dmol.st thomas more c3: starters revise formulae and develop problem solving skills. 123456789...

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C3: StartersRevise formulae and develop

problem solving skills.1 2 3 4 5 6 7 8 9

10

11

12

13

14

15

16

17

18

19

20 21

22 23 24

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Starter 1Solve the equation

for 01sin2cos3 20

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Starter 1Solve the equation

for 01sin2cos3 20

01sin)sin21(3 2

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Starter 1Solve the equation

for 01sin2cos3 20 01sin)sin21(3 2 02sinsin6 2

0)1sin2)(2sin3(

21sin

32sin or

65

6 ,,55.5,87.3 Back

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Starter 2Prove the identity

2244 tansectansec

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Starter 2Prove the identity

44 tansec LHS

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Starter 2Prove the identity

44 tansec LHS)tan)(sectan(sec 2222

)tantan1)(tan(sec 2222 )tan(sec 22

Back

RHS

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Starter 3Prove the identity

tan

cos3cossin3sin

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Starter 3

cos3cossin3sin

LHS

cos3cos)sin()cos(2 2

32

3

)cos()cos(2)sin()cos(2

23

23

23

23

)cos()2cos(2)sin()2cos(2

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Starter 3

cos3cossin3sin

LHS

cos3cos)sin()cos(2 2

32

3

)cos()cos(2)sin()cos(2

23

23

23

23

RHS tan

)cos()2cos(2)sin()2cos(2

Back

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Starter 4Given that andwhere A is acute and B is obtuse, find

43tan A 13

5sin B

)(cos BAec

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Starter 4

By Pythagoras

43tan A 13

5sin B

54

53

cossin

AA

1312

125

costan

BB

1312

125

costan

BB

A is acute

B is obtuse

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Starter 4

)sin(1)(cosBA

BAec

BABA sincoscossin1

135

54

1312

53

1

6520

6536

1

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Starter 4

)sin(1)(cosBA

BAec

BABA sincoscossin1

135

54

1312

53

1

16651

6520

6536

Back

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Starter 5Differentiate

xey x sin2

xey 2sin

xxxy

lncossin

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Starter 5Differentiate

xey x sin2 xeu 2 xv sinx

dxdu e22 xdx

dv cos

xexedxdy xx cossin2 22

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Starter 5Differentiate

xey 2sin xu 2sin uey

xdxdu 2cos2 u

dudy e

xexedxdy xu 2cos22cos2. 2sin

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Starter 5Differentiate

xxxy

lncossin

xxu cossin xv ln

xxdxdu 22 sincos xdx

dv 1

2

1

)(lncossin2cosln

xxxxx

dxdy x

xdxdu 2cos

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Starter 5Differentiate

xxxy

lncossin

xxu cossin xv ln

xxdxdu 22 sincos xdx

dv 1

221

)(ln2sin2cosln

xxxx

dxdy x

xdxdu 2cos

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Starter 5Differentiate

xxxy

lncossin

xxu cossin xv ln

xxdxdu 22 sincos xdx

dv 1

2)(ln22sin2cosln2

xxxxxx

dxdy

xdxdu 2cos

Back

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Starter 6Differentiate

)ln(sec xy

7)3sin1( xy

xexy sin

2cos

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Starter 6Differentiate

1)(cossec xxu uy ln

xxdxdu sin)(cos 2 udu

dy 1

xx

udxdy

2cossin.1

xy ln(sec)

xx

dxdu

2cossin

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Starter 6Differentiate

1)(cossec xxu uy ln

xxdxdu sin)(cos 2 udu

dy 1

xxxx

xx

udxdy tan

coscossin

cossin.1

22

)ln(sec xy

xx

dxdu

2cossin

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Starter 6Differentiate

xu 3sin1 7uy

xdxdu 3cos3 67udu

dy

xxxudxdy 3cos)3sin1(213cos3.7 66

7)3sin1( xy

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Starter 6Differentiate

xu 2cos xev sinxdx

du 2sin2 xdxdv ex sin.cos

x

xx

eexxxe

dxdy

sin2

sinsin ).(cos2cos)2sin2(

xexy sin

2cos

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Starter 6Differentiate

xu 2cos xev sinxdx

du 2sin2 xdxdv ex sin.cos

)cos2cos2sin2(sin xxxedxdy x

xexy sin

2cos

Back

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Starter 7Solve the following equations, giving

exact solutions72 xe

5)12ln( 2 x

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Starter 7Solve the following equations, giving

exact solutions72 xe

7ln2 x7ln2

1x

5)12ln( 2 x5)12ln(2 x

25)12ln( x

25

12 ex

212

5

ex

Back

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Starter 8Show that can in be written in

the form Use the iteration starting with to generate Show that 5.5 is a root of the equation to

one decimal place.

0352 xx35 xx

351 nn xx

50 x 654321 ,,,,, xxxxxx

352 xx

35 xx

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Starter 8Use the iteration starting with to generate Show that 5.5 is a root of the equation to

one decimal place.

351 nn xx

50 x 654321 ,,,,, xxxxxx

2915.55

1

0

xx

Calculator:5 =5Ans+3====

537.5530.5518.5489.54275.52915.5

5

6

5

4

3

2

1

0

xxxxxxx

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Starter 8Show that 5.5 is a root of the equation to

one decimal place. 35)( 2 xxxf

525.0)55.5( f

547.0)45.5( f

Change of sign Root between 5.55 and 5.45Change of sign Root between 5.55 and 5.45Hence, x = 5.5 is a root to 1 decimal place.Back

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Starter 9Sketch the graph Hence, or otherwise, solve

52 xy

552 x

y=2x-5

52 xyy=5

x = 0 or 5 Back

when x = 3

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Starter 10 By differentiating find the coordinates of

the turning point on the curve State the nature of the turning point (i.e.

maximum or minimum).

xxy ln813

0813 2 x

xdxdyFor turning points

813 3 x3x22

2 816x

xdxyd

0816 22

2

x

xdxydwhen x =

3

Hence, minmum point at (3,-61.99)

Back

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Starter 11Solve the following equations, giving

exact solutions422 xxx eee

12)35ln( 2 x

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Starter 11Solve the following equations, giving

exact solutions422 xxx eee

432 xx ee

432 xxTake logs base e

4x

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Starter 11Solve the following equations, giving

exact solutions12)35ln( 2 x12)35ln(2 x6)35ln( x

e to the power of 635 ex

536

ex Back

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Starter 12Complete the table:

ydxdy

x

x

n

a

e

xxxxx

kx

lnsectancossin

aa

e

xxx

xx

nkx

x

x

x

n

ln

tansecsec

sincos

1

2

1

Back

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Starter 13Complete the table:

ydxdy

x

x

n

e

xxxx

xx

5

6

3

2

3

6ln7sec4tan

cos

2sin)13(

3ln)3(5

6

7tan7sec74sec4

sincos3

2cos2)13(6

5

6

1

2

2

12

x

x

x

n

e

xxx

xx

xxnx

Back