core 1 revision day
DESCRIPTION
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 Algebra 11:0011:15 Break 11:15 12:15 Co-ordinate Geometry 12:151:00pm Lunch 1:002:00 Curve Sketching and Indices - PowerPoint PPT PresentationTRANSCRIPT
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Core 1 Revision Day
Let Maths take you Further…
Further Mathematics Support Programme
www.furthermaths.org.uk
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Outline of the Day10:00 11:00 Algebra11:00 11:15 Break11:15 12:15 Co-ordinate Geometry
12:15 1:00pm Lunch
1:00 2:00 Curve Sketching and Indices2:00 3:00 Calculus
3:00pm Home time!
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ALGEBRA
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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
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Question 1
The expression (2x-5)(x+3) is equivalent to:
A) 2x2 + x - 15B) 2x2 - x - 15C) 2x2 + 11x - 15D) 2x2 - 2x - 15E) Don’t know
(2x-5)(x+3) = 2x2 +6x – 5x -15 = 2x2 +x – 15
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Question 2
The discriminant of the quadratic equation 2x2 +5x-1=0
is:
A) 17B) 33C) 27D) -3E) I don’t know
b2 – 4ac = 52 – 4 x 2 x (-1) = 25 + 8 =33
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Question 3Consider the simultaneous equations:
x + 3y = 53x – y =5
The correct value of x for the solution is:A) x=1B) x= -1C) x=2D) x= -2E) I don’t know
3 x (2) 9x –3 y =15 (3) + x + 3y = 5 (1)
10x = 20 x=2
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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.
Worked Example
2 2) 8 29 ( 4) 16 29a x x x
2( 4) 45x Take away 16 since the -4 in the bracket will give us
an extra 16.
2 2) 8 29 0 ( 4) 45 0b x x x 2( 4) 45x
4 45x
4 45x
We must complete the
question
4 5 9 4 3 5x
4 3 5 so c=4 and d=3x
Don’t forget the plus and minus. A very common
error through out A-level!
So a=-4 and b=-45.
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Solve the simultaneous equationsx – 2y = 1,
x2 + y2 = 29.
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Worked Example
Label the equations 1 and 2
1)
2)
Equation 1 1 2x y
2 2
Using this in Equation 2 gives
(1+2y) 29y 2 21 4 4 29 0y y y
25 4 28 0y y
(5 14)( 2) 0y y
142, 5y y
2314 14When , x=1+2 =5 5 5y
Equation 1 does not have any squared terms, so it is easier to expression x in terms of y
You must know how to solve quadratic equations with ease!You could use the formula if you wanted!
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Worked Example
8 2x 14x
Draw a quick sketch to help you.
1The function changes signs at x=3, x= .2 3½1So we need x< or x>32
c) Draw a number line
½ 3¼Since both inequalities must be true we look for where there is red and blue
3x 1 14 2x
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.
xxa 2536)
0372372) 22 xxxxb 0)3)(12( xx
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Question for you to tryQuestion 1
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Question for you to tryQuestion 2
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Question for you to tryQuestion 3
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SolutionsQuestion 1
For part (i) your values of a and b are:
A) a =30, b = 2;B) a = 120, b = 2;C) a = 30, b = 5; D) a = 120, b = 5;E) None of these.
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Worked SolutionQuestion 1
225424550485) i254225
220210 230 a=30 and b=2
)36()36(
363
363)
ii336336
33
336
111
1132
112,
111
qp336)3(363636)36()36( 2
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SolutionsQuestion 2
The formula for r is given by:
A)
B)
C)
D)
E) None of these.
hVr3
hVr3
hVr 3
3Vhr
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SolutionsQuestion 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.
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C1(AQA) Jan 2006Question 1
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C1(AQA) Jan 2007Question 3
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C1(AQA) Jan 2007Question 7
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C1(Edexcel) Jan 2006Question 1
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C1(Edexcel) Jan 2006Question 5
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C1(Edexcel) Jun 2006Question 2
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C1(Edexcel) Jun 2006Question 6
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C1(Edexcel) Jun 2006Question 8
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C1(Edexcel) Jan 2007Question 2
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C1(Edexcel) Jan 2007Question 5
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C1(Edexcel) Jun 2007Question 1
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C1(Edexcel) Jun 2007Question 6
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C1(Edexcel) Jun 2007Question 7
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C1(Edexcel) Jan 2008Question 2
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C1(Edexcel) Jan 2008Question 3
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C1(Edexcel) Jan 2008Question 8
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COORDINATE GEOMETRY
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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
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Gradient = change in y = y2 – y1 change in x x2 – x1
y = mx + c
m = gradientc = y intercept
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Distance between two points
Equation of a circle:
Centre: (a, b) Radius: r
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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
D) The gradient is 1.5 and the y-intercept is 0.3 E) Don’t know
y = 3/10 x + 15/10y = 0.31 x + 1.5
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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 D) isosceles E) Don’t know
The sides are all different lengths
1349)57()14( 22
26251)27()54( 22
525916)52()15( 22
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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 D) The point (−3,−1) lies on the circle
E) Don’t know
The equation represents a circle with centre
(-3, 1) and radius 2. So the statement is incorrect
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Worked Example
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.
y
xOA
C
B(3,4) NOT TO SCALE
Gradient of line AB is 5So gradient of line BC is -1/5.
Equation of BC is: y = -1/5 x + cUsing B(3,4) we get: 4 = -1/5 * 3 + c. So c = 4 + 3/5 = 23/5So Equation of BC is y = -1/5 x + 23/5X coordinate of C is given when y = 0. So 0 = -1/5 x + 23/5. So x = 23.
Gradient of perpendicular = -1 / (gradient of original)
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Worked ExampleA 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 √45b) 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 =452 y2 -6y + 9 =452 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:
Distance = (8,-3)
(1,6)
22 ))3(6()18( 22 97
40.111308149
Watch the minus signs
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Question for you to tryQuestion 1
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Question for you to tryQuestion 2
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Question for you to try (part 1)Question 3 (Part One)
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Question for you to try (part 2)Question 3 (Part Two)
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Solution to Question 1Question 1
The equation of the line is:
A) 3x + 2y = 26B) 3x + 2y = 13C) -3x + 2y = 26D) -3x + 2y = 13.E) Don’t know.
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Solution to Question 1Question 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 + cUse 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.
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Solution to Question 2Question 2
The radius of the circle in (ii) is:
A) √45B) ½ √45C) √85D) ½ √85E) Don’t know.
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Solution to Question 2
i) Gradient of AB =(8-0)/(9-5) = 8/4 =2Gradient of BC = (1-0)/(3-5) = 1/(-2) = -½Product of gradients = 2 x (-½) = -1, so perpendicular.
ii) If AC is diameter then midpoint of AC is centre of the circle.Midpoint of AC =
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.
iii) Let (x,y) be coordinates of D. Midpoint of BD is centre of circle (6,4.5). So
So coordinates of D are (7,9).
Gradient = change in y/change in x
5.4,62
18,2
39
5.4,62
0,2
5
yx .9,7,9,125 yxyx
B(5,0)
D(x,y)
(6,4.5)
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C1(AQA) Jan 2006Question 2
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C1(AQA) Jan 2006Question 5
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C1(AQA) Jun 2006Question 7
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C1(AQA) Jan 2007Question 4
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C1(AQA) Jan 2007Question 2
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C1(AQA) Jun 2007Question 1
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C1(AQA) Jun 2007Question 5
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C1(AQA) Jan 2008Question 1
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C1(AQA) Jan 2008Question 4
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C1(Edexcel) Jan 2006Question 3
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C1(Edexcel) Jun 2006Question 10
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C1(Edexcel) Jun 2006Question 11
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C1(Edexcel) Jun 2007Question 10
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C1(Edexcel) Jun 2007Question 11
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C1(Edexcel) Jan 2008Question 4
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CURVE SKETCHING (AND INDICES)
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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
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Question 1
The vertex of the quadratic graph y=(x-2)2 - 3is :
A) Minimum (2,-3)B) Minimum (-2,3)C) Maximum (2,-3)D) Maximum (-2,-3)E) Don’t know
The graph has a minimum point, since the coefficient of x² is positive. The smallest possible value of (x-2)2 is 0, when x = 2. [When x = 2 y = -3]
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Question 2
The quadratic expression x2 -2x-3 can be written in the form:
A) (x+1)2 - 4B) (x-1)2 - 4C) (x-1)2 - 3D) (x-1)2 - 2E) Don’t know
x2 -2x-3 =(x-1)2 -1 -3 =(x-1)2 - 4
The -1 is present to correct for the +1 we get when multiplying out (x-1)2
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Question 3
The graph of y=x2 -2x-1 has a minimum point at:
A) (1,-1)B) (-1,-1)C) (-1,-2)D) (1,-2)E) Don’t know
y = x2 -2x-1 = (x-1)2 -1 -1 = (x-1)2 - 2So minimum point is (1,-2)
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Worked Exam Questioni) 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.
i) 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.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.Coordinates are (1 +√3,0) and (1 -√3,0)Graph crosses y-axis when x=0.So coordinates are (0,-2)
x
y
(1+√3,0)(1-√3,0)
(1,-3)(0,-2)
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Worked Exam Questioni) 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.
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Question for you to tryQuestion 1
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Question for you to tryQuestion 2
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Question for you to tryQuestion 3
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Question for you to tryQuestion 4
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Solution to Question 1(i)Question 1
Solution to (i) is:
A) 4/27B) 8/81
C) 4/3D) 4/81
E) Don’t know
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Solution to Question 1 (ii)Question 1
Solution to (ii) is:
A)
B)
C)
D)
E) Don’t know
2
8103cba
328103 cba
2
83cb
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Solution to Question 3Question 3
The graph of y=4x2 -24x + 27 is:
A)
B)
x(-1.5,0)(-4.5,0)
(-3,-9)
(0,27)
C)
D)
E) Don’t know
x(1.5,0)(-4.5,0)
(-3,9)
(0,27)
x(4.5,0)(1.5,0)
(3,-9)
(0,27)
x(4.5,0)(1.5,0)
(3,9)
(0,-27)
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C1(AQA) Jan 2006Question 3
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C1(AQA) Jan 2007Question 1
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C1(AQA) Jun 2007Question 3
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C1(AQA) Jan 2008Question 5
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C1(Edexcel) Jun 2006Question 3
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C1(Edexcel) Jun 2006Question 9
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C1(Edexcel) Jan 2007Question 10
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CALCULUS (NON-MEI)
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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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Question 1
The gradient of the curve y=3x2 – 4 at the point (2,8) is :
A) 12B) 6xC) 48D) 8E) Don’t know
If y=3x2 – 4 then dy/dx = 6xx=2
dy/dx = 6x2 = 12So gradient of curve at (2,8) is 12.
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Question 2If
then
A) dy/dx =1.5B) dy/dx = 3/2 - tC) dx/dt = 1.5D) dt/dx = 1.5E) Don’t know
2
23
223tttx
5.123
123
223
2
23
dtdx
ttttx
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Question 3
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Solution to Question 3The 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’.
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Worked ExampleA 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 +cSo y = -x + cWhen 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 + cUsing the point (3,1) gives: 1 = (-1/3) x 3 + c. So x = -2. Equation of normal is y = -1/3 x -2.
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Worked ExampleA 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 havey = 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 = 4So 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).
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Questions for you to try.Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dxii) 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.
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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.
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Questions for you to try.Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dxii) 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) y = -1344x + 2314.B) y = 1344x - 3062.C) y = -21x6 + cD) y = 21x6 + cE) Don’t know
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Questions for you to try.Question 1
A curve has equation y = 10 – 3x7.
i) Find dy/dxii) 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.
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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)
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C1(AQA) Jan 2006Question 7
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C1(AQA) Jan 2007Question 5
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C1(AQA) Jun 2007Question 4
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C1(AQA) Jan 2008Question 2
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C1(Edexcel) Jan 2006Question 9
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C1(Edexcel) Jan 2006Question 10
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C1(Edexcel) Jun 2006Question 5
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C1(Edexcel) Jan 2007Question 1
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C1(Edexcel) Jan 2007Question 8
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C1(Edexcel) Jun 2007Question 3
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C1(Edexcel) Jan 2008Question 5
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POLYNOMIALS (MEI)
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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
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Question 1
Which of the following is a factor of x³ + x² + 2x + 8A) x+2B) x-2C) x+1D) x-1E) Don’t know
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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 thenA) 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.
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Question 2
If x-2 is a factor of f(x)=3x³ – 5x² + ax + 2, then the value of a is:
A) 21B) 3C)-21D)-3E) Don’t know
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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, thenf(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.
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Question 3
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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
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Worked ExampleFind the binomial expansion of (3+x)4, writing each term as simply as possible.
Binomial Expansion
In our example, a=3, b = x and n=4.
nnnnnn babnnnbannbnaaba
1221
)!1(2)1(
!2)1()(
43342241444 3!3
2343!234343)3( xxxxx
432 125410881 xxxx
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Worked ExampleWhen 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.
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Questions for you to try.Question 1
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Questions for you to try.Question 2
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Questions for you to try.Question 3
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Solution to Question 1Question 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.
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Solution to Question 2Question 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.
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Solution to Question 3Question 3
12156452)452)(3)( 2232 xxxxxxxxi12112 23 xxx
quadratic ofnt discrimina theCompute0732254245 is 452 ofnt discrimina The 22 xx
equation quadratic theof roots real no so negative isnt discrimina The0.f(x)equation cubic theofroot real oneonly Hence
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Solution to Question 3 ctd
02212)2(11)2()2(2get to22)( into 2Put x) 23 xfii
101122212112 gives 22f(x) Computing 2323 xxxxxx.22)( ofroot is 2 xSo xf
ax2 bx c , c=-5
x a = 2 bx2
-2 -2ax2 -2c=10
-2c=10, c=-5a=2-2a + b = -1-4 + b =-1, b =3
)532)(2(10112 So 223 xxxxxx)1)(52)(2( xxx
So x=1 and x=-5/2 are other roots of the equation.
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Solution to Question 3 ctd
)1)(52)(2( 22f(x) have We xxx22)1)(52)(2( f(x) So xxx
units. 22by down shift )1)(52)(2( ofgraph is f(x) ofgraph So xxx
y=-22(0,-12)
(3,0)
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C1(MEI) 6th June 2006Question 8
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C1(MEI) 6th June 2006Question 12
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C1(MEI) 16th January 2007Question 4
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C1(MEI) 16th January 2007Question 5
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C1(MEI) 16th January 2007Question 8
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C1(MEI) 7th June 2007Question 4
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C1(MEI) 7th June 2007Question 6
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C1(MEI) January 2008Question 6
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C1(MEI) January 2008Question 7
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C1(MEI) June 2008Question 3
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C1(MEI) June 2008Question 8
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C1(MEI) June 2008Question 11
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EXAM PRACTICE
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That’s all folks!!!