DMO’L.St Thomas More C2: Starters Revise formulae and develop problem solving skills. 123456789 101112131415161718 19 2021 222324 25 26

DMO’L.St Thomas More

C2: Starters

Revise formulae and develop problem solving skills.

1 2 3 4 5 6 7 8 9

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Starter 1

A circle has diameter with end points at (-4,-1) and (0,1).

Find the equation of the circle.


Starter 1


Centre = (-4+0 , -1+1)2 2

= (-2 , 0)


Starter 1


Centre = (-4+0 , -1+1) 2 2= (-2 , 0)

Radius = =

22 )10()42(


Starter 1


Centre = (-4+0 , -1+1) 2 2= (-2 , 0)

Radius = =

22 )10()42(

5Equation of circle:

(x+2)2 + y2 = 5Back


Starter 2

Find the mediator of these the points A(2,6) and B(8,0)


Starter 2


Midpoint of AB = (5,3)


Starter 2


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


Starter 3

Divide 6x3+27x2+14x+8 by x+4


To divide 6x3+27x2+14x+8 by x+4

x+4 6x3 +27x2 +14x +8



6x2

x+4 6x3 +27x2 +14x +8



6x2

x+4 6x3 +27x2 +14x +8

6x3 +24x2



6x2

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2



6x2

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x



6x2 +3x

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x



6x2 +3x

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x



6x2 +3x

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x

2x



6x2 +3x

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x

2x +8



6x2 +3x +2

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x

2x +8



6x2 +3x +2

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x

2x +8

2x +8



6x2 +3x +2

x+4 6x3 +27x2 +14x +8

6x3 +24x2

3x2 +14x

3x2 +12x

2x +8

2x +8

0Back


Divide 2x3+4x2-9x-9 by x+3

Starter 3


To divide 2x3+4x2-9x-9 by x+3

x+3 2x3 +4x2 -9x -9



2x2

x+3 2x3 +4x2 -9x -9



2x2

x+3 2x3 +4x2 -9x -9

2x3 +6x2



2x2

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2



2x2

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x



2x2 -2x

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x



2x2 -2x

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x



2x2 -2x

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x

-3x



2x2 -2x

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x

-3x -9



2x2 -2x -3

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x

-3x -9



2x2 -2x -3

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x

-3x -9

-3x -9



2x2 -2x -3

x+3 2x3 +4x2 -9x -9

2x3 +6x2

-2x2 -9x

-2x2 -6x

-3x -9

-3x -9

0

Back


Simplify each of the following:

4log8log2log3

2log6log5log

3log7log

31

222

55

aaa

4log8log2log3

2log6log5log

21log3log7log

31

222

555

aaa

4log8log2log3

15log2log6log5log

21log3log7log

31

2222

555

aaa 4log8log2log4log8log2log3

15log2log6log5log

21log3log7log

313

31

2222

555

aaaaaa

4log2log8log4log8log2log3

15log2log6log5log

21log3log7log

31

2222

555

aaaaaa

64log4log8log2log3

15log2log6log5log

21log3log7log

31

2222

555

aaaa

C2: Starter 5

Back


Express each of the following in terms of p,q and/or r

zryqxp aaa log,log,log

zyx

xy

a

a

z

yxa

zxy

a

32

3

log

)log

)(log

)(log

3

2

zyx

xy

rqpzyx

a

a

z

yxa

aaazxy

a

32

3

log

)log

)(log

logloglog)(log

3

2

zyx

xy

rqpzyx

rqpzyx

a

a

aaaz

yxa

aaazxy

a

32

3

log

)log

32log3loglog2)(log

logloglog)(log

3

2

zyx

qpyxxy

rqpzyx

rqpzyx

a

aaa

aaaz

yxa

aaazxy

a

32

23

21

23

213

log

loglog)log

32log3loglog2)(log

logloglog)(log

3

2

rqpzyxzyx

qpyxxy

rqpzyx

rqpzyx

aaaa

aaa

aaaz

yxa

aaazxy

a

21

2132

23

21

23

213

32loglog3log2log

loglog)log

32log3loglog2)(log

logloglog)(log

3

2

C2: Starter 6

Back


Solve the equations

C2: Starter 7

Back523

13

1

62

57

273

xx

x

xx


Solve the equations

C2: Starter 8

Back532

12

1

87

255

2564

xx

x

xx 3333.1x

2

1x

43.4x


Find the equation of the circle passing through the points A(-1,1), B(-1,5), C(5,7)

C2: Starter 9



Draw diagram

And then find

the mediators

C2: Starter 9



Mediator of AB

C2: Starter 9



Mediator of AB

y = 3

C2: Starter 9



Mediator of BC

Mid pt of BC = (2,6)

Grad BC = 1/3

Grad = -3

C2: Starter 9



Mediator of BC


Grad BC = 1/3

Grad = -3

y- 6 = -3(x – 2)

C2: Starter 9



Mediator of BC


Grad BC = 1/3

Grad = -3

Y = -3x +12

C2: Starter 9



Find centre using mediators

y = 3

y = -3x + 12

C2: Starter 9




y = 3

y = -3x + 12

Centre = (3,3)

C2: Starter 9




y = 3

y = -3x + 12

Centre = (3,3)

R2 = 42 + 22 = 20

C2: Starter 9



Equation of circle is

(x-3)2+(y-3)2 = 20

C2: Starter 9

Back


Find the sum to 10 terms of

a. 2 + 10 + 50 + 250 + …

b. 100 + 50 + 25 + 12.5 + …

Find the sum to infinity of

c. 4 + 2 + 1 + ½

C2: Starter 10



a. 2 + 10 + 50 + 250 + …

b. 100 + 50 + 25 + 12.5 + …

Find the sum to infinity of

c. 4 + 2 + 1 + ½

C2: Starter 10

Back

4 882 812

199.8047

8


a. Expand (x + 2y)4

b. Find the term in x7 in the expansion of (3x - y)10

c. Find the term that is independent of x in the expansion of (x2 + 5/x)9

C2: Starter 11

x4 + 8x3y + 24x2y2 + 32xy3 + 16y4

-262 440x7y3

1 312 500

Back


a. Expand (x - y)4

b. Find the term in x7 in the expansion of (x + 3y)9


C2: Starter 12

x4 - 4x3y + 6x2y2 - 4xy3 + y4

324x7y2

193 359 375

Back


a. Expand (x - y)5

b. Find the term in x5 in the expansion of (x + 7y)15


C2: Starter 13

x5 - 5x4y + 10x3y2 - 10x2y3 + 5xy4 – y5

(8.48x1011)x5y10

114 688

Back


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

25)sin(2

21 r

)sin( 52

522

210

23.15 cm


a. Use the remainder theorem to show that (x-1) is a factor of 4x3 – 3x2 – 1

b. If y = 2x3 – 3x2 find dy/dx

c. Write log37 -4log3(1/2) as a single logarithm

C2: Starter 15

Rem =f(1) = 4 x 13 – 3x12 – 1 = 4 – 3 – 1 = 0

(x-1) is factor.

dy/dx = 6x2 – 6x

log37 -log3(1/2)4 = log3(7 x 24)

= log3(448)

Back

log37 -log3(1/2)4 = log3(7 x 24)

= log3(448)

log37 -log3(1/2)4 = log3(7 x 24)

= log3(112)


a. Find the remainder when 4x3 – 3x2 – 1 is divided by (x-3)


c. Write 3log32 -4log3(2) as a single logarithm

C2: Starter 16Rem =f(3) = 4 x 33 – 3x32 – 1

= 108 – 27 – 1

= 80

dy/dx = 48x3 – 12x2

log37 -log3(1/2)4 = log3(7 x 24)

= log3(448)

Back

log37 -log3(1/2)4 = log3(7 x 24)

= log3(448)

log323 -log324 = log3(8 ÷ 16)

= log3(1/2)


a. Find the remainder when 4x3 – 3x2 – 1 is divided by (x+1)


c. Write in surd form sin 330o

C2: Starter 17

Rem =f(3) = 4 x (-1)3 – 3x(-1)2 – 1

= -4 – 3 – 1

= -8

dy/dx = 72x2 – 24xBack

sin 330o = -sin 30o = -1/2


a. If y = 12x3 – 6x2 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 = 36x2 – 12x

For turning points

36x2 – 12x = 0

12x(3x - 1) = 0

x = 0 or x = 1/3

For turning points

36x2 – 12x = 0

12x(3x - 1) = 0

x = 0 or x = 1/3

d2y/dx2 = 72x – 12

x = 0 d2y/dx2 < 0 Max at (0,0)

x = 1/3 d2y/dx2 > 0 Min at (1/3,-2/9)

d2y/dx2 = 72x – 12

x = 0 d2y/dx2 < 0 Max at (0,0)

x = 1/3 d2y/dx2 > 0 Min at (1/3,-2/9)

d2y/dx2 = 72x – 12

x = 0 d2y/dx2 < 0 Max at (0,0)

x = 1/3 d2y/dx2 > 0 Min at (1/3,-2/9)

Back



a. 2 + 10 + 18 + 26 + …

b. 200 + 100 + 50 + 25 + …

Without using a calculator find

a. cos 135o

b. sin 330o

c. tan 225o

C2: Starter 19



a. 2 + 10 + 18 + 26 + …380

b. 200 + 100 + 50 + 25 + …399.61


a. cos 135o

b. sin 330o

c. tan 225o

C2: Starter 19



a. 2 + 10 + 18 + 26 + …380

b. 200 + 100 + 50 + 25 + …399.61


a. cos 135o -1/2

b. sin 330o

c. tan 225o

C2: Starter 19



a. 2 + 10 + 18 + 26 + …380

b. 200 + 100 + 50 + 25 + …399.61


a. cos 135o -1/2

b. sin 330o -1/2

c. tan 225o

C2: Starter 19



a. 2 + 10 + 18 + 26 + …380

b. 200 + 100 + 50 + 25 + …399.61


a. cos 135o -1/2

b. sin 330o -1/2

c. tan 225o 1

C2: Starter 19

Back



a. 3 + 12 + 21 + 30 + …

b. 80 + 60 + 45 + 33.75…


a. cos 300o

b. sin 150o

c. tan 120o

C2: Starter 20



a. 3 + 12 + 21 + 30 + … 435

b. 80 + 60 + 45 + 33.75… 301.98


a. cos 300o

b. sin 150o

c. tan 120o

C2: Starter 20



a. 3 + 12 + 21 + 30 + … 435

b. 80 + 60 + 45 + 33.75… 301.98


a. cos 300o 1/2

b. sin 150o

c. tan 120o

C2: Starter 20



a. 3 + 12 + 21 + 30 + … 435

b. 80 + 60 + 45 + 33.75… 301.98


a. cos 300o 1/2

b. sin 150o 1/2

c. tan 120o

C2: Starter 20



a. 3 + 12 + 21 + 30 + … 435

b. 80 + 60 + 45 + 33.75… 301.98


a. cos 300o 1/2

b. sin 150o 1/2

c. tan 120o 3

C2: Starter 20

Back


C2: Starter 21

a. Find f’(x) given;

b. Find f(x) given

c. Solve sin(+60o) = 1/2

xxxxxf

524)('

xxxxf 223)(


a.

b. Find f(x) given


C2: Starter 21

xxxxxf

524)('

222 26)('3)( xxxfxxf xx


a.

b. Find f(x) given


C2: Starter 21

222 26)('3)( xxxfxxf xx

cxxxf

xxfx

xxx

52

2

5

5

2)(

4)('


a.

b. Find f(x) given


= 45 - 60, 135 – 60, 405 – 60

= 75o, 345o

C2: Starter 21

222 26)('3)( xxxfxxf xx

cxxxf

xxfx

xxx

52

2

5

5

2)(

4)('

Back


C2: Starter 22


b. Evaluate

c. Solve tan(+50o) = 1

dxxx )4(2

2

3

xxxxf

323)(


a.

b. Find f(x) given


C2: Starter 22

xxx

xf

xxf xx

222

3)('

3)(

2

2 3

dxxx )4(2

2

3



b. Find f(x) given


C2: Starter 22

xxx

xf

xxf xx

222

3)('

3)(

2

2 3

0)4(2

2

3 dxxx


a.

b. Find f(x) given

Solve tan(+50o) = 1

= -5o, 175o, 355o

C2: Starter 22

xxx

xf

xxf xx

222

3)('

3)(

2

2 3

0)4(2

2

3 dxxx

Back


C2: Starter 23

Find the shaded

area.


C2: Starter 23

Find the shaded

area.

Area under the

line is a triangle

222

22cm


C2: Starter 23

Find the shaded

area.

Area under the

curve

231

3

1

223

11)

2

1

2(

2

12cmdxx

xdx

xx


C2: Starter 22

Find the shaded

area.

Shaded Area

232

3112 cm

Back


C2: Starter 24

Find

a. the coordinates

of A, B and C

b. the shaded area.

C


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


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


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

25.122

55cm

C


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

6

5)5)(1( dxxx

C


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

6

5

26

5)56()5)(1( dxxxdxxx

C


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

26

5 3126

52)56()5)(1( cmdxxxdxxx

C


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

261

31

21 10212 cm

C

Back


Evaluate:

10log

1000log

27

1log

log

1024log

125log

1000

10

3

2

2

5

aa

C2: Starter 25

Back10log

1000log

27

1log

log

1024log

125log

1000

10

3

2

2

5

aa

10log

1000log

27

1log

log

101024log

3125log

1000

10

3

2

2

5

aa

10log

1000log

27

1log

2log

101024log

3125log

1000

10

3

2

2

5

aa

10log

1000log

327

1log

2log

101024log

3125log

1000

10

3

2

2

5

aa

10log

31000log

327

1log

2log

101024log

3125log

1000

10

3

2

2

5

aa

31

1000

10

3

2

2

5

10log

31000log

327

1log

2log

101024log

3125log

aa


Solve the equations:

C2: Starter 26

Back

155 x

221 84 xx

5log)1(log 22 xx

15log5log x

68.15log15log x

68.1 x

15log)22(4log)1( xx

4log15log2)15log24(log x

115log24log4log15log2

x

1 x

32log1

log 22

x

x

xx 321

311 x

311 x

Documents

DMO’L.St Thomas More C2: Starters Revise formulae and develop problem solving skills. 123456789 101112131415161718 19 2021 222324 25 26