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# Core 1 Revision Day - PowerPoint PPT Presentation

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

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Core 1 Revision Day

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www.furthermaths.org.uk

Core 1 Revision Day

Let Maths take you Further…

10:00 11:00Algebra

11:0011:15Break

11:15 12:15Co-ordinate Geometry

12:151:00pmLunch

1:002:00Curve Sketching and Indices

2:003:00Calculus

3:00pmHome time!

• 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

The expression (2x-5)(x+3) is equivalent to:

A) 2x2 + x - 15

B) 2x2 - x - 15

C) 2x2 + 11x - 15

D) 2x2 - 2x - 15

E) Don’t know

(2x-5)(x+3)

= 2x2 +6x – 5x -15

= 2x2 +x – 15

2x2 +5x-1=0

is:

A) 17

B) 33

C) 27

D) -3

E) I don’t know

b2 – 4ac

= 52 – 4 x 2 x (-1)

= 25 + 8 =33

Consider the simultaneous equations:

x + 3y = 5

3x – y =5

The correct value of x for the solution is:

A) x=1

B) x= -1

C) x=2

D) x= -2

E) I don’t know

3 x (2) 9x –3 y =15(3)

+ x + 3y = 5(1) 10x = 20

x=2

Worked Example

Write the expression x2 -8x – 29 in the form (x+a)2 + b, where a and b are constants.

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.

So a=-4 and b=-45.

Don’t forget the plus and minus. A very common error through out A-level!

We must complete the question

Worked Example extra 16.

Solve the simultaneous equations

x – 2y = 1,

x2 + y2 = 29.

1)

2)

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!

Worked Example extra 16.

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.

½

3

¼

½

3

Question for you to try extra 16.

Question 1

Question for you to try extra 16.

Question 2

Question for you to try extra 16.

Question 3

Solutions extra 16.

Question 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.

Worked Solution extra 16.

Question 1

a=30 and b=2

Solutions extra 16.

Question 2

The formula for r is given by:

A)

B)

C)

D)

E) None of these.

Solutions extra 16.

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.

C1(AQA) Jan 2006 extra 16.

Question 1

C1(AQA) Jan 2007 extra 16.

Question 3

C1(AQA) Jan 2007 extra 16.

Question 7

C1(Edexcel) Jan 2006 extra 16.

Question 1

C1( extra 16.Edexcel) Jan 2006

Question 5

C1( extra 16.Edexcel) Jun 2006

Question 2

C1( extra 16.Edexcel) Jun 2006

Question 6

C1( extra 16.Edexcel) Jun 2006

Question 8

C1( extra 16.Edexcel) Jan 2007

Question 2

C1( extra 16.Edexcel) Jan 2007

Question 5

C1( extra 16.Edexcel) Jun 2007

Question 1

C1( extra 16.Edexcel) Jun 2007

Question 6

C1( extra 16.Edexcel) Jun 2007

Question 7

C1( extra 16.Edexcel) Jan 2008

Question 2

C1( extra 16.Edexcel) Jan 2008

Question 3

C1( extra 16.Edexcel) Jan 2008

Question 8

Key Topics extra 16.

For AS-core you should know: extra 16.

• 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

Gradient = extra 16.change in y = y2 – y1

change in x x2 – x1

y = mx + c

c = y intercept

Distance between two points extra 16.

Equation of a circle:

Quick Quiz extra 16.

Question 1 extra 16.

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 + 15/10

y = 0.31 x + 1.5

Question 2 extra 16.

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

Question 3 extra 16.

• 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

So the statement isincorrect

Worked Example extra 16.

y

NOT TO SCALE

B(3,4)

O

x

C

A

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.

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 = -1/5 x + 23/5. So x = 23.

Worked Example extra 16.

A circle has equation (x-2)2 + y2 = 45.

State the centre and radius of this circle.

The circle intersects the line with equation x + y = 5 at two points, A and B. Find algebraically the coordinates of A and B.

Compute the distance between A and B to 2 decimal places.

Centre of circle is (2,0) and radius is √45

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:

Distance =

Watch the minus signs

(1,6)

(8,-3)

Question for you to try extra 16.

Question 1

Question for you to try extra 16.

Question 2

Question for you to try (part 1) extra 16.

Question 3 (Part One)

Question for you to try (part 2) extra 16.

Question 3 (Part Two)

Solution to Question 1 extra 16.

Question 1

The equation of the line is:

A) 3x + 2y = 26

B) 3x + 2y = 13

C) -3x + 2y = 26

D) -3x + 2y = 13.

E) Don’t know.

Solution to Question 1 extra 16.

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.

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.

Solution to Question 2 extra 16.

Question 2

The radius of the circle in (ii) is:

A) √45

B) ½ √45

C) √85

D) ½ √85

E) Don’t know.

Solution to Question 2 extra 16.

Gradient of AB =(8-0)/(9-5) = 8/4 =2

Gradient of BC = (1-0)/(3-5) = 1/(-2) = -½

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 =

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

D(x,y)

(6,4.5)

B(5,0)

C1(AQA) Jan 2006 extra 16.

Question 2

C1(AQA) Jan 2006 extra 16.

Question 5

C1(AQA) Jun 2006 extra 16.

Question 7

C1(AQA) Jan 2007 extra 16.

Question 4

C1(AQA) Jan 2007 extra 16.

Question 2

C1(AQA) Jun 2007 extra 16.

Question 1

C1(AQA) Jun 2007 extra 16.

Question 5

C1(AQA) Jan 2008 extra 16.

Question 1

C1(AQA) Jan 2008 extra 16.

Question 4

C1(Edexcel) Jan 2006 extra 16.

Question 3

C1(Edexcel) Jun 2006 extra 16.

Question 10

C1(Edexcel) Jun 2006 extra 16.

Question 11

C1(Edexcel) Jun 2007 extra 16.

Question 10

C1(Edexcel) Jun 2007 extra 16.

Question 11

C1(Edexcel) Jan 2008 extra 16.

Question 4

CURVE SKETCHING extra 16.(AND INDICES)

Key Topics extra 16.

For AS-core you should know: extra 16.

• 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

Quick Quiz extra 16.

Question 1 extra 16.

The vertex of the quadratic graph y=(x-2)2 - 3

is :

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]

Question 2 extra 16.

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

D) (x-1)2 - 2

E) 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

Question 3 extra 16.

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 - 2

So minimum point is (1,-2)

Worked Exam Question extra 16.

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.

(x-1)2 has its smallest value when x=1, y value at this point is -3.

So minimum point is (1,-3).

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)

y

x

(1-√3,0)

(1+√3,0)

(0,-2)

(1,-3)

Worked Exam Question extra 16.

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.

Intersection when x2 -2x – 2 = x2 +4x – 5.

So -2x-2 = 4x -5, giving 6x – 3 =0. So x= ½.

Intersection when x=½.

Question for you to try extra 16.

Question 1

Question for you to try extra 16.

Question 2

Question for you to try extra 16.

Question 3

Question for you to try extra 16.

Question 4

Solution to Question 1( extra 16.i)

Question 1

Solution to (i) is:

A) 4/27

B) 8/81

C) 4/3

D) 4/81

E) Don’t know

Solution to Question 1 (ii) extra 16.

Question 1

Solution to (ii) is:

A)

B)

C)

D)

E) Don’t know

Solution to Question 3 extra 16.

Question 3

(-3,9)

The graph of y=4x2 -24x + 27 is:

A)

B)

C)

D)

E) Don’t know

(0,27)

(0,27)

x

x

(1.5,0)

(-4.5,0)

(4.5,0)

(1.5,0)

(3,-9)

(3,9)

(0,27)

x

x

(1.5,0)

(4.5,0)

(-1.5,0)

(-4.5,0)

(0,-27)

(-3,-9)

C1(AQA) Jan 2006 extra 16.

Question 3

C1(AQA) Jan 2007 extra 16.

Question 1

C1(AQA) Jun 2007 extra 16.

Question 3

C1(AQA) Jan 2008 extra 16.

Question 5

C1(Edexcel) Jun 2006 extra 16.

Question 3

C1(Edexcel) Jun 2006 extra 16.

Question 9

C1(Edexcel) Jan 2007 extra 16.

Question 10

Key Topics extra 16.

For AS-core you should know: extra 16.

• 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

Quick Quiz extra 16.

Question 1 extra 16.

The gradient of the curve y=3x2 – 4 at the point (2,8) is :

A) 12

B) 6x

C) 48

D) 8

E) Don’t know

If y=3x2 – 4

then dy/dx = 6x

x=2

dy/dx = 6x2 = 12

So gradient of curve at (2,8) is 12.

Question 2 extra 16.

If

then

A) dy/dx =1.5

B) dy/dx = 3/2 - t

C) dx/dt = 1.5

D) dt/dx = 1.5

E) Don’t know

Question 3 extra 16.

Solution to Question 3 extra 16.

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’.

Worked Example extra 16.

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.

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 extra 16.

A curve has equation y = 2x3 -3x2 – 8x + 9.

• Find the equation of the tangent to the curve at the point P (2, -3).

• 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).

Questions for you to try. extra 16.

Question 1

A curve has equation y = 10 – 3x7.

Find dy/dx

Find an equation for the tangent to the curve at the point where x=2.

Determine whether y is increasing or decreasing when x = -3.

Questions for you to try. extra 16.

Question 2

A curve has equation y=x3 + 44x2 + 29x

Find dy/dx

Hence find the coordinates of the points on the curve where dy/dx=0.

Questions for you to try. extra 16.

Question 1

A curve has equation y = 10 – 3x7.

Find dy/dx

Find an equation for the tangent to the curve at the point where x=2.

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 + c

• D) y = 21x6 + c

• E) Don’t know

Questions for you to try. extra 16.

Question 1

A curve has equation y = 10 – 3x7.

Find dy/dx

Find an equation for the tangent to the curve at the point where x=2.

Determine whether y is increasing or decreasing when x = -3.

dy/dx = -21x6

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.

When x=-3, dy/dx = -21 x (-3) 6 = -15309 < 0. So y is decreasing.

Questions for you to try. extra 16.

Question 2

A curve has equation y=x3 + 44x2 + 29x

Find dy/dx

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)

C1(AQA) Jan 2006 extra 16.

Question 7

C1(AQA) Jan 2007 extra 16.

Question 5

C1(AQA) Jun 2007 extra 16.

Question 4

C1(AQA) Jan 2008 extra 16.

Question 2

C1(Edexcel) Jan 2006 extra 16.

Question 9

C1(Edexcel) Jan 2006 extra 16.

Question 10

C1(Edexcel) Jun 2006 extra 16.

Question 5

C1(Edexcel) Jan 2007 extra 16.

Question 1

C1(Edexcel) Jan 2007 extra 16.

Question 8

C1(Edexcel) Jun 2007 extra 16.

Question 3

C1(Edexcel) Jan 2008 extra 16.

Question 5

Key Topics extra 16.

For AS-core you should know: extra 16.

• 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

Quick Quiz extra 16.

Question 1 extra 16.

Which of the following is a factor of x³ + x² + 2x + 8

• x+2

• x-2

• x+1

• x-1

• Don’t know

Solution to Question 1 extra 16.

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

• f(-2) = 0, so x+2 is a factor.

• f(2) =24, so x-2 is not a factor.

• f(-1) = 6, so x+1 is not a factor.

• f(1) = 12, so x-1 is not a factor.

Question 2 extra 16.

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 extra 16.

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 extra 16.

Solution to Question 3 extra 16.

• 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

Worked Example extra 16.

Find 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.

Worked Example extra 16.

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.

Questions for you to try. extra 16.

Question 1

Questions for you to try. extra 16.

Question 2

Questions for you to try. extra 16.

Question 3

Solution to Question 1 extra 16.

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.

Solution to Question 2 extra 16.

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.

Solution to Question 3 extra 16.

Question 3

Solution to Question 3 extra 16.ctd

-2c=10, c=-5

a=2

-2a + b = -1

-4 + b =-1, b =3

So x=1 and x=-5/2 are other roots of the equation.

Solution to Question 3 extra 16.ctd

(3,0)

(0,-12)

y=-22

C1(MEI) 6 extra 16.th June 2006

Question 8

C1(MEI) 6 extra 16.th June 2006

Question 12

C1(MEI) 16 extra 16.th January 2007

Question 4

C1(MEI) 16 extra 16.th January 2007

Question 5

C1(MEI) 16 extra 16.th January 2007

Question 8

C1(MEI) 7 extra 16.th June 2007

Question 4

C1(MEI) 7 extra 16.th June 2007

Question 6

C1(MEI) January 2008 extra 16.

Question 6

C1(MEI) January 2008 extra 16.

Question 7

C1(MEI) June 2008 extra 16.

Question 3

C1(MEI) June 2008 extra 16.

Question 8

C1(MEI) June 2008 extra 16.

Question 11

EXAM PRACTICE extra 16.