Core 1 Revision Day

Core 1 Revision Day

Let Maths take you Further…

Further Mathematics Support Programme

www.furthermaths.org.uk

http://www.dfes.gov.uk/index.htm

http://www.dfes.gov.uk/index.htm

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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

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

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.

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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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(Edexcel) Jan 2006Question 1

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C1(Edexcel) Jun 2006Question 2

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COORDINATE GEOMETRY

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• 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 = change in y = y2 – y1 change in x x2 – x1

y = mx + c

m = gradientc = y intercept

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

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

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 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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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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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

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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) Jun 2006Question 7

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CURVE SKETCHING (AND INDICES)

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• 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]

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

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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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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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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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.

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

Question 3

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.

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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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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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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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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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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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


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.

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


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.

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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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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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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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) 16th January 2007Question 4

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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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EXAM PRACTICE

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That’s all folks!!!

Documents

Core 1 Revision Day