jan 2008

Mark Scheme January 2008

GCE

GCE Mathematics (8371/8373,9371/9373)

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Contents Mark Schemes Page

6663 Core Mathematics 1 5 6664 Core Mathematics 2 17 6665 Core Mathematics 3 25 6666 Core Mathematics 4 31 6674 Further Pure Mathematics FP1 51 6677 Mechanics 1 59 6678 Mechanics 2 63 6679 Mechanics 3 69 6683 Statistics 1 75

6684 Statistics 2 83 6689 Decision Mathematics 1 95

5

January 2008 6663 Core Mathematics C1

Mark Scheme Question Number

Scheme

Marks

1. 323 kxx → or 654 kxx → or kx→− 7 (k a non-zero constant)

33 3x or

64 6x (Either of these, simplified or unsimplified)

xxx 73

2 63 −+ or equivalent unsimplified, such as 1

637

64

33 xxx

−+

+ C (or any other constant, e.g. + K)

M1 A1 A1 B1 (4) 4

M: Given for increasing by one the power of x in one of the three terms.

A marks: ‘Ignore subsequent working’ after a correct unsimplified version of a term is seen.

B: Allow the mark (independently) for an integration constant appearing at any stage (even if it appears, then disappears from the final answer).

This B mark can be allowed even when no other marks are scored.

6

Question Number

Scheme

Marks

2. (a) 2 (b) 9x seen, or ( )3(a) answer to seen, or 33 )2( x seen. 98x

B1 (1) M1 A1 (2) 3

(b) M: Look for 9x first… if seen, this is M1.

If not seen, look for ( )3(a) answer to , e.g. 32 … this would score M1 even if it does not subsequently become 8. (Similarly for other answers to (a)).

In 33 )2( x , the 32 is implied, so this scores the M mark.

Negative answers:

(a) Allow 2− . Allow 2± . Allow ‘2 or 2− ’.

(b) Allow 98x± . Allow ‘ 98x or 98x− ’.

N.B. If part (a) is wrong, it is possible to ‘restart’ in part (b) and to score full marks in part (b).

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Question Number

Scheme

Marks

3. ( )( )

( )( )32

323235

−−

×+−

( )⎟⎟⎠

⎞⎜⎜⎝

⎛ +−=

+−−=

...33710

...3353210

2

( )3713−= ⎟⎟⎠

⎞⎜⎜⎝

⎛ −1

3713 Allow 13 (a = 13)

37− (b = −7)

M1

M1

A1

A1 (4) 4

1st M: Multiplying top and bottom by ( )32 − . (As shown above is sufficient).

2nd M: Attempt to multiply out numerator ( )35 − ( )32 − . Must have at least 3 terms correct.

Final answer: Although ‘denominator = 1’ may be implied, the 3713− must obviously be the final answer (not an intermediate step), to score full marks. (Also M0 M1 A1 A1 is not an option).

The A marks cannot be scored unless the 1st M mark has been scored, but this 1st M mark could be implied by correct expansions of both numerator and denominator.

It is possible to score M1 M0 A1 A0 or M1 M0 A0 A1 (after 2 correct terms in the numerator).

Special case: If numerator is multiplied by ( )32 + instead of ( )32 − , the 2nd M can still be scored for at least 3 of these terms correct:

( )23353210 −+− . The maximum score in the special case is 1 mark: M0 M1 A0 A0.

Answer only: Scores no marks.

Alternative method: )32)(3(35 ++=− ba

33232)32)(3( +++=++ baaba M1: At least 3 terms correct. ba 325 += ba 21 +=− a =… or b =… M1: Form and attempt to solve

simultaneous equations. a = 13, b = −7 A1, A1

8

Question Number

Scheme

Marks

4. (a) ⎟

⎠⎞

⎜⎝⎛ −=

−−

=−−−−

−−−−

=21

147or

147,

)6(843or

86)3(4m

Equation: ))6((214 −−−=− xy or )8(

21)3( −−=−− xy

022 =−+ yx (or equiv. with integer coefficients… must have ‘= 0’)

(e.g. 014714 =−+ xy and 014714 =−− yx are acceptable)

(b) 22 ))3(4()86( −−+−− 22 714 + or 22 7)14( +− or 22 )7(14 −+ (M1 A1 may be implied by 245)

22 714 +=AB or )12(7 222 + or 245

57

M1, A1

M1

A1 (4)

M1

A1

A1cso (3) 7

(a) 1st M: Attempt to use

12

12

xxyy

m−−

= (may be implicit in an equation of L).

2nd M: Attempting straight line equation in any form, e.g ( )11 xxmyy −=−

mxxyy

=−−

1

1 , with any value of m (except 0 or ∞) and either (–6, 4) or (8, –3).

N.B. It is also possible to use a different point which lies on the line, such as the midpoint of AB (1, 0.5). Alternatively, the 2nd M may be scored by using cmxy += with a numerical gradient and substituting (–6, 4) or (8, –3) to find the value of c.

Having coords the wrong way round, e.g. )4(21)6( −−=−− xy , loses the

2nd M mark unless a correct general formula is seen, e.g. )( 11 xxmyy −=− .

(b) M: Attempting to use 212

212 )()( yyxx −+− .

Missing bracket, e.g. 22 714 +− implies M1 if no earlier version is seen. 22 714 +− with no further work would be M1 A0. 22 714 +− followed by ‘recovery’ can score full marks.

9

Question Number

Scheme

Marks

5. (a)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+ −− 12

1

32 xx 1,21

−=−= qp

(b) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++−= −− 12

1

3275 xxxy

5dd

⎟⎠⎞

⎜⎝⎛ =

xy ( )05or x ( 75 −x correctly differentiated)

Attempt to differentiate either px2 with a fractional p, giving )0(1 ≠− kkx p ,

(the fraction p could be in decimal form)

or qx3 with a negative q, giving )0(1 ≠− kkxq .

223

223

3,31221 −−−−

−−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=×−×− xxxx

B1, B1 (2)

B1

M1

A1ft, A1ft (4) 6

(b):

N.B. It is possible to ‘start again’ in (b), so the p and q may be different from those seen in (a), but note that the M mark is for the attempt to differentiate px2 or qx3 .

However, marks for part (a) cannot be earned in part (b).

1st A1ft: ft their px2 , but p must be a fraction and coefficient must be simplified (the fraction p could be in decimal form). 2nd A1ft: ft their qx3 , but q must be negative and coefficient must be simplified.

'Simplified' coefficient means ba where a and b are integers with no common

factors. Only a single + or − sign is allowed (e.g. − − must be replaced by +). Having +C loses the B mark.

10

Question Number

Scheme

Marks

6. (a) Shape: Max in 1st quadrant and 2 intersections on positive x-axis

1 and 4 labelled (in correct place) or clearly stated as coordinates

(2, 10) labelled or clearly stated

(b) Shape: Max in 2nd quadrant and 2 intersections on negative x-axis

−1 and −4 labelled (in correct place) or clearly stated as coordinates

(−2, 5) labelled or clearly stated (c) (a = ) 2 May be implicit, i.e. )2(f +x

Beware: The answer to part (c) may be seen on the first page.

B1

B1

B1 (3) B1

B1

B1 (3)

B1 (1)

7

(a) and (b):

1st B: ‘Shape’ is generous, providing the conditions are satisfied.

2nd and 3rd B marks are dependent upon a sketch having been drawn.

2nd B marks: Allow (0, 1), etc. (coordinates the wrong way round) if the sketch is correct. Points must be labelled correctly and be in appropriate place (e.g. (−2, 5) in the first quadrant is B0).

(b) Special case: If the graph is reflected in the x-axis (instead of the y-axis), B1 B0 B0 can be scored. This requires shape and coordinates to be fully correct, i.e. Shape: Minimum in 4th quadrant and 2 intersections on positive x-axis,

1 and 4 labelled (in correct place) or clearly stated as coordinates, (2, −5) labelled or clearly stated.

11

Question Number

Scheme

Marks

7. (a) )1(1 +p or 1+p (b) ( )( ))()( apa + [(a) must be a function of p]. [ ])1)(1( +++ ppp 2231 pp ++= (*) (c) 1231 2 =++ pp ...0)32( ==+ ppp

23

−=p (ignore p = 0, if seen, even if ‘chosen’ as the answer)

(d) Noting that even terms are the same.

This M mark can be implied by listing at least 4 terms, e.g. ...,21,1,

21,1 −−

21

2008 −=x

B1 (1) M1

A1cso (2) M1 M1 A1 (3)

M1

A1 (2)

8 (b) M: Valid attempt to use the given recurrence relation to find 3x .

Missing brackets, e.g. )1(1 +++ ppp Condone for the M1, then if all terms in the expansion are correct, with working fully shown, M1 A1 is still allowed.

Beware ‘working back from the answer’, e.g. )21)(1(231 2 pppp ++=++ scores no marks unless the recurrence relation is justified.

(c) 2nd M: Attempt to solve a quadratic equation in p (e.g. quadratic formula or completing the square). The equation must be based on 13 =x . The attempt must lead to a non-zero solution, so just stating the zero solution p = 0 is M0.

A: The A mark is dependent on both M marks.

(d) M: Can be implied by a correct answer for their p (answer is p + 1), and can also be implied if the working is ‘obscure’).

Trivialising, e.g. p = 0, so every term = 1, is M0.

If the additional answer 12008 =x (from p = 0) is seen, ignore this (isw).

12

Question Number

Scheme

Marks

8. (a) ( )0)8(2 =−++ kkxx k−8 need not be bracketed

)8(44 22 kkacb −−=−

032404 22 <−+⇒<− kkacb (*)

(b) ...0)4)(8( ==−+ kkk 48 =−= kk

Choosing 'inside' region (between the two k values)

48 <<− k or 84 −>> k

M1

M1

A1cso (3)

M1 A1

M1

A1 (4)

7

(a) 1st M: Using the k from the right hand side to form 3-term quadratic in x ('= 0' can be implied), or…

attempting to complete the square ( )0842

22

=−+−⎟⎠⎞

⎜⎝⎛ + kkkx or equiv.,

using the k from the right hand side. For either approach, condone sign errors.

1st M may be implied when candidate moves straight to the discriminant.

2nd M: Dependent on the 1st M. Forming expressions in k (with no x’s) by using 2b and ac4 . (Usually seen as the discriminant acb 42 − , but separate expressions are fine, and also allow the use of acb 42 + . (For 'completing the square' approach, the expression must be clearly separated from the equation in x). If 2b and ac4 are used in the quadratic formula, they must be clearly separated from the formula to score this mark. For any approach, condone sign errors.

If the wrong statement 042 <− acb is seen, maximum score is M1 M1 A0.

(b) Condone the use of x (instead of k) in part (b). 1st M: Attempt to solve a 3-term quadratic equation in k.

It might be different from the given quadratic in part (a).

Ignore the use of < in solving the equation. The 1st M1 A1 can be scored if −8 and 4 are achieved, even if stated as 4,8 <−< kk . Allow the first M1 A1 to be scored in part (a).

N.B. ‘ 4,8 <−> kk ’ scores 2nd M1 A0 ‘ 4or8 <−> kk ’ scores 2nd M1 A0 ‘ 4and8 <−> kk ’ scores 2nd M1 A1 ‘ 3,2,1,0,1,2,3,4,5,6,7 −−−−−−−=k ’ scores 2nd M0 A0

Use of ≤ (in the answer) loses the final mark.

13

Question Number

Scheme

Marks

9. (a) 24 kxx → or 2

36 kxx → or 1

28 −→ kxx

(k a non-zero constant)

1232 8,4,2)(f −−−= xxxx (+ C) (+ C not required)

At x = 4, y = 1: ( ) C+×−⎟⎠⎞

⎜⎝⎛ ×−×= −12

34844)162(1 Must be in part (a)

C = 3

(b) 29

168)26(16)4(f =+×−=′ ( )m=

Gradient of normal is 92

− ⎟⎠⎞

⎜⎝⎛ −=

m1

Eqn. of normal: )4(921 −−=− xy (or any equiv. form, e.g.

92

41

−=−−

xy )

Typical answers for A1: ⎟⎠⎞

⎜⎝⎛ +−=

917

92 xy ( )01792 =−+ yx ( )8.12.0 && +−= xy

Final answer: gradient ( )291

− or 5.4

1− is A0 (but all M marks are available).

M1

A1, A1, A1

M1

A1 (6) M1 M1

M1 A1 (4)

10

(a) The first 3 A marks are awarded in the order shown, and the terms must be simplified.

'Simplified' coefficient means ba where a and b are integers with no common

factors. Only a single + or − sign is allowed (e.g. + − must be replaced by −). 2nd M: Using x = 4 and y = 1 (not y = 0) to form an eqn in C. (No C is M0)

(b) 2nd M: Dependent upon use of their )(f x′ .

3rd M: eqn. of a straight line through (4, 1) with any gradient except 0 or ∞.

Alternative for 3rd M: Using (4, 1) in cmxy += to find a value of c, but an equation (general or specific) must be seen.

Having coords the wrong way round, e.g. )1(924 −−=− xy , loses the 3rd M

mark unless a correct general formula is seen, e.g. )( 11 xxmyy −=− .

N.B. The A mark is scored for any form of the correct equation… be prepared to apply isw if necessary.

14

Question Number

Scheme

Marks

10. (a) (b) )12)(3( 2 +−+= xxxy 3523 +−+= xxx (k = 3)

(c) 523dd 2 −+= xx

xy

3523 2 =−+ xx or 0823 2 =−+ xx ...0)2)(43( ==+− xxx

2,equiv.)exact (or 34

−== xx

B1

B1

B1

B1 (4)

M1

A1cso (2) M1 A1

M1 M1

A1, A1 (6) 12

(a) The individual marks are independent, but the 2nd, 3rd and 4th B’s are dependent upon a sketch having been attempted. B marks for coordinates: Allow (0, 1), etc. (coordinates the wrong way round) if marked in the correct place on the sketch.

(b) M: Attempt to multiply out 2)1( −x and write as a product with )3( +x , or attempt to multiply out )1)(3( −+ xx and write as a product with )1( −x , or attempt to expand )1)(1)(3( −−+ xxx directly (at least 7 terms). The 2)1( −x or )1)(3( −+ xx expansion must have 3 (or 4) terms, so should not, for example, be just 12 +x .

A: It is not necessary to state explicitly 'k = 3'. Condone missing brackets if the intention seems clear and a fully correct expansion is seen.

(c) 1st M: Attempt to differentiate (correct power of x in at least one term).

2nd M: Setting their derivative equal to 3.

3rd M: Attempt to solve a 3-term quadratic based on their derivative.

The equation could come from 0dd

=xy .

N.B. After an incorrect k value in (b), full marks are still possible in (c).

−4 −3 −2 −1 1 2 3 4

5

10

x

y

15

Question Number

Scheme

Marks

11. (a) )5.1(24302425 −×+=+= dau 6−=

(b) 0)1(5.130)1( =−−=−+ rdna r = 21

(c) { })5.1(1960220

20 −+=S { })5.1(2060221or 21 −+=S { }030

221or 21 +=S

= 315

M1

A1 (2)

M1

A1 (2)

M1 A1ft

A1 (3) 7

(a) M: Substitution of a = 30 and 5.1±=d into (a + 24d). Use of a + 25d (or any other variations on 24) scores M0.

(b) M: Attempting to use the term formula, equated to 0, to form an equation in r (with no other unknowns). Allow this to be called n instead of r. Here, being ‘one off’ (e.g. equivalent to a + nd), scores M1.

(c) M: Attempting to use the correct sum formula to obtain 20S , 21S , or, with their r from part (b), 1−rS or rS . 1st A(ft): A correct numerical expression for 20S , 21S , or, with their r from part (b), 1−rS or rS …. but the ft is dependent on an integer value of r.

Methods such as calculus to find a maximum only begin to score marks after establishing a value of r at which the maximum sum occurs. This value of r can be used for the M1 A1ft, but must be a positive integer to score A marks, so evaluation with, say, n = 20.5 would score M1 A0 A0.

‘Listing’ and other methods (a) M: Listing terms (found by a correct method), and picking the 25th term. (There may be numerical slips).

(b) M: Listing terms (found by a correct method), until the zero term is seen. (There may be numerical slips). ‘Trial and error’ approaches (or where working is unclear or non-existent) score M1 A1 for 21, M1 A0 for 20 or 22, and M0 A0 otherwise.

(c) M: Listing sums, or listing and adding terms (found by a correct method), at least as far as the 20th term. (There may be numerical slips). A2 (scored as A1 A1) for 315 (clearly selected as the answer). ‘Trial and error’ approaches essentially follow the main scheme, beginning to score marks when trying 20S , 21S , or, with their r from part (b), 1−rS or rS . If no working (or no legitimate working) is seen, but the answer 315 is given, allow one mark (scored as M1 A0 A0).

For reference: Sums: 30, 58.5, 85.5, 111, 135, 157.5, 178.5, 198, 216, 232.5, 247.5, 261, 273, 283.5, 292.5, 300, 306, 310.5, 313.5, 315, ……..

17



Scheme Marks

1. a)i) ii) (b) Notes: (a) (b)

f(3) = 33 - 2 x 32 - 4 x 3 + 8 ; = 5 M1; A1 f(–2) = (– 8 – 8 + 8 + 8) = 0 ( B1 on Epen, but A1 in fact) B1 (3) M1 is for attempt at either f(3) or f(–3) in (i) or f(–2) or f(2) in (ii). [(x + 2)]( x 2 – 4 x + 4) (= 0 not required) [must be seen or used in (b)] M1 A1 (x + 2) (x – 2)2 (= 0) ( can imply previous 2 marks) M1 Solutions: x = 2 or – 2 (both) or (–2, 2, 2) [no wrong working seen] A1 (4) [7] No working seen: Both answers correct scores full marks One correct ;M1 then A1B0 or A0B1, whichever appropriate. Alternative (Long division) Divide by (x – 3) OR (x + 2) to get ,2 baxx ++ a may be zero [M1]

5and12 +−+ xx seen i.s.w. (or “remainder = 5”) [A1] 0and442 +− xx seen (or “no remainder”) [B1]

First M1 requires division by a found factor ; e.g )2(),2( −+ xx or what candidate thinks is a factor to get zerobemay),( 2 abaxx ++ . First A1 for [ (x + 2)] ( x 2 – 4 x + 4) or (x – 2)( x 2 – 4) Second M1:attempt to factorise their found quadratic. (or use formula correctly) [Usual rule: .||||),)((2 bcdwheredxcxbaxx =++=++ ] N.B. Second A1 is for solutions, not factors SC: (i) Answers only: Both correct, and no wrong, award M0A1M0A1 (as if B1,B1) One correct, (even if 3 different answers) award M0A1M0A0 (as if B1)

(ii) Factor theorem used to find two correct factors, award M1A1, then M0, A1 if both correct solutions given. ( –2,2,2 would earn all marks) (iii) If in (a) candidate has )4)(2( 2 −+ xx B0, but then repeats in (b), can score M1A0M1(if goes on to factorise)A0 (answers fortuitous) Alternative (first two marks)

02)2()2())(2( 232 =+++++=+++ cxcbxbxcbxxx and then compare with 0842 23 =+−− xxx to find b and c. [M1] 4,4 =−= cb [A1] Method of grouping

)2(4,)2(842 223 ±−=+−− xxxxxx M1; = )2(4)2(2 −−− xxx A1 [= 22 )2)(2(])2)(4( −+=−− xxxx M1 Solutions: 2,2 −== xx both A1

18

Question Number 2. (a) (b) (c) : Notes

3. (a) (b) Notes:

Scheme Marks Complete method, using terms of form ark, to find r M1 [e.g. Dividing ar6 = 80 by ar3 = 10 to find r ; r6 – r3 = 8 is M0] r = 2 A1 (2) Complete method for finding a M1 [e.g. Substituting value for r into equation of form ark = 10 or 80 and finding a value for a. ]

(8a = 10 ) a = 411

45= (equivalent single fraction or 1.25) A1 (2)

Substituting their values of a and r into correct formula for sum. M1 ( ) ( )12

45

11 20 −=

−−

=rraS

n

(= 1310718.75) 1 310 719 (only this) A1 (2) [6]

(a) M1: Condone errors in powers, e.g. ar4 = 10 and/or ar7 = 80, A1: For r = 2, allow even if ar4 = 10 and ar7 = 80 used (just these) (M mark can be implied from numerical work, if used correctly)

(b) M1: Allow for numerical approach: e.g. 310

cr ← 2

10

cr ←

cr10

←10

In (a) and (b) correct answer, with no working, allow both marks. (c) Attempt 20 terms of series and add is M1 (correct last term 655360) If formula not quoted, errors in applying their a and/or r is M0 Allow full marks for correct answer with no working seen.

10

211 ⎟

⎠⎞

⎜⎝⎛ + x = 1 + ⎟⎟

⎠

⎞⎜⎜⎝

⎛1

10 32

21

310

21

210

21

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛ xxx M1 A1

= 1 + 5x ; + 445 (or 11.25)x2 + 315x ( coeffs need to be these, i.e, simplified) A1; A1 (4)

[Allow A1A0, if totally correct with unsimplified, single fraction coefficients)

(1 + 01021 .× )10 = 1 + 5(0.01) + ( )201.0)25.11

445( or + 15(0.01)3 M1 A1√

= 1 + 0.05 + 0.001125 + 0.000015 = 1.05114 cao A1 (3) [7] (a) For M1 first A1: Consider underlined expression only. M1 Requires correct structure for at least two of the three terms:

(i) Must be attempt at binomial coefficients.

[Be generous :allow all notations e.g. 210C , even ⎟

⎠⎞

⎜⎝⎛

210 ; allow “slips”.]

(ii) Must have increasing powers of x , (iii) May be listed, need not be added; this applies for all marks.

First A1: Requires all three correct terms but need not be simplified, allow 110 etc, 2

10C etc, and condone omission of brackets around powers of ½ x Second A1: Consider as B1: 1 + 5 x (b) For M1: Substituting their (0.01) into their (a) result [0.1, 0.001, 0.25, 0.025,0.0025 acceptable but not 0.005 or 1.005] First A1 (f.t.): Substitution of (0.01) into their 4 termed expression in (a) Answer with no working scores no marks (calculator gives this answer)

19

Question Number 4. (a) (b) Notes

Scheme Marks 3 sin2 θ – 2 cos2 θ = 1 3 sin2 θ – 2 (1 – sin2 θ) = 1 (M1: Use of )1cossin 22 =+ θθ M1 3 sin2 θ – 2 + 2 sin2 θ = 1 5 sin2 θ = 3 cso AG A1 (2)

53sin2 =θ , so sinθ = (±)√ 0.6 M1

Attempt to solve both sinθ = +.. and sinθ = – …(may be implied by later work) M1 θ = 50.7685o awrt =θ 50.8° (dependent on first M1 only) A1 θ (= 180º – 50.7685c o ); = 129.23…o awrt 129.2º M1; A1 √ [f.t. dependent on first M and 3rd M] sin θ = – √ 0.6 θ = 230.785o and 309.23152o awrt 230.8º, 309.2º (both) M1A1 (7) [9] (a) N.B: AG; need to see at least one line of working after substituting cos2θ. (b) First M1: Using 5 3sin 2 =θ to find value for sinθ or θ

[Allow such results as 53sin,

53sin =θ=θ …..for M1]

econd M1: Considering the – value for sin .θ (usually later) First A1: Given for awrt 50.8°. Not dependent on second M. Third M1: For (180 – candidate’s 50.8)°, need not see written down Final M1: Dependent on second M (but may be implied by answers) For (180 + candidate’s 50.8)° or (360 – candidate’s 50.8)° or equiv. Final A1: Requires both values. (no follow through) [ Finds k=θ2cos (k = 2/5) and so =θcos (±)...M1, then mark equivalently] NB Candidates who only consider positive value for sinθ can score max of 4 marks: M1M0A1M1A1M0A0 – Very common. Candidates who score first M1 but have wrong sinθ can score maximum M1M1A0M1A√M1A0 SC Candidates who obtain one value from each set, e.g 50.8 and 309.2 M1M1(bod)A1M0A0M1(bod)A0 Extra values out of range – no penalty

20

Question Number 5.

Scheme Marks Method 1 (Substituting a = 3b into second equation at some stage) Using a law of logs correctly (anywhere) e.g. log3 ab = 2 M1 Substitution of 3b for a (or a/3 for b) e.g. log3 3b2 = 2 M1 Using base correctly on correctly derived log3 p= q e.g. 22 33 =b M1 First correct value b = √ 3 (allow 3½) A1 Correct method to find other value ( dep. on at least first M mark) M1 Second answer a = 3b = 3 √ 3 or √27 A1 Method 2 (Working with two equations in log3a and log3b) “Taking logs” of first equation and “separating” ba 333 log3loglog += M1 ( = 1 + b3log ) Solving simultaneous equations to find log 3a or log 3 b M1 [ log 3 a = 1½, log 3 b=½ ] Using base correctly to find a or b M1 Correct value for a or b a = 3 √ 3 or b = √ 3 A1 Correct method for second answer, dep. on first M; correct second answer M1;A1[6] [Ignore negative values]

Notes:

Answers must be exact; decimal answers lose both A marks There are several variations on Method 1, depending on the stage at which a = 3b is used, but they should all mark as in scheme. In this method, the first three method marks on Epen are for (i) First M1: correct use of log law, (ii) Second M1: substitution of a = 3b, (iii) Third M1: requires using base correctly on correctly derived log3 p= q Three examples of applying first 4 marks in Method 1: (i) 2log3log 33 =+ bb gains second M1 2loglog3log 333 =++ bb gains first M1 (2 == bb 33 log,1log ½) no mark yet 3=b ½ gains third M1, and if correct A1 (ii) 2)(log3 =ab gains first M1

23=ab gains third M1 22 33 =b gains second M1 (iii) 23log 2

3 =b has gained first 2 M marks 23log2 3 =⇒ b or similar type of error 13log3 =⇒ b 33 =⇒ b does not gain third M1, as 13log3 =b not derived correctly

21

Question Number 6. (a) (b) Notes:

Scheme Marks N C θ B º 500m 15o 700m ° A BC2 = 7002 +5002 – 2 × 500 × 700 cos 15° M1 A1 ( = 63851.92… ) BC = 253 awrt A1 (3)

BCB

scandidate'15sin

700sin

= M1

sin B = sin 15 × 700 /253c = 0.716.. and giving an obtuse B ( 134.2°) dep on 1st M M1 θ = 180º – candidate’s angle B (Dep. on first M only, B can be acute) M1 θ = 180 – 134.2 = (0)45.8 (allow 46 or awrt 45.7, 45.8, 45.9) A1 (4) [7] [46 needs to be from correct working] (a) If use cos 15º = ….., then A1 not scored until written as BC2 = … correctly Splitting into 2 triangles BAX and CAX, where X is foot of perp. from B to AC Finding value for BX and CX and using Pythagoras M1 BC 2 = 22 )15cos500700()15sin500( oo −+ A1 BC = 253 awrt A1 (b) Several alternative methods: (Showing the M marks, 3rd M dep. on first M))

(i) cosBC

BCBs'xcandidate500x2

700scandidate'500 222 −+= or cc xBCxBC 5002500700

222 −+= M1

Finding angle B M1 dep., then M1 as above (ii) 2 triangle approach, as defined in notes for (a)

tan CBX = valueforBX

valueforAX−700 M1

Finding value for ∠CBX ( 59≈ °) dep M1 )]'75(180[ CBXscandidate ∠+−=θ oo M1 (iii) Using sine rule (or cos rule) to find C first: Correct use of sine or cos rule for C M1, Finding value for C M1 Either B =180° – (15º + candidate’s C) or =θ (15º + candidate’s C) M1 (iv) 700 θcos50015cos BC+=o M2 {first two Ms earned in this case} Solving for θ ; =θ 45.8 (allow 46 or5.7, 45.8, 45.9) M1;A1 Note: S.C. In main scheme, if θ used in place of B, third M gained immediately; Other two marks likely to be earned, too, for correct value of θ stated.

22

Question Number 7 (a) (b) (c)

Scheme Marks Either solving 0 = x(6 – x) and showing x = 6 (and x = 0) B1 (1) or showing (6,0) (and x = 0) satisfies y = 6x – x2 [allow for showing x = 6] Solving 2x = 6x – x2 (x 2 = 4x) to x = …. M1 x = 4 ( and x = 0) A1 Conclusion: when x = 4, y = 8 and when x = 0, y = 0 , A1 (3)

(Area =) ∫ −)4(

)0(2 )6( xx dx Limits not required M1

Correct integration 3

33

2 xx − (+ c) A1

Correct use of correct limits on their result above (see notes on limits) M1

[“3

33

2 xx − ”]4 – [“3

33

2 xx − ”]0 with limits substituted [= 48 – 3226

3121 = ]

Area of triangle = 2 × 8 =16 (Can be awarded even if no M scored, i.e. B1) A1 Shaded area = ± (area under curve – area of triangle ) applied correctly M1

( = 3210)16

3226 =− (awrt 10.7) A1 (6)[10]

Notes

(b) In scheme first A1: need only give x = 4 If verifying approach used: Verifying (4,8) satisfies both the line and the curve M1(attempt at both), Both shown successfully A1 For final A1, (0,0) needs to be mentioned ; accept “clear from diagram”

(c) Alternative Using Area = ± }2;)6{()4(

)0(2 xxx −−∫ dx approach

(i) If candidate integrates separately can be marked as main scheme

If combine to work with = ± )4()4(

)0(2∫ − xx dx, first M mark and third M mark

= (±) [3

23

2 xx − (+ c) ] A1,

Correct use of correct limits on their result second M1, Totally correct, unsimplified ± expression (may be implied by correct ans.) A1 10⅔ A1 [Allow this if, having given - 10⅔, they correct it] M1 for correct use of correct limits: Must substitute correct limits for their strategy into a changed expression and subtract, either way round, e.g [ ] [ ] }{ 0

4 −± If a long method is used, e,g, finding three areas, this mark only gained for correct strategy and all limits need to be correct for this strategy. Final M1: limits for area under curve and triangle must be the same.

S.C.(1) ∫∫ =−−6

0

6

02 2)6( xdxdxxx [ ] ....

33

60

26

0

32 =−⎥

⎦

⎤⎢⎣

⎡− xxx award M1A1

MO(limits)AO(triangle)M1(bod)A0 (2) If, having found ± correct answer, thinks this is not complete strategy and does more, do not award final 2 A marks Use of trapezium rule: M0A0MA0possibleA1for triangle M1(if correct application of trap. rule from x = 0 to x = 4) A0

23

Question Number 8 (a) (b) (c) Notes

Scheme Marks

(x – 6)2 + (y – 4)2 = ; 32 B1; B1 (2)

Complete method for MP : = ( ) ( )22 46612 −+− M1 = 40 or awrt 6.325 A1 [These first two marks can be scored if seen as part of solution for (c)] Complete method for cosθ , sinθ or tanθ M1

e.g. cos θ = 40'

3MPMT

scandidate= (= 0.4743 ) (θ = 61.6835o)

[If TP = 6 is used, then M0] θ = 1.0766 rad AG A1 (4)

Complete method for area TMP ; e.g. = 40321

×× sin θ M1

= 3123 ( = 8.3516..) allow awrt 8.35 A1

Area (sector)MTQ = 0.5 × 32 × 1.0766 (= 4.8446…) M1 Area TPQ = candidate’s (8.3516.. – 4.8446..) M1 = 3.507 awrt A1 (5) [Note: 3.51 is A0] [11] (a) Allow 9 for 32. (b) First M1 can be implied by √40or √31 For second M1:

May find TP = 313)40( 22 =− , then either

..)8859.1(331tan...)8803.0(

4031sin ==== θθ or

MPTP or cos rule

NB. Answer is given, but allow final A1 if all previous work is correct.

(c) First M1: (alternative) 940321

−××

Second M1: allow even if candidate’s value ofθ used. (Despite being given !)

24

Question Number 9 (a) (b) (c) (d) Notes

Scheme Marks (Total area ) = 3xy + 2x2 B1

(Vol: ) x2y = 100 (y = 2

100x

, xy = x

100 ) B1

Deriving expression for area in terms of x only M1 (Substitution, or clear use of, y or xy into expression for area )

(Area =) 22300 xx

+ AG A1 cso (4)

xA

dd = – 2

300x

+ 4x M1A1

Setting xA

dd = 0 and finding a value for correct power of x, for cand. M1

[ x 3 = 75] x = 4.2172 awrt 4.22 (allow exact 3 75 ) A1 (4)

2

2

dd

xA = 4600

3 +x

= positive , > 0; therefore minimum M1;A1(2)

Substituting found value of x into (a) M1 (Or finding y for found x and substituting both in 3xy + 2x2 )

[y = 221724100

. = 5.6228]

Area = 106.707 awrt 107 A1 (2) [12] (a) First B1: Earned for correct unsimplified expression, isw. (b) First M1: At least one power of x decreased by 1, and no “c” term.

(c) For M1: Find 2

2

dd

xA and explicitly consider its sign, state > 0 or “positive”

A1: Candidate’s 2

2

dd

xA must be correct for their

xA

dd , sign must be + ve

and conclusion “so minimum”, (allow QED, √ ). ( may be wrong x, or even no value of x found)

Alternative: M1: Find value of xA

dd on either side of “x = 3 75 ” and consider sign

A1: Indicate sign change of negative to positive for xA

dd , and conclude

minimum. OR M1: Consider values of A on either side of “x = 3 75 ” and compare with”107” A1: Both values greater than “x = 107 ” and conclude minimum. Allow marks for (c) and (d) where seen; even if part labelling confused. Throughout , allow confused notation, such as dy/dx for dA/dx.

25

January 2008

6665 Core Mathematics C3 Mark Scheme

Question Number Scheme Marks

1. 22x 1− 2 1x − 4 22 3 1x x x− + +

4 22 2x x− 2 1x x− + + 2x− 1+ x M1 2a = stated or implied A1 1c = − stated or implied A1

222 1

1xx

x− +

−

2, 0, 1, 1, 0a b c d e= = = − = = 1d = and 0, 0b e= = stated or implied A1 [4]

2. (a)

2 2 2d 2e tan e secd

x xy x xx= + M1 A1+A1

2 2 2d 0 2e tan e sec 0d

x xy x xx= ⇒ + = M1

22 tan 1 tan 0x x+ + = A1

( )2tan 1 0x + = tan 1x = − cso A1 (6)

(b) 0

d 1d

yx

⎛ ⎞=⎜ ⎟

⎝ ⎠ M1

Equation of tangent at ( )0, 0 is y x= A1 (2) [8]

26


3. (a) ( )f 2 0.38= …

( )f 3 0.39= − … M1

Change of sign (and continuity) ⇒ root in ( )2, 3 cso A1 (2) (b) 1 ln 4.5 1 2.50408x = + ≈ M1

2 2.50498x ≈ A1

3 2.50518x ≈ A1 (3) (c) Selecting [ ]2.5045, 2.5055 , or appropriate tighter range, and evaluating at both ends. M1 ( ) 4f 2.5045 6 10−≈ ×

( ) 4f 2.5055 2 10−≈ − ×

Change of sign (and continuity) ( )root 2.5045, 2.5055⇒ ∈

root 2.505⇒ = to 3 dp cso A1 (2) [7] Note: The root, correct to 5 dp, is 2.50524

27


4. (a)

Shape B1 ( )5, 4 B1

( )5, 4− B1 (3) (b) For the purpose of marking this paper, the graph is identical to (a) Shape B1 ( )5, 4 B1

( )5, 4− B1 (3) (c)

General shape – unchanged B1 Translation to left B1 ( )4, 8 B1

( )6, 8− − B1 (4)

In all parts of this question ignore any drawing outside the domains shown in the diagrams above. [10]

( )5, 4− ( )5, 4 y

O x

( )4, 8

( )6, 8− −

y

O x

28

1000

R

t O


5. (a) 1000 B1 (1) (b) 57301000e 500c− = M1

5730 1e2

c− = A1

15730 ln2

c− = M1

0.000121c = cao A1 (4) (c) 229201000e 62.5cR −= = Accept 62-63 M1 A1 (2) (d) Shape B1 1000 B1 (2) [9]

29


6. (a) ( )cos 2 cos 2 cos sin 2 sinx x x x x x+ = − M1

( ) ( )22cos 1 cos 2sin cos sinx x x x x= − − M1

( )2 22cos 1 cos 2(1 cos )cosx x x x= − − − any correct expression A1

34cos 3cosx x= − A1 (4)

(b)(i) ( )

( )

22cos 1 sincos 1 sin1 sin cos 1 sin cos

x xx xx x x x

+ +++ =

+ + M1

( )

2 2cos 1 2sin sin1 sin cos

x x xx x

+ + +=

+ A1

( )

( )2 1 sin

1 sin cosx

x x+

=+

M1

2 2sec

cosx

x= = cso A1 (4)

(c) sec 2x = or 1cos2

x = M1

5,

3 3x π π= accept awrt 1.05, 5.24 A1, A1 (3)

[11]

7. (a) d 6cos 2 8sin 2d

y x xx= − M1 A1

0

d 6d

yx

⎛ ⎞=⎜ ⎟

⎝ ⎠ B1

146

y x− = − or equivalent M1 A1 (5)

(b) ( )2 23 4 5R √= + = M1 A1

4tan , 0.9273

α α= ≈ awrt 0.927 M1 A1 (4)

(c) ( )sin 2 their 0x α+ = M1

2.03, 0.46,1.11, 2.68x = − − A1 A1 A1 (4)

First A1 any correct solution; second A1 a second correct solution; third A1 all four correct and to the specified accuracy or better. Ignore the y-coordinate.

[13]

30


8. (a) 1

33 11 2

2xx y y −⎛ ⎞= − ⇒ = ⎜ ⎟

⎝ ⎠ or 3

12

x− M1 A1 (2)

1

31 1f :

2xx− −⎛ ⎞

⎜ ⎟⎝ ⎠

a Ignore domain

(b) ( ) 3

3gf 41 2

xx

= −−

M1 A1

( )3

3

3 4 1 21 2

xx

− −=

− M1

3

3

8 11 2

xx−

=−

cso A1 (4)

3

3

8 1gf :1 2

xxx−

−a Ignore domain

(c) 38 1 0x − = Attempting solution of numerator = 0 M1

12

x = Correct answer and no additional answers A1 (2)

(d) ( ) ( )

( )

3 2 3 2

23

1 2 24 8 1 6dd 1 2

x x x xyx x

− × + − ×=

− M1 A1

( )

2

23

18

1 2

x

x=

− A1

Solving their numerator = 0 and substituting to find y. M1 0, 1x y= = − A1 (5) [13]

31


Mark Scheme Question Number Scheme Marks

x 0 4

π 2π 3

4π π

1. (a) y 0 1.844321332… 4.810477381… 8.87207 0

awrt 1.84432 B1 awrt 4.81048 or 4.81047 B1

0 can be implied

[2]

Outside brackets awrt 0.39 or 1

2 awrt 0.79× 12 4

π× or 8π

B1

For structure of trapezium rule{ }............. ; M1 (b)

Way 1

( ) }{1Area ; 0 2 1.84432 4.81048 8.87207 02 4

π≈ × × + + + +

Correct expression inside brackets which all must be multiplied by their “outside

constant”.A1

31.05374... 12.19477518... 12.19488π

= × = = (4dp) 12.1948 A1 cao

[4]

4π (or awrt 0.79 ) and a divisor

of 2 on all terms inside brackets.

B1

One of first and last ordinates, two of the middle ordinates

inside brackets ignoring the 2.M1

Aliter (b)

Way 2

}{ 0 1.84432 1.84432 4.81048 4.81048 8.87207 8.87207 04 2 2 2 2Area π + + + +≈ × + + +

which is equivalent to:

( ) }{1Area ; 0 2 1.84432 4.81048 8.87207 02 4

π≈ × × + + + +

Correct expression inside brackets if 1

2 was to be factorised out.

A1

15.52687... 12.19477518... 12.19484π

= × = = (4dp) 12.1948 A1 cao

[4] 6 marks

Note an expression like ( )1Area 2 1.84432 4.81048 8.872072 4

π≈ × + + + would score B1M1A0A0

32


** represents a constant (which must be consistent for first accuracy mark)

2. (a) ( ) ( )1 13 31 1

3 3 3 38 3 8 1 2 18 8x xx ⎛ ⎞ ⎛ ⎞− = − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Takes 8 outside the bracket to give any of

13(8) or 2 .

B1

Expands

13(1 ** )x+ to

give a simplified or an un-simplified

131 ( )(** )x+ ;

M1;

51 2 1 2

2 33 3 3 3 313

( )( ) ( )( )( )2 1 ( )(** ); (** ) (** ) ...

2! 3!x x x

⎧ − − − ⎫= + + + +⎨ ⎬

⎭⎩

with ** 1≠

A correct simplified or an un-simplified

}{.......... expansion with candidate’s followed

through ( )** x

A1

51 2 1 2

2 33 3 3 3 33 3 313 8 8 8

( )( ) ( )( )( )2 1 ( )( ) ( ) ( ) ...

2! 3!x x x⎧ − − − ⎫

= + − + − + − +⎨ ⎬⎭⎩

Award SC M1 if you see

51 2 1 22 33 3 3 3 3( )( ) ( )( )( )

(** ) (** )2! 3!

x x− − −

+

{ }2 351 1

8 64 15362 1 ; ...x x x= − − − − Either { }182 1 ........x− or

anything that cancels to 12 ;

4x−

A1;

2 31 1 52 ; ...4 32 768

x x x= − − − − Simplified 2 351

32 768x x− − A1 [5]

(b) 13 2 31 1 5(7.7) 2 (0.1) (0.1) (0.1) ...

4 32 768≈ − − − −

Attempt to substitute 0.1x = into a candidate’s

binomial expansion.M1

2 0.025 0.0003125 0.0000065104166...= − − − 1.97468099...= awrt 1.9746810 A1 [2]

7 marks

You would award B1M1A0 for 51 2 1 2

2 33 3 3 3 33 313 8 8

( )( ) ( )( )( )2 1 ( )( ) ( ) ( 3 ) ...

2! 3!x x x

⎧ − − − ⎫= + − + − + − +⎨ ⎬

⎭⎩

because ** is not consistent.

If you see the constant term “2” in a candidate’s final binomial expansion, then you can award B1.

Be wary of calculator value of ( )137.7 1.974680822...=

33


Aliter 2. (a)

13(8 3 )x−

Way 2 2 or

13(8) (See note ↓ ) B1

Expands 13(8 3 )x− to give

an un-simplified or simplified

1 23 31

3(8) ( )(8) (** );x−+

M1;

51 23 3 3

83

1 223 31

3

51 233 3 3

( )( )(8) ( )(8) (** ); (8) (** )

2!( )( )( )

(8) (** ) ...3!

x x

x

− −

−

⎧ ⎫−+ +⎪ ⎪⎪ ⎪= ⎨ ⎬− −⎪ ⎪+ +

⎪ ⎪⎭⎩

with ** 1≠

A correct un-simplified or simplified

}{.......... expansion with candidate’s followed

through ( )** x

A1

51 23 3 3

83

1 223 31

3

51 233 3 3

( )( )(8) ( )(8) ( 3 ); (8) ( 3 )

2!( )( )( )

(8) ( 3 ) ...3!

x x

x

− −

−

⎧ ⎫−+ − + −⎪ ⎪⎪ ⎪= ⎨ ⎬− −⎪ ⎪+ − +

⎪ ⎪⎭⎩

Award SC M1 if you see

53

83

1 223 3

51 233 3 3

( )( )(8) (** )

2!( )( )( )

(8) (** )3!

x

x

−

−

−

− −+

{ }2 351 1 1 1 1

3 4 9 32 81 2562 ( )( )( 3 ) ( )( )(9 ) ( )( )( 27 ) ...x x x= + − + − + − +

Anything that

cancels to 12 ;4

x−

or { }182 1 ........x−

A1; 2 31 1 52 ; ...

4 32 768x x x= − − − −

Simplified 2 35132 768x x− − A1

[5]

If you see the constant term “2” in a candidate’s final binomial expansion, then you can award B1.

Be wary of calculator value of ( )137.7 1.974680822...=

34


3. Volume ( )

2

2

1 1d d2 1 2 1

b b

a ax x

x xπ π⎛ ⎞= =⎜ ⎟+⎝ ⎠ +∫ ∫

Use of 2 dV y xπ= ∫ .

Can be implied. Ignore limits.B1

( ) 22 1 db

ax xπ −= +∫

( )1(2 1)

( 1)(2)

b

a

xπ−⎡ ⎤+

= ⎢ ⎥−⎣ ⎦

Integrating to give 1(2 1)p x −± + M1

( ) 112 (2 1)

b

axπ −⎡ ⎤= − +⎣ ⎦

112 (2 1)x −− + A1

( ) 1 12(2 1) 2(2 1)b a

π⎡ ⎤⎛ ⎞ ⎛ ⎞− −

= −⎢ ⎥⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎣ ⎦ Substitutes limits of b and a and

subtracts the correct way round. dM1

2 1 2 12 (2 1)(2 1)

a ba b

π ⎡ ⎤− − + += ⎢ ⎥+ +⎣ ⎦

2( )2 (2 1)(2 1)

b aa b

π ⎡ ⎤−= ⎢ ⎥+ +⎣ ⎦

( )(2 1)(2 1)

b aa bπ −

=+ +

( )(2 1)(2 1)

b aa bπ −+ + A1 aef

[5] 5 marks

Allow other equivalent forms such as

(2 1)(2 1)b a

a bπ π−+ +

or ( )(2 1)(2 1)

a ba bπ− −+ +

or ( )4 2 2 1

b aab a b

π −+ + +

or 4 2 2 1

b aab a b

π π−+ + +

.

Note that π is not required for the middle three marks of this question.

35


Aliter

3. Way 2

Volume ( )

2

2

1 1d d2 1 2 1

b b

a ax x

x xπ π⎛ ⎞= =⎜ ⎟+⎝ ⎠ +∫ ∫

Use of 2 dV y xπ= ∫ .

Can be implied. Ignore limits.B1

( ) 22 1 db

ax xπ −= +∫

Applying substitution d

d2 1 2uxu x= + ⇒ = and changing

limits x u→ so that 2 1a a→ + and 2 1b b→ + , gives

( )22 1

2 1d

2

b

a

u uπ−+

+= ∫

( )2 11

2 1( 1)(2)

b

a

uπ+−

+

⎡ ⎤= ⎢ ⎥−⎣ ⎦

Integrating to give 1pu−± M1

( )2 111

2 2 1

b

auπ

+−

+⎡ ⎤= −⎣ ⎦

112 u−− A1

( ) 1 12(2 1) 2(2 1)b a

π⎡ ⎤⎛ ⎞ ⎛ ⎞− −

= −⎢ ⎥⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎣ ⎦

Substitutes limits of 2 1b + and 2 1a + and subtracts the correct

way round.dM1

2 1 2 12 (2 1)(2 1)

a ba b

π ⎡ ⎤− − + += ⎢ ⎥+ +⎣ ⎦

2( )2 (2 1)(2 1)

b aa b

π ⎡ ⎤−= ⎢ ⎥+ +⎣ ⎦

( )(2 1)(2 1)

b aa bπ −

=+ +

( )(2 1)(2 1)

b aa bπ −+ + A1 aef

[5] 5 marks

Allow other equivalent forms such as

(2 1)(2 1)b a

a bπ π−+ +

or ( )(2 1)(2 1)

a ba bπ− −+ +

or ( )4 2 2 1

b aab a b

π −+ + +

or 4 2 2 1

b aab a b

π π−+ + +

.

Note that π is not required for the middle three marks of this question.

36


4. (i) ( ) ( )2 2ln d 1.ln dx xx x= ⇒∫ ∫( )

12 1

22

dlnd

d 1d

xxx

uux

v v xx

⎧ ⎫= ⇒ = =⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪= ⇒ =⎪ ⎪⎩ ⎭

Use of ‘integration by parts’

formula in the correct direction.

M1

( ) ( ) 12 2ln d ln . dx x

xx x x x= −∫ ∫

Correct expression. A1 ( )2ln 1 dxx x= − ∫ An attempt to multiply x by a

candidate’s ax or 1

bx or 1x . dM1

( )2ln xx x c= − + Correct integration with + c A1 aef [4]

(ii) 2

4

2sin dx xπ

π∫

( )2 2 1

2NB: cos 2 1 2sin or sin 1 cos2x x x x⎡ ⎤= ± ± = ± ±⎣ ⎦ Consideration of double angle formula for cos2x M1

( )2 2

4 4

1 cos2 1d 1 cos2 d2 2

x x x xπ π

π π

−= = −∫ ∫

Integrating to give

sin 2ax b x± ± ; , 0a b ≠ dM1

2

4

12

1 sin 22

x xπ

π⎡ ⎤= −⎣ ⎦ Correct result of anything

equivalent to 1 12 4 sin 2x x− A1

( ) ( )( )2sinsin( )12 2 2 4 2

1 12 2 4 2( 0) ( )

πππ π

π π

⎡ ⎤= − − −⎢ ⎥⎣ ⎦

= − − −⎡ ⎤⎣ ⎦

Substitutes limits of 2π and 4

π and subtracts the correct way

round.ddM1

( )1 1 1

2 4 2 8 4π π= + = + ( )1 1 1 2

2 4 2 8 4 8 8or orπ π π+ + + A1 aef , cso

Candidate must collect their π term and constant term

together for A1

[5]

No fluked answers, hence cso. 9 marks

Note: ( ) ( ) ( ) ( )d

2 2 dln d (their ) ln their . their dx x uxx v v x= −∫ ∫ for M1 in part (i).

18 4Note 0.64269...π + =

37


Aliter 4. (i)

Way 2 ( ) ( )2ln d ln ln 2 d ln d ln 2 dx x x x x x x= − = −∫ ∫ ∫ ∫

ln d 1.ln dx x x x= ⇒∫ ∫1dln

dd 1d

xuu xx

v v xx

⎧ ⎫= ⇒ =⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪= ⇒ =⎪ ⎪⎩ ⎭

1ln d ln . dxx x x x x x= −∫ ∫ Use of ‘integration by parts’


M1

lnx x x c= − + Correct integration of ln x with or without + c A1

ln 2 d ln 2x x c= +∫ Correct integration of ln 2

with or without + c M1

Hence, ( )2ln d ln ln 2x x x x x x c= − − +∫ Correct integration with + c A1 aef

[4]

Note: ( ) ( ) ( )ddln d their ln their . their duxx x v x v x= −∫ ∫ for M1 in part (i).

38


Aliter 4. (i)

Way 3 ( )2ln dx x∫

2d 1d 2

x uux

= ⇒ =

( )2ln d 2 ln dx x u u=∫ ∫

Applying substitution correctly to give

( )2ln d 2 ln dx x u u=∫ ∫

Decide to award 2nd M1 here!

ln d 1.ln du x u u=∫ ∫

1ln d ln . duu x u u u u= −∫ ∫ Use of ‘integration by parts’


M1

lnu u u c= − + Correct integration of lnu with or without + c A1

Decide to award

2nd M1 here! M1

( ) ( )2ln d 2 lnx x u u u c= − +∫

Hence, ( ) ( )2 2ln d lnx xx x x c= − +∫ Correct integration with + c A1 aef

[4]

39


Aliter 4. (ii)

Way 2 2 2

4 4

2 2sin d sin .sin d and sin dx x x x x I x xπ π

π π= =∫ ∫ ∫

dd

dd

sin cossin cos

ux

vx

u x xx v x

= ⇒ =⎧ ⎫⎪ ⎪⎨ ⎬

= ⇒ = −⎪ ⎪⎩ ⎭

{ }2sin cos cos dI x x x x∴ = − + ∫ An attempt to use the correct by parts formula. M1

( ){ }2sin cos 1 sin dI x x x x∴ = − + −∫

{ }2 2sin d sin cos 1 d sin dx x x x x x x= − + −∫ ∫ ∫

{ }22 sin d sin cos 1 dx x x x x= − +∫ ∫ For the LHS becoming 2I dM1

{ }22 sin d sin cosx x x x x= − +∫

{ }2 1

2 2sin d sin cos xx x x x= − +∫ Correct integration A1

( ) ( )2

2 4

4

( ) ( )2 1 12 2 2 2 2 4 4 2

14 4 8

sin d sin( )cos( ) sin( )cos( )

(0 ) ( )

x xπ

π π

π

π π π π

π π

⎡ ⎤∴ = − + − − +⎢ ⎥⎣ ⎦

= + − − +⎡ ⎤⎣ ⎦

∫ Substitutes limits of 2π and 4

π and subtracts the correct way

round.ddM1

1

8 4π= + ( )1 1 1 2

2 4 2 8 4 8 8or orπ π π+ + + A1 aef cso

Candidate must collect their π term and constant term

together for A1

[5]

No fluked answers, hence cso.

1

8 4Note 0.64269...π + =

40


5. (a) 3 24 12x y xy− = ( eqn ∗ )

2

2

8 512 4 12( 8)512 4 96

x y yy y

= − ⇒ − − = −

− − = −

Substitutes 8x = − (at least once) into * to obtain a three term quadratic in y .

Condone the loss of 0.=M1

2

2

4 96 512 024 128 0

y yy y− + =

− + =

( 16)( 8) 0y y− − =

24 576 4(128)2

y± −

=

An attempt to solve the quadratic in y by either factorising or by the formula or by

completing the square.dM1

16 or 8.y y= = Both 16 and 8.y y= =

or ( ) ( )8, 8 and 8, 16 .− −A1

[3]

Differentiates implicitly to include either d dd dor 12y yx xky x± . Ignore d

d ...yx = M1

Correct LHS equation; A1; (b) 2d d d3 8 ; 12 12

d d dy y yx y y xx x x

⎧ ⎫ ⎛ ⎞= − = +⎨ ⎬ ⎜ ⎟

⎝ ⎠⎩ ⎭

Correct application of product rule (B1)

2d 3 12

d 12 8y x yx x y

⎧ ⎫−=⎨ ⎬

+⎩ ⎭ not necessarily required.

Substitutes 8x = − and at least one of their y-values to attempt to find any one of d

d .yx

dM1

One gradient found. A1

( )

( )

d 3(64) 12(8) 96@ 8, 8 , 3,d 12( 8) 8(8) 32d 3(64) 12(16) 0@ 8, 16 , 0.d 12( 8) 8(16) 32

yxyx

−− = = = −

− + −−

− = = =− +

Both gradients of -3 and 0 correctly found. A1 cso [6]

9 marks

41


Differentiates implicitly to include either

2 d dd dor 12x xy ykx y± . Ignore d

d ...xy = M1

Correct LHS equation A1;

Aliter 5. (b)

Way 2 2dx d d3 8 ; 12 12

d d dx xx y y x

y y y⎛ ⎞⎧ ⎫⎪ ⎪= − = +⎜ ⎟⎨ ⎬ ⎜ ⎟⎪ ⎪⎩ ⎭ ⎝ ⎠

Correct application of product rule (B1)

2d 3 12

d 12 8y x yx x y

⎧ ⎫−=⎨ ⎬

+⎩ ⎭ not necessarily required.

Substitutes 8x = − and at least one of their

y-values to attempt to find any one of d

dyx or d

d .xy

dM1

One gradient found. A1

( )

( )

d 3(64) 12(8) 96@ 8, 8 , 3,d 12( 8) 8(8) 32d 3(64) 12(16) 0@ 8, 16 , 0.d 12( 8) 8(16) 32

yxyx

−− = = = −

− + −−

− = = =− +

Both gradients of -3 and 0 correctly found. A1 cso [6]

42


Aliter 5. (b) 3 24 12x y xy− = ( eqn ∗ ) Way 3

2 34 12 0y xy x+ − =

2 312 144 4(4)( )

8x x x

y− ± − −

=

2 312 144 16

8x x x

y− ± +

=

2 312 4 9

8x x x

y− ± +

=

( )122 33 1

2 2 9y x x x= − ± +

A credible attempt to make y the subject

and an attempt to differentiate either 32 x−

or ( )122 31

2 9 .x x+M1

( ) ( )122 33

2d 9 g( )dy k x x xx

−= − ± + A1

( )( ) ( )122 3 23 1 1

2 2 2d 9 ; 18 3dy x x x xx

−= − ± + +

12

2

2 3

d 3 18 3d 2 4(9 )y x xx x x

+= − ±

+

( )( ) ( )122 3 23 1 1

2 2 2d 9 ; 18 3dy x x x xx

−= − ± + + A1

12

d 3 18( 8) 3(64)@ 8d 2 4(9(64) ( 512))yxx

− += − = − ±

+ − Substitutes 8x = − find any one of d

d .yx dM1

3 48 3 482 2 324 (64)

= − ± = − ±

One gradient correctly found. A1

d 3 3 3, 0.d 2 2yx

∴ = − ± = − Both gradients of -3 and 0 correctly found. A1

[6]

43


6. (a) 2 36 & 41 1

OA OB⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

uuur uuur

Finding the difference between OB

uuur and OA

uuur.

M1 ±

3 2 14 6 21 1 2

AB OB OA⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

uuur uuur uuur

Correct answer. A1

[2] An expression of the form

( ) ( )vector vectorλ± M1

(b)

1

2 1: 6 2

1 2l λ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

r or 3 14 21 2

λ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

r

1

2 1: 6 2

1 2l λ

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

r or 3 14 21 2

λ−⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

r

r = ( )theirOA ABλ±uuur uuur

or

r = ( )theirOB ABλ±uuur uuur

or

r = ( )theirOA BAλ±uuur uuur

or

r = ( )theirOB BAλ±uuur uuur

( r is needed.)

A1 aef

[2]

(c) 2

0 1 1: 0 0 0

0 1 1l µ µ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + ⇒ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r r

1 2 2AB = = − +d i j k

uuur, 2 0= + +d i j k & θ is angle

( ) ( )2

2 2 2 2 2 22

1 12 0

2 1cos

. (1) ( 2) (2) . (1) (0) (1)

AB

ABθ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− •⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟• ⎝ ⎠ ⎝ ⎠= =

+ − + + +

d

d

uuur

uuur

Considers dot product between 2d and their .AB

uuur M1

2 2 2 2 2 2

1 0 2cos(1) ( 2) (2) . (1) (0) (1)

θ + +=

+ − + + + Correct followed through

expression or equation. A1

43cos 45 or or awrt 0.79.

3. 2πθ θ= ⇒ = o 445 or or awrt 0.79πθ = o A1 cao

[3]

This means that cosθ does not necessarily have to be the subject of the equation. It could be of the form 3 2 cos 3.θ =

44


6. (d) If l1 and l2 intersect then: 2 1 16 2 01 2 1

λ µ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

: 2 (1): 6 2 0 (2): 1 2 (3)

λ µλλ µ

+ =− =

− + =

ijk

Either seeing equation (2) written

down correctly with or without any other equation or seeing equations (1) and (3) written down correctly.

M1

Attempt to solve either equation (2) or simultaneously solve any two of

the three equations to find …dM1

(2) yields 3

Any two yields 3, 5λλ µ== =

either one of λ or µ correct. A1

1

2 1 5 1 5: 6 3 2 0 5 0 0

1 2 5 1 5l or

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + − = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r r

505

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

or 5 5+i k

Fully correct solution & no incorrect values of λ or µ seen earlier.

A1 cso

[4]

Aliter 6. (d)

Way 2 If l1 and l2 intersect then:

3 1 14 2 01 2 1

λ µ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

: 3 (1): 4 2 0 (2): 1 2 (3)

λ µλλ µ

+ =− =+ =

ijk



M1

(2) yields 2

Any two yields 2, 5λλ µ== =




1

3 1 5 1 5: 4 2 2 0 5 0 0

1 2 5 1 5l or

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= + − = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r r

505

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

or 5 5+i k


A1 cso

[4] 11 marks Note: Be careful! λ and µ are not defined in the question, so a candidate could interchange these or use different scalar parameters.

45


Aliter 6. (d)


2 1 16 2 01 2 1

λ µ−⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠

: 2 (1): 6 2 0 (2): 1 2 (3)

λ µλλ µ

− =+ =

− − =

ijk



M1



(2) yields 3

Any two yields 3, 5λλ µ= −= − =


1

2 1 5 1 5: 6 3 2 0 5 0 0

1 2 5 1 5l or

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r r

505

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

or 5 5+i k


A1 cso

[4]

Aliter 6. (d)


3 1 14 2 01 2 1

λ µ−⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

: 3 (1): 4 2 0 (2): 1 2 (3)

λ µλλ µ

− =+ =− =

ijk



M1

(2) yields 2

Any two yields 2, 5λλ µ= −= − =




1

3 1 5 1 5: 4 2 2 0 5 0 0

1 2 5 1 5l or

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

r r

505

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

or 5 5+i k


A1 cso

[4] 11 marks

46


7. (a) ( ) 1ln 2 ,1

x t yt

⎡ ⎤= + =⎢ ⎥+⎣ ⎦, d 1

d 2xt t

⇒ =+

Must state d 1d 2xt t=

+B1

1Area d

1x

t=

+∫ .

Ignore limits.M1;

ln 4 2

ln 2 0

1 1 1Area( ) d ; dt1 1 2

R xt t t

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠∫ ∫ 1 1 d

1 2t

t t⎛ ⎞ ⎛ ⎞×⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠∫ . Ignore limits. A1 AG

Changing limits, when:

ln 2 ln 2 ln( 2) 2 2 0ln 4 ln 4 ln( 2) 4 2 2

x t t tx t t t= ⇒ = + ⇒ = + ⇒ == ⇒ = + ⇒ = + ⇒ =

changes limits x t→

so that ln 2 0→ and ln 4 2→ B1

Hence, 2

0

1Area( ) d( 1)( 2)

R tt t

=+ +∫

[4]

(b) 1

( 1)( 2) ( 1) ( 2)A B

t t t t⎛ ⎞

= +⎜ ⎟+ + + +⎝ ⎠

( 1) ( 2)A B

t t+

+ + with A and B found M1

1 ( 2) ( 1)A t B t= + + +

Let 1,t = − ( )1 1 1A A= ⇒ = Let 2,t = − ( )1 1 1B B= − ⇒ = −

Finds both A and B correctly. Can be implied.

(See note below)A1

2 2

0 0

1 1 1d d( 1)( 2) ( 1) ( 2)

t tt t t t

= −+ + + +∫ ∫

Either ln( 1)a t± + or ln( 2)b t± + dM1

[ ] 2

0ln( 1) ln( 2)t t= + − +

Both ln terms correctly ft. A1 ( ) ( )ln3 ln 4 ln1 ln 2= − − − Substitutes both limits of 2 and 0

and subtracts the correct way round. ddM1

( )32ln3 ln 4 ln 2 ln3 ln 2 ln= − + = − =

ln 3 ln 4 ln 2− + or ( ) ( )3 14 2ln ln−

or ln3 ln 2− or ( )32ln

A1 aef isw

(must deal with ln 1) [6]

Takes out brackets. Writing down 1 1 1( 1)( 2) ( 1) ( 2)t t t t

= ++ + + +

means first M1A0 in (b).

Writing down 1 1 1( 1)( 2) ( 1) ( 2)t t t t

= −+ + + +

means first M1A1 in (b).

47


( )ln 2x t= + , 11

yt

=+

Attempt to make t =… the subject M1

7. (c) e 2 e 2x xt t= + ⇒ = − giving e 2xt = − A1

Eliminates t by substituting in y dM1 1 1

e 2 1 e 1x xy y= ⇒ =− + −

giving 1

e 1xy =−

A1

[4]

Attempt to make t =… the subject M1 Aliter 7. (c)

Way 2

1 1 11 1 or

1( 1) 1 1 1

yt t ty y y

yy t yt y yt y ty

−+ = ⇒ = − =

−+ = ⇒ + = ⇒ = − ⇒ =

Giving either 1 1t

y= − or 1 yt

y−

= A1

1 1ln 1 2 or ln 2yx xy y

⎛ ⎞ ⎛ ⎞−= − + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ Eliminates t by substituting in x dM1

1ln 1xy

⎛ ⎞= +⎜ ⎟

⎝ ⎠

1 1xey

= +

11xey

− =

1e 1xy =−

giving 1e 1xy =−

A1

[4]

(d) Domain : 0x > 0x > or just > 0 B1 [1] 15 marks

48


Attempt to make 1t + = … the subject M1 Aliter

7. (c) Way 3

e 2 1 e 1x xt t= + ⇒ + = − giving 1 e 1xt + = − A1

Eliminates t by substituting in y dM1

1 11 e 1xy y

t= ⇒ =

+ −

giving 1e 1xy =−

A1

[4]

Attempt to make 2t + =… the subject M1 Aliter 7. (c)

Way 4

1 1 11 2 1 or 2 yt t ty y y

++ = ⇒ + = + + =

Either 12 1ty

+ = + or 12 yty+

+ = A1

1 1ln 1 or ln yx xy y

⎛ ⎞ ⎛ ⎞+= + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ Eliminates t by substituting in x dM1

1ln 1xy

⎛ ⎞= +⎜ ⎟

⎝ ⎠

1 1xey

= +11xey

⇒ − =

1e 1xy =−

giving 1e 1xy =−

A1

[4]

49


8. (a) d d1600 or 1600d dV Vc h k ht t= − = − , Either of these statements M1

( ) d4000 4000dVV hh

= ⇒ = d 4000dVh= or d 1

d 4000hV= M1

d

ddd

d d dd d d

VtVh

h h Vt V t= × =

Either, d 1600 1600 0.4d 4000 4000 4000h c h c h k ht

−= = − = −

or d 1600 1600 0.4d 4000 4000 4000h k h k h k ht

−= = − = −

Convincing proof of ddht

A1 AG

[3]

(b) When 25h = water leaks out such that d 400dVt=

400 400 25 400 (5) 80c h c c c= ⇒ = ⇒ = ⇒ =

From above; 80 0.024000 4000

ck = = = as required Proof that 0.02k = B1 AG

[1]Aliter

(b) 400 4000k h= Way 2

400 4000 25k⇒ = Using 400, 4000 and 25h = 400

20000400 (20000) 0.02k k⇒ = ⇒ = = or 5h = . Proof that 0.02k = B1 AG [1]

(c) d d0.4d 0.4h hk h dtt k h= − ⇒ =

−∫ ∫

Separates the variables with d

0.4hk h−∫ and dt∫ on either side

with integral signs not necessary.

M1 oe

100

0

1 0.02time required d0.020.4 0.02

hh

÷∴ =

÷−∫

100

0

50time required d20

hh

=−∫ Correct proof A1 AG

[2]

50


8. (d) 100

0

50 d20

hh−∫ with substitution 2(20 )h x= −

d 2(20 )( 1)dh xx= − − or d 2(20 )

dh xx= − − Correct d

dhx

B1 aef

2(20 ) 20 20h x h x x h= − ⇒ = − ⇒ = − 50 50d . 2(20 ) d

20h x x

xh= − −

−∫ ∫ M1

20 dx xx

λ −± ∫ or

20 d20 (20 )

x xx

λ −±

− −∫

where λ is a constant

20100 dx x

x−

= ∫

20100 1 dx

x⎛ ⎞= −⎜ ⎟⎝ ⎠∫

ln ; , 0x xα β α β± ± ≠ M1 ( ) ( )100 20lnx x c= − +

100 2000lnx x− A1 change limits: when 0 then 20h x= =

and when 100 then 10h x= =

[ ]100

10

200

50 d 100 2000ln20

h x xh

= −−∫

or ( ) ( )100 100

00

50 d 100 20 2000ln 2020

h h hh

⎡ ⎤= − − −⎣ ⎦−∫ Correct use of limits, ie. putting

them in the correct way round

( ) ( )1000 2000ln10 2000 2000ln 20= − − − Either 10x = and 20x = or 100h = and 0h = ddM1

2000ln 20 2000ln10 1000= − − Combining logs to give... 2000ln 2 1000− 2000ln 2 1000= − or ( )1

22000ln 1000− − A1 aef

[6]

(e) Time required 2000ln 2 1000 386.2943611... sec= − = = 386 seconds (nearest second) = 6 minutes and 26 seconds (nearest second) 6 minutes, 26 seconds B1 [1] 13 marks

51

January 2008 6674 Further Pure Mathematics FP1


Scheme

Marks

1

Integrating factor = 3xe−

3 3( )x xd ye xedx

− −∴ =

3 3 3 31( )3 3

x x x xxye xe dx e e dx− − − −∴ = = − +∫ ∫

= 3 31 ( )3 9

x xx e e c− −− − +

313 9

xxy ce∴ = − − +

Notes: First M for multiplying through by Integrating Factor and evidence of calculus Second M for integrating by parts ‘the right way around’. Be generous – ignore wrong signs and wrong constants. Second M dependent on first. Both As dependent on this M. First A1 for correct expression – constant not required Second A requires constant for follow through. If treated as a second order de with errors then send to review.

B1 M1 M1 A1 A1ft [5]

52

2. 3. (a) (b)

Use (2x+1) as factor to give 2( ) (2 1)( 6 10)f x x x x= + − +

Attempt to solve quadratic to give 6 (36 40)

2x

± −=

Two complex roots are 3 i= ± Notes: First M if method results in quadratic expression with 3 terms (even with remainder). Second M for use of correct formula on their quadratic. Third M for using i from negative discriminant.

Consider ( 3)( 9) (3 5)( 1)( 1)

x x x xx

+ + − − −−

, obtaining 22 20 22( 1)

x xx

− + +−

Factorise to obtain 2( 11)( 1)( 1)

x xx

− − +−

.

Identify x = 1 and their two other critical values Obtain one inequality as an answer involving at least one of their critical values To obtain x < -1, 1 < x < 11 Notes: Second M attempt to factorise quadratic expression with 3 terms (usual rules). Second A don’t require -2 outside but can be part of factors.

M1 A1

M1 A1 M1 A1

(6)

[ 6 ] M1 A1

M1 A1

(4) B1ft M1 A1, A1

(4)[ 8 ]

B1, B1

53

4. (a) (b)

f (0.7) 0.195028497= − and f (0.8) = 0.297206781

Use 0.8 (0.8)0.7 (0.7)

ff

αα

−=

− − to obtain 0.8 (0.7) 0.7 (0.8)

(0.8) (0.7)f f

f fα − +=

−

(=0.739620991) =0.740 Answer required to 3 dp or better

212 2( ) 6 1 sec ( )xf x x′ = + −

Use 2(0.75)0.75(0.75)

fxf

= −′

( = 0.741087218)=0.741 Answer required to 3 dp or better

Notes: (a) Bs for 3dp or better First M for reasonable attempt using fractions and differences. (b) First M attempt to differentiate f(x), term in x is enough. Lose last A if either or both not to 3 dp

M1 A1

(4)

M1 A1

M1 A1 (4)

[ 8 ]

54

Question Number

Scheme

Marks

5. (a) (b)

Method to obtain partial fractions e.g.5 4 ( 1)( 2) ( 2) ( 1)r A r r Br r Cr r+ = + + + + + +And equating coefficients, or substituting values for x. A = 2, B = 1, C =-3 or 32 1

1 2r r r+ ++ −

32 11 2 3

132 1

2 3 432 1

3 4 5

32 11 1

32 11 2

...

....

n

r

n n n

n n n

=

− +

+ +

= + −

+ + −

+ + −

++ + −

+ + −

∑

= 3 322 1 22 , n n+ ++ − − or equivalent

= 7( 1)( 2) 4( 2) 6( 1)2( 1)( 2)

n n n nn n

+ + − + − ++ +

=27 11

2( 1)( 2)n n

n n+

+ +*

Notes:

(a) Require three constants for method. (b) Require first 3 and last 2 of their terms for first method

Second method - dependent on first - for attempt to combine to single fraction. Expansion of (n+1)(n+2) in numerator and correct solution required for final A1

M1 A1 A1 A1 (4) M1 A1, A1 M1 A1 (5)

55

Question Number

Scheme

Marks

6(a)

(b)

(c)

(d)

( i ) Multiply top and bottom by conjugate to give 25

i− −

( ii ) Expand and simplify using i2 = -1 to give 3 4i−

2 25 5 , 5 2z z i z z− = − − = * 2

4arg( )z z π− = − or -450 or 7π/4 or 3150 or -0.7853…or 5.497…

one mark for each point Notes:

(a) -2-i or 2+i OK for method. Attempt to expand required. (b) square root required for method

(c) 2 for correct answer only, tan required for method. 2dp or better. (d) Position of points not clear but both quadrants correct first B1 only.

M1 A1

M1 A1 (4)

M1A1(2)

M1 A1 (2)

B1, B1 ft (2)

[10]

56

Question Number

Scheme

Marks

7 (a)

(b)

Solve auxiliary equation 23 2 0m m− − = to obtain 2 13

m or= −

C.F is 23 x xAe Be− +

Let PI = 2x xλ µ ν+ + . Find 2y xλ µ′ = + , and 2y λ′′ = and substitute into d.e. Giving 71 1

2 2 4, andλ µ ν= − = = −

232 71 1

2 2 4x xy x x Ae Be−∴ = − + − + +

Use boundary conditions: 2= 7

4 A B− + + 231 2

2 3x xy x Ae Be−′ = − + − + and 3 = 1 2

2 3 A B− + Solve to give A=3/4 , B = 3 (

232 7 31 1

2 2 4 4 3x xy x x e e−∴ = − + − + + ) Notes:

(a) Attempt to solve quadratic expression with 3 terms (usual rules)

Both values required for first accuracy. Real values only for follow through Second M 3 term quadratic for PI required Final A1ft for their CF+ their PI dependent upon at least one M

(b) Second M for attempt to differentiate their y and third M for substitution

M1 A1

A1ft

M1A1 A1A1

A1ft(8)

M1A1ft

M1 M1

M1 A1(6)

[14]

57

Question Number

Scheme

Marks

8 (a) (b) (c)

(3 2cos ) 4a aθ+ =

Solve to obtain 12cosθ =

3πθ ±= and points are )

3,4( πa and )

35,4( πa

Use area = 21

2 r dθ∫ to give 2 212 (3 2cos )a dθ θ+∫

Obtain (9 12cos 2cos 2 2)dθ θ θ+ + +∫ Integrate to give11 12sin sin 2θ θ θ+ +

Use limits3π and π , then double or

3π and

35π or theirs

Find a third area of circle =216

3aπ

Obtain required area = 2 238 13 3

3 2a aπ

−

correct shape 5a and 4a marked

2a marked and passes through O

Notes: (a) First A for r=4a second for both values in radians.

Accept 1.0471…and 5.2359….2 dp or better for final A (b) First M for substitution, expansion and attempt to use double angles.

Second M for integrating expression of the form θθ 2coscos cba ++

Lose final A only if 2a missing in last line (c) First B for approximately symmetrical shape about initial line, only 1

loop which is convex strictly within shaded region

M1 M1

A1, A1 (4)

M1

A1

M1 A1

M1

B1 A1 , A1

(8) B1 B1 B1 (3)

[15]

4a 5a 2a O

59

January 2008 6677 Mechanics M1


Scheme

Marks

1(a) (b) 2.(a) (b) (c)

. 5 3 A B I I 1 2 I = 4( 5 – 1) = 16 Ns CLM: 4 x 5 – m x 3 = 4 x 1 + m x 2 ⇒ m = 3.2 or 16 = m (3 + 2) ⇒ m = 3.2 27 = 0 + ½.a.32 ⇒ a = 6 v = 6 x 3 = 18 m s–1 From t = 3 to t = 5, s = 18 x 2 – ½ x 9.8 x 22 Total ht. = s + 27 = 43.4 m, 43 m

M1 A1 (2) M1 A1 DM1 A1 (4) or M1 A1 DM1 A1 (4) 6 M1 A1 (2) M1 A1 f.t. (2) M1 A1 f.t. M1 A1 (4) 8

4

60

Question Number

Scheme

Marks

3.(a) (b) (c) 4.(a) (b) (c)

V Shape ‘V’ 15 Shape for last 22s (with V > 15) 5 Figures t 16 22 ½(15 + 5) x t = 120 ⇒ t = 12 → T = 12 + 16 + 22 = 50 s 120 + ½(V + 5).16 + 22V = 1000 Solve: 30V = 840 ⇒ V = 28 R (// plane): 49 cos θ = 6g sin 30 ⇒ cos θ = 3/5 * R (perp to plane): R = 6g cos 30 + 49 sin θ R ≈ 90.1 or 90 N R (// to plane): 49 cos 30 – 6g sin 30 = 6a ⇒ a ≈ 2.17 or 2.2 m s–2

B1 B1 B1 (3) M1 M1 A1 (3) M1 B1 A1 DM1 A1 (5) 11 M1 A1 A1 (3) M1 A1 DM1 A1 (4) M1 A2,1,0 A1 (4) 11

61

Question Number 5.(a) (b) (c) 6.(a) (b) (c) (d)

Scheme S T M(A): T x 4 = 12g x 2.5 A C B T = 7.5g or 73.5 N R(↑) S + T = 12g ⇒ S = 4.5g or 44.1 N U V A C B 16g 12g M(A) V x 4 = 16g x y + 12g x 2.5 V = 4gy + 7.5g or 39.2y + 73.5 N V ≤ 98 ⇒ 39.2y + 73.5 ≤ 98 ⇒ y ≤ 0.625 = 5/8 Hence “load must be no more than 5/8 m from A” (o.e.) Speed = √(52 + 82) ≈ 9.43 m s–1 Forming arctan 8/5 or arctan 5/8 oe

Bearing = 360 – arctan 5/8 or 270 + arctan 8/5 = 328 At t = 3, p.v. of P = (7 – 15)i + (–10 + 24)j = –8i + 14j Hence –8i + 14j + 4(ui + v j) = 0

⇒ u = 2, v = – 3.5 p.v. of P t secs after changing course = (–8i + 14j) + t(2i – 3.5j) = 7i + ….. Hence total time = 10.5 s

Marks M1 A1 A1 M1 A1 (5) M1 A1 A1 (3) M1 DM1 A1 (3) 11 M1 A1 (2) M1 DM1 A1 (3) M1 A1 M1 DM1 A1 (5) M1 DM1 A1 (3) 13

62

Question Number 7.(a) (b) (c) (d)

Scheme B: 2mg – T = 2m x 4g/9 ⇒ T = 10mg/9 A: T – µ mg = m x 4g/9 Sub for T and solve: µ = 2/3 * When B hits: v2 = 2 x 4g/9 x h Deceleration of A after B hits: ma = µ mg ⇒ a = 2g/3 Speed of A at P: V2 = 8gh/9 – 2 x 2g/3 x h/3 ⇒ V = 3

2 √(gh) Same tension on A and B

Marks M1 A1 A1 (3) M1 B1 A1 DM1 A1 (5) M1 A1 M1 A1 f.t. DM1 A1 (6) B1 (1) 15

63


Mark Scheme Question Number Scheme Marks

1. (a) KE lost is ( )21 2.5 8 80 J2× × = M1 A1 (2)

(b) Work energy 80 20R= × ft their (a) M1 A1 ft 4R = A1 (3) [5] Alternative to (b) ( )2 20 8 2 20 1.6a a= − × × ⇒ = − N2L 2.5 1.6R = × ft their a M1 A1ft 4= A1 (3)

2. (a) ( ) ( )26 6 9 4t t= − + −p i j& ( )1ms− M1 A1 (2) (b) 29 4 0t − = M1 2

3t = DM1 A1 (3) (c) 1 5t = ⇒ =p j& ft their p& B1ft (+/-) ( )2 6 0.5 5− = −i j v j M1

( )14 7 ms−= −v i j M1 A1 (4) [9]

64


3. (a) ( )20000 16 1250F F= = M1 A1 � 550 1000 9.8sinF θ= + × ft their F M1 A1ft Leading to 1

14sinθ = cso A1 (5)

(b) N2L � a1000sin8.91000550 =××+ θ ( 1

14550 1000 9.8 1000a+ × × = ) or 1250 = 1000a

M1 A1

( )( )1.25a = − 2 2 22 16 2 1.25v u as y= + ⇒ = × × M1 102y ≈ accept 102.4, 100 A1 (4) [9] Alternative to (b) Work-Energy 21 1

2 141000 16 1000 9.8 550y y× × − × × = M1 M1 A1 102y ≈ accept 102.4, 100 A1 (4)

4. (a) Triangle Circle S

Mass ratio 126 9π 126 9π− (28.3) (97.7) B1 B1ft

x 7 5 x y 4 5 y 4, 7 seen B1 ( )126 7 9 5 126 9 xπ π× = × + − × ft their table values M1 A1ft

7.58x ≈ (ππ

912645882−− ) awrt 7.6 A1

( )126 4 9 5 126 9 yπ π× = × + − × ft their table values M1 A1ft

y ≈ 3.71 (ππ

912645504

−− ) awrt 3.7 A1 (9)

(b) tan21

yx

θ =−

ft their ,x y M1 A1ft

15θ ≈ ° A1 (3) [12]

65


5. (a) N B 2a 30° a mg R a 3mg A rF Μ(A) 4 cos30 3 sin 30 2 sin 30N a mg a mg a× ° = × °+ × ° M1 A2(1,0)

°= 30tan45 mgN ( = mg

345 = 7.07…m) DM1 A1

→ rF N= , ↑ 4R mg= B1, B1 Using rF Rµ= B1

54 3

mg Rµ=√

for their R M1

516 3

µ =√

awrt 0.18 A1 (10)

[10]

Alternative method: M(B): 30sin430cos430sin3330sin2 aRaFamgamg ×=×+×+× 30sin430cos430sin11 aRaFmga ×=×+

RFmg 22

342

11=+

↑ 4R mg= , Using rF Rµ=

2538 =µ , 5

16 3µ =

√

M1A3(2,1,0) DM1A1 B1 B1 M1 A1

66

6. (a) → 30 2ut= B1 ↑ 247.5 5 4.9ut t− = − M1 A1 247.5 75 4.9t− = − eliminating u or t DM1

( )2 75 47.5 254.9

t += = DM1

5t = cso A1 (6)

(b) 30 2 30 10 3ut u u= ⇒ = ⇒ = M1 A1 (2)

(c) ↑ 5 9.8 34y u t= − = −& M1 requires both M1 A1 → 2 6x u= =& x& and y& A1 ( )22 26 34v = + − DM1

( )134.5 msv −≈ accept 35 A1 (5)

[13]

Alternative to (c)

2 21 12 2 47.5B Amv mv m g− = × × with 2 2 26 15 261Av = + = M1 A(2,1,0)

( )2 261 2 9.8 47.5 1192Bv = + × × = DM1

( )134.5 msBv −≈ accept 35 A1 (5)

BEWARE : Watch out for incorrect use of asuv 222 +=

67


7. (a) 2u u 2m 3m x y LM 4 3 2 3mu mu mx my+ = + M1 A1 NEL 1

2y x u− = B1

Solving to 85

y u= cso M1 A1 (5)

(b) 1110

x u= or equivalent B1

Energy loss ( ) ( )( ) ( )( )2 22 2 81110 5

1 12 22 2 3m u u m u u× − + × − M1 A(2,1,0)

2920 mu= A1 (5)

(c) 8

5 u 3m m s t LM 24

5 3mu ms mt= + M1 A1

NEL 85t s eu− = B1

Solving to ( )25 3s u e= − M1 A1

For a further collision ( )211

10 5 3u u e> − M1

14e > ignore e ≤ 1 A1 (7)

[17]

69


Mark Scheme

Question Number

Scheme

Marks

1.(a) (b) 2.

T or e mgl

λ ×= (even T=m is M1, A0, A0 sp case)

g24.016.0

=×λ

⇒ =λ 49 N or 5g

R(↑) cosT mgθ = or cos mgT

θ =

6.19cos.4.032.0.49 =θ or 4 .cos 2g gθ = or 2 .cosmg mgθ = (ft on their λ )

⇒ cosθ = 21 ⇒ θ = 60° ( or

3π

radians)

m ‘a’ = ± 2

165x

, with acceleration in any form (e.g.2

2

dd

xt

, ddvvx

,ddvt

or

a)

Uses a = ddvvx

to obtain kddvvx= ± k’ 2

32x

Separates variables, k ∫ v dv = k’ ∫ 2

32x

dx

Obtains 221 v = m

x32

(+ C) or equivalent e.g. 20.12 v = –

165x

(+

C) Substituting x = 2 if + used earlier or – 2 if – used in d.e. x = 2, v = ± 8 ⇒ 32 = –16 + C ⇒ C = 48 (or value appropriate to their correct equation)

v = 0 ⇒ x

32 = 48 ⇒ x = 3

2 m (N.B. - 23 is not acceptable for final answer

)

N.B ddx

( 12 m 2v ) = 2

165x

, is also a valid approach.

Last two method marks are independent of earlier marks and of each other

M1 A1 A1 (3) M1 A1ft

A1 (3) 6 B1 M1 dM1 A1 M1 A1 M1 A1 cao 8

Special case sinT mgθ =

giving θ = 30 is M1 A0 A0 unless there is evidence that they think θ is with horizontal – then M1 A1 A0

70

Question Number

Scheme

Marks

3.(a) (b)

Large cone small cone S Vol. )2()2( 2

31 hrπ hr 2

31π hr 2

37π (accept ratios 8 : 1 : 7)

C of M h2

1 , h45 x (or equivalent)

hhr 2

1238 .π – hr 2

31π . h4

5 = hr 237π . x or equivalent

→ x = h28

11 ∗

tan θ = 11 1128 14

2 2 2 28,11

r r rh rx

= = =

6.68≈θ ° or 1.20 radians

(Special case – obtains complement by using tan θ = 2rx

giving 21.4° or .374 radians M1A0A0)

Centres of mass may be measured from another point ( e.g. centre of small circle, or vertex) The Method mark will then require a complete method (Moments and subtraction) to give required value for x ). However B marks can be awarded for correct values if the candidate makes the working clear.

B1 B1, B1 M1

A1 (5)

M1, A1 A1

(3) 8

71

Question Number 4. (a) (b)

Scheme Energy equation with at least three terms, including K.E term

212

mV + ..

+ .. 2 21 2 1 1 2 9. . , . .sin 30, . .

2 16 2 2 16mg a mg amg aa a

+ =

⇒ V = 2ga

Using point where velocity is zero and point where string becomes slack:

2

2

12

1 2 9 3. . , . .sin 302 16 4

mw

mg a amga

=

−

⇒ w =8

3ag

Alternative (using point of projection and point where string becomes slack):

221 112 2 ,

16 8mga mgamw mV− = −

So w = 8

3ag

In part (a) DM1 requires EE, PE and KE to have been included in the energy equation.

If sign errors lead to 2

2gaV = − , the last two marks are M0 A0

In parts (a) and (b) A marks need to have the correct signs In part (b) for M1 need one KE term in energy equation of at least 3 terms with distance 34a

to indicate first method, and two KE terms in energy equation of at least 4 terms with

distance 4a

to indicate second method.

SHM approach in part (b). (Condone this method only if SHM is proved)

Using 2 2 2 2( )v a xω= − with 2 2ga

ω = and 4ax = ± .

Using ‘a’ = 2a to give w =

83ag

.

Marks M1 A1, A1, A1 dM1 A1 (6) M1 A1, A1 A1 (4) M1,A1 A1 A1

10

M1 A1 A1

A1

72

5.(a) (b) (c)

N

Nµ 2

,mv N mgr

µ µ= =

mg

6.08.975

2122

=×

==rgvµ ∗

R(↑) cos , 0.6 sinR R mgα α =m

11

2553.

53

54 mgRmgR =⇒=⎟

⎠⎞

⎜⎝⎛ −⇒

R( ) 2

sin , 0.6 cos mvR Rr

α α± =

v ≈ 32.5 m s–1

In part (b) M1 needs three terms of which one is mg If cos α and sin α are interchanged in equation this is awarded M1 A0 A1

In part (c) M1 needs three terms of which one is 2mv

ror 2mrω

If cos α and sin α are interchanged in equation this is also awarded M1 A0 A1 If they resolve along the plane and perpendicular to the plane in part (b), then attempt at

2

cos sinmvR mgr

α α− = , and 2

0.6 sin cosmvR mgr

α α+ = and attempt to eliminate v

Two correct equations Correct work to solve simultaneous equations Answer In part (c) Substitute R into one of the equations Substitutes into a correct equation (earning accuracy marks in part (b))

Uses 25 25(or )

11 29mg mgR =

Obtain v = 32.5

M1, A1 A1 (3) M1, A1, A1 A1 (4) M1, A1, A1 dM1 A1cao

(5) 12

M1

A1 A1 A1 (4) M1 A1

A1

M1A1 (5)

73

Question 6.(a) (b) (c) (d)

Scheme

Energy equation with two terms on RHS, 21 1 5. sin2 2 2

gamv m mga θ= +

2 (5 4sin )2gav θ⇒ = + ∗

R(\\ string) a

mvmgT2

sin =− θ (3 terms)

)sin65(2

θ+=⇒mgT o.e.

560 sin ,T θ= ⇒ = −

Has a solution, so string slack when α ≈ 236(.4)° or 4.13 radians

At top of small circle, 22

5.21

21 2 mgagammv −= (M1 for energy equation with 3 terms)

gav 232 =⇒ = 14.7a

Resolving and using Force = 2mv

r,

aga

mmgT21

23

.=+ (M1 needs three terms, but any v)

mgT 2=⇒ Use of 2 2 2v u gh= + is M0 in part (a)

Marks M1, A1 A1 cso (3) M1 A1 A1 (3) M1, A1 A1 (3)

M1 A1 A1 M1 A1 A1 (6) 15

74

Question Number 7.(a) (b) (c) (d) (a)

(b) (c) (d)

Scheme (Measuring x from E) 2 2 98( 0.2)x g x= − +&& , and so &&x = −49x SHM period with 2 49ω = so T = 2π

7 Max. acceleration = 49 × max. x = 49 x 0.4 = 19.6 m s–2

String slack when x = –0.2: v2 = 49(0.42 – 0.22)

⇒ v ≈ 2.42 m s–1 7 3

5=

Uses tax ωcos= or use sinx a tω= but not with x = 0 or a± Attempt complete method for finding time when string goes slack –0.2 = 0.4 cos 7t ⇒ cos 7t = – 2

1

t = 212π

≈ 0.299 s

Time when string is slack = (2) 2.42 2 3 0.495

7g×

= ≈ s (2 needed for A

) Total time = 2 x 0.299 + 0.495 ≈ 1.09 s DM1 requires the minus sign. Special case 2 2 98x g x= −&& is M1A1A0M0A0 2 98x x= −&& is M0A0A0M0A0 No use of x&& , just a is M1 A0,A0 then M1 A0 if otherwise correct. Quoted results are not acceptable. Answer must be positive and evaluated for B1 M1 – Use correct formula with their ω , a and x but not x = 0. A1 Correct values but allow x = +0.2 Alternative It is possible to use energy instead to do this part

221

20.60.6

2mv mg

lλ ×

+ × = M1 A1

If they use sinx a tω= with x = 0.2± and add 7

π or 14π this is dM1, A1 if done correctly

If they use tax ωcos= with x = -0.2 this is dM1, then A1 (as in scheme)

If they use tax ωcos= with x = +0.2 this needs their 7π

minus answer to reach dM1, then A1

Marks M1 A1, A1 d M1 A1cso

(5) B1 (1) M1 A1 A1 (3) M1 dM1 A1 A1 M1 A1ft A1 (7) 16

75

January 2008

6683 Statistics S1 Mark Scheme

Question Number

Scheme

Marks

NB (a) (b)

( ) ( )2 2773 724 773 724S 60475 722.1, S 53122 704.4, S 56076 110.810 10 10xx yy xy

×= − = = − = = − =

1st B1 for x∑ and 2nd B1 for y∑ , should be seen or implied. M1 for at least one correct attempt at one of S , S or Sxx yy xy and then using in

the correct formula 1st A1ft for a fully correct expression. (ft their �x and their �y) or 3 correct expressions for yyxyxx S and ,S,S but possibly incorrect values for these placed correctly in r. 2nd A1 for awrt 0.155 If 5.0>r they can score B1g in (b) for saying that it (skills test) is not a good guide to performance but B0h since a second acceptable comment about both tests is not possible. Give B1 for one correct line, B1B1 for any 2. If the only comment is the test(s) are a good guide: scores B0B0 If the only comment is the tests are not good: scores B1B0 (second line) The third line is for a comment that suggests that the interview test is OK but the skills test is not since one is positive and the other is negative. Treat 1st B1 as B1g and 2nd as B1h An answer of “no” alone scores B0B0

1. (a) 773, 724x y= =∑ ∑ B1, B1

2 2

10 56076 773 724(10 60475 773 )(10 53122 724 )

r × − ×=

× − × − o.e. M1 A1ft

0.155357.....r = A1 (5)

(b) Both weak correlation B1g B1h Neither score is a good indication of future performance Interview test is slightly better since correlation is positive (2)

Total 7 marks

76

Question Number

Scheme

Marks

2.

(a) mean is 2757 , 229.7512

= AWRT 230 M1, A1

sd is 2724961 (229.75) , 87.3404512

− = AWRT 87.3 M1, A1

[Accept s = AWRT 91.2] (4)

(b) Ordered list is: 125, 160, 169, 171, 175, 186, 210, 243, 250, 258, 390, 420 ( )1

2 2 186 210 198Q = + = B1

( )11 2 169 171 170Q = + = B1

( )13 2 250 258 254Q = + = B1

(3)(c) 3 3 11.5( ) 254 1.5(254 170), 380Q Q Q+ − = + − = Accept AWRT (370-392) M1, A1 Patients F (420) and B (390) are outliers. B1ft B1ft

(4)

(d) 1 2 3

3 1

2 170 2 198 254 , 0.3254 170

Q Q QQ Q− + − × +

= =− −

& AWRT 0.33 M1, A1

Positive skew. A1ft (3)

Total 14 marks

(a) 1st M1 for using x

n∑ with a credible numerator and n = 12.

2nd M1 for using a correct formula, root required but can ft their mean NB Use of s = 8321.84... 91.22...= is OK for M1A1 here. Answers only from a calculator in (a) can score full marks (b) 1st B1 for median= 198 only, 2nd B1 for lower quartile 3rd B1 for upper quartile S.C. If all 31 Q and Q are incorrect but an ordered list (with > 6 correctly placed) is seen

and used then award B0B1 as a special case for these last two marks. (c) M1 for a clear attempt using their quartiles in given formula,

A1 for any value in the range 370 - 392 1st B1ft for any one correct decision about B or F - ft their limit in range (258, 420) 2nd B1ft for correct decision about both F and B - ft their limit in range (258, 420) If more points are given score B0 here for the second B mark. ( Can score M0A0B1B1 here) (d) M1 for an attempt to use their figures in the correct formula – must be seen

(> 2 correct substitutions) 1st A1 for AWRT 0.33

2nd A1ft for positive skew. Follow through their value/sign of skewness . Ignore any further calculations. “positive correlation” scores A0

77

Question Number Scheme 3.

Width 1 1 4 2 3 5 3 12 Freq. Density 6 7 2 6 5.5 2 1.5 0.5

0.5 ×12 or 6 A1

Total area is (1 6) (1 7) (4 2) ...., 70× + × + × + =

( ) 12

14090.5 78.5their 70

− × × M1

“70 seen anywhere” B1 Number of runners is 12 A1

(5) Total 5 marks

1st M1 for attempt at width of the correct bar (90.5 - 78.5) [Maybe on histogram or in table] 1st A1 for 0.5 12× or 6 (may be seen on the histogram. Must be related to the area of the bar above 78.5 - 90.5.

2nd M1 for attempting area of correct bar 140their 70

×

B1 for 70 seen anywhere in their working 2nd A1 for correct answer of 12.

Minimum working required is 1402 0.5 12 where the 2 should come from 70

× ×

Beware 90.5 - 78.5 = 12 (this scores M1A0M0B0A0) Common answer is 0.5 12× = 6 (this scores M1A1M0B0A0) If unsure send to review e.g. 2 ×0.5 × 12=12 without 70 being seen

Marks M1

78

Question Number Scheme Marks 4.

(a) 41 4061818.5 , 153.910xyS ×

= − = (could be seen in (b)) AWRT 154 M1, A1

241188 19.9

10xxS = − = (could be seen in (b)) A1

(3)

(b) 153.9 , 7.733668....19.9

b = = AWRT 7.73 M1, A1

40.6 4.1( 8.89796....)a b= − × = M1 8.89 7.73y x= + A1

(4)(c) A typical car will travel 7700 miles every year B1ft

(1)(d) 5, 8.89 7.73 5( 47.5 47.6)x y= = + × = − M1

So mileage predicted is AWRT 48000 A1 (2)

Total 10 marks Accept calculations for xyxx S and S in (a) or (b)

(a) M1 for correct attempt or expression for either 1st A1 for one correct 2nd A1 for both correct

(b) Ignore the epen marks for part (b) they should be awarded as per this scheme

1st M1 for their Stheir S

xy

xx

1st A1 for AWRT 7.73 2nd M1 for attempt at correct formula for a (minus required). Ft their b. Quoting a correct formula but making one slip in sub.eg. y =406 is OK 2nd A1 for correct equation with 2dp accuracy. Accept a = 8.89, and b = 7.73 even if not written as final equation. Correct answers only (from calc) score 4/4 if correct to 2dp or 3/4 if AWRT 2dp (c) B1ft for their b 1000× to at least 2 sf. Accept “7.7 thousand” but value is needed (d) M1 for substituting x = 5 into their final answer to (b). A1 for AWRT 48000 (Accept “48 thousands”)

79

5. Diagram may be drawn with ( )CAB ∪⊂ or with the 0 for )( ′∪∩ CAB simply left blank Marks (a) A B 1 2 0 3cc M1 90,3,2,1 A1 1,(0),2 M1A1 90 1 outside A1

3 1 Box B1 (6)

C

2

1 (b) P(none)=0.01 B1ft

(1)(c) P(A but not B)=0.04 M1 A1ft

(2)(d) P(any wine but C)=0.03 M1A1ft

(2)(e) P(exactly two)=0.06 M1A1ft

(2)

(f) P( ) 93P( ) ,

P( ) 96C AC A

A∩

= = or 3231

or AWRT 0.969 M1A1ft,A1

(3)Total 16 marks

(a) 1st M1 for 3 closed, labelled curves that overlap. A1 for the 90, 3, 2 and 1 2nd M1 for one of 1, 0 or 2 correct or a correct sum of 4 values for A, B or C 2nd A1 for all 7 values correct. Accept a blank instead of 0. NB final mark is a B1 for the box not an A mark as on EPEN In parts (b) to (f) full marks can be scored for correct answers or correct ft (b) B1ft Follow through their ‘1’from outside divided by 100 (c) M1 for correct expression eg P( ) P( )A B B∪ − or calculation e.g. 3 + 1 or 4 on top A1 for a correct probability, follow through with their ‘3+1’ from diagram (d) M1 for correct expression or calculation e.g. 1+2+0 or 99-96 or 3 on top

A1 for a correct probability, follow through their ‘2+1+0’ from diagram (e) M1 for a correct expression or calculation e.g. 3+2+1 or 6 on top (f) M1 for a correct expression upto “,” and some correct substitution, ft their values. One of these probabilities must be correct or correct ft. If P(C) on bottom M0 1st A1ft follow through their A C∩ and their A but the ratio must be in (0, 1) 2nd A1 for correct answer only. Answer only scores 3/3, but check working P( A C∩ )/P(C ) is M0

Accept decimals or probs. in Venn diagram

For M marks in (c) to (e) they must have a fraction

80

Question Number Scheme Marks 6 (a) 200 or 200g B1

(1)(b) P(190 210) 0.6X< < = or P(X < 210) = 0.8 or P(X > 210)= 0.2 or diagram (o.e.) M1 Correct use of 0.8 or 0.2 A1

( ) 210 200Zσ−

= ± M1

10 0.8416σ

= 0.8416 B1

11.882129....σ = AWRT 11.9 A1

(5)

(c) 180 200( 180)P X P Zσ−⎛ ⎞< = <⎜ ⎟

⎝ ⎠ M1

( 1.6832)P Z= < − 1 0.9535= − M1 = 0.0465 or AWRT 0.046 A1

(3)Total 9 marks

(a) “mean = 200g” is B0 but “median = 200” or just “200” alone is B1 Standardization in (b) and (c). They must use σ not σσ or 2 . (b) 1st M1 for a correct probability statement (as given or eg P(200<X<210)=0.3 o.e.)

or shaded diagram - must have values on z-axis and probability areas shown 1st A1 for correct use of 0.8 or p = 0.2. Need a correct probability statement. May be implied by a suitable value for z seen (e.g. z = 0.84) 2nd M1 for attempting to standardise. Values for x and µ used in formula. Don’t need z = for this M1 nor a z-value, just mark standardization. B1 for z = 0.8416 (or better) [z = 0.84 usually just loses this mark in (a)] 2nd A1 for AWRT 11.9

(c) 1st M1 for attempting to Standardise with 200 and their sd(>0) e.g. ( )180 200their σ−

±

2nd M1 NB on epen this is an A mark ignore and treat it as 2nd M1 for 1 – a probability from tables provided compatible with their probability statement. A1 for 0.0465 or AWRT 0.046 (Dependent on both Ms in part (c))

81

Question Scheme

7.(a) 1 1 1( 3 0) ,4 4 16

P R B= ∩ = = × = M1, A1

(2) (b)

3

0 3 6 9

2

0 2 4 6

1

0 1 2 3

0

0 0 0 0

B R

0 1 2 3

(c) 7 1,

16 16a b c d= = = = B1, B1 B1

(3)

(d) 1 1 1 1E( ) 1 2 3 4 ...

16 8 8 16T ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= × + × + × + × +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ M1

124

= or exact equivalent e.g. 2.25, 94

A1

(2)

(e) 2

2 2 2 21 1 1 1 9Var( ) 1 2 3 4 ,...16 8 8 16 4

T ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= × + × + × + × + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

M1A1,M1

49 81 374 16 16

= − = or 11516

(o.e.) AWRT 7.19 A1 (4)

Total 14 marks (a) M1 for 1 1

4 4× (c) 1st B1 for 16

7 , 2nd B1 for only one error in b, c, d ( 1 1

16 16 or b c d b c d= = ≠ = = ≠ etc), 3rd B1 all of b, c, d 161=

(d) M1 for attempting P( )t T t=∑ , 3 or more terms correct or correct ft. Must Attempt to sum. NB calculating E(T) and then dividing by a number other than 1 scores M0. (e) 1st M1 for attempt at 2E( )T , 3 or more terms correct or correct ft.

1st A1 for 494

(o.e.) or a fully correct expression (all non-zero terms must be seen)

2nd M1 for subtracting their 2[E( )]T , Must be some attempt to square 94− is M0 but 9

16− could be M1 2nd A1 for correct fraction or AWRT 7.19 Full marks can still be scored in (d) and (e) if a is incorrect

All 0s All 1,2,3s All 4,6,9s

B1 B1 B1 (3)

83

January 2008 Statistics S2 Mark Scheme

Question Number

Scheme Marks

1. (a) (b) (c) (d) Notes 1. (a) (b) (c) (d)

A census is when every member of the population is investigated. There would be no cookers left to sell. A list of the unique identification numbers of the cookers. A cooker B1 Need one word from each group (1) Every member /all items / entire /oe (2) population/collection of individuals/sampling frame/oe enumerating the population on its own gets B0 B1 Idea of Tests to destruction. Do not accept cheap or quick B1 Idea of list/ register/database of cookers/serial numbers B1 cooker(s) / serial number(s) The sample of 5 cookers or every 400th cooker gets B1

B1 B1 B1 B1

(4)

84

2 (a) (b) (c) Notes: 2. (a) (b) (c)

Let X be the random variable the number of faulty bolts

P(X ≤ 2) - P(X ≤ 1) = 0.0355 – 0.0076 or ( ) ( )!2!18

!207.03.0 182

= 0.0279 = 0.0278 1 - P(X ≤ 3) = 1 – 0.1071 = 0.8929

or ( ) ( ) ( ) ( ) ( )( ) ( )2019182173 7.0!1!19!20.7.03.0

!2!18!207.03.0

!3!17!207.03.01 −−−−

46 )1071.0()8929.0(

!6!4!10 = 0.0140.

M1 Either attempting to use P (X ≤ 2) – P (X ≤ 1)

or attempt to use binomial and find p(X = 2). Must have ( ) ( )2 18 20!118!2!

p p− ,

with a value of p A1 awrt 0.0278 or 0.0279. M1 Attempting to find 1 – P(X ≤ 3) A1 awrt 0.893 M1 for k 6 4( ) (1 )p p− . They may use any value for p and k can be any number or nC6p6(1 – p)n-6

A1√ 6 410! ( ) (1 )4!6!

their part b their part b− may write 10C6or 10C4

A1 awrt 0.014

M1 A1

(2) M1 A1

(2)

M1A1√A1

(3) B1 B1

(2)

85

3. (a) (b) (i) (ii) (c)

Events occur at a constant rate. any two of the 3 Events occur independently or randomly. Events occur singly. Let X be the random variable the number of cars passing the observation point. Po(6)

P(X ≤ 4) – P(X ≤ 3) = 0.2851 – 0.1512 or !46e 46−

= 0.1339

1 – P(X ≤ 4) = 1 – 0.2851 or 1 – ⎟⎟⎠

⎞⎜⎜⎝

⎛++++− 1

!16

!26

!36

!46e

2346

= 0.7149 P ( 0 car and 1 others) + P (1 cars and 0 other ) = e-1 x 2e-2 + 1e-1 x e-2 = 0.3679 x 0.2707 + 0.3674 x 0.1353 = 0.0996 + 0.0498 = 0.149 alternative Po(1+2) = Po(3) B1 P(X=1) = 3e-3 M1 A1 = 0.149 A1

B1 M1

A1

M1 A1

(5)B1 M1 A1 A1

(4)

86

Notes 3(a) (b) (i) (ii) (c)

B1 B1 Need the word events at least once. Independently and randomly are the same reason. Award the first B1 if they only gain 1 mark Special case. If they have 2 of the 3 lines without the word events they get B0 B1 B1 Using Po(6) in (i) or (ii)

M1 Attempting to find P(X ≤ 4) – P(X ≤ 3) or 4

4!e λλ−

A1 awrt 0.134 M1 Attempting to find 1 – P(X ≤ 4) A1 awrt 0.715 B1 Attempting to find both possibilities. May be implied by doing 1 2-

2e eλ λλ− × + 2 1-

1e eλ λλ− × any values of 1 2and λ λ

M1 finding one pair of form 1 2-2e eλ λλ− × any values of 1 2and λ λ

A1 one pair correct A1 awrt 0.149 Alternative. B1 for Po(3) M1 for attempting to find P(X=1) with Po(3) A1 3e-3 A1 awrt 0.149

87

4. (a)

(b)

(c) Notes 4. (a) (b) (c)

K(24 + 22 – 2) = 1 K = 1/18

1 – F(1.5) = 1 –181 (1.54 + 1.52 – 2)

= 0.705 or 203288

31 (2 ) 1 2

9( )

0

y y yf y

otherwise

⎧ + ≤ ≤⎪⎪= ⎨⎪⎪⎩

M1 putting F(2) = 1 or F(2) – F(1) = 1 A1 cso. Must show substituting y = 2 and the 1/18 M1 either attempting to find 1 – F(1.5) may write and use F( 2 ) – F ( 1.5) A1 awrt 0.705 M1 attempting to differentiate. Must see either a yn→ yn-1 at least once

A1 for getting 31 (2 )9

y y+ o.e and 1 2y≤ ≤ allow 1< y <2

B1 for the 0 otherwise. Allow 0 for y <1 and 0 for y >2 Allow them to use any letter

M1 A1

(2) M1 A1

(2) M1 A1 B1

(3)

88

Question

Scheme

Marks

5

H0 : p = 0.3; H1 : p > 0.3 Let X represent the number of tomatoes greater than 4 cm : X∼ B(40, 0.3) P(X≥ 18) = 1 – P( X ≤ 17) P(X≥ 18) 1 – P(X≤ 17) = 0.0320 P( X ≥ 17 ) = 1 – P(X≤ 16) = 0.0633 = 0.0320 CR X≥ 18 0.0320 < 0.05 18 ≥ 18 or 18 in the critical region no evidence to Reject H0 or it is significant New fertiliser has increased the probability of a tomato being greater than 4 cm Or Dhriti’s claim is true

B1 B1 B1 M1 A1 M1 B1d cao

(7)

5

B1 for correct H0 . must use p or pi B1 for correct H1 must use p and be one tail. B1 using B(40, 0.3). This may be implied by their calculation M1 attempt to find 1 – P( X ≤ 17) or get a correct probability. For CR method must attempt to find P(X≥ 18) or give the correct critical region A1 awrt 0.032 or correct CR.

M1 correct statement based on their probability , H1 and 0.05 or a correct contextualised statement that implies that.

B1 this is not a follow through .conclusion in context. Must use the words increased, tomato and some reference to size or diameter. This is dependent on them getting the previous M1

If they do a two tail test they may get B1 B0 B1 M1 A1 M1 B0 For the second M1 they must have accept Ho or it is not significant

or a correct contextualised statement that implies that. .

89

6a (i) ii) b)

Let X represent the number of sunflower plants more than 1.5m high X~ Po(10) µ=10 P(8 ≤ X ≤ 13) = P(X ≤ 13) – P(X ≤ 7) = 0.8645 – 0.2202 = 0.6443 awrt 0.644

X~ N(10,7.5)

P(7.5 ≤ X ≤ 13.5) = P 7.5 10 13.5 107.5 7.5

X− −⎛ ⎞≤ ≤⎜ ⎟

⎝ ⎠

= P (-0.913≤ X ≤ 1.278) = 0.8997 – (1 – 0.8186) = 0.7183 awrt 0.718 or 0.719 Normal approx /not Poisson since (n is large) and p close to half. or (np = 10 npq = 7.5) mean ≠ variance or np (= 10) and nq (= 30) both >5. or exact binomial = 0.7148

B1 M1 A1 B1 M1 M1

A1 A1 M1 A1

(10) B1 B1dep

(2)

6a (i) ii)

B1 mean = 10 May be implied in (i) or (ii) M1 Attempting to find P(X ≤ 13) – P(X ≤ 7) A1 awrt 0.644

B1 σ2 = 7.5 May be implied by being correct in standardised formula M1 using 7.5 or 8.5 or 12.5 or 13.5. M1 standardising using 7.5 or 8 or 8.5 or 12.5 or13 or 13.5 and their mean and standard deviation.

A1 award for either 7.5 107.5− or awrt -0.91

A1 award for either 13.5 107.5− or awrt 1.28

M1 Finding the correct area. Following on from their 7.5 and 13.5. Need to do a Prob >0.5 – prob <0.5 or prob <0.5 + prob< 0.5

90

b)

A1 awrt 0.718 or 0.719 only. Dependent on them getting all three method marks. No working but correct answer will gain all the marks first B1 normal second B1 p close to half, or mean≠ variance or np and nq both > 5.They may use a number bigger than 5 or they may work out the exact value 0.7148 using the binomial distribution. Do not allow np> 5 and npq>5

91

7 ai) ii) (b) (c) (d)

A hypothesis test is a mathematical procedure to examine a value of a population parameter proposed by the null hypothesis compared with an alternative hypothesis. The critical region is the range of values or a test statistic or region where the test is significant that would lead to the rejection of H0.

Let X represent the number of incoming calls : X ∼ Po(9) From table P(X≥ 16) = 0.0220 P(x ≤ 3) = 0.0212 Critical region (x ≤ 3 or x ≥ 16) Significance level = 0.0220 + 0.0212 = 0.0432 or4.32%

Ho : λ = 0.45; H1 : λ < 0.45 ( accept : Ho : λ = 4.5; H1 : λ < 4.5) Using X ∼ Po(4.5) P (X ≤ 1 ) = 0.0611 CR X < 0 awrt 0.0611 0.0611 > 0.05. 1 ≥ 0 or 1not in the critical region There is evidence to Accept H0 or it is not significant There is no evidence that there are less calls during school holidays.

B1 B1g B1h

(3)B1 M1 A1 A1 B1

(5) B1

(1) B1 M1 A1 M1

B1cao (5)

Notes 7 ai) ii) (b)

B1 Method for deciding between 2 hypothesis. B1 range of values. This may be implied by other words. Not region on its own B1 which lead you to reject H0 Give the first B1 if only one mark awarded. B1 using Po(9) M1 attempting to find P(X>16) or P(x < 3) A1 0.0220 or P(X>16)

92

(c) (d)

A1 0.0212 or P(x ≤ 3) These 3 marks may be gained by seeing the numbers in part c B1 correct critical region A completely correct critical region will get all 5 marks. Half of the correct critical region eg x ≤ 3 or x ≥ 17 say would get B1 M1 A0 A1 B0 if the M1 A1 A1 not already awarded. B1 cao awrt 0.0432 B1 may use λ or µ. Needs both H0 and H1 M1 using Po(4.5) A1 correct probability or CR only M1 correct statement based on their probability , H1 and 0.05 or a correct contextualised statement that implies that.

B1 this is not a follow through .Conclusion in context. Must see the word calls in conclusion If they get the correct CR with no evidence of using Po(4.5) they will get M0 A0 SC If they get the critical region X ≤ 1 they score M1 for rejecting H0 and B1 for concluding the rate of calls in the holiday is lower.

93

0

0.5

1

1.5

2

2.5

0 1 2 3 4 5

x

f(x)

8. a) b) c) d) e)

Max height of 2 labelled and goes through(2,0) shape must be between 2 and 3 and no other lines drawn (accept patios drawn)

correct shape

3

∫ −3

2dx 2)2x(x =

3

2

2⎥⎦

⎤⎢⎣

⎡− 2x

32x3

= 232

∫ −m

2dx 2)2(x = 0.5

[ ]m22 x4x − = 0.5 m2 – 4m + 4 = 0.5 m2 – 4m + 3.5 = 0

m = 2

24 ±

m =2.71 Negative skew. mean < median < mode .

B1 B1 B1

(3)B1

(1) M1A1 A1

(3) M1 A1 M1 A1

(4)B1 B1dep

(2)

94

Notes 8. (a) (b) (c) (d) (e)

B1 the graph must have a maximum of 2 which must be labelled B1 the line must be between 2 and 3 with not other line drawn except patios. They can get this mark even if the patio cannot be seen. B1 the line must be straight and the right shape. B1 Only accept 3 M1 attempt to find f ( )dx x x∫ for attempt we need to see xn → xn+1. ignore limits A1 correct integration ignore limits

A1 accept 232 or awrt 2.67 or 2.6

•

M1 using f ( )dx x∫ =0.5 A1 m2 – 4m + 4 = 0.5 oe M1 attempting to solve quadratic.

A1 awrt 2.71 or 4 2 2 or 2+ oe

2 2+

First B1 for negative Second B1 for mean < median< mode. Need all 3 or may explain using diagram.

102

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Email [email protected]

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