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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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Further Mathematics Support Programme

www.furthermaths.org.uk

Core 1 Revision Day

Let Maths take you Further…


Outline of the Day

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!


ALGEBRA


Key Topics


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


Quick Quiz


Question 1

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


Question 2

The discriminantof the quadratic equation

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


Question 3

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 Exam Questions


Take away 16 since the -4 in the bracket will give us an extra 16.

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

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

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


Questions for you to try now...


Question for you to try

Question 1


Question for you to try

Question 2


Question for you to try

Question 3


Answers to selected questions


Solutions

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

Question 1

a=30 and b=2


Solutions

Question 2

The formula for r is given by:

A)

B)

C)

D)

E) None of these.


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.


More practice for you.


C1(AQA) Jan 2006

Question 1


C1(AQA) Jan 2007

Question 3


C1(AQA) Jan 2007

Question 7


C1(Edexcel) Jan 2006

Question 1


C1(Edexcel) Jan 2006

Question 5


C1(Edexcel) Jun 2006

Question 2


C1(Edexcel) Jun 2006

Question 6


C1(Edexcel) Jun 2006

Question 8


C1(Edexcel) Jan 2007

Question 2


C1(Edexcel) Jan 2007

Question 5


C1(Edexcel) Jun 2007

Question 1


C1(Edexcel) Jun 2007

Question 6


C1(Edexcel) Jun 2007

Question 7


C1(Edexcel) Jan 2008

Question 2


C1(Edexcel) Jan 2008

Question 3


C1(Edexcel) Jan 2008

Question 8


COORDINATE GEOMETRY


Key Topics


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


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


Quick Quiz


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 + 15/10

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

D) isosceles

E) Don’t know

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

    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 isincorrect


Worked Exam Questions


Worked Example

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 perpendicular = -1 / (gradient of original)

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

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)


Questions for you to try now...


Question for you to try

Question 1


Question for you to try

Question 2


Question for you to try (part 1)

Question 3 (Part One)


Question for you to try (part 2)

Question 3 (Part Two)


Answers to selected questions


Solution to Question 1

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

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.


Solution to Question 2

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

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)


More practice for you.


C1(AQA) Jan 2006

Question 2


C1(AQA) Jan 2006

Question 5


C1(AQA) Jun 2006

Question 7


C1(AQA) Jan 2007

Question 4


C1(AQA) Jan 2007

Question 2


C1(AQA) Jun 2007

Question 1


C1(AQA) Jun 2007

Question 5


C1(AQA) Jan 2008

Question 1


C1(AQA) Jan 2008

Question 4


C1(Edexcel) Jan 2006

Question 3


C1(Edexcel) Jun 2006

Question 10


C1(Edexcel) Jun 2006

Question 11


C1(Edexcel) Jun 2007

Question 10


C1(Edexcel) Jun 2007

Question 11


C1(Edexcel) Jan 2008

Question 4


CURVE SKETCHING (AND INDICES)


Key Topics


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


Quick Quiz


Question 1

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

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

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 Questions


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.

(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

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.


Questions for you to try now...


Question for you to try

Question 1


Question for you to try

Question 2


Question for you to try

Question 3


Question for you to try

Question 4


Answers to selected questions


Solution to Question 1(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)

Question 1

Solution to (ii) is:

A)

B)

C)

D)

E) Don’t know


Solution to Question 3

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)


More practice for you.


C1(AQA) Jan 2006

Question 3


C1(AQA) Jan 2007

Question 1


C1(AQA) Jun 2007

Question 3


C1(AQA) Jan 2008

Question 5


C1(Edexcel) Jun 2006

Question 3


C1(Edexcel) Jun 2006

Question 9


C1(Edexcel) Jan 2007

Question 10


CALCULUS (NON-MEI)


Key Topics


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


Quick Quiz


Question 1

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

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


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’.


Worked Exam Questions


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.

  • 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 now...


Questions for you to try.

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.

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.


Answers to selected questions


Questions for you to try.

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.

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.

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)


More practice for you.


C1(AQA) Jan 2006

Question 7


C1(AQA) Jan 2007

Question 5


C1(AQA) Jun 2007

Question 4


C1(AQA) Jan 2008

Question 2


C1(Edexcel) Jan 2006

Question 9


C1(Edexcel) Jan 2006

Question 10


C1(Edexcel) Jun 2006

Question 5


C1(Edexcel) Jan 2007

Question 1


C1(Edexcel) Jan 2007

Question 8


C1(Edexcel) Jun 2007

Question 3


C1(Edexcel) Jan 2008

Question 5


POLYNOMIALS (MEI)


Key Topics


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


Quick Quiz


Question 1

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

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

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


Worked Exam Questions


Worked Example

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

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 now...


Questions for you to try.

Question 1


Questions for you to try.

Question 2


Questions for you to try.

Question 3


Answers to selected questions


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.


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.


Solution to Question 3

Question 3


Solution to Question 3 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 ctd

(3,0)

(0,-12)

y=-22


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That’s all folks!!!


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