ALL Alevel edexcel maths Mark Schemes Jan 07

Mark Scheme January 2007

GCE

GCE Mathematics (8371/8373, 9371/9373)

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January 2007

Publications Code UA 018759

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Contents

Page

6663 Core Mathematics 1……………………………………………………………………. 1 6664 Core Mathematics 2……………………………………………………………………. 11 6665 Core Mathematics 3………………………………………………………………..... 33 6666 Core Mathematics 4……………………………………………………………………. 39 6674 Further Pure Mathematics…………………………………………………………. 59 6677 Mechanics 1………………………………………………………………………………….68 6678 Mechanics 2………………………………………………………………………………….72 6679 Mechanics 3………………………………………………………………………………… 78 6683 Statistics 1………………………………………………………………………………….. 85 6684 Statistics 2………………………………………………………………………………….. 92 6689 Decision Mathematics 1……………………………………………………………. 98

1

January 2007 6663 Core Mathematics C1

Mark Scheme

Question Scheme Mark Number

1. 234 kxx → or 21

21

2−

→ kxx (k a non-zero constant) M1

.......,12 21

2 −+ xx , ( )01→− A1, A1, B1 (4)

4

Accept equivalent alternatives to 21

−x , e.g. 5.0

21 ,1,1 −x

xx.

M1: 34x ‘differentiated’ to give 2kx , or…

21

2x ‘differentiated’ to give 21

−kx (but not for just 01→− ).

1st A1: 212x (Do not allow just 243 x× )

2nd A1: 21

−x or equivalent. (Do not allow just 2

1

221 −× x , but allow 2

1

1−

x or 21

22 −

x ).

B1: 1− differentiated to give zero (or ‘disappearing’). Can be given provided that at least one of the other terms has been changed. Adding an extra term, e.g. + C, is B0.

2

Question Scheme Marks Number

2. (a) 6√3 (a = 6) B1 (1)

(b) Expanding 2)32( √− to get 3 or 4 separate terms M1

34,7 √− ( )4,7 −== cb A1, A1 (3)

4

(a) 36√± also scores B1.

(b) M1: The 3 or 4 terms may be wrong.

1st A1 for 7, 2nd A1 for 34√− .

Correct answer 347 √− with no working scores all 3 marks.

347 √+ with or without working scores M1 A1 A0.

Other wrong answers with no working score no marks.

3


3. (a) Shape of f(x) B1

Moved up ↑ M1

Asymptotes: y = 3 B1

x = 0 (Allow “y-axis”) B1 (4)

( 3≠y is B0, 0≠x is B0).

(b) 031=+

x No variations accepted. M1

31

−=x (or 33.0− …) Decimal answer requires at least 2 d.p. A1 (2)

6

(a) B1: Shape requires both branches and no obvious “overlap” with the asymptotes (see below), but otherwise this mark is awarded generously. The curve may, e.g., bend away from the asymptote a little at the end. Sufficient curve must be seen to suggest the asymptotic behaviour, both horizontal and vertical. M1: Evidence of an upward translation parallel to the y-axis. The shape of the graph can be wrong, but the complete graph (both branches if they have 2 branches) must be translated upwards. This mark can be awarded generously by implication where the graph drawn is an upward translation of another standard curve (but not a straight line). The B marks for asymptote equations are independent of the graph. Ignore extra asymptote equations, if seen.

(b) Correct answer with no working scores both marks. The answer may be seen on the sketch in part (a).

Ignore any attempts to find an intersection with the y-axis.

e.g. (a) This scores B0 (clear overlap with horiz. asymp.) M1 (Upward translation… bod that both branches have been translated).

B0 M1 B0 M1 B0 M0

No marks unless the original curve is seen, to show upward translation.

−4 −3 −2 −1 1 2 3 4

−4

−2

2

4

6

8

x

y

4


4. 44)2( 22 +−=− xxx or 44)2( 22 ++=+ yyy M: 3 or 4 terms M1

10)2( 22 =+− xx or 10)2( 22 =++ yy M: Substitute M1

0642 2 =−− xx or 0642 2 =−+ yy Correct 3 terms A1

...,0)1)(3( ==+− xxx or ...,0)1)(3( ==−+ yyy M1 (The above factorisations may also appear as )1)(62( +− xx or equivalent).

13 −== xx or 13 =−= yy A1

31 −== yy or 31 =−= xx M1 A1 (7)

(Allow equivalent fractions such as: 26

=x for x = 3).

7 1st M: ‘Squaring a bracket’, needs 3 or 4 terms, one of which must be an 2x or 2y term.

2nd M: Substituting to get an equation in one variable (awarded generously).

1st A: Accept equivalent forms, e.g. 642 2 =− xx .

3rd M: Attempting to solve a 3-term quadratic, to get 2 solutions.

4th M: Attempting at least one y value (or x value).

If y solutions are given as x values, or vice-versa, penalise at the end, so that it is possible to score M1 M1A1 M1 A1 M0 A0.

Strict “pairing of values” at the end is not required.

“Non-algebraic” solutions: No working, and only one correct solution pair found (e.g. x = 3, y = 1): M0 M0 A0 M0 A0 M1 A0 No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M1 A1 M1 A1 Both correct solution pairs found, and demonstrated, perhaps in a table of values: Full marks Squaring individual terms: e.g. 422 += xy M0 104 22 =++ xx M1 A0 (Eqn. in one variable) 3√=x M0 A0 (Not solving 3-term quad.) 77422 √==+= yxy M1 A0 (Attempting one y value)

5


5. Use of acb 42 − , perhaps implicit (e.g. in quadratic formula) M1

( )0)1(890)1(24)3( 2 <++<+−××−− kk A1

178 −<k (Manipulate to get qpk < , or qpk > , or qpk = ) M1

8

17−<k ⎟

⎠⎞

⎜⎝⎛ −<−< 125.2or

812 :equivOr kk A1cso (4)

4

1st M: Could also be, for example, comparing or equating 2b and ac4 . Must be considering the given quadratic equation. There must not be x terms in the expression, but there must be a k term.

1st A: Correct expression (need not be simplified) and correct inequality sign. Allow also 0)1(2432 <+−××−− k .

2nd M: Condone sign or bracketing mistakes in manipulation. Not dependent on 1st M, but should not be given for irrelevant work. M0 M1 could be scored: e.g. where acb 42 + is used instead of acb 42 − . Special cases:

1. Where there are x terms in the discriminant expression, but then division by 2x gives an inequality/equation in k. (This could score M0 A0 M1 A1).

2. Use of ≤ instead of < loses one A mark only, at first occurrence, so an

otherwise correct solution leading to 8

17−≤k scores M1 A0 M1 A1.

N.B. Use of b = 3 instead of 3−=b implies no A marks.

6


6. (a) )34)(34( xx √+√+ seen, or a numerical value of k seen, ( )0≠k . M1 (The k value need not be explicitly stated… see below). xx 92416 +√+ , or k = 24 A1cso (2)

(b) cx→16 or 23

21

cxkx → or 29 cxx → M1

( ) 232

16,2

916d92416 xCxxxxx +++=+√+∫ A1, A1ft (3)

5

(a) e.g. )34)(34( xx √+√+ alone scores M1 A0, (but not 2)34( x√+ alone). e.g xx 91216 +√+ scores M1 A0.

k = 24 or xx 92416 +√+ ,with no further evidence, scores full marks M1 A1.

Correct solution only (cso): any wrong working seen loses the A mark.

(b) A1: Cxx ++2

9162

. Allow 4.5 or 214 as equivalent to

29 .

A1ft: 23

32 xk (candidate’s value of k, or general k).

For this final mark, allow for example 348 as equivalent to 16, but do

not allow unsimplified “double fractions” such as ( )2324 , and do

not allow unsimplified “products” such as 2432× .

A single term is required, e.g. 23

23

88 xx + is not enough.

An otherwise correct solution with, say, C missing, followed by an incorrect solution including + C can be awarded full marks (isw, but allowing the C to appear at any stage).

7


7. (a) 323 cxx → or cx→− 6 or 128 −− →− cxx M1

⎟⎠⎞

⎜⎝⎛ +−+

−−−=

−

xxxCxxxx 86)(

186

33)(f 3

13 A1 A1

Substitute x = 2 and y = 1 into a ‘changed function’ to form an equation in C. M1

141281 =++−= CC A1cso (5)

(b) 22

28623 −−× M1

= 4 A1

Eqn. of tangent: )2(41 −=− xy M1

74 −= xy (Must be in this form) A1 (4)

9

(a) First 2 A marks: + C is not required, and coefficients need not be simplified, but powers must be simplified.

All 3 terms correct: A1 A1 Two terms correct: A1 A0 Only one term correct: A0 A0

Allow the M1 A1 for finding C to be scored either in part (a) or in part (b).

(b) 1st M: Substituting x = 2 into 22 863

xx −− (must be this function).

2nd M: Awarded generously for attempting the equation of a straight line through (2, 1) or (1, 2) with any value of m, however found. 2nd M: Alternative is to use (2, 1) or (1, 2) in cmxy += to find a value for c.

If calculation for the gradient value is seen in part (a), it must be used in part (b) to score the first M1 A1 in (b). Using (1, 2) instead of (2, 1): Loses the 2nd method mark in (a).

Gains the 2nd method mark in (b).

8


8. (a) kx →4 or 21

23

3 kxx → or kxx →− 22 M1

xxxy 4

294

dd 2

1−+= A1 A1 (3)

(b) For x = 4, 8322416)162()443()44( =−+=×−×+×=y (*) B1 (1)

(c) 31694dd

−=−+=xy

M: Evaluate their xy

dd at x = 4 M1

Gradient of normal = 31 A1ft

Equation of normal: )4(318 −=− xy , 203 += xy (*) M1, A1 (4)

(d) ( )20.....:0 −== xy and use 212

212 )()( yyxx −+− M1

22 824 +=PQ or 222 824 +=PQ Follow through from (k, 0) A1ft May also be scored with ( ) ( )22 8 and24 −− . = 8√10 A1 (3)

11

(a) For the 2 A marks coefficients need not be simplified, but powers must be

simplified. For example, 21

323 x× is acceptable.

All 3 terms correct: A1 A1 Two terms correct: A1 A0

Only one term correct: A0 A0

(b) There must be some evidence of the “24” value.

(c) In this part, beware ‘working backwards’ from the given answer.

A1ft: Follow through is just from the candidate’s value of xy

dd .

2nd M: Is not given if an m value appears “from nowhere”. 2nd M: Must be an attempt at a normal equation, not a tangent.

2nd M: Alternative is to use (4, 8) in cmxy += to find a value for c. (d) M: Using the normal equation to attempt coordinates of Q, (even if using x = 0 instead of y = 0), and using Pythagoras to attempt PQ or 2PQ . Follow through from (k, 0), but not from (0, k)…

A common wrong answer is to use x = 0 to give 320

. This scores M1 A0 A0.

For final answer, accept other simplifications of √640, e.g. 2√160 or 4√40.

9


9. (a) Recognising arithmetic series with first term 4 and common difference 3. B1 (If not scored here, this mark may be given if seen elsewhere in the solution). ( )13)1(34)1( +=−+=−+ nndna M1 A1 (3)

(b) { } { } 175,3)110(82

10)1(22

=×−+=−+= dnanSn , M1 A1, A1 (3)

(c) { } 1750)1(382

:1750 <−+< kkSk { } ⎟⎠⎞

⎜⎝⎛ >+

+>+ 175038

21:1750or 1 kkSk M1

( )03492113or 0350053 22 >−+<−+ kkkk M1 A1 (Allow equivalent 3-term versions such as 350053 2 =+ kk ).

0)35)(1003( <+− kk Requires use of correct inequality throughout.(*) A1cso (4)

(d) 3

100 or equiv. seen ⎟⎠⎞

⎜⎝⎛

397or , k = 33 (and no other values) M1, A1 (2)

12

(a) B1: Usually identified by a = 4 and d = 3. M1: Attempted use of term formula for arithmetic series, or… answer in the form (3n + constant), where the constant is a non-zero value. Answer for (a) does not require simplification, and a correct answer without working scores all 3 marks.

(b) M1: Use of correct sum formula with n = 9, 10 or 11. A1: Correct, perhaps unsimplified, numerical version. A1: 175 Alternative: (Listing and summing terms). M1: Summing 9, 10 or 11 terms. (At least 1st, 2nd and last terms must be seen). A1: Correct terms (perhaps implied by last term 31). A1: 175 Alternative: (Listing all sums) M1: Listing 9, 10 or 11 sums. (At least 4, 7, ….., “last”). A1: Correct sums, correct finishing value 175. A1: 175 Alternative: (Using last term).

M1: Using )(2

lanSn += with 9T , 10T or 11T as the last term.

A1: Correct numerical version )314(2

10+ . A1: 175

Correct answer with no working scores 1 mark: 1,0,0.

(c) For the first 3 marks, allow any inequality sign, or equals. 1st M: Use of correct sum formula to form inequality or equation in k, with the 1750.

2nd M: (Dependent on 1st M). Form 3-term quadratic in k. 1st A: Correct 3 terms. Allow credit for part (c) if valid work is seen in part (d).

(d) Allow both marks for k = 33 seen without working. Working for part (d) must be seen in part (d), not part (c).

10


10. (a) (i) Shape or or B1 Max. at (0, 0). B1

(2, 0), (or 2 shown on x-axis). B1 (3)

(ii) Shape B1

(It need not go below x-axis)

Through origin. B1

(6, 0), (or 6 shown on x-axis). B1 (3)

(b) )6()2(2 xxxx −=− M1 0623 =−− xxx Expand to form 3-term cubic (or 3-term quadratic if divided by x), with all terms on one side. The “= 0” M1 may be implied.

...0)2)(3( ==+− xxxx Factor x (or divide by x), and solve quadratic. M1 2 and3 −== xx A1 16:2 −=−= yx Attempt y value for a non-zero x value by M1 substituting back into )2(2 −xx or )6( xx − . 9:3 == yx Both y values are needed for A1. A1 )9,3( and)16,2( −− (0, 0) This can just be written down. Ignore any ‘method’ shown. (But must be seen in part (b)). B1 (7) 13

(a) (i) For the third ‘shape’ shown above, where a section of the graph coincides with the x-axis, the B1 for (2, 0) can still be awarded if the 2 is shown on the x-axis.

For the final B1 in (i), and similarly for (6, 0) in (ii): There must be a sketch. If, for example (2, 0) is written separately from the sketch, the sketch must not clearly contradict this. If (0, 2) instead of (2, 0) is shown on the sketch, allow the mark. Ignore extra intersections with the x-axis.

(ii) 2nd B is dependent on 1st B.

Separate sketches can score all marks.

(b) Note the dependence of the first three M marks. A common wrong solution is (-2, 0), (3, 0), (0, 0), which scores M0 A0 B1 as the last 3 marks.

A solution using no algebra (e.g. trial and error), can score up to 3 marks: M0 M0 M0 A0 M1 A1 B1. (The final A1 requires both y values). Also, if the cubic is found but not solved algebraically, up to 5 marks: M1 M1 M0 A0 M1 A1 B1. (The final A1 requires both y values).

−3 −2 −1 1 2 3 4 5 6 7

−20

−10

10

20

x

y

11


Mark Scheme

Question Number

Scheme Marks

1.

(a) f ´(x) = 3x2 + 6x B1

f ´´(x) = 6x + 6 M1, A1cao

(3)

Notes cao = correct answer only

Examples 1(a) f ´´(x) = 3x2 + 6x B1 1(a) f ´(x) = x2 + 3x B0

M0 A0 f ´´(x) = x + 3 M1 A0 1(a) f ´(x) = 3x2 + 6x B1 1(a) x3 + 3x2 + 5 f ´´(x) = 6x M1 A0 = 3x2 + 6x B1 = 6x + 6 M1 A1 1(a) y = x3 + 3x2 + 5 1(a) f ´(x) = 3x2 + 6x + 5 B0

ddyx

= 3x2 + 3x B0 f ´´(x) = 6x + 6 M1 A1

2

2

dd

yx

= 6x + 3 M1 A0 1(a) f ´(x) = 3x2 + 6x B1

f ´´(x) = 6x + 6 + c M1 A0 1(a) f ´(x) = 3x2 + 6x + c B0

f ´´(x) = 6x + 6 M1 A1

1(a) Acceptable alternatives include 3x2 + 6x1; 3x2 + 3×2x; 3x2 + 6x + 0

Ignore LHS (e.g. use [whether correct or not] of ddyx

and 2

2

dd

yx

)

3x2 + 6x + c or 3x2 + 6x + constant (i.e. the written word constant) is B0

B1

M1 Attempt to differentiate their f ´(x); xn → xn – 1. xn → xn – 1 seen in at least one of the terms. Coefficient of x…. ignored for the method mark. x2 → x1 and x → x0 are acceptable.

M1

Acceptable alternatives include 6x1 + 6x0; 3×2x + 3×2 6x + 6 + c or 6x + 6 + constant is A0

A1 cao

12

Question Number

Scheme Marks

1.

(b) xxxxxx 5

33

4d )53(

3423 ++=++∫ M1, A1

)5141(10845

4

2

1

34

++−++=⎥⎦

⎤⎢⎣

⎡++ xxx M1

= 4315 o.e.

A1 (4)

(7)

Notes o.e. = or equivalent

Examples

1(b) 4

3 54x x x c+ + + M1 A1 1(b)

43 5

4x x x c+ + + M1 A1

4 + 8 + 10 + c – ( 14 + 1 + 5 + c) M1 x = 2, 22 + c

= 3415 A1 x = 1, 6 1

4 + c M0 A0 (no subtraction)

1(b) Attempt to integrate f(x); xn → xn + 1 Ignore incorrect notation (e.g. inclusion of integral sign)

M1

o.e. Acceptable alternatives include

43 5

4x x x+ + ;

4 313 5

4 3x x x+ + ;

4 33 54 3x x x c+ + + ;

4 33 54 3x x x+ +∫

N.B. If the candidate has written the integral (either 4 33 5

4 3x x x+ + or what they think is the

integral) in part (a), it may not be rewritten in (b), but the marks may be awarded if the integral is used in (b).

A1

Substituting 2 and 1 into any function other than x3 + 3x2 + 5 and subtracting either way round. So using their f ´(x) or f ´´(x) or ∫ their f ´(x) dx or ∫ their f ´´(x) dx will gain the M mark (because none of these will give x3 + 3x2 + 5). Must substitute for all x s but could make a slip.

14 8 10 1 54

+ + − + + (for example) is acceptable for evidence of subtraction (‘invisible’

brackets).

M1

o.e. (e.g. 4315 , 15.75, 4

63 )

Must be a single number (so 1422 6− is A0).

A1

Answer only is M0A0M0A0

13

1(b) 2

1

f ( ) dx x∫ = 23 + 3×22 + 5 – (1 + 3 + 5) M0 A0, M0

= 25 – 9 = 16 A0 (Substituting 2 and 1 into x3 + 3x2 + 5, so 2nd M0)

1(b) 2

1

(6 6) dx x+∫ = 22

13 6x x⎡ ⎤+⎣ ⎦ M0 A0 1(b)

22

1

(3 6 ) dx x x+∫ = 23 2

13x x⎡ ⎤+⎣ ⎦ M0 A0

= 12 + 12 – (3 + 6) M1 A0 = 8 + 12 – (1 + 3) M1 A0

1(b) 4

3 54x x x+ + M1 A1

4 4

3 32 12 5 2 1 54 4+ + × − + + M1

(one negative sign is sufficient for evidence of subtraction) = 22 – 6 1

4 = 3415 A1

(allow ‘recovery’, implying student was using ‘invisible brackets’) 1(a) f(x) = x3 + 3x2 + 5

f ´´(x) = 4

3 54x x x+ + B0 M0 A0

(b) 4 4

3 32 12 5 2 1 54 4+ + × − − − M1 A1 M1

= 4315 A1

The candidate has written the integral in part (a). It is not rewritten in (b), but the marks may be awarded as the integral is used in (b).

14

Question Number

Scheme Marks

2.

(a) 5 2 35 4 5 4 3(1 2 ) 1 5 ( 2 ) ( 2 ) ( 2 ) ...

2! 3!x x x x× × ×

− = + × − + − + − +

...8040101 32 +−+−= xxx

B1, M1, A1,

A1

(4)

(b) (1 + x)(1 – 2x)5 = (1 + x)(1 – 10x + …)

= 1 + x – 10x + … M1

≈ 1 – 9x ( ) A1 (2)

(6)

Notes

2(a) 1 – 10x 1 – 10x must be seen in this simplified form in (a).

B1

Correct structure: ‘binomial coefficients’ (perhaps from Pascal’s triangle), increasing powers of x. Allow slips.

Accept other forms: 5C1, 51⎛ ⎞⎜ ⎟⎝ ⎠

, also condone 51

⎛ ⎞⎜ ⎟⎝ ⎠

but must be attempting to use 5.

Condone use of invisible brackets and using 2x instead of –2x. Powers of x: at least 2 powers of the type (2x)a or 2xa seen for a ≥ 1.

M1

40x2 (1st A1) A1 – 80x3 (2nd A1) A1 Allow commas between terms. Terms may be listed rather than added

Allow ‘recovery’ from invisible brackets, so 5 4 3 2 2 35 5 51 1 . 2 1 . 2 1 . 2

1 2 3x x x⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ − + − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

...8040101 32 +−+−= xxx gains full marks.

2 35 4 5 4 31 5 (2 ) (2 ) (2 ) ...2! 3!

x x x× × ×+ × + + + 2 31 10 40 80 ...x x x= + + + + gains B0M1A1A0

Misread: first 4 terms, descending terms: if correct, would score B0, M1, 1st A1: one of 40x2 and –80x3 correct; 2nd A1: both 40x2 and –80x3 correct.

2(a) Long multiplication

(1 – 2x)2 = 1 – 4x + 4x2, (1 – 2x)3 = 1 – 6x + 12x2 – 8x3, (1 – 2x)4 = 1 – 8x + 24x2 – 32x3 {+ 16x4} (1 – 2x)5 = 1 – 10x + 40x2 + 80x3 + …

1 – 10x 1 – 10x must be seen in this simplified form in (a).

B1

Attempt repeated multiplication up to and including (1 – 2x)5 M1

15

40x2 (1st A1) A1 – 80x3 (2nd A1) A1 Misread: first 4 terms, descending terms: if correct, would score B0, M1, 1st A1: one of 40x2 and –80x3 correct; 2nd A1: both 40x2 and –80x3 correct.

16

2(b) Use their (a) and attempt to multiply out; terms (whether correct or incorrect) in x2 or higher can be ignored. If their (a) is correct an attempt to multiply out can be implied from the correct answer, so (1 + x)(1 – 10x) = 1 – 9x will gain M1 A1. If their (a) is correct, the 2nd bracket must contain at least (1 – 10x) and an attempt to multiply out for the M mark. An attempt to multiply out is an attempt at 2 out of the 3 relevant terms (N.B. the 2 terms in x1 may be combined – but this will still count as 2 terms). If their (a) is incorrect their 2nd bracket must contain all the terms in x0 and x1 from their (a) AND an attempt to multiply all terms that produce terms in x0 and x1. N.B. (1 + x)(1 – 2x)5 = (1 + x)(1 – 2x) [where 1 – 2x + … is NOT the candidate’s answer to (a)] = 1 – x i.e. candidate has ignored the power of 5: M0 N.B. The candidate may start again with the binomial expansion for (1 – 2x)5 in (b). If correct (only needs 1 – 10x) may gain M1 A1 even if candidate did not gain B1 in part (a).

M1

N.B. Answer given in question. A1 Example Answer in (a) is 2 31 10 40 80 ...x x x= + + − + (b) (1 + x)(1 + 10x) = 1 + 10x + x M1 = 1 + 11x A0

17

Question Number

Scheme Marks

3. Centre ⎟⎠⎞

⎜⎝⎛ ++−

246,

231 , i.e. (1, 5) M1, A1

r =

2)46())1(3( 22 −+−−

or r2 = (1 – (–1))2 + (5 – 4)2 or r2 = (3 – 1)2 + (6 – 5)2 o.e.

M1

(x – 1)2 + (y – 5)2 = 5 M1,A1,A1 (6)

Notes

Some use of correct formula in x or y coordinate. Can be implied. Use of ( ) ( )( )1 1

2 2,A B A Bx x y y− − → (–2, –1) or (2, 1) is M0 A0 but watch out for use of 12 ( )A A Bx x x+ − etc which is okay.

M1

(1, 5) (5, 1) gains M1 A0.

A1

Correct method to find r or r2 using given points or f.t. from their centre. Does not need to be simplified.

Attempting radius = 2(diameter)

2is an incorrect method, so M0.

N.B. Be careful of labelling: candidates may not use d for diameter and r for radius. Labelling should be ignored. Simplification may be incorrect – mark awarded for correct method. Use of 2 2

1 2 1 2( ) ( )x x y y− − − is M0.

M1

Write down (x ± a)2 + (y ± b)2 = any constant (a letter or a number). Numbers do not have to be substituted for a, b and if they are they can be wrong.

M1

LHS is (x – 1)2 + (y – 5)2. Ignore RHS. A1 RHS is 5. Ignore subsequent working. Condone use of decimals that leads to exact 5.

A1

Or correct equivalents, e.g. x2 + y2 – 2x – 10y + 21 = 0.

Alternative – note the order of the marks needed for ePEN. As above. M1 As above. A1 x2 + y2 + (constant)x + (constant)y + constant = 0. Numbers do not have to be substituted for the constants and if they are they can be wrong.

3rd M1

Attempt an appropriate substitution of the coordinates of their centre (i.e. working with coefficient of x and coefficient of y in equation of circle) and substitute (–1, 4) or (3, 6) into equation of circle.

2nd M1

–2x – 10y part of the equation x2 + y2 – 2x – 10y + 21 = 0. A1 +21 = 0 part of the equation x2 + y2 – 2x – 10y + 21 = 0. A1 Or correct equivalents, e.g. (x – 1)2 + (y – 5)2 = 5.

18

Question Number

Scheme Marks

4. 17log5log =x or x = log517 M1

5log

17log=x A1

= 1.76 A1 (3)

Notes N.B. It is never possible to award an A mark after giving M0. If M0 is given then the marks will be M0 A0 A0.

4 Acceptable alternatives include

17log5log =x ; 10 10log 5 log 17x = ; e elog 5 log 17x = ; ln 5 ln17x = ; x = log517 Can be implied by a correct exact expression as shown on the first A1 mark

1st M1

An exact expression for x that can be evaluated on a calculator. Acceptable alternatives include

5log17log

=x ; 10

10

log 17log 5

x = ; e

e

log 17log 5

x = ; ln17ln 5

x = ; log 17log 5

q

q

x = where q is a number

This may not be seen (as, for example, log517 can be worked out directly on many calculators) so this A mark can be implied by the correct final answer or the right answer corrected to or truncated to a greater accuracy than 3 significant figures or 1.8

Alternative: a numbera number

x = where this fraction, when worked out as a decimal rounds to 1.76.

(N.B. remember that this A mark cannot be awarded without the M mark). If the line for the M mark is missing but this line is seen (with or without the x =) and is correct the method can be assumed and M1 1st A1 given.

1st A1

1.76 cao 2nd A1 N.B. 5 17 = 1.76 and x5 = 17, ∴x = 1.76 are both M0 A0 A0 Answer only 1.76: full marks (M1 A1 A1) Answer only to a greater accuracy but which rounds to 1.76: M1 A1 A0 (e.g. 1.760, 1.7603, 1.7604, 1.76037 etc) Answer only 1.8: M1 A1 A0 Trial and improvement: award marks as for “answer only”.

19

Examples 4. x = 17log5 M0 A0 4. 51.76 = 17 M1 A1 A1

= 1.76 A0 Answer only but clear that x = 1.76 Working seen, so scheme applied 4. 51.8 = 17 M1 A1 A0 4. 51.76 M0 A0 A0

Answer only but clear that x = 1.8 4. 5log 17 = x M1 4. 5log 17 = x M1

x = 1.760 A1 A0 x = 1.76 A1 A1 4. 17log5log =x M1 4. ln 5 ln17x = M1

x = 1.2304...0.69897...

A1 x = 2.833212...1.609437...

A1

x = 1.76 A1 x = 1.76 A1

4. 17log5log =x M1 4. 17log 5 = x M0

x = 2.578901.46497

A1 log5log17

x = A0

x = 1.83 A0 x = 0.568 A0 4. 51.8 = 18.1, 51.75 = 16.7 4. x = 51.76 M0 A0 A0 51.761 = 17 M1 A1 A0

4. 17log5log =x M1 4. 5log

17log=x M1 A1

x = 1.8 A1 A0 x = 1.8 A0 N.B. 4. x5 = 17 M0 A0 4. 5 17 M0 A0

x = 1.76 A0 = 1.76 A0

20

Question Number

Scheme Marks

5.

(a) f(–2) = (–2)3 + 4(–2)2 + (–2) – 6 M1

{ = –8 + 16 – 2 – 6}

= 0, ∴ x + 2 is a factor A1

(2)

(b) x3 + 4x2 + x – 6 = (x + 2)(x2 + 2x – 3) M1, A1

= (x + 2)(x + 3)(x – 1) M1, A1

(4)

(c) –3, –2, 1 B1 (1)

(7)

Notes Line in mark scheme in { } does not need to be seen.

5(a) Attempting f(±2): No x s; allow invisible brackets for M mark Long division: M0 A0.

M1

= 0 and minimal conclusion (e.g. factor, hence result, QED, , ). If result is stated first [i.e. If x + 2 is a factor, f(–2) = 0] conclusion is not needed. Invisible brackets used as brackets can get M1 A1, so f(–2) = –23 + 4×–22 + –2 – 6 { = –8 + 16 – 2 – 6} = 0, ∴ x + 2 is a factor M1 A1, but f(–2) = –23 + 4×–22 + –2 – 6 = –8 – 16 – 2 – 6 = 0, ∴ x + 2 is a factor M1 A0 Acceptable alternatives include: x = –2 is a factor, f(–2) is a factor.

A1

5(b) 1st M1 requires division by (x + 2) to get x2 + ax + b where a ≠ 0 and b ≠ 0 or equivalent with division by (x + 3) or (x – 1).

M1

(x + 2)(x2 + 2x – 3) or (x + 3)(x2 +x – 2) or (x – 1)(x2 + 5x + 6) [If long division has been done in (a), minimum seen in (b) to get first M1 A1 is to make some reference to their quotient x2 + ax + b.]

A1

Attempt to factorise their quadratic (usual rules). M1 “Combining” all 3 factors is not required. A1 Answer only: Correct M1 A1 M1 A1 Answer only with one sign slip: (x + 2)(x + 3)(x + 1) scores 1st M1 1st A12nd M0 2nd A0 (x + 2)(x – 3)(x – 1) scores 1st M0 1st A0 2nd M1 2nd A1

Answer to (b) can be seen in (c).

5(b) Alternative comparing coefficients (x + 2)(x2 + ax + b) = x3 + (2 + a)x2 + (2a + b)x + 2b Attempt to compare coefficients of two terms to find values of a and b

M1

a = 2, b = –3 A1 Or (x + 2)(ax2 + bx + c) = ax3 + (2a + b)x2 + (2b + c)x + 2c Attempt to compare coefficients of three terms to find values of a, b and c.

M1

21

a = 1, b = 2, c = – 3 A1 Then apply scheme as above

5(b) Alternative using factor theorem Show f(–3) = 0; allow invisible brackets M1 ∴x + 3 is a factor A1 Show f(1) = 0 M1 ∴x – 1 is a factor A1

5(c) –3, –2, 1 or (–3, 0), (–2, 0), (1, 0) only. Do not ignore subsequent working. Ignore any working in previous parts of the question. Can be seen in (b)

B1

22

Question Number

Scheme Marks

6. xx sin51)sin1(2 2 =+− M1

03sin5sin2 2 =−+ xx

0)3)(sin1sin2( =+− xx

21sin =x M1, A1

6

5,6

ππ=x

M1, M1,

A1cso (6)

Notes

Use of xx 22 sin1cos −= . Condone invisible brackets in first line if 22 2sin x− is present (or implied) in a subsequent line. Must be using xx 22 sin1cos −= . Using 2 2cos 1 sinx x= + is M0.

M1

Attempt to solve a 2 or 3 term quadratic in sin x up to sin x = … Usual rules for solving quadratics. Method may be factorising, formula or completing the square

M1

Correct factorising for correct quadratic and 21sin =x .

So, e.g. (sin 3)x + as a factor → sin 3x = can be ignored.

A1

Method for finding any angle in any range consistent with (either of) their trig. equation(s) in degrees or radians (even if x not exact). [Generous M mark] Generous mark. Solving any trig. equation that comes from minimal working (however bad). So x = sin–1/cos–1/tan–1(number) → answer in degrees or radians correct for their equation (in any range)

M1

Method for finding second angle consistent with (either of) their trig. equation(s) in radians. Must be in range 0 ≤ x < 2π. Must involve using π (e.g. π ± …, 2π – …) but … can be inexact. Must be using the same equation as they used to attempt the 3rd M mark. Use of π must be consistent with the trig. equation they are using (e.g. if using cos–1 then must be using 2π – … ) If finding both angles in degrees: method for finding 2nd angle equivalent to method above in degrees and an attempt to change both angles to radians.

M1

65,

6ππ c.s.o. Recurring decimals are okay (instead of 1 5and

6 6).

Correct decimal values (corrected or truncated) before the final answer of 6

5,6

ππ is

acceptable.

A1 cso

Ignore extra solutions outside range; deduct final A mark for extra solutions in range.

Special case

Answer only 6

5,6

ππ M0, M0, A0, M1, M1 A1 Answer only 6π M0, M0, A0, M1,

23

M0 A0 Finding answers by trying different values (e.g. trying multiples of π) in 2cos2x + 1 = 5sinx : as for answer only.

24

Question Number

Scheme Marks

7. y = x(x2 – 6x + 5) = x3 – 6x2 + 5x M1, A1

2

53

64

d )56(234

23 xxxxxxx +−=+−∫ M1, A1ft

1

0

23

4

252

4 ⎥⎦

⎤⎢⎣

⎡+−

xxx = 0252

41

−⎟⎠⎞

⎜⎝⎛ +− =

43 M1

2

1

23

4

252

4 ⎥⎦

⎤⎢⎣

⎡+−

xxx = ( )4310164 −+− =

411

− M1, A1(both)

∴total area = 4

1143+ M1

= 27 o.e.

A1cso

(9)

25

Notes

Attempt to multiply out, must be a cubic. M1 Award A mark for their final version of expansion (but final version does not need to have like terms collected).

A1

Attempt to integrate; xn → xn + 1. Generous mark for some use of integration, so e.g. 2 2 2

( 1)( 5) d 52 2 2x x xx x x x x x⎛ ⎞⎛ ⎞

− − = − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∫ would gain method mark.

M1

Ft on their final version of expansion provided it is in the form ...p qax bx+ + . Integrand must have at least two terms and all terms must be integrated correctly.

If they integrate twice (e.g. 1

0∫ and

2

1∫ ) and get different answers, take the better of the two.

A1ft

Substitutes and subtracts (either way round) for one integral. Integral must be a ‘changed’ function. Either 1 and 0, 2 and 1 or 2 and 0. For [ ]10 : – 0 for bottom limit can be implied (provided that it is 0).

M1

M1 Substitutes and subtracts (either way round) for two integrals. Integral must be a ‘changed’ function. Must have 1 and 0 and 2 and 1 (or 1 and 2). The two integrals do not need to be the same, but they must have come from attempts to integrate the same function.

M1

43 and

411

− o.e. (if using 2

1

f ( )x∫ ) or 43 and 11

4 o.e. (if using

1

2

f ( )x∫ or –2

1

f ( )x∫ or

2

1

f ( )x−∫ ) where f(x) = 4 2

3 524 2x xx− + .

The answer must be consistent with the integral they are using (so 2

1

f ( )x∫ = 114

loses this A

and the final A).

411

− may not be seen explicitly. Can be implied by a subsequent line of working.

A1

5th M1 | their value for [ ]10 | + | their value for [ ]21 | Dependent on at least one of the values coming from integration (other may come from e.g. trapezium rules). This can be awarded even if both values already positive.

M1

27 o.e. N.B. c.s.o.

A1 cso

26

Question Number

Scheme Marks

8.

(a) 721400–

dd 2 += −v

vC M1, A1

0721400– 2 =+−v M1

v2 = 4900 dM1

v = 70 A1cso

(5)

(b) 32

2

2800dd −= v

vC M1

v = 70, 2

2

dd

Cv

> 0 {⇒ minimum}

or v = 70, 2

32

d 22800 70 { 0.00816...}d 245

Cv

−= × = = {⇒ minimum}

A1ft

(2)

(c) v = 70, 7702

701400 ×

+=C M1

C = 40 A1 (2) (9)

Notes

8(a) Attempt to differentiate vn → vn – 1. Must be seen and marked in part (a) not part (b). Must be differentiating a function of the form 1av bv− + .

M1

o.e.

( 2 2–14007

v c− + + is A0)

A1

Their 0dd

=vC . Can be implied by their d

dCv

= P + Q → P = ± Q. M1

Dependent on both of the previous Ms.

Attempt to rearrange their ddCv

into the form vn = number or vn – number = 0, n ≠ 0.

dM1

v = 70 cso but allow v = ±70. v = 70 km per h also acceptable. A1cso Answer only is 0 out of 5. Method of completing the square: send to review.

27

8(a) Trial and improvement f(v) = 1400 27v

v+

Attempts to evaluate f(v) for 3 values a, b, c where (i) a < 70, b = 70 and c > 70 or (ii) a, b < 70 and c > 70 or (iii) a < 70 and b, c > 70.

M1

All 3 correct and states v = 70 (exact) A1 Then 2nd M0, 3rd M0, 2nd A0.

8(a) Graph Correct shape (ignore anything drawn for v < 0).

M1

v = 70 (exact) A1 Then 2nd M0, 3rd M0, 2nd A0.

8(b)

Attempt to differentiate their vC

dd ; vn → vn – 1 (including v0 → 0).

M1

2

2

dd

vC must be correct. Ft only from their value of v and provided their value of v is +ve.

Must be some (minimal) indication that their value of v is being used.

Statement: “When v = their value of v, 2

2

dd

vC > 0” is sufficient provided 32800v− > 0 for their

value of v. If substitution of their v seen: correct substitution of their v into 32800v− , but, provided evaluation is +ve, ignore incorrect evaluation. N.B. Parts in mark scheme in { } do not need to be seen.

A1ft

8(c) Substitute their value of v that they think will give Cmin (independent of the method of obtaining this value of v and independent of which part of the question it comes from).

M1

40 or £40 Must have part (a) completely correct (i.e. all 5 marks) to gain this A1.

A1

Answer only gains M1A1 provided part (a) is completely correct..

28

Examples 8(b)

8(b) 32

2

2800dd −= v

vC M1

v = 70, 2

2

dd

Cv

> 0 A1

8(b) 32

2

2800dd −= v

vC M1

> 0 A0 (no indication that a value of v is being used) 8(b) Answer from (a): v = 30

32

2

2800dd −= v

vC M1

v = 30, 2

2

dd

Cv

> 0 A1ft

8(b) 2

32

d 2800d

C vv

−= M1

v = 70, 2

32

d 2800 70d

Cv

−= ×

= 8.16 A1 (correct substitution of 70 seen, evaluation wrong but positive)

8(b) 2

32

d 2800d

C vv

−= M1

v = 70, 2

2

d 0.00408d

Cv

= A0 (correct substitution of 70 not seen)

29

Question Number

Scheme Marks

9.

(a) 662)36(66cos

222

××−+

=PQR 12

⎧ ⎫= −⎨ ⎬⎩ ⎭

M1, A1

3

2π=PQR

A1

(3)

(b) Area = 3

2621 2 π

×× m2 M1

= π12 m2 ( ) A1cso

(2)

(c) Area of ∆ = 3

2sin6621 π

××× m2 M1

= 39 m2 A1cso

(2)

(d) Area of segment = 3912 −π m2 M1

= 22.1 m2 A1

(2)

(e) Perimeter = 26 6 63π⎡ ⎤+ + ×⎢ ⎥⎣ ⎦

m M1

= 24.6 m A1ft (2)

(11)

Notes

9(a) N.B. 2 2 2 2 cosa b c bc A= + − is in the formulae book. Use of cosine rule for PQRcos . Allow A, θ or other symbol for angle. (i) 2 2 2(6 3) 6 6 2.6.6cos PQR= + − : Apply usual rules for formulae: (a) formula not stated, must be correct, (b) correct formula stated, allow one sign slip when substituting.

or (ii) 2 2 26 6 (6 3)cos

2 6 6PQR ± ± ±

=± × ×

Also allow invisible brackets [so allow 2

6 3 ] in (i) or (ii)

M1

Correct expression 2 2 26 6 (6 3)

2 6 6+ −

× ×o.e. (e.g. 36

72− or 1

2− )

A1

23π

A1

30

9(a) Alternative

3sin6

aθ = where θ is any symbol and a < 6. M1

3 3sin6

θ = where θ is any symbol. A1

23π

A1

9(b) Use of 2

1 r2θ with r = 6 and θ = their (a). For M mark θ does not have to be exact. M0 if using degrees.

M1

π12 c.s.o. (⇒ (a) correct exact or decimal value) N.B. Answer given in question

A1

Special case: Can come from an inexact value in (a) PQR = 2.09 → Area = 2

1 ×62 × 2.09 = 37.6 (or 37.7) = π12 (no errors seen, assume full values used on calculator) gets M1 A1. PQR = 2.09 → Area = 2

1 ×62 × 2.09 = 37.6 (or 37.7) = 11.97π = π12 gets M1 A0.

9(c) Use of 2

1 r2sin θ with r = 6 and their (a). θ = 1cos (their )PQR− in degrees or radians Method can be implied by correct decimal provided decimal is correct (corrected or truncated to at least 3 decimal places). 15.58845727

M1

39 c.s.o. Must be exact, but correct approx. followed by 39 is okay (e.g. … = 15.58845 = 39 )

A1cso

9(c) Alternative (using 1

2 bh ) Attempt to find h using trig. or Pythagoras and use this h in 1

2 bh form to find the area of triangle PQR

M1

39 c.s.o. Must be exact, but correct approx. followed by 39 is okay (e.g. … = 15.58845 = 39 )

A1cso

9(d) Use of area of sector – area of ∆ or use of 21

2 ( sin )r θ θ− . M1 Any value to 1 decimal place or more which rounds to 22.1 A1

9(e) 6 + 6 + [6 × their (a)]. M1 Correct for their (a) to 1 decimal place or more A1 ft

31

Question Number

Scheme Marks

10.

(a) {Sn = } a + ar + … + arn – 1 B1

{rSn = } ar + ar2 + … + arn M1

(1 – r)Sn = a(1 – rn) dM1

rraS

n

n −−

=1

)1( ( ) A1cso

(4)

(b) a = 200, r = 2, n = 10,

21)21(200 10

10 −−

=S

M1, A1

= 204,600 A1

(3)

(c) 65

=a , 31

=r B1

raS−

=∞ 1,

31

65

1−=∞S M1

= 45 o.e.

A1

(3)

(d) –1 < r < 1 (or | r | < 1) B1 (1)

(11)

Notes

10(a) Sn not required. The following must be seen: at least one + sign, a, arn – 1 and one other intermediate term. No extra terms (usually arn).

B1

Multiply by r; rSn not required. At least 2 of their terms on RHS correctly multiplied by r. M1 Subtract both sides: LHS must be ±(1 – r)Sn, RHS must be in the form ±a(1 – rpn + q). Only award this mark if the line for Sn = … or the line for rSn = … contains a term of the form arcn + d Method mark, so may contain a slip but not awarded if last term of their Sn = last term of their rSn.

dM1

Completion c.s.o. N.B. Answer given in question A1 cso

10(a) Sn not required. The following must be seen: at least one + sign, a, arn – 1 and one other intermediate term. No extra terms (usually arn).

B1

On RHS, multiply by 11

rr

−−

M1

32

Or Multiply LHS and RHS by (1 – r) Multiply by (1 – r) convincingly (RHS) and take out factor of a. Method mark, so may contain a slip.

dM1

Completion c.s.o. N.B. Answer given in question A1 cso

10(b) Substitute r = 2 with a = 100 or 200 and n = 9 or 10 into formula for Sn. M1

10200(1 2 )1 2−−

or equivalent. A1

204,600 A1

10(b) Alternative method: adding 10 terms (i) Answer only: full marks. (M1 A1 A1) (ii) 200 + 400 + 800 + … {+ 102,400} = 204,600 or 100(2 + 4 + 8 + … {+ 1,024)} = 204,600 M1 for two correct terms (as above o.e.) and an indication that the sum is needed (e.g. + sign or the word sum).

M1

102,400 o.e. as final term. Can be implied by a correct final answer. A1 204,600. A1

10(c) N.B. 1

arS∞ −= is in the formulae book.

31

=r seen or implied anywhere. B1

Substitute 65

=a and their r into 1

ar−

. Usual rules about quoting formula. M1

45 o.e.

A1

10(d) N.B. 1

arS∞ −= for | r | < 1 is in the formulae book.

–1 < r < 1 or | r | < 1 In words or symbols. Take symbols if words and symbols are contradictory. Must be < not ≤.

B1

33


Mark Scheme

Question Number Scheme Marks

1. (a) ( )sin 3 sin 2 sin 2 cos cos 2 sinθ θ θ θ θ θ θ= + = + B1

( )2 22sin cos 1 2sin sinθ θ θ θ= + − B1 B1 3 32sin 2sin sin 2sinθ θ θ θ= − + − M1 33sin 4sinθ θ= − cso A1 (5)

(b) 3

3 3 3 3 3 3 9 3sin 3 3 44 4 4 16 16

θ⎛ ⎞√ √ √ √ √= × − = − =⎜ ⎟⎝ ⎠

or exact

equivalent

M1 A1 (2)

[7]

2. (a) ( ) ( ) ( )( )

2

2

2 , 3 2 3f

2x x

xx

+ − + +=

+ M1 A1, A1

( ) ( )

2 2

2 24 4 3 6 3 1

2 2x x x x x

x x+ + − − + + +

= =+ +

cso A1 (4)

(b) 2

2 1 312 4

x x x⎛ ⎞+ + = + +⎜ ⎟⎝ ⎠

, 0> for all values of x. M1 A1, A1 (3)

(c) ( )( )

2

2

1 32 4f

2

xx

x

⎛ ⎞+ +⎜ ⎟⎝ ⎠=

+

Numerator is positive from (b) ( )22 2 0x x≠ − ⇒ + > (Denominator is positive)

Hence ( )f 0x > B1 (1) [8] Alternative to (b)

( )2 2d 1 31 2 1 0 1d 2 4

x x x x x xx

+ + = + = ⇒ = − ⇒ + + = M1 A1

A parabola with positive coefficient of 2x has a minimum 2 1 0x x⇒ + + > A1 (3) Accept equivalent arguments

34


3. (a) 12sin 2 24 4 2

y x P Cπ π= ⇒ = = × = ⇒ ∈√

√ B1 (1)

Accept equivalent (reversed) arguments. In any method it must be clear

that 1sin4 2π=√

or exact equivalent is used.

(b) d 2cosd

x yy= or d1 2cos

dyyx

= M1 A1

d 1d 2cos

yx y= May be awarded after substitution M1

d 14 d 2

yyx

π= ⇒ =

√ cso A1 (4)

(c) 2m′ = −√ B1

( )2 24

y xπ− = − −√ √ M1 A1

2 24

y x π= − + +√ A1 (4)

[9]

4. (i) ( ) ( )

( ) ( )

2 2

2 22 2

9 2d 9d 9 9

x x xy xx x x

⎛ ⎞+ − −⎜ ⎟= =⎜ ⎟+ +⎝ ⎠

M1 A1

2d 0 9 0 3d

y x xx= ⇒ − = ⇒ = ± M1 A1

1 13, , 3,6 6

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Final two A marks depend on second M only A1, A1 (6)

(ii) ( )1

2 22d 3 1 e 2ed 2

x xyx= + × M1 A1 A1

( )11

ln3 ln3 221 d 3ln 3 1 e 2e 3 4 3 182 d 2

yxx

= ⇒ = + × = × × = M1 A1 (5)

[11]

35


5. (a) ( )22 23 1 2R R= + ⇒ =√ M1 A1

tan 33πα α= ⇒ =√ accept awrt 1.05 M1 A1 (4)

(b) ( ) 1sin their 2

x α+ = M1

5 13 their ,6 6 6

x π π πα ⎛ ⎞+ = ⎜ ⎟⎝ ⎠

A1

11,2 6

x π π= accept awrt 1.57, 5.76 M1 A1 (4)

[8]

The use of degrees loses only one mark in this question. Penalise the first time it occurs in an answer and then ignore.

36


6. (a) ( )ln 4 2y x= −

e 4 2y x= − leading to 12 e2

yx = − Changing subject and removing ln M1 A1

12 e2

xy = − 1 1f 2 e2

x−⇒ −a cso A1

Domain of 1f − is � B1 (4) (b) Range of 1f − is ( )1f 2x− < (and f -1 ( )∈x �) B1 (1) (c) Shape B1 1.5 B1 ln 4 B1

ln 4

2y = B1 (4)

(d) 1 20.3704, 0.3452x x≈ − ≈ − cao B1, B1 (2)

If more than 4 dp given in this part a maximum on one mark is lost. Penalise on the first occasion.

(e) 3 0.354 03019x = − … 4 0.350 926 88x = − … 5 0.352 017 61x = − … 6 0.351633 86x = − … Calculating to at least 6x to at least four dp M1 0.352k ≈ − cao A1 (2) [13] Alternative to (e) 0.352k ≈ − Found in any way

Let ( ) 1g e2

xx x= +

( ) ( )g 0.3515 0.0003, g 0.3525 0.001− ≈ + − ≈ − M1

Change of sign (and continuity) ( )0.3525, 0.3515k⇒ ∈ − − 0.352k⇒ = − (to 3 dp) A1 (2)

37


7. (a) ( ) ( )f 2 16 8 8 16 0− = + − = > B1

( ) ( )f 1 1 4 8 3 0− = + − = − < B1

Change of sign (and continuity) ⇒ root in interval ( )2, 1− − B1ft (3)

ft their calculation as long as there is a sign change

(b) 3d 4 4 0 1d

y x xx= − = ⇒ = M1 A1

Turning point is ( )1, 11− A1 (3) (c) 2, 4, 4a b c= = = B1 B1 B1 (3) (d) Shape B1

ft their turning point in correct quadrant only B1 ft

2 and −8 B1 (3) (e) Shape B1 (1) [13]

38


8. (i) ( ) ( )2 2 2 2sec cosec 1 tan 1 cotx x x x− = + − + M1 A1

2 2tan cotx x= − cso A1 (3) (ii)(a) arccos cosy x x y= ⇒ = B1

sin arcsin2 2

x y x yπ π⎛ ⎞= − ⇒ = −⎜ ⎟⎝ ⎠

B1 (2)

Accept arcsin arcsin cosx y=

(b) arccos arcsin2 2

x x y yπ π+ = + − = B1 (1)

[6] Alternatives for (i) 2 2 2 2sec tan 1 cosec cotx x x x− = = − M1 A1

Rearranging 2 2 2 2sec cosec tan cotx x x x− = − cso

A1 (3)

2 2

2 2 2 2

1 1 sin cosLHScos sin cos sin

x xx x x x

⎛ ⎞−= − =⎜ ⎟

⎝ ⎠

RHS ( )( )2 2 2 22 2 4 4

2 2 2 2 2 2

sin cos sin cossin cos sin coscos sin cos sin cos sin

x x x xx x x xx x x x x x

− +−= − = = M1

2 2

2 2

sin coscos sin

x xx x−

= A1

LHS= or equivalent A1 (3)

39


Mark Scheme Question Number Scheme Marks

** represents a constant

1. ( )2 2

22 5x 1 5xf(x) (2 5x) 2 1 12 4 2

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

Takes 2 outside the

bracket to give any of (2)-2 or 1

4 .B1

M1

⎧ ⎫− − − − −

= + − + + +⎨⎭⎩

2 314

( 2)( 3) ( 2)( 3)( 4)1 ( 2)(* * x); (* * x) (* * x) ...2! 3!

Expands 2(1 * * )x −+ to give an

unsimplified+ −1 ( 2)(* * x) ;

A correct unsimplified }{.......... expansion

with candidate’s ( )* * x

A1

− − −⎧ ⎫− − − − −= + − + + +⎨ ⎬

⎭⎩

5x 5x 5x2 314 2 2 2

( 2)( 3) ( 2)( 3)( 4)1 ( 2)( ); ( ) ( ) ...2! 3!

2 3

14

75x 125x1 5x; ...4 2

⎧ ⎫= + + + +⎨ ⎬

⎩ ⎭

A1;

= + + + +2 31 5x 75x 125x; ...

4 4 16 8

Anything that cancels to +

1 5x ;4 4

Simplified +2 375x 125x

16 8 A1

= + + + +2 31 1 11 51 x; 4 x 15 x ...4 4 16 8

[5] 5 marks

40


Aliter 1. −= − 2f(x) (2 5x)

Way 2 B1

M1

− − −

−

⎧ ⎫− −+ − +⎪ ⎪⎪ ⎪= ⎨ ⎬− − −⎪ ⎪+ +

⎪ ⎪⎭⎩

2 3 4 2

5 3

( 2)( 3)(2) ( 2)(2) (* * x); (2) (* * x)2!

( 2)( 3)( 4) (2) (* * x) ...3!

14 or −2(2)

Expands 2(2 5x)−− to give an

unsimplifed− −+ −2 3(2) ( 2)(2) (* * x) ;

A correct unsimplified }{.......... expansion

with candidate’s ( )* * x

A1

− − −

−

− −⎧ ⎫+ − − + −⎪ ⎪⎪ ⎪= ⎨ ⎬− − −⎪ ⎪+ − +⎪ ⎪⎩ ⎭

2 3 4 2

5 3

( 2)( 3)(2) ( 2)(2) ( 5x); (2) ( 5x)2!

( 2)( 3)( 4) (2) ( 5x) ...3!

⎧ ⎫+ − − +⎪ ⎪= ⎨ ⎬

+ − − +⎪ ⎪⎩ ⎭

21 1 14 8 16

3116

( 2)( )( 5x); (3)( )(25x )

( 4)( )( 125x ) ...

A1;

= + + + +2 31 5x 75x 125x; ...

4 4 16 8

Anything that cancels to +

1 5x ;4 4

Simplified +2 375x 125x

16 8 A1

= + + + +2 31 1 11 51 x; 4 x 15 x ...4 4 16 8

[5]

5 marks Attempts using Maclaurin expansions need to be referred to your team leader.

41


2. (a) Volume ( ) ( )− −

⎛ ⎞ π= π =⎜ ⎟⎜ ⎟+ +⎝ ⎠∫ ∫

1 12 2

1 14 4

2

2

1 1dx dx3 1 2x 9 1 2x

Use of 2V y dx= π∫ .

Can be implied. Ignore limits.B1

= ( )−−

π⎛ ⎞ +⎜ ⎟⎝ ⎠ ∫

12

14

21 2x dx9

Moving their power to the top.(Do not allow power of -1.)

Can be implied. Ignore limits and 9

π

M1

M1

12

14

1(1 2x)9 ( 1)(2)

−

−

⎡ ⎤π +⎛ ⎞= ⎢ ⎥⎜ ⎟ −⎝ ⎠ ⎢ ⎥⎣ ⎦

Integrating to give −± + 1p(1 2x)11

2 (1 2x)−− + A1

−

−

π⎛ ⎞ ⎡ ⎤= − +⎜ ⎟ ⎣ ⎦⎝ ⎠

12

14

112 (1 2x)

9

⎡ ⎤⎛ ⎞⎛ ⎞π − −⎛ ⎞= −⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦12

1 19 2(2) 2( )

π⎛ ⎞= − − −⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠

14 ( 1)

9

π=

12

Use of limits to give exact values of

π12 or π π3 2

36 24or or aef A1 aef

[5](b) From Fig.1, AB = ( )− − = 31 1

2 4 4 units As ≡3

4 units 3cm then scale factor

( )= =

34

3k 4 .

Hence Volume of paperweight = ( ) π⎛ ⎞⎜ ⎟⎝ ⎠

3412

( ) ×34 (their answer to part (a)) M1

V = π

=3 316 cm 16.75516... cm3

π16

3 or awrt 16.8or π64

12 or aefA1

[2] 7 marks

Note: π9 (or implied) is not needed for the middle three marks of question 2(a).

42


Aliter

2. (a) Volume ( ) ( )

1 12 2

1 14 4

2

2

1 1dx dx3 1 2x 3 6x

− −

⎛ ⎞= π = π⎜ ⎟⎜ ⎟+ +⎝ ⎠∫ ∫ Use of 2V y dx= π∫ .

Can be implied. Ignore limits.B1

Way 2

= ( ) ( )12

14

23 6x dx−

−

π +∫ Moving their power to the top.

(Do not allow power of -1.) Can be implied.

Ignore limits and π

M1

M1

( )12

14

1(3 6x)( 1)(6)

−

−

⎡ ⎤+= π ⎢ ⎥−⎣ ⎦

Integrating to give 1p(3 6x)−± +

116 (3 6x)−− + A1

( )

12

14

116 (3 6x)−

−⎡ ⎤= π − +⎣ ⎦

( ) 32

1 16(6) 6( )

⎡ ⎤⎛ ⎞⎛ ⎞− −= π −⎢ ⎥⎜ ⎟⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

( ) 1 1

36 9( )= π − − −⎡ ⎤⎣ ⎦

π=

12

Use of limits to give exact values of

π12 or π π3 2

36 24or or aef A1 aef

[5]

Note: π is not needed for the middle three marks of question 2(a).

43


3. (a) = − = −x 7cos t cos7t , y 7sin t sin7t ,

M1 = − +

dx 7sin t 7sin7tdt

, = −dy 7cos t 7cos7tdt

Attempt to differentiate x and y with respect to t to give

dxdt in the form ± ±A sin t Bsin7t

dydt in the form ± ±Ccos t Dcos7t

Correct dxdt and dy

dt A1

−

∴ =− +

dy 7cos t 7cos7tdx 7sin t 7sin7t

Candidate’s dddd

ytxt

B1

[3]

M1 (b)

When t ,6π

= m(T) =π π

π π

−=− +

76 6

76 6

7cos 7cosdydx 7sin 7sin

;

( )− −= = = − = −

− − −

7 3 7 32 2

7 72 2

7 3 3 awrt 1.737

Substitutes 6t or 30π= o into their dydx expression;

to give any of the four underlined expressions oe

(must be correct solution only)A1 cso

Hence m(N) = −−

13

or 13

awrt 0.58= Uses m(T) to ‘correctly’ find m(N). Can be ft from “their

tangent gradient”.A1 oe.

When π= 6t ,

( )π π= − = − − = =7 3 3 8 376 6 2 2 2x 7cos cos 4 3

( )π π= − = − − = =7 87 16 6 2 2 2y 7sin sin 4

The point ( )4 3, 4

or ( )awrt 6.9, 4B1

N: ( )− = −1

3y 4 x 4 3

Finding an equation of a normal with their point and their normal

gradient or finds c by using y (their gradient)x "c "= + .

M1

A1 oe N: = 1

3y x or = 3

3y x or =3y 3x

Correct simplified EXACT equation of normal.

This is dependent on candidate using correct ( )4 3 , 4

or ( )= + ⇒ = − =1

34 4 3 c c 4 4 0

Hence N: = 1

3y x or = 3

3y x or =3y 3x [6]

9 marks

44


Aliter 3. (a) = − = −x 7cos t cos7t , y 7sin t sin7t ,

M1

Way 2

= − +dx 7sin t 7sin7tdt

, = −dy 7cos t 7cos7tdt

Attempt to differentiate x and y with respect to t to give dx

dt in the form± ±A sin t Bsin7t

dydt in theform± ±Ccos t Dcos7t

Correct dxdt and dy

dtA1

dy 7cos t 7cos7t 7( 2sin4t sin3t) tan4tdx 7sin t 7sin7t 7(2cos4t sin3t)

− − −= = =− + −

Candidate’s dddd

ytxt

B1

[3]

M1 (b)

When t ,6π

= m(T) = π= 46

dy tandx

;

( ) ( )( )

= = − = −−

32

12

2 13 awrt 1.73

2 (1)

Substitutes 6t or 30π= o into their dydx expression;

to give any of the three underlined expressions oe

(must be correct solution only)A1 cso

Hence m(N) = −−

13

or 13

awrt 0.58= Uses m(T) to ‘correctly’ find m(N). Can be ft from “their

tangent gradient”.A1 oe.

When π= 6t ,

( )π π= − = − − = =7 3 3 8 376 6 2 2 2x 7cos cos 4 3

( )π π= − = − − = =7 87 16 6 2 2 2y 7sin sin 4

The point ( )4 3, 4

or ( )awrt 6.9, 4B1

N: ( )− = −1

3y 4 x 4 3

Finding an equation of a normal with their point and their normal

gradient or finds c by using y (their gradient)x "c "= + .

M1

A1 oe N: = 1

3y x or = 3

3y x or =3y 3x

Correct simplified EXACT equation of normal.

This is dependent on candidate using correct ( )4 3 , 4

or ( )= + ⇒ = − =1

34 4 3 c c 4 4 0

Hence N: = 1

3y x or = 3

3y x or =3y 3x [6] 9 marks

45

Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) → ∞ , but obtains M0 if they write y 4 (x 4 3)− = ∞ − . If they write, however, N: x = 4 3 , then they can score M1. Beware: A candidate finding an m(T) = ∞ can obtain A1ft for m(N) = 0, and also obtains M1 if they write y 4 0(x 4 3)− = − or y = 4.

46


4. (a) 2x 1 A B(x 1)(2x 3) (x 1) (2x 3)

−≡ +

− − − −

− ≡ − + −2x 1 A(2x 3) B(x 1) Forming this identity.

NB: A & B are not assigned in this question

M1

Let = 32x , ( )= ⇒ =1

22 B B 4

A1 Let =x 1, ( )= − ⇒ = −1 A 1 A 1 either one of = −A 1 or =B 4 .

both correct for their A, B. A1

giving 1 4(x 1) (2x 3)−

+− −

[3]

(b) & (c) −

=− −∫ ∫dy (2x 1) dx

y (2x 3)(x 1) Separates variables as shown

Can be implied B1

−= +

− −∫ 1 4 dx(x 1) (2x 3)

Replaces RHS with their partial fraction to be integrated. M1

M1 A1 ∴ = − − + − +ln y ln(x 1) 2ln(2x 3) c

At least two terms in ln’sAt least two ln terms correct

All three terms correct and ‘+ c’ A1 [5]

= =y 10, x 2 gives =c ln10 =c ln10 B1 ∴ = − − + − +ln y ln(x 1) 2ln(2x 3) ln10

= − − + − +2ln y ln(x 1) ln(2x 3) ln10 Using the power law for

logarithms M1

⎛ ⎞−= +⎜ ⎟−⎝ ⎠

2(2x 3)ln y ln ln10(x 1)

or

⎛ ⎞−= ⎜ ⎟−⎝ ⎠

210(2x 3)ln y ln(x 1)

Using the product and/or quotient laws for logarithms to obtain a

single RHS logarithmic term with/without constant c.

M1

−=

−

210(2x 3)y(x 1)

−=

−

210(2x 3)y(x 1)

or aef. isw A1 aef

[4] 12 marks

47


Aliter 4.

(b) & (c) −

=− −∫ ∫dy (2x 1) dx

y (2x 3)(x 1) Separates variables as shown

Can be implied B1

Way 2

−= +

− −∫ 1 4 dx(x 1) (2x 3)

Replaces RHS with their partial fraction to be integrated. M1

M1 A1 ∴ = − − + − +ln y ln(x 1) 2ln(2x 3) c

At least two terms in ln’sAt least two ln terms correct

All three terms correct and ‘+ c’ A1 See below for the award of B1 decide to award B1 here!! B1 = − − + − +2ln y ln(x 1) ln(2x 3) c Using the power law for

logarithms M1

⎛ ⎞−= +⎜ ⎟−⎝ ⎠

2(2x 3)ln y ln cx 1

Using the product and/or quotient

laws for logarithms to obtain a single RHS logarithmic term

with/without constant c.

M1

⎛ ⎞−= ⎜ ⎟−⎝ ⎠

2A(2x 3)ln y lnx 1

where =c ln A

or ⎛ ⎞ ⎛ ⎞− −

+⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠= =

2 2(2x 3) (2x 3)ln c lnx 1 x 1lny ce e e e

−

=−

2A(2x 3)y(x 1)

= =y 10, x 2 gives =A 10 =A 10 for B1 award

above

−=

−

210(2x 3)y(x 1)

−=

−

210(2x 3)y(x 1)

or aef & isw A1 aef

[5] & [4]

Note: The B1 mark (part (c)) should be awarded in the same place on ePEN as in the Way 1 approach.

48


Aliter

(b) & (c) −

=− −∫ ∫dy (2x 1) dx

y (2x 3)(x 1) Separates variables as shown Can

be implied B1

Way 3

32

1 2 dx(x 1) (x )−

= +− −∫ Replaces RHS with their partial

fraction to be integrated. M1

M1 A1 3

2ln y ln(x 1) 2ln(x ) c∴ = − − + − + At least two terms in ln’s

At least two ln terms correctAll three terms correct and ‘+ c’ A1

[5] = =y 10, x 2 gives ( )1

2c ln10 2ln ln40= − = ( )12c ln10 2ln= − orc ln40= B1 oe

3

2ln y ln(x 1) 2ln(x ) ln40∴ = − − + − + 23

2ln y ln(x 1) ln(x ) ln10= − − + − + Using the power law for logarithms M1

23

2(x )ln y ln ln40

(x 1)⎛ ⎞−

= +⎜ ⎟−⎝ ⎠ or

23

240(x )ln y ln

(x 1)⎛ ⎞−

= ⎜ ⎟−⎝ ⎠

Using the product and/or quotient laws for logarithms to obtain a

single RHS logarithmic term with/without constant c.

M1

23

240(x )y

(x 1)−

=−

23

240(x )y

(x 1)−

=−

or aef. isw A1 aef

[4]

Note: Please mark parts (b) and (c) together for any of the three ways.

49


5. (a) sin x cos y 0.5+ = ( eqn ∗ )

dy dycos x sin y 0dx dx

⎧ ⎫= − =⎨ ⎬

⎩ ⎭ ( eqn # )

Differentiates implicitly to include dysin ydx

± . (Ignore ( )dydx = .) M1

dy cos xdx sin y

= cos xsin y A1 cso

[2]

(b) dy cos x0 0 cos x 0dx sin y

= ⇒ = ⇒ =

Candidate realises that they need to solve ‘their numerator’ = 0

…or candidate sets =dd 0yx in their

(eqn #) and attempts to solve the resulting equation.

M1

giving 2 2x or xπ π= − = both 2 2x ,π π= − or 90x = ± o or awrt x 1.57= ± required here

A1

When 2x π= − , ( )2sin cos y 0.5π− + = When 2x π= , ( )2sin cos y 0.5π + =

Substitutes either their 2x π= or π= − 2x into eqn ∗ M1

cos y 1.5⇒ = ⇒ y has no solutions cos y 0.5⇒ = − ⇒ 2 2

3 3y orπ π= − Only one of 2 2

3 3y or or 120π π= − o

or 120°− or awrt -2.09 or awrt 2.09 A1

In specified range ( ) ( ) ( )2 22 3 2 3x, y , and ,π π π π= − Only exact coordinates of

( ) ( )2 22 3 2 3, and ,π π π π− A1

Do not award this mark if candidate states other

coordinates inside the required range.

[5] 7 marks

50


6. x x ln2y 2 e= =

(a) xln2dy ln2.edx

= xln2dy ln2.edx

= M1

Way 1

Hence x xdy ln2.(2 ) 2 ln2dx

= = AG x2 ln2 AG A1 cso

[2]Aliter

(a) ( )xln y ln 2= leads to ln y x ln2= Takes logs of both sides, then uses the power law of logarithms…

Way 2

1 dy ln2y dx

= … and differentiates implicitly to

give dy1y dx ln2=

M1

Hence xdy y ln2 2 ln2dx

= = AG x2 ln2 AG A1 cso

[2]

M1 (b)

2(x )y 2= 2(x )dy 2x. 2 .ln2dx

⇒ =

2(x )Ax 22(x )2x. 2 .ln2

or 2x.y.ln2 if y is defined A1

When x = 2, 4dy 2(2)2 ln2dx

= Substitutes x = 2 into their dy

dx

which is of the form ± 2( )2 xk

or 2(x )Ax 2

M1

dy 64ln2dx

= 44.3614...= 64ln2 or awrt 44.4 A1

[4] 6 marks

51


Aliter 6. (b) ( )2xln y ln 2= leads to 2ln y x ln2=

Way 2

M1

1 dy 2x.ln2y dx

=

1 dy Ax.ln2y dx

=

1 dy 2x.ln2y dx

= A1

When x = 2, 4dy 2(2)2 ln2dx

= Substitutes x = 2 into their dy

dx

which is of the form ± 2( )2 xk or 2(x )Ax 2

M1

dy 64ln2

dx= 44.3614...= 64ln2 or awrt 44.4 A1

[4]

52


7. = = + + ⇒ =

= = + − ⇒ =

uuur uuur

uuur uuurOA 2 2 OA 3

OB 4 OB 18

a i j k

b i j k

( )BC 2 2= ± + +i j kuuur

⇒ =uuurBC 3

( )AC 4= ± + −i j kuuur

⇒ =uuurAC 18

(a) OC 3 3 3= = + −c i j k

uuur 3 3 3+ −i j k B1 cao

[1]

M1

A1

(b) 2 1OA OB 2 1 2 2 4 0

1 4

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

uuur uuur or…

1 2BO BC 1 2 2 2 4 0

4 1

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

uuur uuur or…

1 2AC BC 1 2 2 2 4 0

4 1

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

uuur uuur or…

2 1AO AC 2 1 2 2 4 0

1 4

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

uuur uuur

An attempt to take the dot product between eitherOA and OB

uuur uuur

OA and ACuuur uuur

, AC and BCuuur uuur

or OB and BCuuur uuur

Showing the result is equal to zero.

and therefore OA is perpendicular to OB and hence OACB is a rectangle.

perpendicular and OACB is a rectangle A1 cso

Using distance formula to find

either the correct height or width. M1

Multiplying the rectangle’s height by its width. M1 Area 3 18 3 18 9 2= × = =

exact value of 3 18 , 9 2 , 162 or aef A1

[6]

(c) ( )12OD 3 3 3= = + −d i j k

uuur ( )1

2 3 3 3+ −i j k B1 [1]

53


using dot product formula M1

(d) ( )51 1

2 2 2DA = ± + +i j kuuur

& ( )3 3 32 2 2DC = ± + −i j k

uuur

or ( )BA 5= ± + +i j kuuur

& ( )OC 3 3 3= ± + −i j kuuur

Identifies a set of two relevant vectors

Correct vectors ± A1 Way 1

dM1

( ) ( ) ( )

0.5 1.50.5 1.5 3 3 152.5 1.5 14 4 4cos D

27 327 27. 42 2

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟•⎜ ⎟ ⎜ ⎟

+ −⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠= ± = ± = ±

Applies dot product formula on multiples

of these vectors.Correct ft.

application of dot product formula.

A1

1 1D cos3

− ⎛ ⎞= −⎜ ⎟⎝ ⎠

Attempts to find the

correct angle D rather than 180 D° − .

ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]Aliter using dot product formula and direction vectors

M1 (d) d ( )BA 5= ± + +i j k

uuur & d ( )OC = ± + −i j k

uuur

Identifies a set of two direction vectors

Correct vectors ± A1 Way 2

dM1

( ) ( ) ( )

1 11 11 5 1 1 5 1cos D

33 . 27 3 . 27

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




A1

1 1D cos3

− ⎛ ⎞= −⎜ ⎟⎝ ⎠



ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]

54


Aliter using dot product formula and similar triangles M1

(d) ( )dOA 2 2= + +i j kuuur

& d ( )OC = + −i j kuuur

Identifies a set of two

direction vectorsCorrect vectors A1

Way 3

dM1

( )1

2

2 12 11 1 2 2 1 1cos D

9 . 3 9 . 3 3

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




A1

1 1D 2 cos3

− ⎛ ⎞= ⎜ ⎟

⎝ ⎠

Attempts to find the correct angle D by

doubling their angle for 1

2 D .ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]

Aliter using cosine rule (d) 51 1

2 2 2DA = + +i j kuuur

, 3 3 32 2 2DC = + −i j k

uuur, AC 4= + −i j kuuur

Way 4

M1 27DA

2=

uuur , 27DC

2=

uuur, AC 18=

uuur

Attempts to find all the lengths of all

three edges of ADC∆

All Correct A1

dM1

( )

2 2227 27 18

2 2 1cos D327 272

2 2

⎛ ⎞ ⎛ ⎞+ −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

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

Using the cosine rule formula with correct

‘subtraction’.Correct ft application

of the cosine rule formula

A1

1 1D cos3

− ⎛ ⎞= −⎜ ⎟⎝ ⎠



ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]

55


Aliter using trigonometry on a right angled triangle (d) 51 1

2 2 2DA = + +i j kuuur

OA 2 2= + +i j kuuur

AC 4= + −i j kuuur

Way 5

M1

Let X be the midpoint of AC

27DA2

=uuur

, 12

3DX OA2

= =uuur uuur

, 1 12 2AX AC 18= =

uuur uuur

(hypotenuse), (adjacent) , (opposite)

Attempts to find two out of the three

lengths in ADX∆

Any two correct A1

dM1

1821

2 272

sin( D) = , 321

2 272

cos( D) = or 1821

2 32

tan( D) =

Uses correct sohcahtoa to find 1

2 DCorrect ft application

of sohcahtoa A1

eg. 18

1 232

D 2 tan−⎛ ⎞⎜ ⎟=⎜ ⎟⎝ ⎠



2 D .ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]

Aliter using trigonometry on a right angled similar triangle OAC (d) OC 3 3 3= + −i j k

uuur OA 2 2= + +i j kuuur

AC 4= + −i j kuuur

M1

Way 6

OC 27=uuur

, OA 3=uuur

, AC 18=uuur

(hypotenuse), (adjacent), (opposite)

Attempts to find two out of the three

lengths in OAC∆

Any two correct A1

dM1

12

18sin( D)27

= , 12

3cos( D)27

= or 12

18tan( D)3

=

Uses correct sohcahtoa to find 1

2 DCorrect ft application

of sohcahtoa A1

eg. 1 18D 2 tan3

− ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠



2 D .ddM1

D 109.47122...= o 109.5° or awrt109° or 1.91c A1

[6]

56


Aliter

7. (b) (i) ( )( )

OC 3 3 3

AB 5

= = ± + −

= ± − − −

c i j k

i j k

uuur

uuur

Way 2

2 2 2 2 2 2OC (3) (3) ( 3) (1) (1) ( 5) AB= + + − = + + − =uuur uuur A complete method of

proving that the diagonals are equal.

M1

As OC AB 27= =

uuur uuur Correct result. A1

then the diagonals are equal, and OACB is a rectangle. diagonals are equal and OACB is a rectangle A1 cso

[3]

= = + + ⇒ =

= = + − ⇒ =

uuur uuur

uuur uuurOA 2 2 OA 3

OB 4 OB 18

a i j k

b i j k

( )BC 2 2= ± + +i j kuuur

⇒ =uuurBC 3

( )AC 4= ± + −i j kuuur

⇒ =uuurAC 18

( )

( )

OC 3 3 3 OC 27

AB 5 AB 27

= = ± + − ⇒ =

= ± − − − ⇒ =

c i j k

i j k

uuur uuur

uuur uuur

Aliter

7. (b) (i)

2 2 2

2 2 2

2 2 2

2 2 2

( ) ( ) ( )or ( ) ( ) ( )or ( ) ( ) ( )or ( ) ( ) ( )

OA AC OCBC OB OCOA OB ABBC AC AB

+ =

+ =

+ =

+ =

or equivalent

Way 3

M1 ( )2

2 2(3) ( 18) 27⇒ + = A complete method of

proving that Pythagoras holds using their values.

Correct result A1

and therefore OA is perpendicular to OB or AC is perpendicular to BC and hence OACB is a rectangle.

perpendicular and OACB is a rectangle A1 cso

[3] 14 marks

57


8. (a)

x 0 1 2 3 4 5 y 1e 2e 7e 10e 13e 4e

or y 2.71828… 7.38906… 14.09403… 23.62434… 36.80197… 54.59815…Either 7e , 10e and 13e

or awrt 14.1, 23.6 and 36.8or e to the power

awrt 2.65, 3.16, 3.61(or mixture of decimals and e’s)

At least two correct B1 All three correct B1 [2]

(b)

( ) }{≈ × × + + + + +7 10 131 2 41I 1 ; e 2 e e e e e2

Outside brackets 1 1

2×

For structure of trapezium rule{ }............. ;

B1; M1

= × = =1 221.1352227... 110.5676113... 110.62

(4sf) 110.6 A1 cao

[3]

Beware: In part (b) candidates can add up the individual trapezia: (b) ( ) ( ) ( ) ( ) ( )≈ + + + + + + + + +7 7 10 0 13 31 2 2 41 1 1 1 1

2 2 2 2 2I .1 e e .1 e e .1 e e .1 e e .1 e e1 1

58


M1 (c)

1 12 21

2dtt (3x 1) .3.(3x 1)dx

−= + ⇒ = +

… or = + ⇒ =2 dtt 3x 1 2t 3dx

12A(3x 1)−+ or =

dd

tt Ax

123

2 (3x 1)−+ or =dt2t 3dx A1

so 1

2

dt 3 3dx 2t2.(3x 1)

= =+

dx 2tdt 3

⇒ = Candidate obtains either dtdx or dx

dt in terms of t …

(3x 1)I e dx+∴ = ∫ t dxe .dtdt

= ∫ t 2te . .dt3

= ∫

… and moves on to substitute this into I to

convert an integral wrt x to an integral wrt t.

dM1

t2

3I t e dt∴ = ∫ t23 t e∫ A1

change limits: when x = 0, t = 1 & when x = 5, t = 4

changes limits x t→ so that 0 1→ and 5 4→

B1

Hence 4

t23

1

I te dt= ∫ ; where 23a 1, b 4, k= = =

[5]

(d) dudt

t tdvdt

u t 1

e v e

= ⇒ =⎧ ⎫⎪ ⎪⎨ ⎬

= ⇒ =⎪ ⎪⎩ ⎭

Let k be any constant for the first three marks of this

part.

M1

( )t t tk t e dt k t e e .1 dt= −∫ ∫

Use of ‘integration by parts’ formula in the

correct direction.Correct expression with a

constant factor k. A1

( )t tk t e e c= − + Correct integration

with/without a constant factor k

A1

( ) ( ){ }∴ = − − −∫4

t 4 4 1 123

1

2te dt 4e e e e3

Substitutes their changed limits into the integrand

and subtracts oe. dM1 oe

4 42

3 (3e ) 2e 109.1963...= = = either 42e or awrt 109.2 A1 [5] 15 marks

• Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark • ddM1 denotes a method mark which is dependent upon the award of the previous two method marks.

59

January 2007 6674 Further Pure Mathematics FP1

Mark Scheme Question Number

Scheme

Marks

1.

(a) Method for finding z : z = 2

6842 −±− , =

2642 i±−

[Completing the square: ( ) izz 161,0161 2 ±−==++ M1,A1

z = – 1 i4± )4,1( ±=−= ba

(b)

Notes (a) First A1 is unsimplified but requires i -1 i8± only scores M1 unless intermediate step seen when M1A1 possible Correct answer with no working is full marks SC: If M0 awarded, k ± 4 i, k + 4 i , k – 4 i scores B1 (Epen M0A0A1) Use of z = a + i b (i) z2 – 2a z + =+ 22 ba z2 +2z + 17 = 0 and compare coefficients M1 =+ 22 ba 17 and a = – 1; z = – 1 i4± A1, A1

(ii) (a + i b)2 + 2(a + i b) + 17 = 0 and compare coefficients M1 41,1720)1(2 22 ±=−=−=+−=+ bandaabaandab A1, A1 (b) Must be a conjugate pair. Allow: Coords marked at points or “correct” numbers on axes.(allow “graduations”) (Ignore any lines drawn)

M1, A1

A1 (3)

B1 √ (1) [4]

60

2.

Attempt to arrange in correct form x

xyxdx

dy cos2=+

Integrating Factor: = e∫ dx

x2

, [ (= e xln2 = e2ln x ) = 2x

[ xxxydxdyx cos22 =+ implies M1M1A1]

∴ dxx

xxyx cos.22 ∫= or equiv.

[ I.F. y = dxRHSscandidateFI )'(..∫ ]

By Parts: ( =)2 yx ∫− dxxxx sinsin i.e. =)( 2 yx )(cos,sin cxxx ++

22

cossinxc

xx

xxy ++=

Notes: First M: At least two terms divided by x. “By parts” M: Must be complete method, e.g xxx dcos2"∫ requires two applications Because of functions involved, be generous with sign, but ∫± dxxxx cossin is M0 (S.C. “Loop” integral like

),""1,cos methodincompletedespitepartsbyofnsapplicatiotwoifMallowdxxe x∫

Final A f.t. for dividing all terms by candidates I.F., providing “c” used.

M1 M1,A1 M1√ M1 A1, A1cao A1√ [8]

61

3.

(a) ( ))35()35.

351

1

2

ii

ipi

zz

−−

++

=

= )34(

3355 pipi +−+ [Multiply out and attempt use of i2 = - 1]

= ipp34

3534

35 −+

+ or ipp34

3334

35 −−

+

(b) For idczz

+=1

2 using 4

tan π=

cd :

[ pp 3535 +=− ] 4=⇒ p

Notes:

In (a) if 2

1

zz used treat as MR. Can score (a)M1M1A0 (b)M1A0

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

+

−+

+

+41)(

153

135

)(22

bip

pp

pa

Allow A1 if answer “all over” 34, real and imag. collected up)

1 + )35()( iibaip ++= : M1 compare real and imag. is first M mark

If denominator in (a) incorrect, both marks in (b) still available

In (b), if use arg z2 - arg z1= 4π :

M1 for 45

3arctanarctan π=−p [ 3258.1....5404.0

4arctan =+=

πp ]

Allow A1 for p = 4 without further work or for that shown in brackets, i.e. assume

values retained on calculator (no penalty because it looks as though not exact)

M1 M1 A1 (3) M1 A1 (2) [5]

62

4.

Working from RHS:

(a) Combining 1

11+

−rr

[ )1(

1+rr

]

Forming single fraction : )1(

)1()1)(1(+

−+++−rr

rrrrr

= )1(1

)1(1)1( 32

++−

=+

+−rr

rrrr

rr AG

Note: For A1, must be intermediate step, as shown

Working from LHS:

(a) )1(

1)1)(1()1(

1)1( 2

++−+

=+

+−rrrrr

rrrr =

)1(11+

+−rr

r M1

Splitting )1(

1+rr

into partial fractions M1

Showing = 1

111)1(

1)1( 2

+−+−=

++−

rrr

rrrr no incorrect working seen A1

Notes:

In first method, second M needs all necessary terms, allowing for sign errors

In second method first M is for division:

Second method mark is for method shown (allow “cover up” rule stated)

If long division, allow reasonable attempt which has remainder constant or linear

function of r.

Setting 1)1(

1)1( 2

++=

++−

rB

rA

rrrr is M0

If 3 or 4 constants used in a correct initial statement,

M1 for finding 2 constants; M1 for complete method to find remaining constant(s)

M1 M1 A1cso (3)

63

(b) ∑∑∑ ⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+−nnn

rrr

111 1111

= ,)(,2

)1( nnn−

+ + ………….. ……. +

⎥⎦

⎤⎢⎣

⎡+

−++−+− )1

11(............)31

21()

211(

nn=

Simplification of method of differences: 1

11+

−n

{ = ])1(

11[2

)1(+

−+−

nnn }

Attempt single fraction: = )1(2

2)1)(1(+

+−+n

nnnn (dep. prev. M1)

= )1(2)1(2

)1( 32

++

++

nnnor

nnn

Alternative: Using Difference method on whole expression:

[0 + 1 – 21 ] + [1 +

21 –

31 ] + [ 2 +

31 –

41 ] …….. [

1111+

−+−nn

n ]

= ( 1 + 2 + 3 ………. + n – 1 ), + [(1 – 1

1+n

)] any form

= 2

)1( −nn ’ {+ 1+n

n }

= )1(2

2)1)(1(+

+−+n

nnnn [Attempt single fraction]

= )1(2)1(2

)1( 32

++

++

nnnor

nnn

Notes:

First M mark is for use of method of differences and attempt at some simplification

First A mark is for simplified result of this method (no more than 2 terms)

Second M mark for attempt at forming single fraction, dependent on first M mark

In alternative first B1 need not be added but need to see 1 2 …….. (n – 1)

B1, B1 M1 A1 depM1 A1 (6) [9] M1 B1, + [A1] B1, depM1 A1

64

Question Number

Scheme

Marks

5.

(a) [ ( )2−>x ]: Attempt to solve )2)(1(312 +−=− xxx ]0734[ 2 =−+ xx

47,1 −= orx

[( )]2−<x : Attempt to solve )2)(1(312 +−−=− xxx Solving 631 +=+ xx (2x2 + 3x – 5 = 0)

25

−=x

(b) 147

<<− x One part

Both correct and enclosed

25

−<x { Must be for x < –2 and only one value}

Notes: “Squaring” in (a) If candidates do not notice the factor of (x – 1)2 they have quartic to solve;

Squaring and finding quartic = 0 [ 0353625188 234 =+−−+ xxxx ]

Finding one factor and factorising 0)35268)(1( 23 =−++− xxxx M1 Finding one other factor and reducing other factor to quadratic, likely to be 0)35348()1( 22 =++− xxx M1 Complete factorisation 0)74)(52()1( 2 =++− xxx M1 [SecondM1 implies the first, if candidate starts there or cancels (x – 1)2] x = 1 B1, x = – 7/4 A1, x = – 5/2 A1 x = 1 allowed anywhere, no penalty in (b) In (b) correct answers seen with no working is independent of (a) (graphical calculator) mark as scheme. Only allow the accuracy mark if no other interval, in both parts

usedtimefirstpenaliseused≤

M1 B1, A1 M1 M1dep A1 (6) M1 A1 B1 √ (3) [9]

65

6.

(a) f(2.0) = – 0.30685……. = – 0.3069 AWRT 3 d.p. f(2.5) = 0.41629……… = 0.4163 both correct 4 d.p. States change of sign, so root (between 2 and 2.5) Note: B1 gained if candidate’s 2 values do show a change of sign and statement made

(b) 5.0x|)5.2(f||)2(f|

|)2(f|)2(+

+=α or |)5.2(||)0.2(|

5.22

ff

=−−

αα or equivalent

Or |)5.2(f|

5.0|)2(f|

xx −= and x found

= 2.212 AWRT (c) f(2.25) = 0.06093……. (≥ 3 d.p.) [ Allow ln.2.25 + 2.25 – 3]

)444.1(9

139414.1)25.2(f,11)(f alloworor

xx &=′+=′

208.2....20781.2,(2.25)f

)25.2(f25.2 ==′

−=α AWRT

(d) f(2.2075) =, { –6.3…. x 10 – 4 }

f(2.2085) = , { 8.1…. x 10 – 4 } ∴ Correct values ( .)fs.1≥ , (root in interval) so root is 2.208 to 3 d.p. Notes:

c) First M in (c) is just for 11+

x

If no intermediate values seen B1M1A1M1A0 is possible for 2.209 or 2.21, otherwise as scheme (B1 eased to award this if not evaluated)

(d) A1 requires values correct ( 1≥ s.f.) and statement (need not say change of sign) M can be given for candidate’s f(2.2075) and f(2.2085)

Allow N-R applied at least twice more, but A1 requires 2.20794 or better and statement MR in (c) 2.5 instead of 2.25 (Answer 2.203) award on ePen B0M1A0M1A1

M1 A1 B1 (3) M1 A1 (2) B1 M1,A1 M1A1 (5) M1 A1 (2) [12]

66

7.

(a) ⇒= − 2xy txx

ty

dd2

dd 3−−= [Use of chain rule; need

dtdx ]

⇒2

4

2

23

2

2

dd6,

dd2

dd

⎟⎠⎞

⎜⎝⎛+−=

−−

txx

txx

ty

( )byd.e.given 4x÷ 31dd6

dd2

2

2

42

2

3 −=⎟⎠⎞

⎜⎝⎛−

xtx

xtx

x

becomes 3)3( 2

2

2

2

=+−=− ydt

ydydt

yd AG

(b) Auxiliary equation: 012 =+m and produce Complementary Function y = … ( tBtAy sincos) += Particular integral: y = 3 ∴ General solution: ( 3sincos) ++= tBtAy

(c) =2

1x

3sincos ++ tBtA

1)34(0,21

=+=⇒== AAtx

Differentiating (to include )dd

tx : – tBtA

dtdxx cossin2 3 +−=−

0)00(0,0dd

=+=⇒== BBttx

t

xsotx cos3

1cos312 +

=+=∴

(d) (Max. value of x when cos t = – 1) so max x = 2

1 or AWRT 0.707

Notes: (See separate sheet for several variations) (a) Second M1 is for attempt at product rule. (be generous)

Final A1 requires all working correct and sufficient “substitution” work (b) Answer can be stated; M1 is implied by correct C.F. stated (allow θ for t) A1 f.t. for candidates CF + PI Allow m2 + m = 0 and m2 -1 = 0 for M1. Marks for (b) can be gained in (c)

(b) Second M : complete method to find other constant (This may involve solving two equations in A and B)

M1 A1√, M1A1 A1 cso (5) M1 A1cao B1 A1√ (4) B1 M1 M1 A1 cao (4) B1 (1) [14]

67

8.

(a) θθθ 3cossin4cos == rx

θθθθ

224 sincos12cos4 −=ddx any correct expression

Solving 0=θd

dx [ 0)sin3cos(cos40 222 =−⇒= θθθθd

dx ]

3

1tan23cos

21sin === θθθ oror

6πθ =⇒ AG

23

6cos

6sin4 2 ==

ππr AG

(b) θ

π

π

drA ∫=4

6

2

21 = θθθ

π

π

d44

6

2 cossin16.21

∫

θθθθθθθ 2sincos2)cossin4(cos2cossin8 2222242 == = (cos2θ + 1) θ2sin2

= 2

4cos12sin2cos 2 θθθ −+ =Answer AG

(c) Area = ⎟⎠

⎞⎜⎝

⎛

⎟⎠

⎞⎜⎝

⎛⎥⎦⎤

⎢⎣⎡ −+

4

6

3

84sin

22sin

61

π

π

θθθ (ignore limits)

= ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−+−⎟⎠

⎞⎜⎝

⎛−+

83

2sin

123sin

61

8sin

82sin

61 33

ππππππ (sub. limits)

= ⎟⎟⎠

⎞⎜⎜⎝

⎛−+−⎟

⎠⎞

⎜⎝⎛ +

163

12163

861 ππ =

24,

61 π

+ both cao

Notes: (a) So many ways x may be expressing e.g

θθθθθθ 4sin)2/1(2sin),2cos1(2sin,cos2sin2 2 ++

leading to many results for θd

dx

Some relevant equations in solving ,0)sin41([ 2 =− θ ,0)3cos4( 2 =−θ 0)tan31( 2 =− θ , cos3 0=θ ]

Showing that 6πθ = satisfies

θddx = 0, allow M1A1 providing

θddx correct

Starting with x = r sin θ can gain M0M1M1 in (a) (b) First M1 for use of double angle formula for sin 2A Second M1 for use of 1cos22cos 2 −= AA Answer given: must be intermediate step, as shown, and no incorrect work

(c) For first M, of the form θθθ 4sin2

2sin 3 ba ±+ (Allow if two of correct form)

On ePen the order of the As in answer is as written

M1 M1A1 M1

A1 cso A1 cso (6) M1 M1 A1 cso (3) M1A1 M1 A1,A1 (5) [14]

68

January 2007 6677 Mechanics M1

Mark Scheme


1. (a) sin 30 24P ° = M1 A1 48P = A1 3 (b) cos30Q P= ° M1 A1 41.6≈ accept 24 3√ , awrt 42 A1 3 6 2. (a) Μ(C) 80 120 0.5x× = × M1 A1 0.75x = cso A1 3 (b) Using reaction at C = 0 B1 Μ(D) 120 0.25 1.25W× = × ft their x M1 A1 24 (N)W = A1 4 (c) ¡ 24 120 144X = + = (N) ft their W M1 A1ft 2 (d) The weight of the rock acts precisely at B. B1 1 10

3. (a) ( ) ( )15 4 3 24

− − +=

i j i ja 3 1.5= −i j M1 A1 2

(b) N2L 6 3m= = −F a i j ft their a M1 A1

( ) ( )2 26 3 6.71 N= + ≈√F accept 45√ , awrt 6.7 M1 A1 4

(c) ( ) ( )6 3 2 3 1.5 6= + + −v i j i j ft their a M1 A1ft

( )121 7 ms−= −i j A1 1 9

69


4. (a) CLM ( )0.3 0.3 2 0.6 5u = × − + × M1 A1 8u = M1 A1 4 (b) ( )0.6 5 3 NsI = × = M1 A1 2 (c) ( )10

35 1.5v u at a a= + ⇒ = × = M1 A1

N2L 100.6 23

R = × = M1 A1 4 10

5. (a) 2 2 2 22 0 21 2 9.8v u as h= + ⇒ = − × × M1 A1 ( )22.5 mh = A1 3 (b) 2 2 2 22 0 2 9.8 24v u as v= + ⇒ = + × × or equivalent M1 A1 ( )470.4=

( )122 msv −≈ accept 21.7 A1 3

(c) 470.4 21 9.8v u at t= + ⇒ − = −√ or equivalent M1 A2 (1, 0)

− 1 each error

( )4.4 st ≈ accept 4.36 A1 4 10

70


6. (a) R �R P 20° 30g Use of F Rµ= B1 cos 20P Rµ° = M1 A1 ¡ sin 20 30R P g+ ° = M1 A1 ( )cos 20 30 sin 20P g Pµ° = − ° M1

0.4 30cos 20 0.4sin 20

gP ×=

°+ ° M1

( )110 N≈ accept 109 A1 8 (b) ¡ 150sin 20 30R g+ ° = M1 A1

( )242.7R ≈

N2L 150cos 20 30R aµ°− = M1 A1

150cos 20 0.4 242.730

a ° − ×≈ M1

= 1.5 (ms-2) accept 1.46 A1 6 14

71


7. (a) N2L Q 2 2g T a− = M1 A1 N2L P 3 sin 30 3T g a− ° = M1 A1 4 (b) 2 3 sin 30 5g g a− ° = M1 a = 0.98 (ms-2) cso A1 2 (c) ( )2T g a= − or equivalent M1

( )18 N≈ accept 17.6 A1 2 (d) The (magnitudes of the) accelerations of P and Q are equal B1 1 (e) ( )2 2 22 2 0.98 0.8 1.568v u as v= + ⇒ = × × = M1

( )11.3 msv −≈ accept 1.25 A1 2 (f) N2L for P 3 sin 30 3g a− ° =

( ) 12

a g= − M1 A1

2 21 12 20 1.568 4.9s ut at t t= + ⇒ = −√ or equivalent M1 A1

( )0.51 st = accept 0.511 A1 5 16

A maximum of one mark can be lost for giving too great accuracy.

72


Mark Scheme

73

74

75

76

77

78


Mark Scheme

Question Number

Scheme

Marks

1.

(a) Maximum speed when accel. = 0 (o.e.)

(b) ( ) =− x30121

xvv

dd (acceln = … + attempt to integrate)

Use of xvv

dd : ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

230

121

2

22 xxv ( + c)

Substituting x = 30, v = 10 and finding c (= 12.5), or limits 2

1212 525 xxv −+= (o.e.)

(a) Allow “acceln > 0 for x < 30, acceln < 0 for x > 30” Also “accelerating for x < 30, decelerating for x > 30” But “acceln < 0 for x > 30” only is B0 (b) 1st M1 will be generous for wrong form of acceln (e.g. dv/dx)! 3rd M1 If use limits, they must use them in correct way with correct values Final A1. Have to accept any expression, but it must be for v2 explicitly (not 1/2v2), and if in separate terms, one can expect like terms to be collected. Hence answer in form as above, or e.g. ( )2

121 60300 xx −+ ; also 2

121 )30(100 x−−

B1 (1) M1 ↓ M1 A1 ↓ M1 A1 (5)

79

2.

A a G h

Height of cone = αtan

a = 3a

Hence h = a4

3

θθ ⇒==34tan

43 aa = 53.1°

1st M1 (generous) allow any trig ratio to get height of cone (e.g. using sin) 3rd M1 For correct trig ratio on a suitable triangle to get θ or complement (even if they call the angle by another name – hence if they are aware or not that they are getting the required angle)

M1 A1 ↓ M1 ↓ M1 A1 (5)

80

3

(a) E.P.E. = 26.321 x

amg =

2

36.3

21

⎟⎠⎞

⎜⎝⎛ a

amg

= 0.2 mga

(b) Friction = mgµ ⇒ work done by friction = ⎟⎠⎞

⎜⎝⎛

34amgµ

Work-energy: gam 2.21 = mgdµ + 0.2 mga (3 relevant terms)

Solving to find µ : 6.0=µ (b) 1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction). 2nd M1: “relevant” terms, i.e. energy or work terms! A1 f.t. on their work done by friction

M1 A1 A1 (3) M1 A1 M1 A1√ ↓ M1 A1 (6)

81

4.

(a) Energy: )cos1(3. 2

21

21 θ+=− mgamvagm

)cos21(2 θ−= agv (o.e.)

(b) avmmgT

2

cos =+ θ

Hence mgT )cos31( θ−= (*) (c) Using T = 0 to find cos θ Hence height above A = a3

4 Accept 1.33a (but must have 3+ s.f.)

(d) agv 3

12 = (o.e.) f.t. using 31cos =θ in v2

consider vert motion: ( ) ghv 2sin 2 =θ (with v resolved) 9

82sin =θ (or θ = 70.53, sin θ = 0.943) and solve for h (as ka) h = a27

4 or 0.148a (awrt)

OR consider energy: ( ) 2

212

21 cos mvmghvm =+θ (3 non-zero terms)

Sub for v, θ and solve for h h = a27

4 or 0.148a (awrt)

M1 A1 A1 (3) M1 A1 A1 cso (3) M1 A1 (2)

B1√ M1 A1 ↓ M1 A1 M1 A1 ↓ M1 A1

82

Question Number

Scheme

Marks

5.

(a) mgT =θcosb 2sin ωθ mrTT =+↔ (3 terms) r = h tanθ

( )θ

θωθθ cos

sinsin1cos

g 2hmm=+ (eliminate r)

⎟⎠⎞

⎜⎝⎛ +

=θθω

sinsin12

hg (*) (solve for 2ω )

(b) ⎟⎠⎞

⎜⎝⎛ += 1

sin12

θω

hg >

hg2 ( θsin < 1) ⇒ ω >

hg2 (*)

(c) ⎟⎠⎞

⎜⎝⎛ +

=θθ

sinsin13

hg

hg ⇒ 2

1sin =θ

mgT =θcos ⇒ mgT3

32= or 1.15mg (awrt)

(a) Allow first B1 M1 A1 if assume different tensions (so next M1 is effectively for eliminating r and T. (b) M1 requires a valid attempt to derive an inequality for ω . (Hence putting sin θ = 1 immediately into expression of ω 2 [assuming this is the critical value] is M0.)

B1 M1 A1 B1 ↓ M1 ↓ M1 A1 (7) M1 A1 (2) M1 A1 ↓ M1 A1 (4)

83

6.

(a) Moments: π xVxxy =∫ d2

1

2 or ∫∫ =2

1

22

1

2 dd xyxxxy

xy d2

1

2∫ = xx

d412

1 4∫ = 2

1312

1⎥⎦⎤

⎢⎣⎡−

x (=

967 ) (either)

xxy d2

1

2∫ = xx

d412

1 3∫ = 2

128

1⎥⎦⎤

⎢⎣⎡−

x (=

323 ) (both)

Solving to find x (= 7

9 ) ⇒ required dist = 72

79 1=− m (*)

(b) H S T

Mass ( )ρ3

21

32

⎟⎠⎞

⎜⎝⎛π , ( )ρ

967π H + S

⎥⎦⎤

⎢⎣⎡= πρ)(

121 ⎥⎦

⎤⎢⎣⎡= πρ)(

325

Dist of CM from base m1619 m

75 x

Moments: ⎥⎦⎤

⎢⎣⎡= πρ)(

121 ⎟

⎠⎞

⎜⎝⎛1619 + ( )ρ

967π

⎟⎠⎞

⎜⎝⎛

75 = ⎥⎦

⎤⎢⎣⎡ πρ)(

325 x

x = 3029 m or 0.967 m (awrt)

Allow distances to be found from different base line if necessary

M1 M1 A1 A1 ↓ M1 A1 cso (6) B1, M1 B1 B1 M1 A1 A1 (7)

84

7.

(a) A T = )05.0(8.0λ = 0.25g

0.8 2.390.05

)g25.0)(8.0(==λ (*)

0.05 x

(b) T = )05.0(8.02.39

+x

mg – T = ma (3 term equn)

0.25g – =+ )05.0(8.02.39 x 0.25 x&& (or equivalent)

x&& = – 196 x

SHM with period 714

22 ππωπ

== s (*)

(c) 14=v √{(0.1) })05.0( 22 − = 1.21(24…) ≈ 1.21 m s–1 (3 s.f.) Accept 7√3/10

(d) Time T under gravity = )1237.0(g

..21.1 s=

Complete method for time T ′ from B to slack.

[↑ e.g. +28π t, where 0.05 = 0.1sin 14t

OR T ′ , where – 0.05 = 0.1 cos 14T ′ ] T'' = 0.1496s Total time = T + T ′ = 0.273 s

(b) 1st M1 must have extn as x + k with k ≠ 0 (but allow M1 if e.g. x + 0.15), or must justify later

For last four marks, must be using x&& (not a)

(c) Using x = 0 is M0 (d) M1 – must be using distance for when string goes slack. Using x = – 0.1 (i.e. assumed end of the oscillation) is M0

M1 A1 (2) M1 M1 A1 A1 ↓ M1 A1 cso (6)

M1 A1√ A1 (3) B1√ M1 A1 A1 A1 (5)

85

January 2007

6683 Statistics S1 Mark Scheme

Question Scheme Marks number

1. (a) (£) 17 Just 17 B1 (1)

(b) 212 and 61t m= =∑ ∑ (Accept as totals under each column in qu.) B1, B1

61 212248510tmS ×

= − , = 1191.8 awrt 1190 or 119 (3sf) M1, A1

983.6 (awrt 984) and 1728.9 (awrt 1730)tt mmS S= = (or 98.4 and 173) A1, A1 (6)

(c) 1191.8983.6 1728.9

r =×

M1, A1f.t.

= 0.913922… awrt 0.914 A1 (3)

(d) 0.914 (Must be the same as (c) or awrt 0.914) B1f.t. (|r| <1)

e.g. linear transformation, coding does not affect coefficient (or recalculate) dB1 (2)

(e) 0.914 suggests longer spent shopping the more spent. (Idea more time, more spent) B1

0.178 different amounts spent for same time. B1 (2)

(f) e.g. might spend short time buying 1 expensive item OR might spend a long time

checking for bargains, talking, buying lots of cheap items. B1g (1)

15 marks

(b) M1 for one correct formula seen, f.t. their ,t m∑ ∑ [Use 1st A1 for 1 correct, 2nd A1 for 2 etc]

(c) M1 for attempt at correct formula, 24852101 5478×

scores M1A0A0

A1ft f.t. their values for ttS etc from (b) but don’t give for ttS = 5478 etc (see above)

Answer only (awrt 0.914) scores 3/3, 0.913 (i.e. truncation) can score M1A1ft by implication.

(d) 2nd B1 dependent on 1st B1 Accept ∑ ∑ ∑ →=== 914.06725,8541,261 2 tmmm

(e) One mark for a sensible comment relating to each coefficient

For 0.178 allow “little or no link between time and amount spent”. Must be in context.

Just saying 0.914 is strong +ve correlation between amount spent and time shopping and

0.178 is weak correlation …scores B0B0.

(f) B1g for a sensible, practical suggestion showing that other factors might affect the amount spent.

E.g. different day (weekend vs weekday) or time of day (time spent queuing if busy)

86


2. (a) 0.03 D (0.0105) Correct tree shape M1 A 0.35 D A, B and C and 0.35 and 0.25 A1 0.06 D (0.015)

0.25 B D (x3) and 0.03, 0.06, 0.05 A1 (3)

D (May be implied by seeing

0.05 D (0.02) P ( )A D∩ etc at the ends)

C

D

(b)(i) P( ) 0.35 0.03A D∩ = × , = 0.0105 or 212000

M1, A1

P(C) = 0.4 (anywhere) B1

(ii) P(D) = (i) + 0.25x 0.06 + (0.4x0.05) M1

= 0.0455 or 912000

A1 (5)

(c) P( ) 0.4 0.05P( ) , = P( ) (ii)C DC D

D∩ ×

= M1, A1ft

= 0.43956… or 4091

0.44 or awrt 0.440 A1 (3)

[Correct answers only score full marks in each part] 11 marks

(a) M1 for tree diagram, 3 branches and then two from each. At least one probability attempted.

(b) 1st M1 for 0.35x0.03. Allow for equivalent from their tree diagram.

B1 for P(C) = 0.4, can be in correct place on tree diagram or implied by 0.4x0.05 in P(D).

2nd M1 for all 3 cases attempted and some correct probabilities seen, including +. Can ft their tree.

Condone poor use of notation if correct calculations seen. E.g. P( | )C D for P ( )C D∩ .

(c) M1 for attempting correct ratio of probabilities. There must be an attempt to substitute some

values in a correct formula. If no correct formula and ration not correct ft score M0.

Writing P(D|C) and attempting to find this is M0.

Writing P(D|C) but calculating correct ratio – ignore notation and mark ratios.

A1ft must have their 0.4 x0.05 divided by their (ii).

If ratio is incorrect ft (0/3) unless correct formula seen and part of ratio is correct then M1.

87


3. (a) N.B. Part (a) doesn’t have to be in a table, could be a list P(X = 1) = …etc B1, B1, B1

x 1 2 3 4 5 6

P(X = x) 136

336

536

736

936

1136

0.0278, 0.0833, 0.139, 0.194, 0.25, 0.306 (Accept awrt 3 s.f) (3)

(b) P(3) + P(4) + P(5) =, 127or

3621 or awrt 0.583 M1, A1 (2)

(c) E(X) = ...3632

361

+×+ , = 16136

or 17364.472 or 4& M1, A1 (2)

(d) ...3632

361)(E 22 +×+=X , = 791

36 or full expression or 36

3521 or awrt 21.97 M1, A1

Var(X) = 2791 161

36 36⎛ ⎞− ⎜ ⎟⎝ ⎠

, = 1.9714… * M1, A1c.s.o. (4)

(e) Var(2 – 3X) = 29 1.97 or ( 3) 1.97× − × , = 17.73 awrt 17.7 or 1442555 M1, A1 (2)

13 marks

(a) 1st B1 for x = 1, … 6 and at least one correct probability N.B. 3 91 136 12 36 4and = =

2nd B1 for at least 3 correct probabilities

3rd B1 for a fully correct probability distribution.

(b) M1 for attempt to add the correct three probabilities, ft their probability distribution

(c) M1 for a correct attempt at E(X). Minimum is as printed. Exact answer only scores M1A1.

[Division by 6 at any point scores M0, no ISW. Non-exact answers with no working score M0.]

(d) 1st M1 for a correct attempt at 2E( )X . Minimum as printed. 79136

or awrt 21.97 scores M1A1.

2nd M1 for ( )22their E( ) their E( )X X− .

2nd A1 cso needs awrt 1.97 and 2791 161

36 36⎛ ⎞− ⎜ ⎟⎝ ⎠

or 25551296

or any fully correct expression seen.

Can accept at least 4 sf for both. i.e. 21.97 for 79136

, 4.472 for 16136

, 20.00 for 2161

36⎛ ⎞⎜ ⎟⎝ ⎠

.

(e) M1 for correct use of Var(aX + b) formula or a full method.

NB 23 1.97− × followed by awrt 17.7 scores M1A1 BUT 23 1.97− × alone, or followed by

– 17.7, scores M0A0.

88


4. (a) Positive skew (both bits) B1 (1)

(b) 19.5 + (60 29) 1043−

× , = 26.7093…. awrt 26.7 M1, A1 (2)

(N.B. Use of 60.5 gives 26.825… so allow awrt 26.8)

(c) 3550 29.5833...120

µ = = or 12729 awrt 29.6 B1

2 2 2138020 138020 or =120 120

σ µ σ µ= − − M1

16.5829... or ( 16.652...)sσ = = awrt 16.6 (or s = 16.7) A1 (3)

(d) 3(29.6 26.7)16.6−

M1A1ft

= 0.52…. awrt 0.520 (or with s awrt 0.518) A1 (3) (N.B. 60.5 in (b) ...awrt 0.499[or with s awrt 0.497]) (e) 0.520 > 0 correct statement about their (d) being >0 or < 0 B1ft So it is consistent with (a) ft their (d) dB1ft (2) (f) Use Median B1 Since the data is skewed or less affected by outliers/extreme values dB1 (2) (g) If the data are symmetrical or skewness is zero or normal/uniform distribution B1 (1)

(“mean =median” or “no outliers” or “evenly distributed” all score B0) 14 marks

(b) M1 for (19.5 or 20) + (60 29) 1043−

× or better. Allow 60.5 giving awrt 26.8 for M1A1

Allow their 0.5n [or 0.5(n+1)] instead of 60 [or 60.5] for M1. (c) M1 for a correct expression for 2 2, , or s sσ σ . NB 2 2274.99 and 277.30sσ = = Condone poor notation if answer is awrt16.6 (or 16.7 for s) (d) M1 for attempt to use this formula using their values to any accuracy. Condone missing 3. 1st A1ft for using their values to at least 3sf. Must have the 3. 2nd A1 for using accurate enough values to get awrt 0.520 (or 0.518 if using s) NB Using only 3 sf gives 0.524 and scores M1A1A0 (e) 1st B1 for saying or implying correct sign for their (d). B1g and B1ft. Ignore “correlation” if seen. 2nd B1 for a comment about consistency with their (d) and (a) being positive skew, ft their (d) only This is dependent on 1st B1: so if (d)>0, they say yes, if (d)<0 they say no. (f) 2nd B1 is dependent upon choosing median.

89


5. (a) Time is a continuous variable or data is in a grouped frequency table B1 (1) (b) Area is proportional to frequency or A ∝ f or A = kf B1 (1) (c) 98.026.3 ×=× M1 dM1 1 child represented by 0.8 A1 cso (3)

(d) (Total) = 240.8

, = 30 M1, A1 (2)

7 marks

(b) 1st B1 for one of these correct statements.

“Area proportional to frequency density” or “Area = frequency” is B0

(c) 1st M1 for a correct combination of any 2 of the 4 numbers: 3.6, 2, 0.8 and 9

e.g. 2

0.8or 0.83.6or 26.3 × etc BUT e.g.

26.3 is M0

2nd M1 dependent on 1st M1 and for a correct combination of 3 numbers leading to 4th.

May be in separate stages but must see all 4 numbers

A1cso for fully correct solution. Both Ms scored, no false working seen and comment required.

(d) M1 for 240.8

seen or implied.

90


6. (a) Used to simplify or represent a real world problem

Cheaper or quicker or easier (than the real situation) or more easily modified (any two lines) To improve understanding of the real world problem B1 Used to predict outcomes from a real world problem (idea of predictions) B1 (2) (b) (3 or 4) Model used to make predictions. (Idea of predicted values based B1 on the model) (4 or 3) (Experimental) data collected B1 (7) Model is refined. B1 (3) 5 marks

(a) 1st B1 For one line

2nd B1 For a second line

Be generous for 1st B1 but stricter for B1B1

(b) 1st & 2nd B1 These two points can be interchanged.

Idea of values from (experimental) data and predicted values based on the model.

1st B1 for predicted values from model e.g. “model used to gain suitable data”

2nd B1 for data collected. Idea of experimental data but “experiment” needn’t be explicitly seen

3rd B1 This should be stage 7. Idea of refinement or revision or adjustment

91


7. (a) P(X < 91 ) = P(Z < 91 10015−

) Attempt standardisation M1

= P(Z < - 0.6) A1

= 1 - 0.7257 M1

= 0.2743 awrt 0.274 A1 (4)

(b) 1 – 0.2090 = 0.7910 0.791 B1

P(X > 100+k) = 0.2090 or P(X < 100+k) = 0.7910 (May be implied) M1

Use of tables to get z = 0.81 B1

15

100100 −+ k ,=0.81 (ft their z = 0.81, but must be z not prob.) M1, A1ft

k = 12 A1 cao (6)

10 marks

(a) 1st M1 for attempting standardisation. 2

(91 )or

µσ σ

−± . Can use of 109 instead of 91.Use of 90.5 etc is M0

1st A1 for – 0.6 (or + 0.6 if using 109)

2nd M1 for 1 – probability from tables. Probability should be > 0.5)

(b) 1st B1 for 0.791 seen or implied.

1st M1 for a correct probability statement, but must use X or Z correctly. Shown on diagram is OK

2nd B1 for awrt 0.81 seen (or implied by correct answer - see below) (Calculator gives 0.80989…)

2nd M1 for attempting to standardise e.g. 100 10015k+ −

or 15k

10015

X − scores 2nd M0 until the 100+ k is substituted to give k, but may imply 1st M1 if k= 112.15 seen

1st A1ft for correct equation for k (as written or better). Can be implied by k = 12.15 (or better)

2nd A1 for k = 12 only.

Answers only

k = 112 or 112.15 or better scores 3/6 (on EPEN give first 3 marks)

k = 12.15 or better (calculator gives 12.148438…) scores 5/6 (i.e loses last A1 only)

k = 12 (no incorrect working seen) scores 6/6

NB Using 0.7910 instead of 0.81 gives 11.865 which might be rounded to 12. This should score no

more than B1M1B0M1A0A0.

92

January 2007 6684 Statistics S2

Mark Scheme

Question Number

Scheme

Marks

1. (a)

(b) (i)

(ii)

A random variable;

a function of known observations (from a population). data OK Yes No

B1 B1

(2)

B1

(1) B1

(1)

Total 4

2. (a)

(b)

P(J≥10) = 1 – P(J ≤ 9) or =1-P(J<10) = 1 – 0.9919 implies method = 0.0081 awrt 0.0081

P ( K ≤1) = P(K = 0) + P(K = 1) both, implied below even with ‘25’ missing

= (0.73)25 + 25(0.73)24(0.27) clear attempt at ‘25’ required = 0.00392 awrt 0.0039 implies M

M1

A1 (2)

M1

M1

A1

(3) Total 5

93

Question Number

Scheme

Marks

3. (a)

(b)

(c)

(d)

Let W represent the number of white plants. W ~ B(12,0.45) use of P(W = 5) = P(W≤5) – P(W≤4) 12C50.4550.557 or equivalent award B1M1 = 0.5269 – 0.3044 values from correct table implies B = 0.2225 awrt 0.222(5)

P(W ≥7) = 1 – P(W ≤6) or =1-P(W<7) = 1 – 0.7393 implies method = 0.2607 awrt 0.261

P( 3 contain more white than coloured)=!7!3!10 (0.2607)3(1 – 0.2607)7 use of B,n=10

= 0.256654… awrt 0.257

mean = np = 22.5 ; var = npq = 12.375

P(W > 25) ≈ P ⎟⎟⎠

⎞⎜⎜⎝

⎛ −>

375.125.225.25Z ± standardise with σ and µ; ±0.5 c.c.

≈ P(Z > 0.8528..) awrt 0.85

≈1 – 0.8023 ‘one minus‘

≈0.1977 awrt 0.197 or 0.198

B1 M1

A1 (3)

M1

A1 (2)

M1A1∫

A1 (3)

B1B1

M1;M1

A1

M1

A1

(7)

Total 15

94

Question Number

Scheme

Marks

4. (a)

(b)

(c)

(d)

(e)

λ > 10 or large µ ok The Poisson is discrete and the normal is continuous. Let Y represent the number of yachts hired in winter P (Y<3) = P(Y ≤ 2) P(Y ≤ 2) & Po(5) = 0.1247 awrt 0.125 Let X represent the number of yachts hired in summer X~Po(25). N(25,25) all correct, can be implied by standardisation below

P(X > 30) ≈ ⎟⎠⎞

⎜⎝⎛ −

>5

255.30ZP ± standardise with 25 & 5; ±0.5 c.c.

≈ P(Z > 1.1) 1.1 ≈ 1 – 0.8643 ‘one minus’ ≈ 0.1357 awrt 0.136 no. of weeks = 0.1357 x 16 ANS (d)x16

= 2.17 or 2 or 3 ans>16 M0A0

B1

(1) B1

(1)

M1

A1 (2)

B1

M1;M1

A1

M1

A1 (6)

M1

A1∫ (2)

Total 12

95

Question Number

Scheme

Marks

5.

(a)

(b)

(c)

(d)

(e)

f(x) = ⎪⎩

⎪⎨⎧ <<

−otherwise. , 0

, , 1 βααβ

x function including inequality, 0 otherwise

22

α β+= ,

853

=−−αβα

or equivalent

24534=+

=+βα

βα

3(4 ) 5 242 12

β ββ− + ==

attempt to solve 2 eqns

β = 6 α = -2 both

E(X) = 752

0150=

+cm 75

Standard deviation = 2)0150(121

−

= 43.30127…cm 25 3 or awrt 43.3

P(X < 30) + P(X > 120) = 15030

15030

+ 1st or at least one fraction, + or double

=60 2 or or 0.4 or equivalent fraction

150 5

B1,B1

(2)

B1,B1

M1

A1 (4)

B1 (1)

M1

A1 (2)

M1,M1

A1

(3)

Total 12

96

Question Number

Scheme

Marks

6. (a)

(b)

(c)

H0 : p = 0.20, H1: p < 0.20 Let X represent the number of people buying family size bar. X ~ B (30, 0.20) P(X ≤ 2) = 0.0442 or P(X ≤ 2) = 0.0442 awrt 0.044 P(X ≤ 3) = 0.1227 CR X ≤ 2 0.0442 < 5%, so significant. Significant There is evidence that the no. of family size bars sold is lower than usual.

H0 : p = 0.02, H1: p ≠ 0.02 4λ = etc ok both Let Y represent the number of gigantic bars sold. Y~ B (200, 0.02) ⇒Y~ Po (4) can be implied below P(Y = 0) = 0.0183 and P ( Y ≤ 8) = 0.9786⇒P(Y ≥ 9) = 0.0214 first, either Critical region Y = 0 U Y ≥ 9 0Y ≤ ok N.B. Accept exact Bin: 0.0176 and 0.0202 Significance level = 0.0183 + 0.0214 = 0.0397 awrt 0.04

B1,B1

M1A1

M1

A1 (6)

B1

M1

B1,B1

B1,B1

B1 (1)

Total 13

97

Question Number

Scheme

Marks

7. (a)

(b)

(c)

(d)

(e)

(f)

1 – F(0.3) = 1 – (2 × 0.32 – 0.33) ‘one minus’ required = 0.847 F(0.60) = 0.5040 F(0.59) = 0.4908 both required awrt 0.5, 0.49 0.5 lies between therefore median value lies between 0.59 and 0.60.

f(x) = ⎩⎨⎧ ≤≤+−

otherwise. , 0,10 ,43 2 xxx attempt to differentiate, all correct

xxxxxx d43d)f(1

0231

0 ∫∫ +−= attempt to integrate f( )x x

= 1

0

34

34

43

⎥⎦

⎤⎢⎣

⎡+

− xx sub in limits

=127

or 0.583& or 0.583 or equivalent fraction

046d

)df(=+−= x

xx

attempt to differentiate f(x) and equate to 0

2 or 0.6 or 0.6673

x = &

mean < median < mode, therefore negative skew. Any pair, cao

M1 A1

(2)

M1A1

B1 (3)

M1A1

(2)

M1

M1

A1

(3)

M1

A1

(2)

B1,B1 (2)

Total 14

98

99

100

101

102

103

Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone 01623 467467 Fax 01623 450481 Email [email protected] Order Code UA 018759 January 2007 For more information on Edexcel qualifications, please visit www.edexcel.org.uk/qualifications Alternatively, you can contact Customer Services at www.edexcel.org.uk/ask or on 0870 240 9800 Edexcel Limited. Registered in England and Wales no.4496750 Registered Office: One90 High Holborn, London, WC1V 7BH

ALL Alevel edexcel maths Mark Schemes Jan 07

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