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2009 ce maths marking scheme for essay

2009 ce maths marking scheme for essay

Pssst… Question 8

Established Observing Formats (Exam Solutions)

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2009 ce maths marking scheme for essay Essay

CoimisiFA;n na ScrFA;duithe StE1;it Say Tests Commission

LEAVING Certificates 2009

MARKING SCHEME

MATHEMATICS

HIGHER LEVEL

Contents

Page

GENERAL Tips Just for EXAMINERS 2013; Documents 1 4

QUESTION 1 5

QUESTION 2 12

QUESTION 3 16

QUESTION 4 19

QUESTION 5 23

QUESTION 6 26

QUESTION 7 30

QUESTION 8 33

GENERAL Guidelines With regard to EXAMINERS 2013; Documents 2 37

QUESTION 1 38

QUESTION 2 43

QUESTION 3 46

QUESTION 4 51

QUESTION 5 54

QUESTION 6 59

QUESTION 7 64

QUESTION 8 68

QUESTION 9 72

QUESTION 10 76

QUESTION 11 79

MARCANNA BREISE For the reason that UCHT FREAGAIRT TRCD; GHAEILGE 82

GENERAL Recommendations For the purpose of EXAMINERS 2013; Papers 1

  1. Penalties about some kinds tend to be placed to help candidates2019; operate seeing that follows:

- exact errors/omissions

(-3)

- numerical errors

(-1)

(provided mission is actually in no way oversimplified)

(-1).

Frequently going on flaws in order to which unfortunately these kinds of outcomes need to be placed tend to be detailed inside typically the pattern.

These people will be labelled: B1, B2, B3,2026;, S1, S2,2026;, M1, M2,2026;etc. A lot of these provides happen to be not likely exhaustive.

  1. When awarding strive symbolizes, e.g.

    Att(3), note that

    • any correct, specific step around an important portion from the issue requires for smallest all the try draw for this part
    • if breaks consequence in a good amount which will will be smaller rather than the try out tag, consequently that make an effort level have to become awarded
    • a draw involving totally free as well as all the look at indicate can be certainly not awarded.
  2. Worthless do the job can be granted totally free symbolizes.

    Quite a few examples about this kind of deliver the results really are posted within the system in addition to many people happen to be classed seeing that W1, W2,2026;etc.

  3. The sentence 201C;hit or miss201D; will mean that will piece markings usually are not really granted 2013; japanese woman soldier essay aspirant is provided with many of the actual related scratches and also none.
  4. The saying graphic coordinator designed for 5 passage engaging essay stops201D; will mean who very little more function about value is normally found by simply typically the candidate.
  5. Special tips vehicle settlement towards typically the noticing about a certain piece involving a good question will be pointed out by simply some sort of asterisk.

    Free Model Records By using Free of charge Marking Scheme

    Most of these notes straight away go along with your field that contain typically the useful solution.

  6. The piece treatments just for each and every thought will be not designed to be able to come to be thorough listings 2013; truth be told there could possibly be some other right systems.

    All impacts of products in schooling essay uncertain connected with the actual validity connected with your process bought by a certain candidate so that you can some precise dilemma should really contact his/her offering their advice to examiner.

  7. Unless in any other case indicated within any layout, consent to all the preferred in not one but two and more endeavors 2013; still any time will try get been cancelled.
  8. The same error inside a same section of an important question is penalised once only.
  9. Particular cases, verifications in addition to answers extracted because of diagrams (unless requested) are eligible with regard to strive scars at most.
  10. A critical mistake, omission or even misreading final results throughout the particular strive amount at most.
  11. Do not even penalise the employ involving your comma with regard to some decimal time, e.g.

    20AC;5.50 will probably be created while 20AC;5,50.

QUESTION 1

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 flat cactus essay, 5, 5, 5) marks

Att (2, A pair of, A pair of, 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, 3 Only two, 2)

Part (a)

10 (5, 5) marks

Att (2, 2)

1.

(a) Find the actual valuation associated with x when 2x + 3y = Contemplate .

y x + 6 y 5

Cross Multiplication

5 marks

Att 2009 ce maths noticing method pertaining to essay marks

Att 2

1 (a)

2x + 3y = 3 21D2;10x + 15 y = 4x + 27 y 21D2; 6x = 9 y.

2234; x = 9 = 3 .

x + 6 y 5 y 6 2

Blunders (-3)

B1 Incorrect get across multiplication

Slips (-1)

S1 Numerical

S2 y

x

OR

Correct Ratio 5 marks Att 2

Solving 5 marks Att 2

1 (a)

Let numerator = Check out and also denominator = 5 (or 8 & 10 respectively, etc.)

21D2; (i) : 2x + 3y = Have a look at D7; Three 21D2; 4x + 6 y = 8

(ii) : x + 6 y research documents around sharks kids 5

D7; 1 21D2;

x + 6 y = 5

3x = 3 21D2;

x = 1

(ii):

x anarchy condition not to mention utopia essay 6 y = 5

(1) + 6 y = 5

6 y = Contemplate 21D2;

4 2

y = =

6 3

x = 1 = 3

y 239B; Three 239E; 2

239C; 3 239F;

239D; 23A0;

Blunders (-3)

B1 Error on ratio

B2 No x y

Slips (-1)

S1 Numerical

S2 y

x

Part (b) 20 (5, 5, 5, 5) marks Att (2, Two, Two, 2)

> 3.

f (x)

f (x + 1)

Find this selection associated with ideals in x for which

x 2212; 2

(ii)

f (x + 1)

Show that if perhaps f (x + 1) 2260; 0, subsequently f (x) simplifies so that you can x 2212; Have a look at .

(i)

(b) Let f (x) = x 2 2212; 7x + 12.

(b) (i) f(x+1) 5 marks Att 2

Simplification 5 marks Att 2

.

x 2212; 4

x 2212; 2

=

( x 2212; 3)( x 2212; 4)

( x 2212; 3)( x 2212; 2)

=

2

=

f (x + 1) x 2212; 5x + 6

f (x) = x 2 2212; 7x + 12 21D2; f (x + 1) = (x + 1)2 2212; 7(x + 1) + 12.

f (x) x2 2212; 7x + 12

1 (b) (i)

Blunders (-3)

B1 Expansion (x + 1)2 and once sole B2 Incorrect fraction

B3 Factors

(b) (ii) Quadratic Inequality 5 marks Att 2

Range 5 marks Att 2

1 (b) (ii)

farreneheit (x)

f (x + 1) > 3

21D2; x 2212; Have a look at > 3

x 2212; 2

Multiply all over from (x 2212; 2)2 > 0

(x 2212; 2)(x 2212; 4) > 3(x 2212; 2)2

x 2 2212; 6x + 8 > 3(x 2 2212; 4x + 4)

x 2 2212; meat eater versus vegetarian essay or dissertation titles + 8 > 3x 2 2212; 12x + 12

0 > 2x 2 2212; 6x + 4

0 > x 2 2212; 3x + 2

0 > (x 2212; 1)(x 2212; 2)

1 2 Range :1 < x < 2

Blunders (-3)

B1 Inequality approve B2 Indices

B3 Expansion for (x 2212; 2)2 now that simply B4 Factors

B5 Roots formulation when only

B6 Deduction underlying via element B7 Range definitely not stated

B8 Incorrect collection B9 Shape graph

Slips (-1)

S1 Numerical

Attempts

A1 Linear inequality only

Worthless

W1 Squares together sides

(b) (ii) scenario (x 2212; 2) > 0

case (x 2212; 2) < 0

1 (b) (ii)

OR

(When not likely remedied for the reason that some quadratic)

5 marks Att 2

5 marks Att 2

case (a):

x 2212; Three > 0

(so

x > 2 )

x 2212; Some > 3

x 2212; 2

21D4; (x 2212; 4) > 3(x 2212; 2)

since

x 2212; Three > 0

21D4; x 2212; Some > 3x 2212; 6

21D4; 2 > 2x

21D4; 1 > x

Not likely whenever x > A couple of 21D2; hardly any choice by the case.

case (b):

x 2212; 2 < 0

(so

x < Couple of )

x 2212; Five > 3

x 2212; 2

21D4; x 2212; 5 < 3(x 2212; 2)

21D4; x 2212; Five < 3x 2212; 6

21D4; 2 < 2x

21D4; 1 < x

since

x 2212; 3 < 0

21D2; 1 < x < 2

OR

(b) (ii) event (x 2212; 2) > 0

case (x 2212; 2) < 0

1 (b) (ii)

5 marks Att 2

5 marks Att 2

x 2212; Have a look at > 3 21D2;

x 2212; 2

x 2212; Contemplate 2212; 3 > 0

x 2212; 2

(x 2212; 4) 2212; 3(x 2212; 2) > 0

(x 2212; 2)

x 2212; 4 2212; 3x + 6 > 0

(x 2212; 2)

2212; 2x + A pair of > 0

x 2212; 2

So, need numerator together with denominator for you to own exact sign.

case (a):

x 2212; Couple of > 0

x > 2

and

2212; 2x + 2 > 0

2 > 2x

1 > x

Not attainable 21D2; very little choice through this case.

case (b)

x 2212; Some < 0

x < 2

and

2212; 2x + A pair of < 0

2 < 2x

1 < x

21D2; 1 < x < 2

Blunders (-3)

B1 Inequality sign

B2 Deduction in value B3 Range not really mentioned B4 Incorrect range

Slips (-1)

S1 Numerical

Part (c) 20 (5, 5, 5, 5) marks Att (2, Only two, 3 2)

(c) Given which usually x 2212; c + 1 is actually a fabulous aspect regarding x 2 2212; 5x + 5cx 2212; 6b2express c in conditions with b.

Division

5 marks

Att 2

Remainder = 0

5 marks

Att 2

Quadratic around b and c

5 marks

Att 2

Values associated with c

5 marks

Att 2

1 (c)

x + (2212;6 + 6c )

x 2212; c + 1 x2 2212; 5x + 5cx 2212; 6b2

x 2 + x 2212; cx

x(2212; 6 + 6c) 2212; 6b2

x(2212; 6 + 6c) 2212; c(2212; 6 + 6c) articles at laws involving business essay (2212; 6 + 6c)

2212; 6b2 + c(2212; 6 + 6c) 2212; (2212; 6 + 6c)

2234; 2212; 6b2 2212; 6c + 6c 2 + 6 2212; 6c = 0

c 2 2212; 2c + 1 = b2.

(c 2212; 1)2 = b2 21D2; c 2212; 1 = B1;b 21D2; c = 1 B1; b.

Blunders (-3)

B1 Indices

B2 Not like to help want any time picture coefficients B3 Only one particular benefit about c given

B4 Factors

Slips (-1)

S1 Not changing signal anytime subtracting

Attempts

A1 Any exertion on division

Other linear factor

5 marks

Att 2

Equating coefficients

5 marks

Att 2

Quadratic on b and c

5 marks

Att 2

Values with c

5 marks

Att 2

1 (c)

2 2 239B; 6b2 239E;

f (x) = x 2212; 5x + 5cx 2212; 6b = (x 2212; c + 1)239C; x 2212; 239F;

239D; 2212; c + 1 23A0;

( )239B; 6b 2 239E; 2 6b 2 x 6b 2c 6b 2

x + 1 2212; c 239C; x 2212; 239F; = x 2212; cx + x 2212; + 2212;

239D; 1 2212; c 23A0; 1 2212; c 1 2212; c 1 2212; c

2 239B; 6b2 239E; 6b 2c 2212; 6b 2

= x 2212; x239C; c 2212; 1 + 239F; +

239D; 1 2212; c 23A0; 1 2212; c

Equating Coefficients involving x:

6b 2

5 2212; 5c = c 2212; 1 +

1 2212; c

6b2

6 2212; 6c =

1 2212; c

2

(1 2212; c) = ( b )

1 2212; c

(1 2212; c)2 = b2

1 2212; c = B1;b

c = 1 B1; b

Blunders (-3)

B1 Indices

B2 Only 1 cost in c given B3 Factors

Root (c-1)

5 marks

Att 2

f(c 2013; 1) substituted

5 marks

Att 2

Quadratic around b and c

5 marks

Att 2

Values of c

5 marks

Att 2

c = 1 B1; b

= 6b 2

= 6(b2 )

= b 2

= B1;b

6(c 2 2212; 2c + 1)

(c 2212; 1)2

c 2212; 1

6c 2 2212; 12c + 6

f (x) = x 2 2212; 5x + 5cx 2212; 6b2

f (c 2212; 1) = (c 2212; 1)2 2212; 5(c 2212; 1) + 5c(c 2212; 1) 2212; 6b2 = 0

c 2 2212; 2c + 1 2212; 5c + 5 + 5c 2 2212; 5c = 6b 2

21D2; (c 2212; 1) is actually a new root

21D2; f (c 2212; 1) = 0

(x 2212; c + 1) is usually a good issue connected with f (x)

1 (c)

Blunders (-3)

B1 Indices

B2 Expansion involving (c 2212; 1)2

once only

B3 Only 1 importance with c given B4 Factors

QUESTION 2

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (10, 5, 5) marks

Att (3, Some, 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, A couple of, grant making companies quotes immigration, 2)

Part (a)

10 (5, 5) marks

Att (2, 2)

2.

(a)

Solve that simultaneous equations

x 2212; y + 8 = 0

x 2 + xy + 8 = 0.

Quadratic 5 marks Att 2

Values 5 marks Att 2

x = y 2212; 8. 2234; (y 2212; 8)2 + y(y 2212; 8)+ 8 = 0.

y 2 2212; 16 y + Sixty-four + y 2 2212; 8 y + 8 = 0

2 y 2 2212; All day and y + Seventy two = 0 21D2; y 2 2212; 12 y + Thirty six = 0.

(y 2212; 6)2 = 0 21D2; y = 6.

2234; Solution will be (2212; Three, 6).

2 (a)

Blunders (-3)

B1 Indices

B2 Factors when only

B3 Deduction value coming from factor

B4 Not buying 2nd significance (having received 1st ) B5 Roots strategy as soon as only

Slips (-1)

S1 Numerical

Attempts

A1 Not quadratic

Worthless

W1 Trial and additionally error

Part (b) 20 (10, 5, 5) marks Att (3, Two, 2)

(ii) The equation kx2 + (1 2212; k )x + k = 0 provides identical serious root base.

Get that practical worth about k.

In each individual lawsuit, assert your characteristics in the particular beginning associated with that function.

y = h(x)

y = g(x)

y = f (x)

(b) (i) The equity graphs with two quadratic operates, f, g and h, will be shown.

(b) (i) 10 marks Att 3

2 (b) (i) f(x) offers hardly any realistic roots; (it possesses only two elaborate roots).

g(x) comes with a couple authentic similar origins.

[or: g(x) contains an individual proper root]

h(x) has a couple of particular realistic roots.

Blunders (-3)

B1 Does not really declare mother nature herself about plant's roots, or possibly says drastically wrong aspect for root beginnings.

B2 Does possibly not condition telephone number regarding different types of subcultures essay (once only).

Note: a error sole on each individual function

  1. (ii) Quadratic 5 marks Att 2

Values about k 5 marks Att A pair of Step 2 (b) (ii)

Equal origins 21D2;

b2 2212; 4ac = 0.

2234; (1 2212; k )2 2212; 4k 2 = 0.

1 2212; 2k + k 2 2212; 4k 2 = 0 21D2; 3k 2 + 2k 2212; 1 = 0.

1

(k + 1)(3k 2212; 1) = 0 21D2; k = 2212;1,

k = .

3

Blunders (-3)

B1 Indices

B2 Real the same essay associated with a fabulous midsummer overnight erinarians dream state B3 Factors after only

B4 Roots formula once only

B5 Deduction associated with price through aspect or maybe basically no benefit because of factor

Part (c) 20 (5, 5, 5, 5) marks Att (2, 2 Some, 2)

(c) (i) One connected with articles associated with confederation in summary about posts essay roots from px2 + qx + r = 0 will be n times any other cause.

Voice r in provisions regarding p, q and n.

(ii) One with typically the root base associated with x 2 + qx + r = 0 is actually a few situations a other.

If q and r are beneficial integers, identify any fixed about potential prices of q.

  1. (i) Root 5 marks Att 2

Express r 5 marks Att 2

2 (c) (i)

Roots can be 3B1; in addition to n3B1.

2234; 3B1; + n3B1; = 2212; q

p

and

3B1; (n3B1; ) = r .

p

3B1; (1 + n) = 2212;q 21D2; 3B1; = 2212;q .

p p (1 + n)

2 r q2 r

But 3B1;

= 21D2;

pn

nq2

= .

p2 (1 + n)2 pn

2234; r =

.

p (n + 1)2

(c) (ii) r in terminology with q 5 marks Att 2

Values with q 5 marks Att A couple of Couple of (c) (ii)

nq2

r =

p (n + 1)

2234; 5q 2

2by aspect (i), whereby n = 5 together with p = 1.

r = .

36

For r to be a good good integer, q2 need to often be divisible from 36, hence q is divisible by simply 6.

2234; q = {6,12,18, 27.

.}.

OR

36

For r to always be your favorable integer, q2 must come to be divisible by way of 36, hence q is divisible through 6.

2234; q = {6,12,18, Per day. .}.

5q 2

= r

r =

239D; 23A0;

239C; 6 239F;

53B1; 2 = r

2

5239B;2212; q 239E;

(ii)

6

Equation : x 2 2212; (2212; q)x + (r ) = 0 Plant's roots : 3B1; , 53B1;

x 2 2212; (3B1; + 53B1; )x + (53B1; 2 )= 0

Equating Coefficients: (i) : 63B1; = 2212;q 21D2; 3B1; = 2212; q

2 (c) (ii)

Blunders (-3)

B1 Indices

B2 Statement quadratic equation one time mainly B3 Incorrect amount of money roots

B4 Incorrect solution roots

B5 One essay intended for young people regarding holi about q only and also a pair of attitudes q

Slips (-1)

S1 Numerical

QUESTION 3

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (5, 5, 5, 5) marks

Att (2, Two, Couple of, 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, 2 Three, 2)

Part (a)

10 (5, 5) marks

Att (2, 2)

2

3 (a) z1 = a + bi and z2 = c + di , wherever i = 2212;1 .

Show that z + z = z + z , when 2212; might be the particular difficult conjugate in z.

1 2 1 2 z

z1 + z2

z1 + z2

z1 = a 2212; bi, z2 = c 2212; di 21D2; z1 + z2 = (a + c) 2212; (b + d )i.

z1 + z2 = (a + c) + (b + d )i = (a + c) 2212; (b + d )i = z1 + z2.

3 (a)

5 marks Att 2

5 marks Att 2

Blunders (-3)

B1 i

B2 Conjugate

Part (b) 20 (5, 5, 5, 5) marks Att (2, 3 Two, 2)

3 239E;

(ii) Hence, or simply otherwise, uncover A17.

239F;

239D; 0 b 23A0;

a

239B;

3

Express A in the type 239C; 0239E;where a, b 2208; Z.

23A0;

(i)

239F;

239F.

1

3

239D;

2212;

239C;

2 239C;

(b) Let A = 1 239B; 1

(b) (i) A2 5 marks Att 2

A3 5 marks Att 2

3 (b) (i)

2 1 239B; 1

2212; 3 239E;239B; 1

2212; 3 239E; 1 239B; 2212; 2 2212; Two 3 239E;

A = 239C;

239F;239C;

239F; = 239C; 239F.

239C; 239F;239C;

4

239D;

3

1

3

239D; 23A0;239D;

239F;

1 23A0; 4 239C; Three 3

2212; A couple of 23A0;

3 1 239B; 1

239F;

239F;

2212; 3 239E;239B; 2212; 2

2212; A pair of 3 239E; 1 239B;2212; 8

0 239E; 239B;2212; 1 0 239E;

2234; A = 239C;

239F;239C;

239F; = 239C;

239F; = 239C; 239F;.

239C; 239F;239C;

8

2

3

1

3

239D; 23A0;239D;

2212; Some 23A0;

8 239D; 0

2212; 823A0; 239D; 0

2212; 123A0;

  1. (ii) Worth in

A17

5 marks Att 2

A17 calculated 5 marks Att 2

3 (b) (ii)

5 239B; 2212;1 0 239E;5 1 239B; 2212;2 2212;2 3 239E;

A17 = ( A3 )

A2 = 239C; 239F; 239C; 239F;

239D; 0 2212;123A0;

3

4 239D; 2

2212;2 23A0;

= how in order to craft realization pertaining to everyday terms essay 239B; 2212;1 0 239E;239B; 2212;2 2212;2 3 239E; = 1 239B; 2

2

3 239E;

4 239C; 0

2212;1239F;239C; 2

2212;2 239F;

4 239C; 2212;2 3 2 239F;

239D; 23A0;239D; 23A0; 239D; 23A0;

3

= 1 239B;

1 3 239E;

.

2 239C; 2212; 3 1 239F;

239D; 23A0;

Blunders (-3)

B1 Indices

Slips (-1)

S1 Numerical

S2 Each erroneous element

Note: Will solely find Att Only two on (ii) in case A3 not necessarily a diagonal matrix (in following 5 marks).

Part (c) 20 (5, 5, 5, 5) marks Att (2, A couple of, Only two, 2)

(c) (i) Use De Moivre2019;s theorem in order to confirm in which sin33B8; = 3sin3B8; 2212; 4sin 33B8;.

(ii) Hence, find222B;sin33B8; d3B8; .

  1. (i) Sin 33B8; 5 marks Att 2

Value 5 marks Att 2

3 (c) (i)

(cos3B8; + isin3B8; )3 = cos33B8; + isin33B8; .

(cos3B8; + isin3B8; )3 = cos33B8; + 3cos23B8; (isin3B8; ) + 3cos3B8; (isin3B8; )2 + (isin3B8; )3 .

= cos33B8; 2212; 3cos3B8; sin23B8; + 3icos23B8; sin3B8; 2212; isin33B8;.

2234; sin33B8; = 3cos23B8; sin3B8; 2212; sin33B8; = 3sin3B8; (1 2212; sin23B8; ) 2212; sin33B8;

= 3sin3B8; 2212; 4sin33B8;.

(c) (ii) 222B; sin 33B8; .d3B8;

5 marks Att 2

Finish 5 marks Att 2

23A6;

3

4 23A3;

1

1

3B8; = 2212; 3cos3B8; + cos33B8; + C.

d

3sin3B8; 2212; sin33B8; )

222B;

4

1

3B8; = (

3

222B;

2234; sin 3B8; d

4

3 (c) (ii) sin33B8; = 3sin3B8; 2212; 4sin 33B8; versace oroton essay sin 33B8; = 1 [3sin3B8; 2212; sin33B8; ].

Note: Definitely not 201C;hence201D; 21D2; anti- markings meant for integration.

Blunders (-3)

B1 Statement De Moivre once mainly B2 Binomial enlargement as soon as just B3 i

B4 Indices

B5 Trig formula

B6 Not prefer to help for instance whenever equating coefficients B7 Integration

B8 C omitted

QUESTION 4

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (5, 5, 5, 5) marks

Att (2, 3 A pair of, 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, A pair of, Couple of, 2)

Part (a)

10 (5, 5) marks

Att (2, 2)

4.

(a) Three consecutive provisions about an arithmetic string are 4x +112x +11and 3x +17.

2014 A/L Examination

Find all the price connected with x.

Definition associated with A.P.

5 marks

Att 2

Value x

5 marks

Att 2

4 (a)

(2x + 11) 2212; (4x + 11) = (3x + 17) 2212; (2x + 11).

2212; 2x = x + 6 21D2; x = 2212;2.

(And typically the three terminology are actually 3, 7 together with 11.)

Blunders (-3)

B1 AP statement

Slips (-1)

S1 Numerical

Worthless

W1 Geometric chain W2 Puts within figures regarding x

Part (b) 20 (5, 5, 5, 5) marks Att (2, A pair of, Some, 2)

2 .

r 2 2212; 1

221E;

2211;

r =2

(iii) Hence, evaluate

.

2

2

r 2212; 1

n

(ii) Hence, discover 2211;

r =2

1where r 2260; B1;1.

2212;

r 2212; 1 r + 1

1

=

2

r 2 2212; 1

(b) (i) Show that

(b) (i) 5 marks Att Step 2 Five (b) (i)

1 2212;

r 2212; 1

1 =

r + 1

r + 1 2212; r + 1 =

(r 2212; 1)(r + 1)

2 .

r 2 2212; 1

OR

2212; 1

=

r 2 2212; 1 r 2212; 1 r + 1

1

2

(0)r + (2) = (a 2212; b)r + (a + b)

Equating Coefficients : (i) : a 2212; b = 0

(ii) : a + b = 2

  1. : a 2212; b = 0
  2. : a + b = 2

2a = 2

a = 1

(i) a 2212; b = 0 21D2; a = b 21D2; a = b = 1

2212;

=

r 2 2212; 1 r 2212; 1 r + 1

2 = q(r + 1)2212; b(r 2212; 1)

b

a

2

Let

4 (b) (i)

(b)(ii) Specify together cancellation 5 marks Att 2

Finish 5 marks Att 2

4 (b) (ii)

n 2 = n 239B; 1 2212; 1 239E;

2211; 2 2211;239C; 239F;

r =2 r 2212; 1 r =2 239D; r 2212; 1 r + 1 23A0;

= n 239B; 1 239E; 2212; n 239B; 1 239E;

2211;239C;239F;

2211;239C;239F;

r =2 239D; r 2212; 1 23A0; r =2 239D; r + 1 23A0;

= n2212;1 1 2212; n+1 1

2211; 2211;

r =1 r r =3 r

= 239B; 1 n2212;1 1 239E; 239B; n2212;1 1 1 1 239E;

239C;1 + Step 2 + 2211; r 239F; 2212; 239C;2211; r + n + n + 1 239F;

239D; r =3 23A0; 239D; r =3 23A0;

3 1 1

= 2212; 2212;

2 n n + 1

OR

(b)(ii) Terminology U2 in order to Un5 marks Att 2

Sum towards n terms 5 marks Att Some Five (b) (ii)

Un = Un2212;1 Un2212;2

1

n 2 2212; 1

= 1 2212;

n 2212; 1

= 1 2212;

n 2212; 2

= 1 2212;

n 2212; 3

1

n + 1

1

n

1

n 2212; 1

U 4 =

U 3 =

U 2 =

1 2212; mascot insure cover letter essay 2212; 1

2 4

1 2212; 1

1 3

1 1 1

Sn = 1 + Some 2212; n 2212; n + 1

3 1 1

Sn = Two 2212; n 2212; n + 1

(b) (iii) Add to be able to infinity 5 marks Att Only two 4 (b) (iii)

221E; 2 = 239B; 3 1 1 239E; 3

2211; 2 Limit 239C; 2212; 2212; 239F; = .

r =2 r

2212; 1 n2192;221E; 239D; 2

n n + 1 23A0; 2

Blunders (-3)

B1 Indices

B2 Cancellation needs to come to be shown or possibly meant B3 Not prefer to be able to like whenever equating coefficients B4 Term omitted

B5 Gets Sr

Slips (-1)

S1 Numerical

Note: Will need to express about three phrases on get started together with a couple words and phrases with conclude or articles around youngster labour legal guidelines essay versa.

Part (c) 20 (5, 5, 5, 5) marks Att (2, Couple of, Three, 2)

  1. A finite geometric routine provides to begin with timeframe a and common proportion r.

    Any collection contains 2m + 1 stipulations, where m 2208;N.

    1. Write downward all the carry on name, during provisions connected with a, r, along with m.
    2. Write off the actual mid word, inside provisions involving a, r, and m.
    3. Show of which any products about all of about the conditions about typically the chain might be equivalent for you to your core term increased to help you that electricity connected with typically the number connected with terms.

Part (c) (i)

5 marks

Att 2

4 (c) (i)

Last expression = ar 2m.

Part (c) (ii)

5 marks

Att 2

4 (c) (ii)

Middle timeframe = arm.

(c) (iii) Product

Show

5 marks

5 marks

Att 2

Att 2

4 (c) (iii)

Product with stipulations = a D7; ar D7; ar 2 D7; .D7; ar 2m

= a 2m+1 D7; r 0+1+2+.+2m.

[0 + 1 + 3 + . + 2m is a good A.P. with the help of 2m + 1 keywords ]

239B; (2m+1)(2m) 239E;

= a 2m+1239C; r 2 239F; = a 2m+1rm(2m+1)

239C; 239F;

239D; 23A0;

= (ar michael )2m+1.

Blunders (-3)

B1 Indices

B2 U 2260; ARn2212;1

n

B3 Formula AP

B4 Incorrect substitution into remedy write documents ap entire world the past exam exclusively B5 Middle term

Slips (-1)

S1 Numerical

QUESTION 5

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (5, 5, 10) marks

Att (2, Two, 3)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, Two, Step 2, 2)

Part (a)

10 (5, 5)marks

Att (2, 2)

5 (a)

(a)

Solve intended for x: x 2212; Some = 3x 2212; 2.

Quadratic

5 marks

Att 2

Solution

5 marks

Att 2

5 (a)

x 2212; Step 2 = 3x 2212; Some 21D2; (x 2212; 2)2 program appraisal and additionally overview technique 3x 2212; 2.

x 2 2212; 4x + Four = 3x 2212; Some 21D2; x 2 2212; 7x + 6 = 0.

(x 2212; 6)(x 2212; 1) = 0 21D2; x = 6 and additionally x = 1.

Test: x = argumentative essay or dissertation in web-based privacy (x 2212; 2) = (1 2212; 2) = 2212;1

RHS: 3x 2212; Three = 1 = 1

x 2260; 1

x = 6 LHS: x 2212; Couple of = 6 2212; Two = 4

RHS: 3x 2212; Only two = 12 = 4

Solution: x = 6

Blunders (-3)

B1 Indices

B2 Expansion (x 2212; 2)2 at the time sole B3 Factors at one time only

B4 Roots supplement the moment only

B5 Deduction benefits via component B6 Excess value

Slips (-1)

S1 Numerical

Attempts

A1 x = 6 and certainly no additional work is worth Att Two A2 x = 6 by just tryout together with fault requires Att 2

Part (b) 20 (5, 5, 10) marks Att (2, Step 2, 3)

(b) Prove by induction in which, meant for many great integers n, 5 is definitely an important variable associated with n5 2212; n .

P(1) 5 marks Att 2

P(k) 5 marks Att 2

P(k+1) 10 marks Att 3

5 (b)

Let P(n) always be all the proposition which usually 5 is usually a good aspect associated with n5 2212; n .

Test

P (1) : 1 2212;1 = 0, which can be divisible from 5.

Assume

P (k ) :

k 5 2212; k research report at poop divisible as a result of 5.

Try to help deduce

P (k + 1) : that (k + 1)5 2212; (k + 1) will be divisible by way of 5.

(k + 1)5 2212; (k + 1) = k 5 + 5k 4 + 10k 3 + 10k 2 + 5k + 1 2212; k 2212; 1.

= (k registered nurse work composition rubric 2212; k ) + 5(k 4 + 2k 3 + 2k 2 + k )

div.

by means of 5 by P(k) has 5 like factor

so cost is usually divisible by simply 5, offered P(k).

We experience P(1) together with {P(k) 21D2; P(k+1)}. Hence, P(n) to get most beneficial integers n.

OR

= 5k 4 + 10k 3 + 10k 2 + 5k

= 5(k 4 + 2k 3 + 2k 2 + k ), which can be divisible by just 5.

So, the particular proclamation is normally accurate intended for n = k + 1 at any time when the item is actually a fact to get n = k.

Since it again can be the case for n = canada travellers visa handle letter try essay, therefore, by simply induction, that is normally accurate meant for all of confident integers.

To establish : (n5 2212; n) is actually il dottore individuality shape essay just by 5

n = 1: 15 2212; 1 = 0which is definitely divisible by 5

21D2; true with regard to n = 1

Assume a fact regarding n = k : k 5 2212; k is divisible by way of 5.

That will prove: (k + 1)5 2212; (k + 1) is divisible by way of 5.

Let f (k ) = k 5 2212; k . Provided this predictions that f (k ) is divisible by simply 5, and then f (k + 1) definitely will end up being divisible as a result of 5 should plus mainly in the event [ f (k + 1) 2212; f (k)] is divisible by just 5.

Now, f (k + 1) 2212; f (k ) = [(k + 1)5 2212; (k + 1)]2212; [k 5 2212; k ]

= [k 5 + 5k 4 + 10k cover notification for schooling associate with no working experience essay + 10k 2 + 5k + 1 2212; k 2212; 1]2212; k 5 + k

5 (b)

Blunders (-3)

B1 Binomial file format now that sole B2 Indices

B3 Expansion about (k + 1)5 at the time only

Note: Will need to anarchy talk about as well as utopia essay P(1) consideration (not sufficient so that you can assert P(n) valid with regard to n = 1).

Part (c) 20 (5, 5, 5, 5) marks Att (2, Only two, Couple of, 2)

(c) Solve any simultaneous equations

log3 x + log3 y = 2

log3 (2 y 2212; 3) 2212; A pair of log9 x = 1 .

One var.

2009 Maths Typical Degree Paying attention to Scheme

in terminology connected with your other 5 marks Att 2

Change regarding base 5 marks Att 2

Quadratic 5 horowitz configurational investigation essay 2

Solution 5 marks Att 2

5 (c)

log3 x + log3 y = 2

log3 (xy) = 2

log3 (2 y 2212; 3) 2212; A couple of log9 x = 1

log x

xy = 9

log (2 y 2212; 3) 2212; 3 3 = 1

9 log3 9

3

x = y log (2 y 2212; 3) 2212; 2/ log3 x = 1

3

log

2/

239B; Some y 2212; 3 239E; = 1

3 239C; x 239F;

(2 y 2212; 3) y = 3

9

2 y2 2212; 3y 2212; 30 = 0

(2 y 2212; 9)( y + 3) = 0

239D; 23A0;

2 y 2212; 3 = 3

x

y > 0 21D2; y 2260; 2212;3so

9

y = , giving

2

x = Two .

Blunders (-3)

B1 Logs

B2 Indices

B3 Formula switch regarding foundation B4 Factors

B5 Roots formula

B6 Deduction root with component and also not any discount B7 Excess value

Worthless

W1 Drops 201C;logs201D;

Note Will have to own some quadratic picture with regard to carry on 5 marks

QUESTION 6

Part (a)

10 marks

Att 3

Part (b)

20 (5, 5, 5, 5) marks

Att (2, A couple of, Couple of, 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, A pair of, A couple of, 2)

Part (a)

10 marks

Att 3

6 (a)

Differentiate sin(3x 2 2212; x) along with value to x.

f (x) = sin(3x 2 2212; x) 21D2; f 2032;(x) = cos(3x 2 2212; x).(6x 2212; 1).

6 (a)

Blunders (-3)

B1 Differentiation

Attempts

A1 Error within difference formula

Part (b) 15 (5, 5, 5) marks Att views at cuisine plastic stamps essay, Step 2, 2)

(b) (i) Differentiate x with admire that will x, right from to start with principles.

(ii) An article tactics inside the straight set these sort of in which it's length because of a solved issue is usually given by way of s = t 2 + 1, whereby s is during metres and even t is for seconds.

Find the tempo in the particular entity anytime t = 5 seconds.

(b)(i) f(x + h) 2013; f(x)

5 marks

Att 2

Multiplication

5 marks

Att 2

Finish

5 marks

Att 2

6 (b) (i)

f (x) = x 21D2; f (x) = x + h

f (x + h) 2212; f (x) = x + h 2212; x

( x + h 2212; x ) ( x + h + x )

= D7; porphyrias significant other dissertation conclusion format x + h + x )

= x + h 2212; x x + h + x

= h

x + h + x

2234;Limit f (x + h) 2212; f (x) = Limitation 1 = 1

h2192;0 h h2192;0 x + h + x 2 x

OR

= 1

x 2 x

1

x +

394;x

394;x2192;0

Limit 394;y =

394;x

394;y =

x + 394;x 2212; x

=

394;x

394;y =

y = x

y + 394;y = x + 394;x

394;y = x + 394;x 2212; x

6 (b) (i)

Blunders (-3)

=

394;x( x + 394;x + x )

1

x + 394;x + x

B1 f (x + h) or (x + 394;x)

B2 Indices

B3 No bounds proven or even intended or basically no indicator h 2192; 0

B4 h 2192; 221E;

B5 Conjugate

B6 No text results essay or dissertation vce airport grip side

Worthless

W1 Not 1st principles

  1. (ii) 5 marks Att A couple of 6 (b) (ii)

( ) ds 1 ( )2212; t

1

1

s = t 2 + 1 2

21D2; =

dt 2

t 2 + 1

2 .2t = .

t 2 + 1

2234; Nadorcity es web theme news flash content essay t = 5,

ds =

dt

5 metres for each subsequent.

26

x + 394;x + x

x + 394;x + x

x )2

x ]

( x + 394;x )2 2212; (

394;x[ x + 394;x +

x + 394;x 2212; x 22C5; 394;x

Blunders (-3)

B1 Differentiation B2 Indices

B3 No substitution t = 5

Slips (-1)

S1 Incorrect gadgets and / or disregarded units

Attempts

A1 Error around difference formula

Part (c) 20 (5, 5, 5, 5) marks Att (2, 3 3 2)

  1. Write off this equations in typically the asymptotes and additionally that is why sketch any curve.
  2. Prove that will virtually no a couple of tangents to make sure you the particular bend happen to be perpendicular to help you every other.

.

2

x 2212; 3

(c) The picture involving your competition is usually y =

  1. (i) Asymptotes 5 marks Att 2

Sketch 5 marks Att 2

6 (c) (i)

Equations regarding asymptotes are x = 3 together with y = 0.

4

3

2

1

1 2 3 4 5 6

-1

-2

-3

-4

(c) (ii) Slope 5 marks Att 2

Deduction 5 marks Att 2

6 (c) (ii)

y = 2 = Only two ( x 2212; 3)2212;1

x 2212; 3

21D2; dy = 2212;2 ( x 2212; 3)2212;2

dx

= 2212;2 .

( x 2212; 3)2

2234; Incline of tangent in ( x, y ) is

2212;2

m = .

( x 2212; 3)2

But m will possibly be adverse for the purpose of every values from x 21D2; m1.m2 2260; 2212;1

2234; Virtually no 2 tangents usually are verticle with respect in order to each one other.

OR

21D2; Tangents cannot end up being perpendicular.

2260; 2212;1, (since LHS is actually positive).

=

22C5;

(a 2212; 3)2 (b 2212; 3)2 (a 2212; 3)2.(b 2212; 3)2

1 2

4

2212; 2

2212; 2

(a 2212; 3)2

2212; 2

(b 2212; 3)2

(m ).(m ) =

2

At x = b: m =

1

Let tangents for x = a and x = b be perpendicular

2212; 2

At x = a: m =

2

2212; 2

(x 2212; 3)

dy

m = =

dx

y = 2(x 2212; 3)2212;1

6 (c) (ii)

Blunders (-3)

B1 Indices

B2 Asymptote

B3 Differentiation

B4 Slope 2260; dy

dx

B5 m1m2 2260; 2212;1

B6 Incorrect deductions or simply absolutely no deduction

Slips

S1 Curve not even approaching asymptotes.

Attempts

A1 Error for differentiation formula

Worthless

W1 Integration

QUESTION 7

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (5, 5, 5, 5) marks

Att (2, Some, A pair of, 2)

Part (c)

20 (5, 5, 5, kearsley united kingdom essays marks

Att (2, Some, Couple of, 2)

Part (a)

10 (5, 5) marks

Att (2, 2)

7 (a) (a) The picture of some sort of curve is definitely x2 2212; y 2 = 30.

See dy in phrases of x and y.

dx

Differentiate 5 marks Att 2

Isolate

dy 5 marks Att 2

dx

dy = x .

dx y

x 2 2212; y 2 = Twenty-five 21D2; 2x 2212; A pair of y dy = 0 21D2;

dx

7 (a)

OR

dy = x

dx y

23A5;

23A2;

23A3;

= x y

1

2

2

x 2212; 25)

y = 2212;(

1

2

2

x 2212; 25)

y = (

y = 2212; x 2 2212; 25

OR

x 2 2212; y 2 = 25

y 2 = x 2 2212; 25

y = x 2 2212; 25

7 (a)

Blunders (-3)

x 2 2212; 24 23A6;

23A1; x 23A4;

= 2212;

23A6;

= x

x 2 2212; 25

= x y

23A2;

dx 23A3; 2

dx 2

23A4;

2

1

2212;

2

x 2212; 25) .2x23A5;

23A1;1

= 2212; (

dy

1

2

2212;

2

x 2212; 25) .2x

= (

dy 1

B1 Differentiation B2 Indices

Attempts

A1 Error through differentiation formula

A2 dy = 2x 2212; A couple of y dy and uses not one but two dy terms in 1st 5 marks.

dx dx dx

Worthless

W1 No differentiation W2 Integration

Part (b) 20 (5, 5, 5, 5) marks Att (2, Only two, Three, 2)

  1. Find dy in conditions associated with t.

dx

  1. Find any equation for all the tangent to the actual curve from a stage assigned just by t = 2.

2

t 2212; 2

and y =

2

6

3t

t 2212; 2

x =

(b) A shape can be described from any parametric equations

, just where t 2260; B1; 2.

  1. (i)

dx , dy

5 marks Att 2

dt dt

dy 5 marks Att 2

dx

.

=

=

2212; 3t 2 2212; 6 3t 2 + 6 t 2 + 2

.

(t 2 2212; 2)2

dy dt 2212; 12t (t 2 2212; 2)2 12t 4t

= .

=

dx dt dx

dy

.

(t 2 2212; 2)2

dt

2212; 12t

= 6(t 2 2212; 2)2212;1 21D2; dy = 2212;6(t 2 2212; 2)2212;2.2t =

y =

.

2212; 3t 2 2212; 6

(t 2 2212; 2)2

=

(t 2 2212; 2)2

=

dt

dy 3(t 2 let this men and women get surf essay 2)2212; 3t.2t

21D2;

3t w not 2 2212; 2

6

t 2 2212; 2

x =

7 (b) (i)

(b)(ii) Incline, point 5 marks Att 2

Equation 5 marks Att 2

7 (b) (ii)

6 6

t = 2

21D2; x = = 3 as well as t = 2 21D2; y = = 3.

2234; Phase is

(3, 3).

2 2

Slope connected with tangent within t = 2 is

8 = Have a look at .

6 3

2234; Equation connected with tangent: y 2212; 3 = 4 ( x 2212; 3)

3

21D2; 4x 2212; 3y 2212; 3 = 0.

Blunders (-3)

B1 Differentiation B2 Indices

B3 Error during acquiring dy

dx

B4 Equation associated with tangent B5 Error with downward slope formula.

Slips (-1)

S1 Numerical

Attempts

A1 Error with difference formula

Part (c) 20 (5, 5, 5, 5) Att (2, Step 2, Only two, 2)

  1. The purpose f (x) = x3 2212; 3x2 + 3x 2212; 4 includes mainly 1 substantial root.
    1. Show the fact that this underlying issues is amongst Three and additionally 3.

Anne in addition to Craig are usually each individual employing typically the Newton-Raphson system to help you estimate that actual.

Anne is definitely starting off with the help of Two seeing that some sort of initially approximation not to mention Craig can be starting up with the help of 3.

    1. Show which usually Anne2019;s getting into approximation is definitely better to help you this heart as compared with Barry2019;s. (That is without a doubt, display this all the main is definitely a smaller amount than 2B7;5.)
    2. Show, having said that, which Barry2019;s subsequent approximation might be closer so that you can this cause when compared with Anne2019;s.
  1. (i) 5 marks Att 2

f (x) = x3 2212; 3x2 + 3x 2212; Check out .

f (2) = 8 2212; 12 + 6 2212; 4 = 2212;2 < 0.

f (3) = 25 2212; 30 + 9 2212; Some = 5 2009 ce maths paying attention to structure meant for essay 0.

2234;root fabrications in between Some along with 3.

(c) (ii) 5 marks Att 2

f (2.5) = (2.5)3 2212; 3(2.5)2 + 3(2.5) 2212; 4

= 15.625 2212; 18.75 + 7.5 2212; 4

= 0.375

f (2) < 0 along with f (2.5) > 0.

2234; underlying cause will be between A pair of along with 2.5. Hence, heart is usually short to help Three compared to to help 3.

(c) (iii) Formula + Differentiation 5 marks Att 2

Finish 5 marks Att 2

7 (c) (iii)

x = x

n

f (x )

2212; n , where

f (x) = x3 2212; 3x 2 + 3x 2212; 4 and

f 2032;(x) = 3x 2 2212; 6x + 3

n+1

n f 2032;(x )

Ann:

f (2) = 2212;2

and

f 2032;(2) = 3.

x

= Three 2212; f (2) = 3 2212; 2212; A pair of = A couple of Only two = 2.666.

Barry:

f (3) = 5 and

f 2032;(3) = 12.

2 f 2032;(2)

x = 3 2212; f (3) = 3 2212; 5

3

= 2 7

3

= 2.583.

2 f 2032;3

12 12

Both about these types of will be over the actual main, which means typically the lessen you is more detailed (i.e.

Barry2019;s).

Blunders (-3)

B1 Indices

B2 Incorrect deductions via f(2) and additionally f (3) or simply not any reduction in price B3 No f(2.5)

B4 Newton 2013;Raphson formula B5 Differentiation

B6 Incorrect discount and absolutely no deductions right from job around (iii)

QUESTION 8

Part (a)

10 marks

Att 3

Part (b)

20 (5, 5, 5, 5) marks

Att (2, A pair of, 3 2)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, 3 A couple of, 2)

Part (a)

10 marks

Att 3

8 (a) Find 222B;239B; 6x + 3 + 1 239E; .

239C; 2 239F;dx

239D; x 23A0;

239F;dx = 3x 2 + 3x 2212; 1 + C.

x

x 2 23A0;

1 239E;

6x + 3 +

239C;

239D;

239B;

222B;

8 (a)

Blunders (-3) B1 Integration B2 Indices

B3 No c

Attempts

A1 Only c correct

Worthless

W1 Differentiation just for integration

Part (b) 20 (5, 5, 5, 5) marks Att (2, Three, Only two, 2)

1 + ex dx .

ln8

222B;ex

ln3

(ii)

3C0;

4

(i) 222B;sin3x sinx dx

2212;3C0;

4

(b) Evaluate

Integration 5 marks Att 2

Value 5 marks Att 2

8 (b) (i)

3C0; 3C0; 3C0;

4 4 4

222B;sin3x sinx dx = 1 222B;(cos2x 2212; cos4x)dx = 1 23A1;1 sin2x 2212; 1 sin4x23A4;

2

2 23A2;23A3; 2

4

23A5;23A6;

3C0;

2212;3C0; 2212;3C0;

4 4

= 1 23A1;239B; 1 sin 3C0;

2212; 1 sin3C0; 239E; 239B; 1

2212;

4

239B; 2212; 3C0; 239E; 1 239E;23A4;

2 23A2;239C; 2 2 4

239F; 2212; 239C; 2 sin239C;

2 239F; 2212; Several sin(2212; 3C0; )239F;23A5;

23A3;239D; 23A0; 239D; 239D; 23A0; 23A0;23A6;

= 1 23A1;239B; 1 2212; 0239E; 239B; 1

0239E;23A4; = 1 .

2 23A2;239C; 2

239F; 2212; 239C;2212; 2 2212; 239F;23A5; 2

23A3;239D; 23A0; 239D; 23A0;23A6;

Integration 5 marks Att 2

Value 5 marks Att 2

8 (b) (ii)

Let u = 1 + ex.

2234; du = exdx.

ln8

1+ eln8 1

222B;ex

ln3

1 + ex

dx =

222B;

1+eln3

u 2 cipr degree assignments = 8 and

eln3 = 3.

9 1 23A1; 3 23A4; 9

222B;

= u 2 du = 23A2; 2 u 2 23A5; = Only two [27 2212; 8] = 38 .

23A2; 3 23A5; 3 3

4 23A3; 23A6; 4

8 (b) (ii)

Using x limits:

ln8

OR

x=ln 8

222B; ex

ln3

1 + ex dx

3

= u 2

2

3

x=ln 3

ln 8

= Only two (1 +

3

3

ex 2

)

ln 3

3

= essay shakespeare sonnet 16 theme 23A1;(1 +

23A2;

3 23A3;

eln 8 )2 2212; (1 +

3

) 23A4;

e

ln 3 2

23A5;23A6;

= Only two [(9)3 2212; (4)3 ]

2 2

3

= A couple of [27 2212; 8] = Two (19) = 38

3 3 3

8 (b) (ii)

ln8

222B; ex

ln3

dx

= 222B; 1 + ex .e xdx

1 + ex

OR

Let u = ex

du = ex

dx

= 222B; (1 + u)2 du

1

du = exdx

= Only two (1 + u)3

x=ln 8

3

= Three (1 +

3

2

3

)

ex 2

x=ln 3

ln 8

ln 3

3

= Two 23A1;(1 +

23A2;

3 23A3;

eln 8 )2 2212; (1 +

3

) 23A4;

e

ln 3 2

23A5;23A6;

= Three [(1 + 8)3 2212; (1 + 3)3 ]

2 2

3

= 3 [(9)3 2212; (4)3 ] = Step 2 [27 2212; 8] = Step 2 (19) = 38

2 2

3 3 3 3

* Erroneous replacement together with unable to help surface finish yields endeavor for most

Blunders (-3)

B1 Trig solution B2 Integration

B3 Differentiation B4 Limits

B5 Erroneous purchase on employing confines B6 Not establishing replaced rules B7 Not necessarily adjusting limits

B8 Indices

B9 Logs

B10

eln a 2260; a

Slips (-1)

S1 Numerical S2 Trig value

S3 Answer never tidied up

Worthless

W1 Differentiation rather for integration with the exception of where several other work benefits attempts

Part (c) 20 (5, 5, 5, 5) marks Att (2, A pair of, 2 2)

(c) Use integration options to make sure you grow that standard method pertaining to your quantities in a cone.

Diagram + slope 5 marks Att 2

Correct subst.

Download A/L Newspapers, Noticing themes in addition to analysis reports

into sound formula 5 marks Att 2

Integration 5 marks Att 2

Volume 5 marks Att 2

8 (c)

y = mx

21D2; y = r x.

h

h

y-axis

r

Volume about cone = 3C0; 222B; y 2 dx,

0

where y = x.

h

h

y = mx

h 2

r

2

1 r 2 [ 3 ]

1 r 2 3

V = 3C0; 222B; 3 x dx =

0

h

V = 1 3C0; r 2h.

3

3C0; x

3 h 2

0 = 3 3C0; l 2 h

r

900

h

x-axis

Blunders (-3)

B1 Integration B2 Slope connected with line

B3 Equation about line

B4 Volume formula granted the item will be quadratic B5 Limits

B6 No Limits

B7 Incorrect arrangement inside making use of restrictions B8 Indices

Slips (-1)

S1 Numerical

Attempts

A1 Uses v = 3C0;y

Worthless

W1 Differentiation as a substitute for integration

CoimisiFA;n na ScrFA;duithe StE1;it Think Assessments Commission

LEAVING Official document 2009

MARKING SCHEME

MATHEMATICS - Daily news 2

HIGHER LEVEL

GENERAL Guidelines For the purpose of EXAMINERS 2013; Cardstock 2

  1. Penalties in several models are utilized to make sure you candidates2019; perform for the reason that follows:

- statistical errors/omissions

(-3)

- mathematical errors

(-1)

(provided chore is certainly definitely not oversimplified)

(-1).

Frequently developing glitches to be able to of which such bank charges will have to get utilized usually are stated on the particular pattern.

Many really are labelled: B1, B2, B3,2026;, S1, S2,2026;, M1, M2,2026;etc. These kind of directories happen to be definitely not exhaustive.

  1. When awarding endeavor marks, e.g.

    Att(3), word that

    • any correct, relevant step for a good area regarding an important thought merits from the very least typically the make an effort symbol intended for in which part
    • if rebates outcome through the tag which will might be cheaper as opposed to any effort bench mark, next that look at recognise have to be awarded
    • a bench mark relating to totally free as well as the actual look at indicate can be hardly ever awarded.
  2. Worthless give good results might be accorded nothing grades.

    Numerous ideas connected with this type of do the job tend to be ranked for a pattern as well as individuals really are branded while W1, W2,2026;etc.

  3. The term 201C;hit or maybe miss201D; means which usually general scratches are certainly not honored 2013; this applicant is provided with most for typically the pertinent represents or perhaps none.
  4. The words 201C;and stops201D; will mean of which simply no further operate in benefit is actually proven by just typically the candidate.
  5. Special notices car to help you all the paying attention to of a selected part from any topic are usually said by simply any asterisk.

    These kind of hints straight away observe all the box containing all the important solution.

  6. The pattern choices with regard to every dilemma really are never expected to help always be inclusive email lists 2013; presently there can often be additional accurate methods.

    All examiner not sure associated with the particular validity of a approach put into practice by way of a fabulous specified prospect towards a new unique issue should really speak to his/her counseling examiner.

  7. Unless or else pointed out throughout the actual scheme, settle for this 2009 ce maths paying attention to scheme regarding essay of not one but two and even more efforts 2013; quite possibly when ever attempts currently have already been cancelled.
  8. The same error throughout any same section of your query is without a doubt penalised once only.
  9. Particular situations, verifications together with responses received through diagrams (unless requested) are considered meant for strive scratches within most.
  10. A really serious mistake, omission and also misreading benefits with this look at draw for most.
  11. Do never penalise that work with of a new comma to get the decimal time, e.g.

    20AC;5.50 might become composed mainly because 20AC;5,50.

QUESTION 1

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (10, 10) marks

Att (3, 3)

Part (c)

20 (10, 5, 5) marks

Att (3, 3 2)

Part (a)

10(5, 5) marks

Att (2, 2)

239B; 2t 1 2212; t 2 239E;

1 (a) Show which, to get most of character with t 2208; 3rd there’s r, typically the stage 239C; , 239F; fabrications on the

239C;1 + t 2 1 + t 2 239F;

239D; 23A0;

circle x 2 + y 2 = 1.

Part (a) Substitution

5 marks

Att 2

Finish

5 marks

Att 2

1 (a)

4t 2 (1 2212; t 2 )2 4t 2 + 1 2212; 2t 2 + t 4 1 + 2t 2 + t 4 (1 + t 2 )2

x 2 + y 2 = ( ) + ( ) = ( ) = ( ) = ( ) = 1.

1 + t 2 2 1 + t 2 2 1 + t 2 2 1 + t 2 2 1 + t 2 2

Blunders (-3)

B1 Incorrect squaring (apply now that if perhaps exact same model regarding error) B2 Incorrect factors

B3 Incorrect conclusion

Slips (-1)

S1 Arithmetic error

Attempts (2, 2 marks)

A1 Some perfect alternative intended for x or y

A2 Effort in providing t 2 in phrases with y

Part (b) 20 (10, 10) marks Att (3, 3)

(b) (i) Find typically the picture about any tangent to help you the particular group of friends x2 + what rhymes with speak to essay 2 = 10 in a time (3,1).

(ii) Find all the principles of k 2208; s pertaining to which unfortunately the brand x 2212; y + k = 0 is normally a fabulous tangent to be able to all the radius (x 2212; 3)2 + ( y + 4)2 = 50.

Part (b) (i) 10 marks Att 3 1 (b) (i)

Equation about tangent: xx1 + yy1 = r 2

or

21D2; 3x + y = 10.

Centre with eliptical (0,0) 21D2; Incline size = Formula regarding tangent: y-1 = -3(x-3)

1 21D2; Pitch Tangent = -3

3

Blunders (-3)

B1 Error on slope formula

B2 Slope from tangent certainly not verticle with respect to help the actual size B3 Error within situation for lines formula

B4 Error with emesa dissertation regarding tangent blueprint B5 Incorrect center regarding circle

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 Equation from tangent components A2 Slope from height only

A3 Equation for set by means of several substitution

Part (b) (ii) 10 marks Att 3

2 21D2; 7 + k = 10 21D2; 7 + k = B1;10.

2234; k = 3 as well as k = 2212;17.

2234; 3 + 3 + k = 5

2

1 (b) (ii) Centre (3, 2212; 4)and radius = 50 = 5 2.

Since a fabulous tangent, perpendicular individuals from middle of the town (3, 2212; 4) that will x 2212; y + k = 0 means radius.

OR

Part (b) (ii) 10 marks Att 3

21D2; (2 + 2k )2 2212; 4.2(k 2 + 8k 2212; 25)= 0

21D2; k 2 + 14k 2212; Fifty-one = 0

21D2; (k 2212; 3)(k + 17) = 0

21D2; k = 3, k= -17

y = x + k

(x 2212; 3)2 + ((x + k ) + 4)2 = 50 2x 2 + (2 + 2k )x + (8k + 25) = 0

One issue connected with contact

Blunders (-3)

B1 Incorrect middle about circle

B2 Error with verticle with respect mileage strategy B3 Incorrect radius

B4 One importance about k only B5 Incorrect squaring B6 Error on factors

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 Centre and radius correct

A2 Some best replacement towards verticle with respect formula

A3 Some appropriate alternative from y = times + k or identical right into circle

Part (c) 20 (10, 5, 5) marks Att (3, A couple of, 2)

(c) Two forums intersect from p(2, 0) and also q(2212; 2 8).

QUESTION 7

a extended distance by that centre of

each range to help you the common chord [ pq] is 20. Come across the equations involving any 2 circles.

Part (c) Initial picture within f and g 10 marks Att 3

Equation for an individual variable 5 marks Att 2

Finish 5 marks Att 2

1 (c)

Slope

pq = 8 2212; 0 = 2212;2 21D2; slope st

2212; 2 2212; 2

= 1 .

2

q (2212; Step 2, 8)

r

t (2212; g, 2212; f )

2234; Check out + f

0 + g

= 1 21D2; g = 2 f

2

+ 8.

20

s (0, 4)

st 2 = 20 21D2; (0 + g )2 + (4 + f )2 = 20 21D2; g 2 + f 2 + 8 f = 4

21D2; (2 f + 8)2 + f 2 + 8 f = Several 21D2; 5 f 2 + 45 f

+ 62 = 0.

p (2, 0)

2234; f 2 + 8 f

+ 12 = 0 21D2;

( f + 2)( f

+ 6) = 0.

f = 2212;2 21D2; g = 4

or f

= 2212;6 21D2; g 2212; 4.

2234;Centres can be (2212; Four, 2) and

(4, 6),

r = 40 .

Circles are:(x + 4)2 + (y 2212; 2)2 = 40

and

(x 2212; 4)2 + (y 2212; 6)2 = 40.

or x 2 + y 2 answers that will court case understand dressed in for the woman nerves 8x 2212; 5 y 2212; 20 = 0

and

x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0

OR

Part (c) 1st situation around f and g 10 marks Att 3

Equation within a person variable 5 marks Att 2

Finish 5 marks Att 2

21D2; 2212;2g 2212; f 2212; 4 = B1;10 21D2; 2g + f + 14 = 0 not to mention 2g + f 2212; 6 = 0

Distance via (0,4) (= midpoint pq) to help you (2212;g, 2212;f) 21D2; (0 + g ) 2 +(4 + f )2 = 20

Solving relating to g 2 +(4 + f )2 = 20 along with 2g + farreneheit 2212; 6 =0 gives g = Four together with f = -2

Solving relating to g 2 +(4 + f )2 = 20 plus 2g + antigone as well as ismene compare and also compare composition thesis + 17 =0 gives g = -4 together with f = -6

Eq.

1: x 2 + y 2 + 8x 2212; Have a look at y + c = 0

(2, 0) concerning ring 21D2; c = -20 21D2; x 2 + y 2 + 8x 2212; Five y 2212; 20 = 0 Eq2: Exact same way 21D2; c = 12 21D2; x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0

20

5

2212; 2g 2212; f topical composition meaning Contemplate =

y = 2212;2(x 2212; 2) or perhaps 2x + y 2212; Contemplate =0

Perp.

distance (-g, -f) to help you pq:

= 2212;2

0 2212; 8

2 2212; 2212;2

Slope pq =

Equation pq:

OR

Part (c) First equation socio political dissertation significance spanish f and g 10 marks Att 3

Equation during you variable 5 marks Att 2

Finish 5 marks Att 2

1 (c)

(2,0) 2208;Circle 21D2; 24 + 0 + 2g(2) + Couple of f (o) + c = 0

21D2; 4g + c = 2212;4 21D2; c = 2212;4g 2212; 4

q (2212; Couple of, 8)

r

t (2212; g, 2212; f )

(2212;2,8) 2208; Eliptical 21D2; 2212;4g + 06 f

+ c + 68 = 0 20

21D2; 2212;4g + 12 f 2212; 4g 2212; Some + 68 = 0 21D2; g = 2( f

+ 4)

s (0, 4)

s(midpoint) = (0,4)

20

But

g 2 + (4 + f ) Step 2 =

21D2; g 2 + (4 + f )2 = 20

21D2; (2( f

+ 4))2 + (4 + f )2 = 20 21D2; 5( f

+ 4)2 = 20

p (2, 0)

21D2; ( f

+ 4)2 = 4 21D2;

f + Five = B1;2 21D2; f

= 2212;6

and 2212; 2

f = 2212;6 21D2; g = 2212;4 21D2; c = 12

f = 2212;2 21D2; g = 4 21D2; c = 2212;20

Circles :

x 2 + y 2 + 8x 2212; Some y 2212; 20 = 0

x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0

OR

Part (c) First of all equation on f and g 10marks Att 3

Equation on 1 variable 5 marks Att 2

Finish 5 marks Att 2

20

pq = ( 2 + Only two )2 + ( 0 2212; 8 )2 = 80 21D2; ps =

pt = 20 + 20 = Fourty 21D2; pt = 40

2234; p(2,0) simply because center associated with a new range having radius 40

21D2; (x 2212; 2)2 + y 2 = 40

But (2212;g,2212; f ) at circle

21D2; (2212;g 2212; 2)2 + (0 + f ) Two = 40

st is a new chord.

8 2212; 2

21D2; 3 + f = 1 21D2; g = 2 f + 8

0 + g 2

2234;(2212;2 f 2212; 8 2212; 2) Some + f 2 = 40

21D2; 5 f 2 + 30 f + 60 = 0 21D2; f 2 + 8 f + 12 = 0

21D2; ( f + 6)( f + 2) = 0 21D2; f = 2212;2 and additionally f = 2212;6

f = 2212;6 21D2; g = 2212;4 21D2; c = 12 f = 2212;2 21D2; g = 5 21D2; c = 2212;20 Circles

x 2 easy snapshot works tips designed for fundraisers y 2 + 8x 2212; Four y 2212; 20 = 0

x 2 + y 2 2212; 8x 2212; 12 y + 12 = 0

2

= 2212;2 21D2; slope st = 1

2212; Some 2212; 2

Slope pq =

2

Blunders (-3)

B1 Error around way away formulation B2 Error throughout core stage formula

B3 Error on verticle with respect distance formula

B4 Incorrect request about Pythagoras supplement B5 Error inside downward slope formula

B6 Error in squaring B7 Error inside factors

B8 Equation involving you ring only

Slips (-1)

S1 Arithmetic error

Attempts (3,2,2 marks)

A1 Mid point and mountain pq A2 c expressed on words and phrases associated with g A3 Radius only

QUESTION 2

Part (a)

10 (5, 5) marks

Att (2, 2)

Part (b)

20 (10, 10) marks

Att (3, 3)

Part (c)

20 (10, 10) marks

Att (3, 3)

Part (a)

10 (5, 5) marks

Att (2, 2)

2 (a)

2192; 2192; 2192; 2192; 2192; 2192; 2212;2192;

(a) If a = A couple of i + j , b = 2212; i + 5 j , obtain the actual appliance vector on your guidance connected with ab .

Part (a)

2212;2192;

ab .

5 marks Att 2

Finish 5 marks Att 2

2 (a)

2212;2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192;

ab = b2212; a = 2212; i + 5 j 2212; A pair of i 2212; j = 2212;3 i + Contemplate j .

2212;2192; 2192; 2192;

9 + 16

ab = 2212; 3 i + Have a look at j = = 5.

2212;2192; 2192; 2192;

Unit vector = ab

= 2212; 3 i + Five j

= 2212; 3 2192;+ Have a look at 2192;

2212;2192; 5

ab

i j .

5 5

Blunders (-3)

2212;2192;

B1 Error during ab

2192; 2192;

= b 2212; a

B2 Error on method for anna hazare and all the lokpal statement essay regarding vector B3 Answer definitely not said within best form

Slips (-1)

S1 Arithmetic error

Attempts (2,2 marks)

A1 Norm solution having many substitution

2212;2192;

A2 ab

2192; 2192;

= b 2212; a

and stops

Part (b) 20 (10, 10) marks Att (3, 3)

(ii) Hence demonstrate that will abqc is some sort of parallelogram.

2192;

2192; 2192;

2192;

Express q in phrases associated with a , b and c .

(i)

2212;2192; 2212;2192; 2212;2192; 2212;2192;

(b) In all the triangle abc, p is the factor about a facet [bc] equivalent explanation numbers essay level q lies out of any triangle such that pq = pb+ pc2212; pa .

(b) (i) 10 marks Att 3 2 (b) (i)

2212;2192; 2212;2192;

2212;2192; 2212;2192;

2192; 2192; 2192; 2192;

2192; 2192; 2192; 2192;

pq = pb+ pc2212; pa 21D2; q2212; p = b2212; p+ c 2212; p2212; a+ p .

2192; 2192; 2192; 2192;

2234; q = b + c 2212; a .

Blunders (-3)

2212;2192;

B1 pq or comparative shown incorrectly

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

2212;2192;

A1 pq or corresponding listed correctly

  1. (ii) 10 marks Att 3 A pair of (b) (ii)

2192; 2192; 2192; 2192;

2192; 2192; 2192; 2192;

2212;2192; 2212;2192;

By area (i) : q = b + c 2212; a 21D2;

2234; abqc is a fabulous parallelogram.

q2212; b = c 2212; a 21D2; bq = ac .

Blunders (-3)

2192; 2192; 2192;

B1 c 2212; a 2260; ac

2192; 2192; 2192;

B2 q 2212; b 2260; bq

B3 No bottom line and / or wrong conclusion

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

2192; 2192; 2192; 2192;

A1 q 2212; b = c 2212; a

Part (c) 20 (10, 10) marks Att (3, 3)

239D; 23A0;

2192;22A5; 2192;22A5;

239C; r 2212; s 239F; = r 2212; what is definitely galvanometer essay of which pertaining to virtually all vectors r and s

(ii)

2192;

2192;22A5;

2192;

2192;22A5;

k l 2212; q = p 2212; q .

Find all the value in that scalar k such that

2192; 2192; 2192;

p = 12 i + 5 j and q = 3 i + Have a look at j .

2192; 2192; 2192;

(i)

(c)

Part (c) (i) 10 marks Att 3 3 (c) (i)

2192; 22A5; 2192; 2192; 22A5; 2192;

2192; 2192; 2192; 2192;

2192; 2192; 2192; 2192;

k delaware 2212; q = p 2212; q

21D2; k 2212; 5 i + 12 j 2212; 3 i 2212; 3 j

= 2212; 5 i + 12 j 2212; 3 i + Five j .

2192; 2192; 1 2

2234; k 2212; 8 + 8 = 13 2212; 5 21D2; 128k = 8 21D2; 8 2k = 8 21D2; k = 21D2; k = .

i j

2 2

Blunders (-3)

2192;

B1 p 22A5; incorrect

B2 Error on formula with regard to typic about vector B3 k not for surd form

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 Norm with q

2192;

A2 p 22A5; only

Part (c) (ii) 10 marks Att 3 Only two (c) (ii)

2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192; 2192;

Let

r = a we + b j and define altruism essay = c we + d n .

2234; r 2212; s = (a 2212; c) i + (b 2212; d ) j .

239B;2192;2192;239E; 22A5; 2192; 2192;

239C; r 2212; s 239F; = 2212;(b 2212; d ) i + (a 2212; c) j

239D; 23A0;

2192;22A5; 2192;22A5;

2192; 2192; 239B;

2192; 2192;239E;

2192; 2192; 239B;2192;

2192;239E; 22A5;

r 2212; s

= 2212;b as i + a n 2212; 239C;2212; d my spouse and i + c n 239F; = 2212;(b 2212; d ) i + (a 2212; c) j = 239C; r 2212; s 239F; .

239D; 23A0; 239D; 23A0;

Blunders (-3)

2192;

B1 r 22A5; incorrect

B2 No summary and drastically wrong conclusion

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 One related verticle with respect correct

2192; 2192; 2192; 2192;

A2 r - s expressed on conditions in i and j

2192; 2192;

A3 Numerical valuations just for r and s fully did the trick outside 2018;correctly2019.

QUESTION 3

Part (a)

10 marks

Att 3

Part (b)

20 (10, 10) marks

Att (3, 3)

Part (c)

20 (5, 5, 5, 5) marks

Att (2, A pair of, 3 2)

Part (a)

10 marks

Att 3

3 (a)

Find typically the formula of that path in which contains that time (1, 0) and additionally tickets through

the point of intersection with your creases 2x 2212; y + 6 = 0 and additionally 10x + 3y 2212; Couple of = 0.

Part (a) 10 marks Att 3

3 (a)

6x 2212; 3y + Eighteen = 0

10x + 3y 2212; Couple of = 0

16x + 04 = 0

21D2; x = 2212;1 and

y = 4.

(1, 0) and

(2212; 1, 4) 21D2; m = 0 2212; 5 = 2212;2.

1 + 1

2234; Picture of brand : y 2212; 0 = 2212;2(x 2212; 1)

21D2; 2x + y 2212; Three = 0.

Blunders (-3)

B1 Error in incline formula

B2 Error for picture from set formula

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 One co-ordinate about time in intersection A2 2x 2013; gym + 6 + 3BB; (10x + 3y 2013; 2) = 0

Part (b) 20 (10, 10) marks Att (3, 3)

Prove which this quantify involving a person about your ways somewhere between a pair of wrinkles by using slopes m1 and additionally m2 is given by

tan3B8; = m1 2212; m2 .

1 + m1m2

Find the equations from this not one but two wrinkles who forward through your level (6,1) along with generate a strong point of view with 450 by means of your set x + 2y = 0.

(ii)

(b) (i)

Part (b) (i) 10 marks Att 3 3 (b) (i)

Slope L1 = m1

and

slope L2 = m2 .

Let 3B8;1 and also 3B8;2

be this great ways produced by

L1 not to mention L2

respectively using the particular positive

sense connected with any x-axis.

Then tan3B8;1 = m1 and

tan3B8;2 = m2

L1 L2

Case 1: (3B8;1 > 3B8;2 )

3B8;1 = 3B8; + 3B8; 2

3B8;

21D2; 3B8; = 3B8;1 2212; 3B8; 2.

tan3B8;

= tan(3B8;1

2212; 3B8; 2 ) =

tan3B8;1 2212; tan3B8; 2

1 + tan3B8;1tan3B8; 2

3B8;2 3B8;1

horizontal

m 2212; m

slope L2 = m2

slope L1= m1

2234; tan3B8; =12 .

1 + m1m2

Case 2: (3B8;1 < 3B8;2 )

3B8;2 = 3B8;2032; + 3B8;1 21D2;

3B8;2032; = 2212;(3B8;1

2212; 3B8;2 )

L2L1

tan 3B8;2032; = 2212; tan(3B8;1

2212; 3B8;2 ) =

tan 3B8;1 2212; color 3B8;2 3B8;2019;

1 + tan 3B8; tan 3B8;

1 2

= 2212; m1 2212; m2 .

1 + m1m2

3B8;13B8;2

horizontal

In this kind of case, that some other opinion involving hope by ariel dorfman essay lines

is 3B8; = 180B0; 2212; 3B8; 2032;giving tan3B8; = 2212; tan3B8; 2032; .

slope L1= m1

slope L2 = 2009 ce maths marking scheme meant for essay One case that will end up established with regard to comprehensive marks

Blunders (-3)

B1 Error on expressing 3B8; in terms and conditions of

3B8;1 and even 3B8;2

B2 Error in growth from Tan(3B8;1 -- 3B8;2 )

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 3B8;1 = 3B8; + 3B8;2 and also stops

Part 2009 ce maths paying attention to system designed for essay (ii) 10 marks Att 3 3 (b) (ii)

x + 3 y = 0 seems to have mountain = 2212; 1 .

2

tan450 = B1; m1 2212; m2

1 + m1m2

1

m1 + 2

, where

m2 = 2212; 1 .

2

2234; 1 = B1;

1 2212; 1 m1

2

21D2; A pair of 2212; m1 = B1;(2m1 + 1)

2 2212; m1 = essays for all the patriot act + 1 21D2;

1

m1 = 3

or 2 2212; m1 = 2212;2m1 2212; 1 21D2;

m1 = 2212;3.

y 2212; 1 = 1 (x 2212; 6)

3

and

y 2212; 1 = 2212;3(x 2212; 6)

x 2212; 3y = 3

and

3x + y = 19.

Blunders (-3)

B1 Error throughout slope

B2 Product involving fields 2260; -1 B3 One picture only

Slips (-1)

S1 Arithmetic error

Attempts (3 marks)

A1 Slope of times +2y = 0 A2 Tan 45 0 =1

Part (c) 20 (5, 5, 5, 5) marks Att (2, A couple of, Two, 2)

  1. f is this change for better (x, y) 2192; (x2032;, y2032;), where x2032; = 2212;x + Two y and y2032; = 2x 2212; y .
    1. L is the collection ax + by + c = 0.

      Prove that will f (L) is without a doubt a fabulous line.

    2. The line y = mx is it has the have look below f. Acquire all the a pair of field research case study worth of m.
  2. (i) x and y in phrases with x2032; and y2032; 5 marks Att 2

Substitution 5 marks Att 2

Finish 5 marks Att 2

3 (c) (i)

x2032; = 2212;x + Couple of y

2 y2032; = 4x 2212; A couple of y

x2032; + Only two y2032; = 3x 21D4;

x = 1 (x2032; + Couple of y2032;).

3

y = 2x 2212; y2032; 21D2;

y = Three (x2032; + A pair of y2032;) 2212; y2032;

3

21D2; y = 1 (2x2032; + y2032;).

3

(2234; Any inverse relationship will be some performance plus as a result f is obviously bijective 3A0;0 2192; 3A0;0 .)

The set

f (L)

is a set in place in all of items (x2032;, y2032;) intended for which usually (x, y) 2208; L .

ax + by + c = 0

21D4; a (x2032; + Three y2032;) + b (2x2032; + y2032;) + c = 0

3 3

21D4; (a + 2b)x2032; + (2a + perfect keep on case study 2017 + 3c = 0.

2234; f (L) all the form

is any set, (since it comprise associated with your place in most of elements extremely rewarding the picture of

px + qy + r = 0 ).

OR

(c) (i) Use f to vector form 5 marks Att 2

Substitution 5 marks Att 2

Finish 5 marks Att 2

= {f( f ) + tf (m) | t 2208; R}, since f is linear.

This will be any tier, considering f (m) 2260; 0(as det( f ) = 2212;3 2260; 0 21D2; f is invertible).

239D; a 23A0;

239D; 2212;c h 23A0;

2234; f(L) will be the set {f (c + tm) | t 2208; R}

239B;2212; b 239E;

  

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