 – 6x -8 = 4
So 6x2 – 6x -12 =0, so x2 – x - 2 =0.
Thus (x-2)(x+1) = 0, which implies x=2 or x=-1.
WE NEED THE COORDINATES
P is where x=2, so Q is the point where x=-1.
When x = -1, y = 2(-1)3 -3(-1)2 – 8(-1) + 9 = -2 -3 +8 +9 =12.
So coordinates of Q are (-1,12).
122
Questions for you to try.
Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dx
ii) Find an equation for the tangent to the curve at the point where x=2.
iii) Determine whether y is increasing or decreasing when x = -3.
123
Questions for you to try.
Question 2
A curve has equation y=x3 + 44x2 + 29x
i) Find dy/dx
ii) Hence find the coordinates of the points on the curve where dy/dx=0.
124
125
Questions for you to try.
Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dx
ii) Find an equation for the tangent to the curve at the point where x=2.
iii) Determine whether y is increasing or decreasing when x = -3.
The equation of the tangent in part (ii) is:
A)
B)
C)
D)
E)
y = -1344x + 2314.
y = 1344x - 3062.
y = -21x6 + c
y = 21x6 + c
Don’t know
126
Questions for you to try.
Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dx
ii) Find an equation for the tangent to the curve at the point where x=2.
iii) Determine whether y is increasing or decreasing when x = -3.
i) dy/dx = -21x6
ii) When x = 2, dy/dx = - 21 x 26 = -1344.
So y = -1344 x + c.
When x = 2, y = 10 – 3 (2) 7 = - 374.
So (2,-374) is point on the curve.
Using this point in y = -1344 x + c gives -374 = -1344 x 2 + c.
So c = 2314.
Equation of tangent is y = -1344x + 2314.
iii. When x=-3, dy/dx = -21 x (-3) 6 = -15309 < 0. So y is decreasing.
127
Questions for you to try.
Question 2
A curve has equation y=x3 + 44x2 + 29x
i) Find dy/dx
ii) Hence find the coordinates of the points on the curve where dy/dx=0.
i) dy/dx = 3x2 + 88x + 29
ii) dy/dx = 0 when 3x2 +88x + 29 = 0.
So 3x2 +88x + 29 = 0 implies (3x+1)(x+29) = 0, so x=-29 or x =-1/3.
We need the coordinates.
When x = -1/3, y = (-1/3)3 + 44(-1/3)2 + 29(-1/3) = 1/9 + 44/9 -29/3 = -14/3.
So one point has coordinate (-1/3, -14/3).
When x = -29, y = (-29)3 + 44(-29)2 + 29(-29) = 11774.
So second point has coordinate (-29,11774)128
129
C1(AQA) Jan 2006
Question 7
130
C1(AQA) Jan 2007
Question 5
131
C1(AQA) Jun 2007
Question 4
132
C1(AQA) Jan 2008
Question 2
133
Question
9
C1(Edexcel) Jan
2006
134
C1(Edexcel) Jan 2006
Question 10
135
C1(Edexcel) Jun 2006
Question 5
136
C1(Edexcel) Jan 2007
Question 1
137
C1(Edexcel) Jan 2007
Question 8
138
C1(Edexcel) Jun 2007
Question 3
139
C1(Edexcel) Jan 2008
Question 5
140
POLYNOMIALS (MEI)
141
142
For AS-core you should know:
•
•
•
•
•
•
How to add, subtract and multiply polynomials.
How to use the factor theorem.
How to use the remainder theorem.
The curve of a polynomial of order n has at most (n – 1) stationary points.
How to find binomial coefficients.
The binomial expansion of (a + b)n.
QUICK QUIZ
143
146
Question 1
Which of the following is a factor of x³ + x² + 2x + 8
A) x+2
B) x-2
C) x+1
D) x-1
E) Don’t know
Solution to Question 1
The solution is (A).
If (x-a) is a factor of f(x), then f(a)=0.
If f(x) = x³ + x² + 2x + 8 then
A) f(-2) = 0, so x+2 is a factor.
B) f(2) =24, so x-2 is not a factor.
C) f(-1) = 6, so x+1 is not a factor.
D) f(1) = 12, so x-1 is not a factor.
Question 2
If x-2 is a factor of f(x)=3x³ – 5x² + ax + 2, then the
value of a is:
A) 21
B) 3
C)-21
D)-3
E) Don’t know
Solution to Question 2
The correct answer is (d).
If x-a is a factor of f(x), then f(a)=0
If f(x)=3x³ – 5x² + ax + 2, then
f(2) = 3(2)³ – 5(2)² + a(2) + 2 = 2a+6.
Since 2 is a factor f(2)=0, so 2a+6 =0, so a=-3.
Question 3
Solution to Question 3
• The correct answer is C
• The graph of y=(x-a)(x-b)(x+c) crosses the x-axis
at (a, 0), (b, 0) and (-c, 0).
• Since two of the intersections are with the
positive x-axis and one with the negative x-axis,
the graph must be either A or C.
• Since y is positive for large positive values of x,
the correct graph is C
153
Worked Example
Find the binomial expansion of (3+x)4, writing each term as simply as possible.
Binomial Expansion
(a  b) n  a n  nan1b 
n(n  1) n2 2
n(n  1)  2 n1 n
a b 
ab  b
2!
(n  1)!
In our example, a=3, b = x and n=4.
(3  x) 4  34  4  341  x 
4  3 4  2 2 4  3  2 4 3 3
3 x 
3 x  x4
2!
3!
 81 108x  54x 2  12x3  x 4
154
Worked Example
When x3 + 3x +k is divided by x-1, the remainder is 6. Find the value of k.
Remainder Theorem:
If f(x) is divided by x-a, then the remainder is f(a)
If f(x) = x3 + 3x +k , then f(1) = 13 + 3x1 +k = 6.
So 4+k =6, so k=2.
155
156
Questions for you to try.
Question 1
157
Questions for you to try.
Question 2
158
Questions for you to try.
Question 3
159
160
Solution to Question 1
Question 1
i) We need the discriminant to be greater than or equal to zero.
b2 -4ac = 52 -4x1xk =25-4k.
For one or more real roots we need 25-4k≥0.
So 25/4 ≥ k.
ii) 4x2 +20 x + 25 = (2x+5)(2x+5)
So 4x2 +20 x + 25 =0 implies (2x+5)(2x+5)=0, so x=-5/2.
161
Solution to Question 2
Question 2
f(x) = x3 + ax2 +7
Put x=-2, to get f(-2) = (-2)3 + a(-2)2 +7 = 0
So -8 + 4a +7 =0. Thus 4a=1, a = ¼.
162
Question 3
Solution to Question 3
i)(x  3)(2x 2  5x  4)  2x3  5x2  4x  6x 2 15x 12
 2 x3  x 2  11x  12
Computethediscriminant of quadratic
T hediscriminant of 2 x 2  5x  4 is 52  4  2  4  25  32  7  0
T hediscriminant is negativeso no real rootsof thequadratic equation
Henceonlyone real rootof thecubic equationf(x)  0.
163
Solution to Question 3 ctd
ii)Put x 2 into f ( x)  22 toget 2(2)3  (2)2 11(2) 12  22  0
So x  2 is root of f ( x)  22.
Computingf(x) 22 gives 2x3  x 2 11x 12  22  2x3  x 2 11x  10
ax2
bx
x
a=2
bx2
-2
-2ax2
c , c=-5
-2c=10
So 2x3  x 2 11x  10  ( x  2)(2x 2  3x  5)
 ( x  2)(2 x  5)(x  1)
So x=1 and x=-5/2 are other roots of the equation.
164
-2c=10, c=-5
a=2
-2a + b = -1
-4 + b =-1, b =3
Solution to Question 3 ctd
We havef(x) 22  ( x  2)(2 x  5)(x  1)
So f(x)  ( x  2)(2 x  5)(x  1)  22
So graph of f(x)is graph of ( x  2)(2 x  5)(x  1) shift down by 22 units.
(3,0)
(0,-12)
165
y=-22
166
C1(MEI) 6th June 2006
Question 8
167
C1(MEI) 6th June 2006
Question 12
168
C1(MEI) 16th January 2007
Question 4
169
C1(MEI) 16th January 2007
Question 5
170
C1(MEI) 16th January 2007
Question 8
171
C1(MEI) 7th June 2007
Question 4
172
C1(MEI) 7th June 2007
Question 6
173
Question 6
C1(MEI) January 2008
174
Question 7
C1(MEI) January 2008
175
Question 3
C1(MEI) June 2008
176
Question 8
C1(MEI) June 2008
177
Question 11
C1(MEI) June 2008
178
EXAM PRACTICE
179
That’s all folks!!!
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