2010 Jun As Mark Scheme 31165

GCE MARKING SCHEME

MATHEMATICS AS/Advanced SUMMER 2010

INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 2010 examination in GCE MATHEMATICS. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

Paper

Page

C1

1

C2

7

C3

12

C4

17

FP1

21

FP2

27

FP3

31

C1 1.

(a)

(i)

(ii)

Gradient of AC = increase in y increase in x Gradient of AC = 4/3

M1 (or equivalent)

A correct method for finding the equation of AC using candidate’s gradient for AC Equation of AC: y – 4 = 4/3[x – (– 6)] (or equivalent) (f.t. candidate’s gradient for AC) Equation of AC: 4x – 3y + 36 = 0 (convincing)

A1

M1 A1 A1

(iii)

M1⎫ ⎧Gradient of BD = increase in y ⎩ increase in x ⎭ (to be awarded only if corresponding M1 is not awarded in part (i)) Gradient of BD = – 3/4 (or equivalent) A1 An attempt to use the fact that the product of perpendicular lines = –1 (or equivalent) M1 Gradient AC × Gradient BD = –1 ⇒ AC, BD perpendicular A1

(iv)

⎧A correct method for finding the equation of BD using the ⎫ ⎩candidate’s gradient for BD M1⎭ (to be awarded only if corresponding M1 is not awarded in part (ii)) Equation of BD: y – 11 = – 3/4[x – (– 7)] (or equivalent) (f.t. candidate’s gradient for BD) A1

Note: Total mark for part (a) is 9 marks

(b)

(i)

An attempt to solve equations of AC and BD simultaneouslyM1 x = – 3, y = 8 (convincing) A1

(ii)

A correct method for finding the length of BE BE = 15

1

M1 A1

2.

(a)

5√7 – √3 = (5√7 – √3)(√7 + √3) √7 – √3 (√7 – √3)(√7 + √3)

M1

Numerator: 5 × 7 + 5 × √7 × √3 – √7 × √3 – 3 A1 Denominator: 7–3 A1 5√7 – √3 = 8 + √21 (c.a.o.) A1 √7 – √3 Special case If M1 not gained, allow B1 for correctly simplified numerator or denominator following multiplication of top and bottom by √7 – √3 (b)

3.

(a)

√15 × √20 = 10√3 √75 = 5√3 √60 = 2√3 √5 (√15 × √20) – √75 – √60 = 3√3 √5

B1 B1 B1 (c.a.o.) B1

dy = 2x – 8 dx An attempt to differentiate, at least one non-zero term correct) An attempt to substitute x = 3 in candidate’s expression for dy dx Value of dy at P = – 2 dx Gradient of normal =

M1 m1

(c.a.o.) A1

–1 m1 candidate’s value for dy dx Equation of normal to C at P: y – (– 5) = 1/2(x – 3) (or equivalent) (f.t. candidate’s value for dy provided M1 and both m1’s awarded) A1 dx (b)

Putting candidate’s expression for dy = 4 M1 dx x-coordinate of Q = 6 A1 y-coordinate of Q = – 2 A1 c = – 26 A1 (f.t. candidate’s expression for dy and at most one error in the dx enumeration of the coordinates of Q for all three A marks provided both M1’s are awarded)

2

4.

5.

6.

7.

(a)

(1 + x)6 = 1 + 6x + 15x2 + 20x3 + . . . All correct Three correct

B2 B1

(b)

An attempt to substitute x = – 0·01(or x = – 0·1) in the expansion of part (a) (f.t. candidate’s coefficients from part (a)) M1 (0·99)6 ≈ 1 – 6 × 0·01 + 15 × 0·0001 – 20 × 0·000001 (At least three correct, f.t. candidate’s coefficients from part (a)) A1 6 (correct to four decimal places) (0·99) = 0·94148 = 0·9415 (c.a.o.) A1

(a)

a=2 b=3 c = – 25

(b)

6x2 + 36x – 17 = 3 [a(x + b)2 + c] + k (k ≠ 0, candidate’s a, b, c) M1 Least value = 3c + 4 (candidate’s c) A1

(a)

An expression for b2 – 4ac, with at least two of a, b or c correct M1 b2 – 4ac = k 2 – 4 × 2 × 18 A1 2 m1 Candidate’s expression for b – 4ac < 0 –12 < k < 12 (c.a.o.) A1

(b)

Finding critical values x = – 0·5, x = 0·6 B1 A statement (mathematical or otherwise) to the effect that x ≤ – 0·5 or 0·6 ≤ x (or equivalent) (f.t. only x = ± 0·5, x = ± 0·6) B2 Deduct 1 mark for each of the following errors the use of < rather than ≤ the use of the word ‘and’ instead of the word ‘or’

(a)

y + δy = – (x + δx)2 + 5(x + δx) – 9 Subtracting y from above to find δy δy = – 2xδx – (δx)2 + 5δx Dividing by δx and letting δx → 0 dy = limit δy = – 2x + 5 dx δx→ 0 δx

(b)

dy = 3 × 1 × x–2/3 + (– 2) × 12 × x–3 B1, B1 dx 4 3 Either 8–2/3 = 1 or second term = (–) 24 (or equivalent fraction) B1 4 512 (c.a.o.) B1 dy = 1 (or equivalent ) dx 64

B1 B1 B1

3

B1 M1 A1 M1 (c.a.o.) A1

8.

(a)

Use of f (–2) = 0 – 96 + 4k + 26 – 6 = 0 ⇒ k = 19 Special case Candidates who assume k = 19 and show f (–2) = 0 are awarded

M1 A1 B1

(b)

f (x) = (x + 2)(12x2 + ax + b) with one of a, b correct M1 f (x) = (x + 2)(12x2 – 5x – 3) A1 2 f (x) = (x + 2)(4x – 3)(3x + 1)(f.t. only 12x + 5x – 3 in above line) A1 Special case Candidates who find one of the remaining factors, (4x – 3) or (3x + 1), using e.g. factor theorem, are awarded B1

(c)

Attempting to find f (1/2) M1 Remainder = – 25 A1 4 If a candidate tries to solve (c) by using the answer to part (b), f.t. when candidate’s expression is of the form (x + 2) × two linear factors

4

9.

(a)

y

(1, 6)

(–4, 0)

(6, 0)

O

x

Concave down curve with maximum at (1, a), a ≠ 3 Maximum at (1, 6) Both points of intersection with x-axis

B1 B1 B1

(b) y

(–1, 3)

x

(– 6, 0)

O

(4, 0)

Concave down curve with maximum at (b, 3), b ≠ 1 Maximum at (–1, 3), Both points of intersection with x-axis

5

B1 B1 B1

10.

(a)

dy = 3x2 – 6 dx 2 Putting derived dy = 0 dx x = – 2, 2 (both correct) (f.t. candidate’s dy) dx

B1 M1 A1

Stationary points are (– 2, 11) and (2, – 5) (both correct) (c.a.o.) A1

A correct method for finding nature of stationary points yielding either (– 2, 11) is a maximum point or (2, – 5) is a minimum point (f.t. candidate’s derived values) M1 Correct conclusion for other point (f.t. candidate’s derived values) A1

6

C2

1.

1 1·25 1·5 1·75 2

1·414213562 1·341640787 1·290994449 1·253566341 1·224744871

(5 values correct) (3 or 4 values correct)

Correct formula with h = 0·25 I ≈ 0·25 × {1·414213562 + 1·224744871 + 2 2(1·341640787 + 1·290994449 + 1·253566341)} I ≈ 10·41136159 ÷ 8 I ≈ 1·301420198 I ≈ 1·301 (f.t. one slip)

B2 B1 M1

A1

Special case for candidates who put h = 0·2 1 1·414213562 1·2 1·354006401 1·4 1·309307341 1·6 1·274754878 1·8 1·247219129 2 1·224744871 (all values correct) B1 Correct formula with h = 0·2 M1 I ≈ 0·2 ×{1·414213562 + 1·224744871 + 2(1·354006401 + 1·309307341 + 2 1·274754878 + 1·247219129)} I ≈ 13·00953393 ÷ 10 I ≈ 1·300953393 I ≈ 1·301 (f.t. one slip) A1 Note: Answer only with no working earns 0 marks 2.

(a)

12(1 – sin 2θ ) – 5 sin θ = 10 (correct use of cos2θ = 1 – sin 2θ ) M1 An attempt to collect , form and solve quadratic equation in sin θ, either by using the quadratic formula or by getting the expression into the form (a sin θ + b)(c sin θ + d), with a × c = coefficient of sin2θ and b × d = constant m1 2 12 sin θ + 5 sin θ – 2 = 0 ⇒ (4 sin θ – 1)(3 sin θ + 2) = 0 ⇒ sin θ = 1, sin θ = – 2 (c.a.o.) A1 4 3 θ = 14·48°, 165·52° B1 θ = 221·81°, 318·19° B1 B1 Note: Subtract 1 mark for each additional root in range for each branch, ignore roots outside range. sin θ = +, –, f.t. for 3 marks, sin θ = –, –, f.t. for 2 marks sin θ = +, +, f.t. for 1 mark

7

3.

(b)

2x = – 58°, 122°, 302° (at least one value) B1 x = 61°, 151°, (both values) B1 Note: Subtract a maximum of 1 mark for additional roots in range, ignore roots outside range.

(c)

sin φ + 2 sin φ cos φ = 0 or tan φ + 2 tan φ cos φ = 0 M1 or sin φ⎧ 1 + 2⎫ = 0 ⎩cos φ ⎭ (both values) A1 sin φ = 0 (or tan φ = 0) , cos φ = –1 2 φ = 0°, 180° (both values) A1 φ = 120° A1 Note: Subtract a maximum of 1 mark for each additional root in range for each branch, ignore roots outside range. Special Case: No factorisation but division throughout by sin φ (or tan φ) to yield 1 + 2 cos φ = 0 (or equivalent) M1 φ = 120° A1

(a)

1 × AC × 11 × sin 110° = 31 2 (substituting the correct values in the correct places in the area formula) AC = 5·998 (AC = 6) BC2 = 62 + 112 – 2 × 6 × 11 × cos 110° (substituting the correct values in the correct places in the cos rule, (f.t. candidate’s value for AC) BC = 14·22 (f.t. candidate’s value for AC)

(b)

4.

sin XZY = sin 60° 2 2√3 – 1 sin XZY = 2 × sin 60° 2√3 – 1 sin XZY = 6 + √3 11

3 × x3/2 – 6 × x–3 – x + c 3/2 –3

M1 A1

M1 A1

(substituting the correct values in the correct places in the sine formula) M1 m1 A1

B1, B1, B1 (–1 if no constant term present)

8

5.

6.

(a)

Sn = a + [a + d] + . . . + [a + (n – 1)d] (at least 3 , one at each end) B1 Sn = [a + (n – 1)d] + [a + (n – 2)d] + . . . + a Either: 2Sn = [a + a + (n – 1)d] + [a + a + (n – 1)d] + . . . + [a + a + (n – 1)d] Or 2Sn = [a + a + (n – 1)d] + (n times) M1 2Sn = n[2a + (n – 1)d] Sn = n[2a + (n – 1)d] (convincing) A1 2

(b)

n[2 × 4 + (n – 1) × 2]= 460 M1 2 Either: Rewriting above equation in a form ready to be solved 2n2 + 6n – 920 = 0 or n2 + 3n – 460 = 0 or n(n + 3) = 460 or: n = 20, n = – 23 A1 n = 20 (c.a.o.) A1

(c)

a + 4d = 9 (a + 5d) + (a + 9d) = 42 An attempt to solve the candidate’s two linear equations Simultaneously by eliminating one unknown d=4 (c.a.o.) a=–7 (f.t. candidate’s value for d)

(a)

(b)

r = – 0·6 S∞ = 40 1 – (– 0·6) S∞ = 25 (i)

(ii)

7.

B1 B1 M1 A1 A1

B1 M1 (c.a.o.) A1

B1 ar3 = 8 ar2 + ar3 + ar4 = 28 B1 An attempt to solve these equations simultaneously by eliminating a M1 3 2 r = 8 ⇒ 2r – 5r + 2 = 0 (convincing) A1 r2 + r3 + r4 28 r = 0·5 (r = 2 discarded, c.a.o.) B1 a = 64 (f.t. candidate’s value for r, provided | r | < 1) B1

3

Area = ⌠⎧3x + x3 ⎫ dx ⌡⎩ 5 ⎭

(use of integration)

3x2 + x4 2 4×5 Area = (27/2 + 81/20) – (3/2 + 1/20)

(correct integration)

Area = 16

9

M1

B1, B1

(an attempt to substitute limits) m1 (c.a.o.) A1

8.

(a)

(b)

Let p = logax Then x = ap xn = apn ∴loga xn = pn ∴loga xn = pn = n logax

(relationship between log and power) B1 (the laws of indices) B1 (relationship between log and power) (convincing) B1

Either: (2y – 1) log 10 6 = log 10 4 (taking logs on both sides and using the power law) M1 A1 y = log 10 4 + log 10 6 2 log 10 6 y = 0·887 (f.t. one slip, see below) A1 Or: (rewriting as a log equation) 2y – 1 = log 6 4 y = log 6 4 + 1 2 y = 0·887 (f.t. one slip, see below) Note: an answer of y = – 0·113 from y = log 10 4 – log 10 6 2 log 10 6 earns M1 A0 A1 an answer of y = 1·774 from y = log 10 4 + log 10 6 log 10 6 earns M1 A0 A1

(c)

log a4 = 1 ⇒ 4 = a1/2 2 a = 16

M1 A1 A1

(rewriting log equation as power equation) M1 A1

10

9.

(a)

A(4 , –1) A correct method for finding radius Radius = √10

(b)

(i)

(ii)

10.

B1 M1 A1

Either: An attempt to substitute the coordinates of P in the equation of C M1 Verification that x = 7, y = – 2 satisfy equation of C and hence P lies on C A1 Or: An attempt to find AP2 M1 2 A1 AP = 10 ⇒ P lies on C A correct method for finding Q M1 Q(1, 0) (f.t. candidate’s coordinates for A) A1

(c)

An attempt to substitute (2x – 4) for y in the equation of the circle M1 x2 – 4x + 3 = 0 (or 5x2 – 20x + 15 = 0) A1 x = 1, x = 3 (correctly solving candidate’s quadratic, both values) A1 Points of intersection are (1, – 2), (3, 2) (c.a.o.) A1

(a)

(i) (ii)

(b)

An attempt to eliminate θ Use of R2 – r2 = (R + r)(R – r) r = 2K – R L

L = Rθ – rθ K = 1 R2θ – 1 r2θ 2 2

B1 B1

11

M1 m1 A1

C3 1.

0 0·2 0·4 0·6 0·8

0·5 0·401312339 0·310025518 0·231475216 0·167981614

(3 values correct) (5 values correct)

B1 B1

Correct formula with h = 0·2 I ≈ 0·2 × {0·5 + 0·167981614 + 4(0·401312339 + 0·231475216) 3 + 2(0·310025518)} I ≈ 0·2 × 3·819182871 ÷ 3 I ≈ 0·254612191 I ≈ 0·2546

M1

(f.t. one slip)

A1

Note: Answer only with no working earns 0 marks 2.

3.

(a)

e.g. θ = π 2 cos θ + cos 4θ = 1 (choice of θ and one correct evaluation) cos 2θ + cos 3θ = –1 (both evaluations correct but different)

B1 B1

(b)

2 (sec2θ – 1) = sec θ + 8 (correct use of tan 2θ = sec2θ – 1) M1 An attempt to collect , form and solve quadratic equation in sec θ, either by using the quadratic formula or by getting the expression into the form (a sec θ + b)(c sec θ + d), with a × c = coefficient of sec2θ and b × d = constant m1 2 2 sec θ – sec θ – 10 = 0 ⇒ (2 sec θ – 5)( sec θ + 2) = 0 ⇒ sec θ = 5 , sec θ = – 2 2 (c.a.o.) A1 ⇒ cos θ = 2 , cos θ = – 1 5 2 θ = 66·42°, 293·58° B1 θ = 120·0°, 240·0° B1 B1 Note: Subtract 1 mark for each additional root in range for each branch, ignore roots outside range. cos θ = +, –, f.t. for 3 marks, cos θ = –, –, f.t. for 2 marks cos θ = +, +, f.t. for 1 mark

(a)

d(y4) = 4y3 dy dx dx 2 d(4x y) = 4x2 dy + 8xy dx dx 3 d(3x – 5x) = 9x2 – 5 dx dy = 9x2 – 5 – 8xy dx 4y3 + 4x2

B1 B1 B1 (c.a.o.) B1

12

(b)

4.

dx = 4 – 2 sin 2t, dt dy = 3 cos 3t dt Use of dy = dy ÷ dx dx dt dt Substituting π for t in expression for dy 12 dx dy = 1 dx √2

f (x) = 4x3 – 2x – 5 An attempt to check values or signs of f (x) at x = 1, x = 2 f (1) = – 3 < 0, f (2) = 23 > 0 Change of sign ⇒ f (x) = 0 has root in (1, 2) x0 = 1·2 x1 = 1·227601026 (x1 correct, at least 5 places after the point) x2 = 1·230645994 x3 = 1·230980996 x4 = 1·231017841 = 1·23102 (x4 correct to 5 decimal places) An attempt to check values or signs of f (x) at x = 1·231015, x = 1·231025 f (1·231015) = – 1·197 × 10– 4 < 0, f (1·231025) = 4·218 × 10– 5 > 0 Change of sign ⇒ α = 1·23102 correct to five decimal places

B1 B1 M1 m1 A1

M1 A1 B1

B1 M1 A1 A1

Note: ‘Change of sign’ must appear at least once. 5.

(a)

(i)

(ii)

(iii)

dy = 13 × (7 + 2x)12 × f (x), (f (x) ≠1) dx dy = 26 × (7 + 2x)12 dx 5 or 1 or 5 dy = dx √1 – (5x)2 √1 – (5x)2 √1 – 5x2 dy = 5 dx √1 – 25x2 dy = x3 × f (x) + e4x × g(x) dx dy = x3 × f (x) + e4x × g(x) (either f (x) = 4e4x or g(x) = 3x2) dx dy = x3 × 4e4x + e4x × 3x2 (all correct) dx

13

M1 A1 M1 A1 M1 A1 A1

(b)

6.

(a)

d (tan x) = cos x × m cos x – sin x × k sin x (m = ±1, k = ±1) dx cos2 x d (tan x) = cos x × cos x – sin x × (– sin x ) dx cos2 x 2 d (tan x) = cos x + sin2 x dx cos2 x d (tan x) = 1 = sec2 x (convincing) 2 dx cos x

(i)

(ii)

(iii)

(b)

⌠(7x – 9)1/2 dx = k × (7x – 9)3/2 + c (k = 1, 7, 1/7) ⌡ 3/2 1/2 1 ⌠(7x – 9) dx = /7 × (7x – 9)3/2 + c ⌡ 3/2 x/6 x/6 ⌠e dx = k × e + c (k = 1, 6, 1/6) ⌡ ⌠ex/6dx = 6 × ex/6 + c ⌡ ⌠ 4 dx = 4 × k × ln | 5x – 1 | + c (k = 1, 5, 1/5) ⌡5x – 1 ⌠ 4 dx = 4 × 1/5 × ln | 5x – 1 | + c ⌡5x – 1

⌠(3x – 4)–3 dx = k × (3x – 4)–2 ⌡ –2

(k = 1, 3, 1/3)

4

M1 A1

A1

M1 A1 M1 A1 M1 A1

M1

4

⌠8 × (3x – 4) dx = ⎡8 × /3 × (3x – 4) ⎤ ⌡ ⎣ –2 ⎦ –3

1

–2

2

A1

2

Correct method for substitution of limits

M1

4

⌠8 × (3x – 4)–3 dx = 5 = 0·3125 ⌡ 16 2

14

(f.t. for k = 1, 3 only)

A1

7.

(a)

Trying to solve either 3x + 1 ≤ 5 or 3x + 1 ≥ – 5 3x + 1 ≤ 5 ⇒ x ≤ 4/3 3x + 1 ≥ – 5 ⇒ x ≥ – 2 (both inequalities) 4 Required range: – 2 ≤ x ≤ /3 (f.t. one slip)

M1 A1 A1

Alternative mark scheme (3x + 1)2 ≤ 25 (forming and trying to solve quadratic) M1 4 A1 Critical points x = – 2 and x = /3 Required range: – 2 ≤ x ≤ 4/3 (f.t. one slip in critical points) A1 (b)

(i) y y = f (x)

x

O

Correct graph

B1

(ii) y y = f (x – 3) + 2 (0, 5)

(3, 2) x

O

Translation of graph of f (x) = | x | with vertex at (± 3, ±2) Coordinates of vertex = (3, 2) Crosses y-axis at (0, 5)

15

M1 A1 A1

8.

(a)

(b)

g´(x) = 3 × f (x) + 2 f (x) ≠ 1 4x2 + 9 g´(x) = 3 × 8x + 2 4x2 + 9 g´(x) = 24x + 8x2 + 18 = 2(2x + 3)2 4x2 + 9 4x2 + 9 (i)

(ii)

9.

(a)

(b)

10.

M1 A1 (convincing)

A1

At stationary point, 2(2x + 3)2 = 0 4x2 + 9 or 3 × 8x + 2 = 0 4x2 + 9 2(2x + 3)2 = 0 only when x = – 3/2 4x2 + 9

M1

g´(x) > 0 either side of x = – 3/2 (or at all other points) Stationary point is a point of inflection

M1 A1

A1

y – 5 = ln (3x – 2) B1 An attempt to express candidate’s equation as an exponential equation M1 x = (ey – 5 + 2) (f.t. one slip) A1 3 (f.t. one slip) A1 f -1(x) = (ex – 5 + 2) 3 D(f -1) = [5, ∞) B1

(a)

R(f) = [1, ∞) R(g) = [– 3, ∞)

B1 B1

(b)

gf(x) = 2√(x + 4)2 – 3. gf(x) = 2x + 5

M1 A1

(c)

fg(x) = √(2x2 – 3 + 4) [fg(x)]2 = 172 x2 = 144 x = ±12

(correct composition) (candidate’s fg(x)) (f.t. one numerical slip) (c.a.o.)

16

B1 M1 A1 A1

C4

1.

(a)

f (x) ≡A + B + C (correct form) M1 2 x (x – 2) (x – 2) 8 – x – x2 ≡ A(x – 2)2 + Bx(x – 2) + Cx (correct clearing of fractions and genuine attempt to find coefficients) m1 A = 2, C = 1, B = – 3 (2 coefficients, c.a.o.) A1 (third coefficient, f.t. one slip in enumeration of other 2 coefficients) A1

(b)

2 f ´(x) = –2 + 3 – x2 (x – 2)2 (x – 2)3 f ´(1) = 3

2.

(at least one of first two )B1 (third term) B1 (f.t. candidates values for A, B, C) (c.a.o.) B1

⎧4x dy + 4y⎫ ⎩ dx ⎭ ⎧– 3y2 dy ⎫ ⎩ dx ⎭

10x + 4x dy + 4y – 3y2 dy = 0 dx dx

B1 B1

dy = 1 (o.e.) (c.a.o.) B1 dx 4 Use of gradnormal × gradtangent = –1 M1 Equation of normal: y – (– 2) = – 4(x – 1) ⎧f.t. candidate’s value for dy ⎫ A1 ⎩ dx ⎭ 3.

(a)

2(2 cos2θ – 1) = 9 cos θ + 7

(correct use of cos 2θ = 2 cos 2θ – 1) M1

An attempt to collect , form and solve quadratic equation in cos θ, either by using the quadratic formula or by getting the expression into the form (a cos θ + b)(c cos θ + d), with a × c = coefficient of cos2θ and b × d = constant m1 2 4 cos θ – 9 cos θ – 9 = 0 ⇒ (4 cos θ + 3)(cos θ – 3) = 0 ⇒ cos θ = –3, (cos θ = 3) (c.a.o.) A1 4 θ = 138·59°, 221·41° B1 B1 Note: Subtract (from final two marks) 1 mark for each additional root in range from 4 cos θ + 3 = 0, ignore roots outside range. cos θ = – , f.t. for 2 marks, cos θ = +, f.t. for 1 mark

17

(b)

(i)

R = 13 B1 Correctly expanding sin (x – α) and using either 13 cos α = 5 or 13 sin α = 12 or tan α = 12 to find α 5 (f.t. candidate’s value for R) M1 α = 67·38° (c.a.o) A1 Least value of 1 = 1 . 5 sin x – 12 cos x + 20 13 × (±1) + 20 (f.t. candidate’s value for R) M1 (f.t. candidate’s value for R) A1 Least value = 1 33 Corresponding value for x = 157·38° (o.e.) (f.t. candidate’s value for α ) A1

(ii)

4.

π/3

Volume = π⌠sin2x dx ⌡

B1

π/6

Use of sin2x = (±1 ± cos 2x) M1 2 Correct integration of candidate’s (±1 ± cos 2x) A1 2 Correct substitution of correct limits in candidate’s integrated expression M1 Volume = π2 = 0·822(467...) 12

5.

(c.a.o.)

⎧1 – x ⎫1/2 = 1 – x – x2 ⎩ 4⎭ 8 128

⎧1–x⎫ ⎩ 8⎭ 2 ⎧– x ⎫ ⎩ 128 ⎭

| x | < 4 or – 4 < x < 4 √3 ≈ 1 – 1 – 1 2 8 128 √3 ≈ 111 . 64

A1

B1 B1

B1 (f.t. candidate’s coefficients) B1 (convincing) B1

18

6.

(a)

(b)

7.

(a)

(b)

Use of dy = dy ÷ dx and at least one of dx = – 2 , dy = 4 correct dx dt dt dt t2 dt dy = – 2t2 (o.e.) dx Equation of tangent at P: y – 4p = – 2p2⎧x – 2⎫ ⎩ p⎭ (f.t. candidate’s expression for dy) dx (convincing) y = – 2p2x + 8p

M1

Substituting x = 2, y = 3 in equation of tangent 4p2 – 8p + 3 = 0 p=1,3 (both values, c.a.o.) 2 2 Points are (4, 2), (4/3, 6) (f.t. candidate’s values for p)

M1 A1 A1

⌠ x3 ln x dx = f (x) ln x – ⌠f (x) g(x) dx ⌡ ⌡ 4 f (x) = x , g(x) = 1 4 x 3 4 ⌠ x ln x dx = x ln x – x4 + c ⌡ 4 16

M1

A1

m1 A1

A1

A1, A1 (c.a.o.)

A1

⌠x(2x – 3)4 dx = ⌠ f (u) × u4 × k du ⌡ ⌡ (f (u) = pu + q, p≠ 0, q ≠ 0 and k = 1/2 or 2) M1 ⌠x(2x – 3)4 dx = ⌠(u + 3) × u4 × du ⌡ ⌡ 2 2 5 4 6 5 ⌠(au + bu ) du = au + bu ⌡ 6 5

A1 (a ≠ 0, b ≠ 0)

B1

Either: Correctly inserting limits of –1, 1 in candidate’s au6 + bu5 6 5 or: Correctly inserting limits of 1, 2 in candidate’s a(2x – 3)6 + b(2x – 3)5 m1 6 5 2

⌠x(2x – 3)4 dx = 3 ⌡ 10

(c.a.o.)

1

19

A1

8.

9.

10.

(a)

dV = – kV2 dt

B1

(b)

⌠dV = –⌠k dt (o.e.) 2 ⌡ ⌡V – 1 = – kt + c V c = –1 12000 V = 12000 = 12000 12000kt + 1 at + 1

M1 A1 (c.a.o.)

A1

(convincing)

A1

(c)

Substituting t = 2 and V = 9000 in expression for V M1 a=1 A1 6 Substituting t = 4 in expression for V with candidate’s value for a M1 V = 7200 (c.a.o) A1

(a)

(2i – 2j + k).(i – 4j + 8k) = 18 B1 |2i – 2j + k| = √9, |i – 4j + 8k| = √81 (one correct) B1 Correctly substituting in the formula a.b = |a| × |b| × cosθ M1 θ = 48·2° (c.a.o.) A1

(b)

(i) (ii)

(c)

2 – λ = –1 + μ – 2 – 2λ = – 4 + μ 1 + 7λ = – 2 – μ (o.e.) (comparing coefficients, at least one equation correct) M1 (at least two equations correct) A1 Solving two of the equations simultaneously m1 (f.t. for all 3 marks if candidate uses his/her expression for AB) λ = – 1, μ = 4 (o.e.) (c.a.o.) A1 Correct verification that values of λ and μ satisfy third equation A1 Position vector of point of intersection is 3i – 6k (f.t. one slip) A1

AB = – i – 2j + 7k B1 Use of a + λAB, a + λ(b – a), b + λAB or b + λ(b – a) to find vector equation of AB M1 r = 2i – 2j + k + λ (– i – 2j + 7k) (o.e.) (f.t. if candidate uses his/her expression for AB) A1

Assume that positive real numbers a, b exist such that a + b < 2√ab . Squaring both sides we have:(a + b)2 < 4ab ⇒ a2 + b2 + 2ab < 4ab a2 + b2 – 2ab < 0 ⇒ (a – b)2 < 0 This contradicts the fact that a, b are real and thus a + b ≥ 2√ab

20

B1 B1 B1

FP1 1. f ( x + h) − f ( x) =

1 1 − 2 1 + ( x + h) 1 + x 2

1 + x 2 − 1 − x 2 − 2 xh − h 2 2 2 = [1 + ( x + h) ](1 + x ) − 2 xh − h 2 2 2 = [1 + ( x + h) ](1 + x ) f ′( x) =

lim − 2 xh − h 2 h → 0 h[1 + ( x + h) 2 ](1 + x 2 )

− 2x 2 2 = (1 + x )

2.

z−

M1A1

A1 A1

M1 A1

5z 5(2 + i) = 2−i− z 2−i

B1

5(2 + i) (2 + i)(2 − i)

M1

5(4 + 4i − 1) 5 =2–i–

A1

2−i− =

2

[Award A1 for either a correct numerator or denominator] = – 1 – 5i

A1

Modulus = 26 (5.10) tan −1 ( y / x) = 1.37 (78.7°) Argument = 4.51 ( 258.7°)

B1

21

B1 B1

3.

(a)

(b)

Determinant = 2(10 − 5λ ) + λ (4λ − 5) + 3(5 − 8) 2 = 4λ − 15λ + 11 Singular when 4λ2 − 15λ + 11 = 0 λ= 1, 11/4

M1 A1 M1 A1

When λ = 3,

⎡2 3 3⎤ ⎢1 2 3⎥ ⎢ ⎥ ⎢4 5 5⎥⎦ ⎣ A= (i)

(ii)

Determinant = 2

B1

⎡− 5 7 − 3⎤ ⎢ 0 −2 2 ⎥ ⎢ ⎥ ⎢⎣ 3 − 3 1 ⎥⎦ Cofactor matrix =

M1

3⎤ ⎡− 5 0 ⎢ 7 − 2 − 3⎥ ⎢ ⎥ ⎢⎣ − 3 2 1 ⎥⎦ Adjugate =

A1

3⎤ ⎡− 5 0 1⎢ 7 − 2 − 3⎥⎥ ⎢ 2 −1 1 ⎥⎦ A = ⎢⎣ − 3 2

A1

3 ⎤⎡ 2 ⎤ ⎡ x⎤ ⎡− 5 0 ⎢ y ⎥ = 1 ⎢ 7 − 2 − 3⎥ ⎢− 1⎥ ⎢ ⎥ 2⎢ ⎥⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ − 3 2 1 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎡1 ⎤ ⎢2⎥ ⎢ ⎥ ⎢ − 2⎥ = ⎣ ⎦

22

M1

A1

4.

α + β = −2,αβ = 3

B1

Consider ⎛ 1 1 1 1 ⎞ α − 2 + β − 2 = α + β − ⎜⎜ 2 + 2 ⎟⎟ β α β ⎠ ⎝α

M1

α +β −

[(α + β ) − 2αβ ] 2

α 2β 2 [4 − 6] 16 −2− =− 9 9 =

A1

⎛ ⎛1 1⎞ 1 ⎞⎛ 1 ⎞ 1 ⎜⎜ α − 2 ⎟⎟⎜ β − 2 ⎟ = αβ + 2 2 − ⎜⎜ + ⎟⎟ β ⎠⎝ α ⎠ α β ⎝α β ⎠ ⎝

M1

=

⎛α + β − ⎜⎜ α β ⎝ αβ = 1 2 34 3+ + = 9 3 9 =

αβ +

1

2

2

A1

⎞ ⎟⎟ ⎠

A1 A1

16 34 x2 + x + =0 9 9 The required equation is

5.

2 The statement is true for n = 1 since 4 − 1 = 15 Let the statement be true for n = k, ie 2k 42 k − 1 is divisible by 15 or 4 − 1 = 15 N Consider 42 ( k +1) − 1 = 16 × 42 k − 1 = 16(1 + 15 N ) − 1

= 16 × 15 N + 15

B1

B1 M1 M1 A1 A1

This is divisible by 15 so true for n = k ⇒ true for n = k + 1, hence proved by induction. A1

23

6.

(a)

A B A(r + 2) + Br 1 = + = r ( r + 2) r r + 2 r ( r + 2)

Let

A = 1 / 2, B = −1 / 2

M1 A1A1

(b)

Sn =

1 ⎡1 1 1 1⎤ + + + ... + ⎥ ⎢ 2 ⎣1 2 3 n⎦ 1 ⎡1 1 1 1 1 ⎤ − ⎢ + + ... + + + 2 ⎣3 4 n n + 1 n + 2 ⎥⎦

3 1 1 − − = 4 2(n + 1) 2(n + 2) n + 2 + n +1 3 − = 4 2(n + 1)(n + 2)

A1 A1

3 2n + 3 − = 4 2(n + 1)(n + 2)

7.

(a)

ln f ( x) = − 2 x ln x

M1A1

1 f ′( x) = −2 ln x − 2 x × f ( x) x

M1A1

f ′(x) (b)

M1A1

−2 x = − 2(ln x + 1) × x

A1

At a stationary point, f ′( x) = 0 so lnx = – 1 x = 1/e (= 0.368) 2 1/ e y = (e ) = 2.09

M1 A1

It is a maximum because f(0.3) = 2.06 and f(0.4) = 2.08 [Accept f ′(0.3) = 0.84 and f ′(0.4) = −0.11]

24

A1 M1A1

8.

(a)

Rotation matrix ⎡0 − 1 0 ⎤ = ⎢⎢1 0 0⎥⎥ ⎢⎣0 0 1⎥⎦

B1

⎡1 0 − 3⎤ ⎢0 1 1 ⎥ ⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ Translation matrix =

B1

⎡1 0 − 3⎤ ⎡0 − 1 0⎤ ⎢0 1 1 ⎥ ⎢1 0 0⎥ ⎢ ⎥⎢ ⎥ ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1⎥⎦ T=

M1

⎡0 − 1 − 3⎤ ⎢1 0 1 ⎥⎥ ⎢ ⎢0 0 1 ⎥⎦ =⎣

(b)

Fixed points satisfy ⎡0 − 1 − 3⎤ ⎡ x ⎤ ⎡ x ⎤ ⎢1 0 1 ⎥⎥ ⎢⎢ y ⎥⎥ = ⎢⎢ y ⎥⎥ ⎢ ⎢⎣0 0 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ ⎢⎣ 1 ⎥⎦ –y –3=x x+1=y (x,y) = (−2,−1)

(c)

M1

A1 M1A1

Let ( x, y ) → ( x′, y′) under T.

x′ = − y − 3 y′ = x + 1 We are given that x′ + 2 y′ = 3 so that − y − 3 + 2( x + 1) = 3 y = 2x – 4 [Special case: x + 2y = 3 giving y = 2x + 10 – award 3/4]

25

M1A1 M1 A1

u + iv =

9.

(a)

1 1 − x − iy

M1

1 − x + iy 2 2 = (1 − x) + y u=

v=

(b)

(c)

A1

1− x (1 − x) 2 + y 2

A1

y (1 − x) 2 + y 2

A1

Dividing, v y = u 1− x =1 The equation of the locus is v = u.

M1 A1

Fixed points satisfy 1 z= 1− z or

z2 − z + 1 = 0

M1A1

Solutions are

1± − 3 1± i 3 = 2 2 z=

M1

⎛1 3⎞ ⎜ ,± ⎟ ⎜2 2 ⎟⎠ ⎝ corresponding to points

A1

Alternative solution: Fixed points satisfy y 1− x ;y= x= 2 2 (1 − x) + y (1 − x) 2 + y 2

M1A1

x = 1 – x giving x = ½

A1

3 1 + y2 = 1 ± 2 4 giving y =

A1

26

FP2

1.

3 xdx 2 and [0,2] → [0, 2 2 ]

u = x x ⇒ du =

I=

2 3

2 2

∫ 0

B1 B1

du

M1

9 − u2 2 2

2⎡ u ⎤ = ⎢sin −1 ( )⎥ 3⎣ 3 ⎦0 = 0.821

2.

3.

A1 A1

(a)

r=5 θ = tan −1 (4 / 3) = 0.927 (53.1°)

(b)

First root = ( 51 / 3 ,0.309 ) = 1.63 + 0.520i Second root = (51 / 3 ,0.309 + 2π / 3) = – 1.26 + 1.15i Third root = (51 / 3 ,0.309 + 4π / 3) = – 0.364 – 1.67i

Substituting for sin and cos, 5 × 2t 5(1 − t 2 ) − =1 1+ t2 1+ t2 10t − 5 + 5t 2 = 1 + t 2 2t 2 + 5t − 3 = 0 Solving, t = ½, – 3. x ⎛ x⎞ 1 tan⎜ ⎟ = ⇒ = 0.464 + nπ (26.6° + 180n°) 2 ⎝2⎠ 2 x = 0.927 + 2nπ (53.1° + 360n°) x ⎛x⎞ tan⎜ ⎟ = −3 ⇒ = −1.25 + nπ (– 71.6° + 180n°) 2 ⎝2⎠ x = – 2.50 + 2nπ ( – 143° + 360n°)

27

B1 B1 M1 A1A1 M1 A1 M1 A1

M1A1 A1 M1A1 M1A1 A1 M1 A1

4.

(a)

3x 2 A Bx + C ≡ + 2 2 ( x + 2)( x + 2) x + 2 x + 2

Let

A( x 2 + 2) + ( Bx + C )( x + 2) ( x + 2)( x 2 + 2) A = 2, B = 1, C = – 2

=

(b)

I = 2∫

2

1

M1 A1A1A1

2 xdx 2 dx dx +∫ 2 − 2∫ 2 1 1 x+2 x +2 x +2

[

]

M1 2

2 2 ⎡ −1 ⎛ x ⎞⎤ 1 A1A1A1 = 2[ln( x + 2)] + ln( x 2 + 2) 1 − ⎟⎥ ⎢ tan ⎜ 2 2⎣ ⎝ 2 ⎠ ⎦1 1 2 ⎛ −1 ⎛ 2 ⎞ −1 ⎛ 1 ⎞ ⎞ ⎜⎜ tan ⎜ A1 = 2 ln 4 − 2 ln 3 + (ln 6 − ln 3) − ⎟ − tan ⎜ ⎟ ⎟⎟ 2 2⎝ ⎝ 2⎠ ⎝ 2 ⎠⎠ = 0.441 cao A1 2 1

5.

cos 5θ + i sin 5θ = (cosθ + isinθ )5

B1

= 5 cos 4 θ .i sin θ + 10 cos 2 θ .i 3 sin 3 θ + i5 sin 5 θ + real

M1A1

Considering imaginary ,

sin 5θ = 5 cos 4 θ sin θ − 10 cos 2 θ sin 3 θ + sin 5 θ sin 5θ = 5 cos 4 θ − 10 cos 2 θ (1 − cos 2 θ ) + (1 − cos 2 θ ) 2 sin θ = 5 cos 4 θ − 10 cos 2 θ + 10 cos 4 θ + 1 − 2 cos 2 θ + cos 4 θ = 16 cos 4 θ − 12 cos 2 θ + 1 As θ → 0, cosθ → 1 so the required limiting value is 5. [FT their previous expression]

28

A1 A1

A1 M1A1

6.

(a)

f ′( x) =

( x − 1) 2 − 2 x( x − 1) ( x − 1) 4

M1A1

Stationary points occur when ( x − 1) 2 = 2 x( x − 1) x = −1, y = −1 / 4 (b)

m1 A1

The asymptotes are x = 1 and y = 0.

B1B1

(c)

G1G1

y

0

(d)

x 1

Consider x =2 ( x − 1) 2

M1

2 x2 − 5x + 2 = 0 x = ½, 2

A1 A1

f −1 ( A) = [0,1 / 2] ∪ [ 2, ∞)

A1A1

29

7.

(a)

(b)

g (− x) = f (− x) + f ( x) = g ( x) , therefore even h(− x) = f (− x) − f ( x) = −h( x) therefore odd The result follows from the fact that 1 1 f ( x ) = g ( x ) + h( x ) 2 2 (i) g ( x) = ln(1 + sin x) + ln(1 − sin x) = ln cos 2 x = 2lncosx (ii)

h( x) = ln(1 + sin x) − ln(1 − sin x) ⎛ 1 + sin x ⎞ = ln⎜ ⎟ ⎝ 1 − sin x ⎠ ⎛ (1 + sin x) 2 ⎞ ⎟⎟ = ln⎜⎜ 2 ⎝ cos x ⎠ ⎛ 1 + sin x ⎞ = 2 ln⎜ ⎟ ⎝ cos x ⎠

B1 B1 B1 M1 A1 A1 M1 A1 A1 A1

= 2 ln(sec x + tan x)

8.

(a)

Writing the equation in the form x 2 = −8 y we note that, in the usual notation, a = 2 and x,y are interchanged with M1 the negative sign indicating that the graph is below the x-axis. The focus is (0, – 2) and the directrix y = 2.

(b)

(i)

(ii)

The result follows since (4 p ) 2 + 8(−2 p 2 ) = 0

dy − 4 p = = −p dx 4

y + px = 2 p This es through (λ,2) if 2 + pλ = 2 p 2 2

or

B1 B1

The equation of the tangent is y + 2 p 2 = − p( x − 4 p) (iii)

A1A1

M1 A1 M1

2 p − pλ − 2 = 0 2

M1 The 2 roots p1 , p2 satisfy p1 p2 = −1 And since the gradient of the tangent, m, satisfies m = − p , it follows A1 that m1m2 = −1 which is the condition for perpendicularity.

30

FP3

1.

(a)

f ′( x) = 2 cosh 2 x − 14 cosh x + 8

M1

= 2(2 cosh 2 x − 1) − 14 cosh x + 8

A1

= 2(2 cosh x − 7 cosh x + 3) 2

(b)

We solve 2 cosh 2 x − 7 cosh x + 3 = 0 (2 cosh x − 1)(cosh x − 3) = 0 coshx = ½, 3

M1 M1 A1

coshx = ½ has no solution (so only one stationary point given by coshx = 3) A1

(c)

2.

x = cosh −1 3 = 1.76

A1

f ′′( x) = 2(4 cosh x sinh x − 7 sinh x) f ′′(1.76) = 28 (exact value 20√2) Positive therefore minimum.

B1 B1 B1

dx = coshudu ; [0,3] → [ 0, θ ] where θ = sinh −1 3 θ

I=

∫

B1B1

2

sinh u cosh udu

M1

sinh 2 u + 1

0

θ

= ∫ sinh 2 udu

A1

0

θ

1 = ∫ (cosh2u - 1) du 20

A1

θ

1 ⎡1 ⎤ sinh 2u − u ⎥ ⎢ 2 ⎣2 ⎦0 = 3.83

A1

=

A1

31

3.

(a)

(b)

(c)

Putting f ( x) = x x and taking logs, ln f ( x) = x ln x f ′( x) = 1 + ln x f ( x) f ′( x) = x x (1 + ln x)

2y

A1

(i)

The Newton-Raphson formula is x ( x n − 2) xn+1 = xn − x n M1A1 xn n (1 + ln xn ) Starting with 1.5, one application of the formula gives 1.56 A1

(ii)

f (1.555) − 2 = −0.01326...; f (1.565) − 2 = .01564... The change of sign indicates that α = 1.56 correct to 2 dps. M1A1 d ⎛⎜ e dx ⎜⎝

ln2 ⎞ ⎟ = − ln22 e x ⎟ x ⎠ Putting x = 1.5 gives – 0.489 (1.56 gives – 0.444)

(i)

ln2 x

dy dy 3 x 2 3 ( = 3x 2 ⇒ = x) dx dx 2 y 2

Arc length =

B1 B1

The iteration converges because this is less than 1 in modulus. B1 Successive values are 1.56 1.559437401 M1 1.559687398 1.559576282 1.559625664 The required value is 1.5596. A1 [Award A1 only if calculations done to reasonable accuracy]

(iii)

4.

M1

1

∫

0

1+

B1

9x dx 4

M1A1 1

⎡ 2 4 ⎛ 9 x ⎞3 / 2 ⎤ = ⎢ × ⎜1 + ⎟ ⎥ 4 ⎠ ⎦⎥ ⎣⎢ 3 9 ⎝ 0 =

8 ⎛⎜ ⎛ 13 ⎞ ⎜ ⎟ 27 ⎜⎝ ⎝ 4 ⎠

3/ 2

M1A1

⎞ − 1⎟ ⎟ ⎠

A1

= 1.44

A1

32

5.

(a)

(i)

f(0) = 0

B1

cosh x f ′( x) = ; f ′(0) = 1 1 + sinh x sinh x( 1 + sinh x) − cosh 2 x ;f ′′( 0 ) = −1 f ′′( x) = ( 1 + sinh x)2 f ′′′( x) =

cosh x(1 + sinh x) 2 − 2(1 + sinh x) cosh x(sinh x − 1) ; f ′′′(0) = 3 (1 + sinh x) 4

(ii)

(b)

(a)

B1B1 B1B1

The Maclaurin series is x 2 3x 3 x2 x3 0+ x− + + ... = x − + + ... M1A1 2 6 2 2 Both even and odd powers of x are present or equivalent argument based on series. B1

x 2 x3 x− + = 10 x 2 2 2 2 x − 21x + 2 = 0

M1 A1

x = 0.096 [FT from (a)]

6.

B1B1

A1

Consider x= rcosθ = cos 2 θ + 2 sin θ cos θ M1 dx = −2 cos θ sin θ + 2 cos 2 θ − 2 sin 2 θ A1 dθ dx At the required point, = 0 , giving M1 dθ sin 2θ = 2 cos 2θ A1 A1 tan 2θ = 2 1 A1 θ = tan −1 2 = 0.55 2 r = 1.9 A1 [Treat consideration of y = rsinθ as a special case and award B2B2 for (2.18,1.34)]

33

(b)

1 2

Area =

1 2

=

1 2

=

π /2

∫ (cos θ + 2 sin θ ) dθ 2

M1

0

π /2

∫ (cos

2

θ + 4 sin θ cos θ + 4 sin 2 θ )dθ

A1

0

π /2

1

1

∫ ( 2 + 2 cos 2θ + 2 sin 2θ + 2 − 2 cos 2θ )dθ

M1A1

0

π /2

1 ⎡ 5θ 3 ⎤ = ⎢ − cos 2θ − sin 2θ ⎥ 2⎣ 2 4 ⎦0

A1

⎛ 5π ⎞ = 2.96 ⎜1 + ⎟ 8 ⎠ ⎝

A1

π /2

7.

(a)

In =

∫ cos

n −1

M1

xd(sin x)

0

[

]

π /2

= cos n−1 x sin x 0 π /2

= (n − 1) ∫

0

π /2

+∫

0

sin x.(n − 1) cos n−2 x sin xdx

(1 − cos 2 x) cos n−2 xdx

= (n − 1) I n−2 − (n − 1) I n

A1

M1A1 A1

⎛ n −1⎞ In = ⎜ ⎟ I n−2 ⎝ n ⎠ (b)

(i)

(ii)

3 3 1 I2 = × I0 4 4 2 3 π /2 = ∫ dx 8 0 3π = (0.589) 16

I4 =

Integral =

∫

π /2

0

M1 A1 A1

cos5 x(1 − cos 2 x)dx = I 5 − I 7

⎛ 4 2 6 4 2 ⎞ π /2 = ⎜ × − × × ⎟ ∫ cos xdx ⎝5 3 7 5 3⎠ 0 ⎛4 2 6 4 2⎞ = ⎜ × − × × ⎟ ×1 ⎝5 3 7 5 3⎠ 8 = 105

GCE Mathematics C1-C4 & FP1-FP3 Mark Scheme - Summer 2010 / JSM

34

M1A1 M1 A1 A1

INTRODUCTION The marking schemes which follow were those used by WJEC for the Summer 2010 examination in GCE MATHEMATICS. They were finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conferences were held shortly after the papers were taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conferences was to ensure that the marking schemes were interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conferences, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about these marking schemes.

Paper

Page

M1 M2 M3

1 5 9

S1 S2 S3

15 19 22

Mathematics M1 Notes:

1.

cao = correct answer only, oe = or equivalent, si = seen or implied, ft = follow through (c) = candidate's value acceptable (a)

Use of v2 = u2 + 2as with u = (±)2.1, a = (±)9.8, s = (±)15.4 v2 = 2.12 + 2 ×9.8 ×15.4 v = 17.5 (ms-1)

(b)

Use of v = u + at with v = 17.5(c), a = (±)9.8, u = (±)2.1 17.5 = 2.1 + 9.8t 11 t = 7

M1 A1 cao A1 oe M1 A1 cao A1

2. (a) v ms-1

2·7

0

ts 15

105

120

attempt at v-t graph with one correct section and axes M1 second correct section A1 completely correct graph with labels A1 (b)

Distance = 0.5(90 + 120) × 2.7

attempt to calculate total area M1 any correct value for an area B1 cao A1

= 283.5 (m) (c) R a

75g

2.7 a = = (0 ⋅18) 15 Apply N2L to woman R – 75g = 75a R = 75(9.8 + 0.18) = 748.5 (N) 1

B1 all forces, dim correct M1 correct equation A1 ft a A1

3. R

5 13 12 cosα = 13

F

a

sinα =

52g

Resolve perpendicular to plane Use of

M1

R = 52gcosα F = μR

m1

12 = 0.2 × 52 × 9.8 × 13 = 94.08 (N)

si A1

Apply N2L to object down slope 52gsinα - F = 52a 5 52 × 9.8 × - 94.08 = 52a 13 a = 1.96 (ms-2) 4.

12 ms-1

(a)

Dim correct, all forces M1 A1

cao A1 7 ms-1

A

B

2 kg

3 kg

vA

vB

attempt at conservation of momentum equation 2 × 12 + 3 × 7 = 2vA + 3vB 2vA + 3vB = 45

M1 A1

attempt at restitution equation vB - vA = -0.6(7 – 12) -3vA + 3vB = 9

M1 A1

attempt to solve simultaneously dep. 5 vA = 36 vA = 7.2 (ms-1) vB = 10.2 (ms-1) (b)

Use of Impulse = change in momentum I = 3(10.2 – 7) = 9.6 (Ns)

2

Both M’s m1 cao A1 cao A1 M1 ft sensible results only A1

5. R

a T

A

F

a 6g T B 4g

(a)

Apply N2L to B/A 4g – T = 4a

M1 A1

Apply N2l to other particle T – F = 6a

M1 A1

Resolve vertically, particle A R = 6g F = μR = 0.4 × 6g = 2.4g attempt to solve equations simultaneously 4g – 2.4g = 10a a = 0.16g = 1.568 (ms-2) T = 32.928 (N) (b)

6.

si B1 B1 m1 cao A1 cao A1

Light strings enable the assumption that tension is constant throughout the string to be used. B1

Attempt to resolve in direction of 12 N force Y = 12 - 5√3 sin 60° - 3√2 sin 45° Y = 1.5

M1 A1

Attempt to resolve in perpendicular direction X = 5√3 cos 60° - 3√2 cos 45° X = 1.33

M1 A1

Resultant R = (1.5) + 1.332 = 2.(0048) (N) 2

M1 ft A1

⎛ 1.33 ⎞ θ = tan-1 ⎜ ⎟ = 41.6° ⎝ 1.50 ⎠ Dir of R is 41.6° to the right with the 12 N force

3

M1 ft A1

7.

TB

TA C A

0·3

B

0·4 6g

10g

Moments about A

8.

9.

1.4 TB = 0.7 × 6g + 0.3 × 10g TB = 50.4 (N) Resolve vertically dim correct, all forces TA + TB = 16g TA = 106.4 (N)

dim. correct equation, all forces M1 any correct moment B1 A1 cao A1 oe M1 A1 ft TB A1

Use of s = ut + 0.5at2 with s = 95, t = 5 95 = 5u + 0.5 × a × 25

M1 A1

Use of v = u + at with t = 7. v = 29.8 29.8 = u + 7a

M1 A1

attempt to solve simultaneously 10.8 = 4.5a a = 2.4 u = 13

m1

(a)

Lamina ABCD XYZ Decoration

cao A1 cao A1

Area 80 9 89

from AD 4 3 x

from AB 5 3 y one correct pair of distances B1 all four correct B1 correct areas B1

Moments about AD 89 x = 80 × 4 + 9 × 3 x = 3.90 (cm)

M1 ft A1 cao A1

Moments about AB 89y = 80 × 5 + 9 × 3 y = 4.80 (cm)

M1 ft A1 cao A1

(b)

⎛ x ⎞ θ = tan-1 ⎜⎜ 10 − y ⎟⎟ ⎝ ⎠ ⎛ 3 .9 ⎞ = tan-1 ⎜⎜ 10 − 4.8 ⎟⎟ ⎝ ⎠

correct triangle M1

ft A1

= 36.9°

4

Mathematics M2 1.

(a)

∫ 3 − 4t dt

=

Attempt to integrate a v v

=

3t – 2t (+C)

v

=

-2t2 + 3t – 1

When t = 0, v = -1 Therefore C = -1 (b)

(c)

M1

2

A1 use of initial conditions m1 A1

When P is at rest, v = 0 -2t2 + 3t -1 = 0 2t 2 – 3t + 1 = 0 (2t – 1)(t – 1) = 0 t = 0.5, 1

M1

ft C ≠ 0

Attempt to integrate v

A1 M1

1

Distance required =

∫ − 2t

2

+ 3t − 1 dt

limits oe m1

1 2

1

3 ⎡ 2 ⎤ = ⎢− t 3 + t 2 − t ⎥ 2 ⎣ 3 ⎦1

ft v A1

1 = 24

cao A1

2

2.

(a)

dr dt v = 6t i + (13 – 4t) j v =

Speed =

A1

(6t ) 2 + (13 − 4t ) 2

When t = 2, speed = (b)

used M1

M1

144 + 25 = 13

velocity is perpendicular to (2i – j) when v.(2i – j) = 0 12t – 13 + 4t = 0 13 t = 16

(c)

cao A1

Acceleration of P = a =

Magnitude =

M1

method for dot product M1

dv dt

used M1

a = 6 i – 4j

(d)

ft A1

independent of t

36 + 16 =

52

Let θ be the required angle. Use of a.b = ⏐a⏐⏐b⏐cosθ with b = v when t = 2

ft v A1

ft A1

M1

52 × 13 cos θ

ft A1

72 – 20 = 52 × 13 cos θ θ = 56.3°

cao A1

(6i – 4j).(12i + 5j) =

5

3.

(a)

Use of Hooke's Law T =

λx

M1

l

294 x 2 x = 0.2 (m) 3 × 9.8 =

(b)

A1 cao A1

Use of loss in potential energy = mgh Loss in PE = 3 × 9.8 × (0.8 + 0.2) = 29.4 (J) Gain in KE = 0.5 × 3v2 = 1.5 v2 (J) 1 λ × x2 × Use of gain in elastic energy = 3 energies 2 l 1 294 × 0.2 2 × Gain in EE = 2 2 = 2.94 (J) Use of conservation of energy 29.4 = 1.5 v2 + 2.94 v = 4.2 (ms-1)

4.

M1 si

ft x A1 B1 M1 ft x A1

M1 ft A1 cao A1

(a) a

N T R

α

1500g

P M1 v 30 ×1000 T= (= 3750 N) si A1 8 N2L up slope all forces, dim correct equation M1 T – R – 1500gsinα = 1500a -1 each error A2 6 3750 – 600 – 1500 × 9.8 × = 1500a 49 cao A1 a = 0.9 (ms-2)

Use of T =

(b)

At maximum attainable speed, a = 0 Apply N2L to particle up the slope

T = R + mgsinα 30000 6 = 600 + 1500 × 9.8 × v 49 -1 v = 12.5 (ms )

6

used M1 M1 A1

cao A1

5.

(a)

Initial vertical speed = Vsin30° = (0.5V) B1 2 2 Use of v = u + 2as with u = Vsin30°, v = 0, a = (±)9.8, s = (±)4.9 M1 A1 0 = 0.25V2 + 2(-9.8)(4.9) V = 19.6 A1 (b) Use of s = ut + 0.5at2 with s = (±)39.2, a = (±)9.8, u = (±)0.5×19.6(c) -39.2 = 9.8t – 4.9t t 2 – 2t – 8 = 0 t = 4s

(c)

M1 A1 attempt to solve quadratic m1 A1

2

initial horizontal velocity = 19.6cos30° = (16.97) Use of v = u + at with u = 9.8(c), a = (±)9.8, t = 3 v = 9.8 – 9.8 × 3 v = -19.6

B1 M1 A1

Speed = 16 ⋅ 97 2 + 19 ⋅ 6 2 = 25.9 (ms-1)

M1 A1

6. R

v2 40 F 60g 30°

Let the maximum speed be v ms-1. On the point of slipping, friction is limiting and F = μR = 0.25R Resolving vertically Rcos30° = mg + Fsin30° R 3 R − = 60 g 2 8 R(4√3 – 1) = 480g R = 793.495 (N)

M1

dim correct equation M1 A1

Apply N2L horizontally Rsin 30° + Fcos30° = ma v2 a = 40 3 3 2 R 1 + R = v 2 4 2 2 1.5v2 = 568.54 v = 19.5 (ms-1) 7

all forces, dim correct M1 A1 M1

cao A1

7.

(a)

Use of conservation of energy 0.5m × 132 = 0.5mv2 + mg(2.5)(1 - cosθ)

M1 KE A1 PE A1

v2 = 169 – 2 × 9.8 × 2.5(1 - cosθ) v2 = 120 + 49cosθ v = √(120 + 49cosθ) When cosθ = 0.5

(b)

v2 = 120 + 24.5 = 144.5 v = 12(.02)

Apply N2L towards centre mv 2 T - mg cosθ = r

T = 3 × 9.8 cosθ +

ft A1 all forces, dim correct M1 A1

3 × (120 + 49cosθ ) 2.5

T = 144 + 88.2 cosθ (c)

cao A1

substitution m1 cao A1

Minimum value of cosθ is -1. Therefore T > 0 for all values of θ. M1 A1 Therefore P describes complete circles.

8

Mathematics M3 1. a

N

R

T

720g

(a)

P v 81×1000 T= v

Use of T =

M1 si A1

Apply N2L

to car dim correct equation

M1

T – R = ma 81000 dv − 90v = 720 v dt Divide by 90 and multiply by v throughout dv 900 − v 2 = 8v dt (b)

A1

A1

Attempt to separate variables 8v ∫ 900 − v 2 dv = ∫ dt

M1 A1

Integrating -4 ln⏐900 – v2⏐ = t + C

t = -4 ln ⏐900 – v2⏐ - C

[

Required time = − 4 ln 900 − v 2

]

20

5

correct ln term all correct

A1 A1

subtraction of t values M1 correct limits oe

A1

⎡ ⎛ 900 − 25 ⎞⎤ ⎟⎟⎥ = 4 ⎢ln⎜⎜ ⎣ ⎝ 900 − 400 ⎠⎦ ⎡ ⎛ 875 ⎞⎤ = 4 ⎢ln⎜ ⎟⎥ ⎣ ⎝ 500 ⎠⎦ = 4 ln (1.75) = 2.24 (s)

cao A1

9

2.

(a)

(b)

At equilibrium 12g =

0 ⋅ 75 λ = 1764 (N)

(d)

λ (0 ⋅ 05 + x ) 0 ⋅ 75

= 12 x

A1

M1 ft λ A1

x = -(14)2 x

Therefore is SHM (with ω = 14).

A1

Amplitude = 0.05 (m) 2π π Period = = s ω 7

B1 B1

Maximum speed = aω = 0.05 × 14 = 0.7 (ms-1)

used M1 ft a A1

Use of v2 = ω2(a2 – x2) with ω = 14, a = 0.05(c), x = 0.03

v2 = 142(0.052 – 0.032) = 142 × 0.042 v = 0.56 (ms-1) (e)

use of Hook's Law M1

Consider a displacement x from the equilibrium position. Apply N2L 12g – T = 12 x 12g -

(c)

λ × 0 ⋅ 05

M1 ft a A1 cao A1

Displacement from Origin = x x = (-)0.05cos(14t) When t = 1.6 x = (-) 0.05 cos(14 × 1.6) x = (-)0.046 (m)

10

M1 ft a A1 cao A1

3.

Auxiliary equation 4m2 – 12m + 9 = 0 (2m – 3)2 = 0 m = 1.5 (twice) Complementary function x = (A + Bt)e1.5t

dx =a dt -12a + 9(at + b) = 18t – 87 9a = 18 a = 2 -24 + 9b = -87 b = -7

For PI, try x = at + b,

B1 ft B1 M1

A1 comparing coefficients m1 both A1

General solution x = (A + Bt)e1.5t + 2t -7 Use of initial conditions t = 0, x = 5,

B1

dx = 10 dt

A–7 = 5 A = 12

ft B1 in general solution M1 cao A1

dx = (A + Bt)(1.5)e1.5t + Be1.5t + 2 dt 1.5A + B + 2 = 10 B = -10

correct diff. ft B1

cao A1

x = (12 – 10t)e1.5t + 2t – 7

11

4.

When the string jerks tight, each particle begins to move in direction PQ with equal speeds v.

P

P

60° 0·3

v

60° 0·6

0·3

J 0·6 u

J 8 ms C

–1

C

Q

Just before jerk

Q

v

Just after jerk

1 2 3 sin ∠Q = 2

cos ∠Q =

si B1

Use of impulse = change in momentum J = 3v Applied to P J = 5 × 8sin 60° - 5v Applied to Q

M1 B1 A1

Attempt to solve simultaneously 3 - 5v 3v = 40 × 2 5 3 v = = 4.33 (ms-1) 2 Speed of particle P is 4.33 ms-1.

m1

cao A1

Magnitude of impulsive tension = J = 3v 15 3 = 12.99 (Ns) = 2

cao A1 units B1

Perpendicular to PQ, there is no impulse Speed of particle Q perpendicular to PQ = 8cos60° = 4 ms-1

⎛5 3⎞ ⎟ 4 2 + ⎜⎜ ⎟ 2 ⎝ ⎠ = 5.89 ms-1

Speed of particle Q =

12

B1

2

M1 cao A1

5.

(a)

(b)

Use of N2L 150g – 10v2 = 150a dv 15g – v2 = 15v ds

M1 A1

Attempt to separate variables 15v dv ∫ v 2 −15 g = − ∫ ds 15 ln v 2 −15 g = − s (+ C ) 2

M1

A1

A1 correct ln A1 all correct A1

Use of boundary conditions s = 0, v = 30 15 ln 900 −15 g = C 2 15 753 ln 2 s = 2 v −15 g (c)

v = 14

m1

cao A1 used M1

⎞ 15 ⎛ 753 ⎟ ln⎜⎜ 2 2 ⎝ 14 −15 × 9 ⋅ 8 ⎟⎠ s = 20.49

s =

(d)

Removing ln 753 ⎛2 ⎞ exp ⎜ s ⎟ = 2 v −15 g ⎝ 15 ⎠ ⎛ 2 ⎞ v2 – 15g = 753exp ⎜ − s ⎟ ⎝ 15 ⎠ ⎛ 2 ⎞ v2 = 15g + 753exp ⎜ − s ⎟ ⎝ 15 ⎠ ⎛ 2 ⎞ v2 = 147 + 753exp ⎜ − s ⎟ ⎝ 15 ⎠

13

cao A1 M1 ft A1

cao A1

6. Q A

S

θ

80g

R

20g

B

(a)

Use of Friction = μ × Normal reaction Q = 0.3 S Attempt at taking mom. about B 4 , 20g × 2.5 sinθ + 80g × 3sinθ = 294 + 1411.2 = 4.9S = S=

(b)

4 S + 3Q 4S + 0.9S 1705.2 348 (N)

Resolve vertically Q + R = 80g + 20g R = 100g – 0.3 × 348 R = 875.6 N

F

si M1 A1 dim correct equation M1 -1 each error A2 cao A1 4 , dim correct M1 A1

Resolve horizontally F = S (= 348) Use of

F ≤ μR 348 μ ≥ 875 ⋅ 6 μ ≥ 0.397

B1 M1

= 0 ⋅ 39744 cao

14

A1

Mathematics S1 1.

(a) (b)

P(A∩B) = 0.6 × 0.3 P(A∪B) = 0.6 + 0.3 – 0.6 × 0.3 = 0.72 EITHER P (B ′) = 1 – P(B) = 0.7 P( A ∪ B ′ ) = P(A) + P( B ′ ) – P(A)P( B ′) = 0.6 + 0.7 – 0.6 × 0.7 = 0.88

B1 M1 A1 B1 M1 A1

OR

B 0.12

0.18 A

2.

3.

0.42

Correct Venn diagram Required prob = 1 – 0.12 = 0.88

B1 M1A1

(a)

E(Y) = 3 × 4 – 1 = 11 Var(Y) = 9 × 2 =18

M1A1 M1A1

(b)

E (Y 2 ) = Var (Y ) + {E (Y )}2 = 18 + 121 = 139 [FT 1 arithmetic slip in (a)]

(a)

Mean = 6 (i) (ii)

(b)

si B1 3

6 = 0.0892 3! Prob = 1 – 0.7149 = 0.2851 [FT on mean] Prob = e −6 ×

M1 A1

Prob of no arrivals = e −0.1t Attempting to solve e −0.1t = 0.25 − 0.1t log e = log 0.25 log 0.25 t=− = 13.86 0.1log e [Award 2 marks for 14 using tables]

15

M1A1 M1A1

B1 M1 A1 A1

4.

(a)

Prob wins on 1st throw = 0.8 × 0.3 = 0.24

M1A1

(b)

Prob wins on 2nd throw = 0.8 × 0.7 × 0.8 × 0.3 = 0.1344 [FT from (a) if M1 awarded in (a)]

M1A1

(c)

Prob wins = 0.24 + 0.24 × 0.56 + 0.24 × 0.56 2 + ... 0.24 = 6/11 (0.55) = 1 − 0.56

M1A1 M1A1

[For candidates who solve for Bill first, award M0A0 for (a), M1A1 for 0.168 in (b), M1A1 for 0.3 + 0.3 × 0.56 + ... and M1A1 for 0.3/(1 – 0.56) = 15/22 (0.68) in (c)]

5.

P(correct ans) = 0.6 × 1 + 0.4 × 0.25 = 0 .7 [Award M1 if 1 and 0.25 reversed] 0.6 Reqd prob = [FT denominator from (a) if answer < 1] 0.7 6 = cao 7

M1A1 A1

(a)

∑p

M1A1

(b)

(i)

(a)

(b)

6.

x

= 16k = 1 so k = 1/16

E( X ) =

1 (1 × 1 + 3 × 3 + 5 × 5 + 7 × 7 ) 16

= 5.25 [Accept 84k] (ii)

(i)

(ii)

1 1 1⎞ ⎛1⎞ 1⎛ 1 E ⎜ ⎟ = ⎜1 × + 3 × + 5 × + 7 × ⎟ 3 5 7⎠ ⎝ X ⎠ 16 ⎝ 1

)

16

M1

M1A1 A1

Possibilities are 1,5 and 3,3 1 (1 × 5 + 5 × 1 + 3 × 3) Prob = 256 [Award M1 for 2 or 3 ] 19 (0.074) [Accept 19 / k 2 ] = 256 Possibilities are 1,1 ; 3,3 ; 5,5 ; 7,7 1 2 1 + 32 + 5 2 + 7 2 Prob = 256 [Award M1 for 3 or 4 ] = 0.328 (21/64) [Accept 84 / k 2 ]

(

B1

A1

= 0.25 [Accept 4k] (c)

B1B1

si B1 M1A1

A1 si B1 M1 A1

7.

(a)

(b)

Number of 6s obtained, X, is B(50,0.2) ⎛ 50 ⎞ (i) Prob = ⎜⎜ ⎟⎟ × 0.212 × 0.8 38 = 0.1033 ⎝ 12 ⎠ or 0.8139 – 0.7107 or 0.2893 – 0.1861 = 0.1032

M1A1

(ii)

M1A1

P(at least 10) = 0.5563 or 1 – 0.4437 = 0.5563

Prob of 2 6s = 0.22 = 0.04 X is now B(200, 0.04) which is approx P(8) [FT p from previous line] P( 5 ≤ X ≤ 10 ) = 0.8159 – 0.0996 or 0.9004 – 0.1841 = 0.7163 cao

8.

(a)

1

∫ kx(1 − x 0

2

) dx = 1

B1

B1 B1 B1B1 B1

M1 1

⎡ x2 x4 ⎤ Integral = k ⎢ − ⎥ 4 ⎦0 ⎣2 [Limits must appear somewhere for M1] k = 4

B1

A1

so k = 4 (b)

E(X) =

1

∫ x.4 x(1 − x 0

2

) dx

M1A1

1

⎡ 4x3 4x5 ⎤ = ⎢ − ⎥ 5 ⎦0 ⎣ 3 [Limits not required until 2nd line] 8 = 15

17

A1

A1

(c)

(i)

F(x) =

∫

x

0

4t (1 − t 2 )dt

M1

[Limits not required for M1]

[

]

x

= 2t 2 − t 4 0 [Limits must appear here] = 2x 2 − x 4 (ii)

(iii)

Prob = F(0.75) – F(0.25) = 2 × 0.75 2 − 0.75 4 − (2 × 0.25 2 − 0.25 4 ) = 0.6875 [FT if M1 awarded in (c)(i) and answer sensible] The median m satisfies 2m 2 − m 4 = 0.5 [FT from (c)(i) if M1 awarded there] 2m 4 − 4m 2 + 1 = 0 4± 8 m2 = 4 m = 0.541

18

A1 A1

M1 A1

M1 A1 m1 cao A1

Mathematics S2

1.

(a)

(b)

2.

120 − 106 = 1.75 (Accept ±) 8 Prob = 0.0401

z=

Distribution of total weight T is N(1060, 640) 1000 − 1060 z= = −2.37 640 P(T < 1000) = 0.0089 [No FT on incorrect variance]

(a)

Under H0, mean = 15 P(X ≤ 9) = 0.0699 (Accept 0.0778 from Normal app) P(X ≥ 22) = 0.0531 (Accept 0.0465 from Normal app) Sig level = 0.0699 + 0.0531 = 0.123 [FT one slip but treat Normal apps as incorrect here]

(b)

X is now Po(150) which is approx N(150,150) [FT their mean] 169.5 − 150 z= 150 [Award M1A0 for incorrect continuity correction] = 1.59 Prob from tables = 0.0559 p-value = 0.1118 Insufficient evidence to reject H0 (Accept ‘Accept H0’). [No c/c gives z = 1.63, prob = 0.0516 and p-value = 0.1032 Incorrect c/c gives z = 1.67. prob = .0475 and pv = 0.095]

19

M1A1 A1 M1A1A1 M1A1 A1

si B1 B1 B1 M1A1

B1 M1A1 A1 A1 B1 B1

3.

4.

(a)

11.5 + 11.7 + 11.6 (= 11.6) 3 0 .2 SE of X = (= 0.115…) 3 95% conf limits are 11.6 ± 1.96 × 0.2/√3 [M1 correct form, A1 1.96] giving [11.4,11.8]

x=

B1 B1 M1A1 cao A1

(b)

H 0 : μ = 12; H1 : μ > 12 B1 12.1 + 12.2 + 12.4 + 12.1 (=12.2) B1 y= 4 12.2 − 12 A1 Test stat = 0 .2 2 / 4 [Award M1 only if there is division by 4 in the denominator] [FT on slip in calculating y ] = 2.0 1 p-value = 0.0228 A1 Strong evidence for thinking that μ exceeds 12. B1

(c)

SE of y − x =

(a)

0 .2 2 0 .2 2 + (= 0.152..) 3 4 90% confidence limits are 12.2 – 11.6 ± 1.645 × 0.152… giving [0.35,0.85]

E(X) = 3 ⇒ np = 3 Using Var(X) = E ( X 2 ) − [ E ( X )]2 np(1 – p) = 11.1 – 9 = 2.1 Solving, 1 – p = 0.7 so p = 0.3 n = 10

(b)

M1A1 m1A1 cao A1 B1 M1 A1 m1A1 A1

E(Y) = 6 E(XY) = 3 × 6 = 18 E (Y 2 ) = 3.6 + 6 2 = 39.6

B1 B1 M1A1

E ( X 2Y 2 ) = 11.1 × 39.6 = 439.56 Var(XY) = 439.56 – 182 = 115.56

M1A1 m1A1

20

5.

(a)

A = 0.5 × PQ cos θ .PQ sin θ = 8 sin θ cos θ = 4 sin 2θ

B1

(b)

P ( A ≤ 2) = P(4 sin 2θ ≤ 2) = P(2θ ≤ sin −1[1 / 2]) = P(θ ≤ π / 12) [Accept 15°] π / 12 = π /4 = 1/3 4 f (θ ) = [Only award if quoted or used in (c)]

M1

(c)

π

E(A) = = =

6.

π /4

4

0

π

∫

8

π

× 4 sin 2θdθ

M1 A1 B1 M1

[− cos 2θ ]π0 / 4

A1

8

A1

π

(a)

H 0 : p = 0.75 versus H1 : p < 0.75

(b)

(i)

(ii)

A1 A1

Under H 0 , X (No germ) is B(50,0.75) and Y (No not germ) is B(50,0.25) Sig level = P(X < 30) ⏐ H 0 )

B1 B1 (si) B1 M1

= P(Y > 20⏐ H 0 ) = 0.0063 [Award B1B0M1A0 if Normal approx used]

A1

Required prob = P(X ≥ 30 |p = 0.5) = P(Y ≤ 20) = 0.1013

M1 A1

[Award M1A0 if Normal approx used] (c)

X is now B(200,0.75) which is approx N(150,37.5) 140.5 − 150 z= 37.5 = – 1.55

B1B1 M1 A1

[Award M1A0 for incorrect continuity correction] p-value = 0.0606 A1 [No c/c gives p = 0.0516, incorrect c/c gives p = 0.0436] Insufficient evidence to doubt the statement on the packet

21

B1

Mathematics S3 1.

(a) (b)

2.

B1

si M1A1 M1A1 A1

(c)

No because 0.5 lies within the interval.

B1

(a)

H 0 : μ = 1; H1 : μ < 1 99.6 x= = 0.996 100 99.24 99.6 2 s2 = = 0.000387878… − 99 99 × 100 [Accept division by 100 giving 0.000384] 0.996 − 1 Test stat = [M0 if square root omitted] 0.000387878 / 100 = –2.03 ( –2.04) p-value = 0.021 Strong evidence to reject H0 (or accept H1) [FT on p-value]

B1

(b)

3.

140 = 0.56 250 0.56 × 0.44 ESE = (= 0.031394..) 250 99% confidence limits are 0.56 ± 2.576 × 0.031394.. giving [0.48,0.64]

pˆ =

(c)

We assume that the sample mean is normally distributed

(a)

(i)

1 3 2 3 × × ×3 = 6 5 4 20 1 3 2 6 P(20p,10p,5p) = × × × 6 = 6 5 4 20 1 2 1 1 P (20p,5p,5p) = × × × 3 = 6 5 4 20 3 2 1 1 P (10p,10p,10p) = × × = 6 5 4 20 3 2 2 6 P (10p,10p,5p) = × × × 3 = 6 5 4 20 3 2 1 3 P (10p,5p,5p) = × × × 3 = 6 5 4 20

P (20 p,10 p,10 p) =

B1 B1

M1A1 A1 A1 B1 B1 M1A1 A1 A1 A1 A1 A1

The sampling distribution of T is t P(T = t)

20p 3/20

25p 6/20

30p 2/20

35p 6/20

40p 3/20 M1A1

22

4.

5.

(a)

Σx = 18 ; Σx 2 = 312 si UE of μ = 1.5 312 182 − UE of σ 2 = 11 11 × 12 = 25.909…

B1 B1 M1 A1

(b)

DF = 11 si B1 At the 95% confidence level, critical value = 2.201 B1 The 95% confidence limits are 25.909.. M1A1 1.5 ± 2.201 12 [Only award M1 if t-distribution used and square root present] [FT values from (a)] giving [ – 1.7,4.7] A1

(c)

The claim is justified since all the possible means within the interval are less than 5 in modulus. [FT from confidence interval] B1

(a)

H 0 : μ x = μ y ; H1 : μ x ≠ μ y

B1

(b)

x = 1.1013..; y = 1.1506..

B1B1

92.4 82.6 2 − = 0.01932.. (0.01906..) 74 74 × 75 102.2 86.32 2 sy = − = 0.03915... (0.03863) 74 74 × 75 s x2 =

[Accept division by 75] 0.01932.. 0.03915 SE = + (= 0.02792.., 0.02773) 75 75 1.1506 − 1.1013 [M0 if no square root] Test stat = 0.02792 = 1.77 (accept 1.8 or 1.79) Prob from tables = 0.0384 (0.0375, 0.0367) p-value = 0.0768 (0.075, 0.734) [FT from line above] (c)

B1 B1

M1A1 M1 A1 A1 B1

Insufficient evidence to state there is a difference in mean life-times. B1

23

6.

(a)

(b)

(c)

(d)

E(X) = − θ + 1 − 3θ = 1 − 4θ Var(X) = θ + 1 − 3θ − (1 − 4θ ) 2 = θ + 1 − 3θ − 1 + 8θ − 16θ 2 = 2θ (3 − 8θ ) 1 − E( X ) [M1A0 if E omitted] 4 1 − (1 − 4θ ) = 4 =θ Var ( X ) Var(U) = 16 2θ (3 − 8θ ) = 16n

E(U) =

N is B(n,2θ) ; E(N) = 2nθ si 2nθ = θ [B0 if E omitted] E (V ) = 2n Var(N) = 2nθ(1 – 2θ) si Var( N ) Var(V) = 4n 2 θ (1 − 2θ ) = 2n 1 ⎛θ 3θ ⎞ +θ 2 ⎟ Var(V) – Var(U) = ⎜ − θ 2 − n⎝2 8 ⎠

(a)

A1

M1 A1 B1 B1 B1 M1 A1 M1 A1

∑ x = 210, ∑ x

B2

2

=9100, ∑ y = 14.92, ∑ xy = 554.4

[B1 for 2 or 3 correct] S xy = 554.4 − 210 × 14.92 / 6 = 32.2

S xx = 9100 − 210 / 6 = 1750 32.2 b = = 0.0184 1750 14.92 − 210 × 0.0184 a= 6 = 1.84 [Working must be seen for marks to be awarded] 9100 SE of a = 0.02 (= 0.0186..) 6 × 1750 The 90% confidence interval for α is given by 1.84 ± 1.645 × 0.0186 (1.81, 1.87) 2

(b)

M1

(> 0) 8n [FT from previous results] U is better because Var(U) < VarV) =

7.

θ

B1 M1 A1

GCSE Mathematics M1-M3 & S1-S3 Mark Scheme (Summer 2010)/JSM

24

B1

B1 B1 M1A1 M1 A1 M1A1 M1A1 cao A1

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