DMO’L.St Thomas More C2: Starters Revise formulae and develop problem solving skills

DMO’L.St Thomas More Starter 1 A circle has diameter with end points at (-4,-1) and (0,1). Find the equation of the circle.

DMO’L.St Thomas More Starter 1 A circle has diameter with end points at (-4,-1) and (0,1). Centre = (-4+0, -1+1) 2 = (-2, 0)

DMO’L.St Thomas More Starter 1 A circle has diameter with end points at (-4,-1) and (0,1). Centre = (-4+0, -1+1) 22 = (-2, 0) Radius = =

DMO’L.St Thomas More Starter 1 A circle has diameter with end points at (- 4,-1) and (0,1). Centre = (-4+0, -1+1) 2 2 = (-2, 0) Radius = = Equation of circle: (x+2) 2 + y 2 = 5 Back

DMO’L.St Thomas More Starter 2 Find the mediator of these the points A(2,6) and B(8,0)

DMO’L.St Thomas More Starter 2 Find the mediator of these the points A(2,6) and B(8,0) Midpoint of AB = (5,3)

DMO’L.St Thomas More Starter 2 Find the mediator of these the points A(2,6) and B(8,0) Midpoint of AB = (5,3) Gradient of AB = - 6 / 6 = -1 Equation of Mediator is y – 3 = 1(x – 5) x – y -2 = 0 Back

DMO’L.St Thomas More Starter 3 Divide 6x 3 +27x 2 +14x+8 by x+4

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 x+46x36x3 +27x 2 +14x+8

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 x+46x36x3 +27x 2 +14x+8

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x 2x2x

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x 2x2x+8

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x+2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x 2x2x+8

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x+2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x 2x2x+8 2x2x

DMO’L.St Thomas More To divide 6x 3 +27x 2 +14x+8 by x+4 6x26x2 +3x+2 x+46x36x3 +27x 2 +14x+8 6x36x3 +24x 2 3x23x2 +14x 3x23x2 +12x 2x2x+8 2x2x 0 Back

DMO’L.St Thomas More Divide 2x 3 +4x 2 -9x-9 by x+3 Starter 3

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 x+32x32x3 +4x 2 -9x-9

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 x+32x32x3 +4x 2 -9x-9

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x -3x

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x -3x-9

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x-3 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x -3x-9

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x-3 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x -3x-9 -3x-9

DMO’L.St Thomas More To divide 2x 3 +4x 2 -9x-9 by x+3 2x22x2 -2x-3 x+32x32x3 +4x 2 -9x-9 2x32x3 +6x 2 -2x 2 -9x -2x 2 -6x -3x-9 -3x-9 0 Back

DMO’L.St Thomas More Simplify each of the following: C2: Starter 5 Back

DMO’L.St Thomas More Express each of the following in terms of p,q and/or r C2: Starter 6 Back

DMO’L.St Thomas More Solve the equations C2: Starter 7 Back

DMO’L.St Thomas More Solve the equations C2: Starter 8 Back

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Draw diagram And then find the mediators C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Mediator of AB C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Mediator of AB y = 3 C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Mediator of BC Mid pt of BC = (2,6) Grad BC = 1 / 3 Grad  = -3 C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Mediator of BC Mid pt of BC = (2,6) Grad BC = 1 / 3 Grad  = -3 y- 6 = -3(x – 2) C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Mediator of BC Mid pt of BC = (2,6) Grad BC = 1 / 3 Grad  = -3 Y = -3x +12 C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Find centre using mediators y = 3 y = -3x + 12 C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Find centre using mediators y = 3 y = -3x + 12 Centre = (3,3) C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Find centre using mediators y = 3 y = -3x + 12 Centre = (3,3) R 2 = = 20 C2: Starter 9

DMO’L.St Thomas More Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7) Equation of circle is (x-3) 2 +(y-3) 2 = 20 C2: Starter 9 Back

DMO’L.St Thomas More Find the sum to 10 terms of a … b … Find the sum to infinity of c ½ C2: Starter 10

DMO’L.St Thomas More Find the sum to 10 terms of a … b … Find the sum to infinity of c ½ C2: Starter 10 Back

DMO’L.St Thomas More a.Expand (x + 2y) 4 b.Find the term in x 7 in the expansion of (3x - y) 10 c.Find the term that is independent of x in the expansion of (x / x ) 9 C2: Starter 11 x 4 + 8x 3 y + 24x 2 y xy y x 7 y Back

DMO’L.St Thomas More a.Expand (x - y) 4 b.Find the term in x 7 in the expansion of (x + 3y) 9 c.Find the term that is independent of x in the expansion of (x / x ) 12 C2: Starter 12 x 4 - 4x 3 y + 6x 2 y 2 - 4xy 3 + y 4 324x 7 y Back

DMO’L.St Thomas More a.Expand (x - y) 5 b.Find the term in x 5 in the expansion of (x + 7y) 15 c.Find the term that is independent of x in the expansion of (x / x ) 8 C2: Starter 13 x 5 - 5x 4 y + 10x 3 y x 2 y 3 + 5xy 4 – y 5 (8.48x10 11 )x 5 y Back

DMO’L.St Thomas More If  =  /5 and the radius is 10 cm find, to 3 significant figures, the area of the shaded segment. C2: Starter 14 Area of segment Back 55 2525

DMO’L.St Thomas More a.Use the remainder theorem to show that (x-1) is a factor of 4x 3 – 3x 2 – 1 b.If y = 2x 3 – 3x 2 find dy / dx c.Write log log 3 ( 1 / 2 ) as a single logarithm C2: Starter 15 Rem =f(1) = 4 x 1 3 – 3x1 2 – 1 = 4 – 3 – 1 = 0  (x-1) is factor. dy / dx = 6x 2 – 6x log 3 7 -log 3 ( 1 / 2 ) 4 = log 3 (7 x 2 4 ) = log3(448) Back log 3 7 -log 3 ( 1 / 2 ) 4 = log 3 (7 x 2 4 ) = log3(448) log 3 7 -log 3 ( 1 / 2 ) 4 = log 3 (7 x 2 4 ) = log 3 (112)

DMO’L.St Thomas More a.Find the remainder when 4x 3 – 3x 2 – 1 is divided by (x-3) b.If y = 12x 4 – 4x 3 find dy / dx c.Write 3log log 3 (2) as a single logarithm C2: Starter 16 Rem =f(3) = 4 x 3 3 – 3x3 2 – 1 = 108 – 27 – 1 = 80 dy / dx = 48x 3 – 12x 2 log 3 7 -log 3 ( 1 / 2 ) 4 = log 3 (7 x 2 4 ) = log3(448) Back log 3 7 -log 3 ( 1 / 2 ) 4 = log 3 (7 x 2 4 ) = log3(448) log log = log 3 (8 ÷ 16) = log 3 ( 1 / 2 )

DMO’L.St Thomas More a.Find the remainder when 4x 3 – 3x 2 – 1 is divided by (x+1) b.If y = 24x 3 – 12x 2 find dy / dx c.Write in surd form sin 330 o C2: Starter 17 Rem =f(3) = 4 x (-1) 3 – 3x(-1) 2 – 1 = -4 – 3 – 1 = -8 dy / dx = 72x 2 – 24x Back sin 330 o = -sin 30 o = -1 / 2

DMO’L.St Thomas More a.If y = 12x 3 – 6x 2 find dy / dx b.Find the two turning points and use the second differential to determine which maximum and which is minimum. C2: Starter 18 dy / dx = 36x 2 – 12x For turning points 36x 2 – 12x= 0 12x(3x - 1) = 0 x = 0 or x = 1 / 3 For turning points 36x 2 – 12x= 0 12x(3x - 1) = 0 x = 0 or x = 1 / 3 d 2 y / dx 2 = 72x – 12 x = 0 d 2 y / dx 2 < 0 Max at (0,0) x = 1 / 3 d 2 y / dx 2 > 0 Min at ( 1 / 3,- 2 / 9 ) d 2 y / dx 2 = 72x – 12 x = 0 d 2 y / dx 2 < 0 Max at (0,0) x = 1 / 3 d 2 y / dx 2 > 0 Min at ( 1 / 3,- 2 / 9 ) d 2 y / dx 2 = 72x – 12 x = 0 d 2 y / dx 2 < 0 Max at (0,0) x = 1 / 3 d 2 y / dx 2 > 0 Min at ( 1 / 3,- 2 / 9 ) Back

DMO’L.St Thomas More Find the sum to 10 terms of a … b … Without using a calculator find a.cos 135 o b.sin 330 o c.tan 225 o C2: Starter 19

DMO’L.St Thomas More Find the sum to 10 terms of a …380 b … Without using a calculator find a.cos 135 o b.sin 330 o c.tan 225 o C2: Starter 19

DMO’L.St Thomas More Find the sum to 10 terms of a …380 b … Without using a calculator find a.cos 135 o -1 /  2 b.sin 330 o c.tan 225 o C2: Starter 19

DMO’L.St Thomas More Find the sum to 10 terms of a …380 b … Without using a calculator find a.cos 135 o -1 /  2 b.sin 330 o -1 / 2 c.tan 225 o C2: Starter 19

DMO’L.St Thomas More Find the sum to 10 terms of a …380 b … Without using a calculator find a.cos 135 o -1 /  2 b.sin 330 o -1 / 2 c.tan 225 o 1 C2: Starter 19 Back

DMO’L.St Thomas More Find the sum to 10 terms of a … b … Without using a calculator find a.cos 300 o b.sin 150 o c.tan 120 o C2: Starter 20

DMO’L.St Thomas More Find the sum to 10 terms of a … 435 b … Without using a calculator find a.cos 300 o b.sin 150 o c.tan 120 o C2: Starter 20

DMO’L.St Thomas More Find the sum to 10 terms of a … 435 b … Without using a calculator find a.cos 300 o 1 / 2 b.sin 150 o c.tan 120 o C2: Starter 20

DMO’L.St Thomas More Find the sum to 10 terms of a … 435 b … Without using a calculator find a.cos 300 o 1 / 2 b.sin 150 o 1 / 2 c.tan 120 o C2: Starter 20

DMO’L.St Thomas More Find the sum to 10 terms of a … 435 b … Without using a calculator find a.cos 300 o 1 / 2 b.sin 150 o 1 / 2 c.tan 120 o  3 C2: Starter 20 Back

DMO’L.St Thomas More C2: Starter 21 a.Find f’(x) given; b. Find f(x) given c. Solve sin(  +60 o ) = 1 /  2

DMO’L.St Thomas More a. b. Find f(x) given c. Solve sin(  +60 o ) = 1 /  2 C2: Starter 21

DMO’L.St Thomas More a. b. Find f(x) given c. Solve sin(  +60 o ) = 1 /  2 C2: Starter 21

DMO’L.St Thomas More a. b. Find f(x) given c. Solve sin(  +60 o ) = 1 /  2  = , 135 – 60, 405 – 60 = 75 o, 345 o C2: Starter 21 Back

DMO’L.St Thomas More C2: Starter 22 a.Find f’(x) given; b. Evaluate c. Solve tan(  +50 o ) = 1

DMO’L.St Thomas More a. b. Find f(x) given c. Solve tan(  +50 o ) = 1 C2: Starter 22

DMO’L.St Thomas More a. Find f’(x) given; b. Find f(x) given c. Solve tan(  +50 o ) = 1 C2: Starter 22

DMO’L.St Thomas More a. b. Find f(x) given Solve tan(  +50 o ) = 1  = -5 o, 175 o, 355 o C2: Starter 22 Back

DMO’L.St Thomas More C2: Starter 23 Find the shaded area.

DMO’L.St Thomas More C2: Starter 23 Find the shaded area. Area under the line is a triangle

DMO’L.St Thomas More C2: Starter 23 Find the shaded area. Area under the curve

DMO’L.St Thomas More C2: Starter 22 Find the shaded area. Shaded Area Back

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C b. the shaded area. C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Triangle ACM M C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Triangle ACM M C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Under curve BC M C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Under curve BC M C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Under curve BC M C

DMO’L.St Thomas More C2: Starter 24 Find a.the coordinates of A, B and C A = (1,0)B = (5,0) C = (6,5) b. the shaded area. Shaded area M C Back

DMO’L.St Thomas More Evaluate: C2: Starter 25 Back

DMO’L.St Thomas More Solve the equations: C2: Starter 26 Back