1

GCE ‘A’ Level H2 Maths Nov 2007 Paper 1

1) 2x2 – x – 19x2 + 3x + 2 – 1

= 2x2 – x – 19 – (x2 + 3x + 2)

x2 + 3x + 2

= x2 – 4x – 21x2 + 3x + 2 Shown

2x2 – x – 19x2 + 3x + 2 > 1

2x2 – x – 19x2 + 3x + 2 – 1 > 0

x2 – 4x – 21x2 + 3x + 2 > 0

(x + 3)(x – 7)(x + 2)(x + 1) > 0

+ – + – + –3 –2 –1 7

∴x < –3 or –2 < x < –1 or x > 7 2i) Rg = [0, ∞) ⊄ IR \{3} = Df Hence fg does not exist. Rf = IR \{0} ⊆ IR = Dg Hence gf exists.

gf(x) = g( 1

x – 3 ) = 1

(x – 3)2

∴gf : x → 1

(x – 3)2 , x ∈IR, x ≠ 3

(ii) Let y = 1

x – 3 , x ≠ 3

x – 3 = 1y

x = 1y + 3

∴f –1 : x → 1x + 3, x ∈IR , x ≠ 0

3a)

(b) ww* + 2w = 3 + 4i (a + ib)(a – ib) + 2(a+ib) = 3 +4i a2 + b2 + 2a + 2ib = 3 + 4i Comparing imag. parts: 2b = 4 ⇒ b = 2 Comparing real parts: a2 + 4 + 2a = 3

a2 + 2a + 1 = 0 (a + 1)2 = 0 a = –1 ∴w = –1 + 2i

4) 4 dIdt = 2 – 3I

4 ∫ 1

2 – 3I dI = ∫ dt

– 43 ∫

–32 – 3I dI = ∫ dt

– 43 ln | 2 – 3I | = t + c

ln | 2 – 3I | = – 34 (t + c)

| 2 – 3I | = e–3(t+c)/4

2 – 3I = ± e–3t/4 e–3c/4 = Ae–3t/4

I = 13 (2 – Ae–3t/4)

I = 2 when t = 0 ⇒ 2 = 13 (2 – A)

⇒ A = –4

∴I = 23 (1 + 2e–3t/4)

For large values of t, I → 23

5) y = 2x + 7x + 2

= 2(x + 2) + 3

x + 2

= 2 + 3

x + 2

∴A = 2, B = 3

y = 1x

translate left by 2 units

y = 1

x + 2

stretch // y–axis by factor 3

y = 3

x + 2

translate up by 2 units

y = 2 + 3

x + 2 = 2x + 7x + 2

6i) →OA •

→OB

= ⎝⎜⎛

⎠⎟⎞1

–12

• ⎝⎜⎛

⎠⎟⎞2

41

= 2 – 4 + 2 = 0

Hence OA ⊥ OB.

(ii) →

OM = 2→OA +

→OB

3

= 13 (2

⎝⎜⎛

⎠⎟⎞1

–12

+ ⎝⎜⎛

⎠⎟⎞2

41

)

= 13 ⎝⎜⎛

⎠⎟⎞4

25

= ⎝⎜⎛

⎠⎟⎞4/3

2/35/3

(iii) Area of triangle OAC

= 12 |

→OA ×

→OC |

= 12 ⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞1

–12

× ⎝⎜⎛

⎠⎟⎞–4

22

= 12 ⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞–2 – 4

–(2 + 8)2 – 4

= 12 ⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞–6

–10–2

= ⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞3

51

= 32 + 52 + 12 = 35 7i) A second root is z = re–iθ

A quadratic factor of P(z) is (z – reiθ)(z – re–iθ) = z2 – z(reiθ + re–iθ) + reiθ re–iθ = z2 – rz(cos θ + i sin θ + cos θ – i sin θ) + r2 eiθ–iθ = z2 – 2rz cos θ + r2 Shown (ii) z6 = –64 = 64eiπ = 26 ei(π+2kπ)

z = 2e i(π6 +

kπ3 )

, k = –3, –2, –1, 0, 1, 2 = 2e–5iπ/6, 2e–iπ/2, 2e–iπ/6, 2eiπ/6, 2eiπ/2, 2e5iπ/6 (iii) z6 + 64

= (z2 – 4z cos π6 + 22)(z2 – 4z cos

π2 + 22)(z2 – 4z cos

5π6 + 22)

= (z2 – 4z 3

2 + 4)(z2 + 4)(z2 +

4z 3

2 + 4)

= (z2 – 2 3 z + 4)(z2 + 4)(z2 + 2 3 z + 4)

–2

3

13

x = –2

y = 2(0,

72 )

(– 72 , 0)

2

8) →AB =

⎝⎜⎛

⎠⎟⎞–2

31

– ⎝⎜⎛

⎠⎟⎞1

24

= ⎝⎜⎛

⎠⎟⎞–3

1–3

Equation of l is r = ⎝⎜⎛

⎠⎟⎞1

24

+ λ⎝⎜⎛

⎠⎟⎞–3

1–3

Substitute eqn of l into eqn of p:

⎝⎜⎜⎛

⎠⎟⎟⎞

1 – 3λ2 + λ4 – 3λ

• ⎝⎜⎛

⎠⎟⎞3

–12

= 17

3 – 9λ – 2 – λ + 8 – 6λ = 17 16λ = –8

λ = – 12

∴point of intersection

= (1 + 32 , 2 –

12 , 4 +

32 )

= (2.5, 1.5, 5.5) (ii) sin θ

=⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞–3

1–3

• ⎝⎜⎛

⎠⎟⎞3

–12

32+12+32 32+12+22

= ⎪⎪⎪

⎪⎪⎪–9 – 1 – 6

19 14=

16 19 14

θ = 78.8°

(iii) Dist. = ⎪⎪⎪

⎪⎪⎪

⎝⎜⎛

⎠⎟⎞1

24

• ⎝⎜⎛

⎠⎟⎞3

–12

– 17

32 + 12 + 22

= ⎪⎪⎪

⎪⎪⎪3 – 2 + 8 – 17

14=

814

= 2.14

9i)

From GC α = 0.619, β = 1.512 (ii) Let the sequence converge to a value x. i.e. xn → x as n → ∞

Then x satisfies x = 13 ex

i.e. 3x = ex

i.e. ex – 3x = 0 Hence x = either α or β (iii)

If x1 = 0 or 1, the sequence converges to the value α = 0.619 If x1 = 2, the sequence diverges. (iv) xn+1 – xn

= 13 e

xn – xn

= 13 (e

xn – 3xn)

From the graph, this is < 0 if α < xn < β but > 0 if xn < α or xn > β Hence xn+1 < xn if α < xn < β, xn+1 > xn if xn<α or xn>β (v) Since 0 < α = 0.619, if x1 = 0, the sequence increases in value and approaches the value α = 0.619. Since α < 1 < β, if x1 = 1, the sequence decreases in value and approaches the value α = 0.619. Since 2 > β = 1.512, if x1 = 2, the sequence increases in value and approaches ∞.

10) a + 3d = ar ⇒ d = 13 (ar – a)

a + 5d = ar2 ⇒ d = 15 (ar2 – a)

∴15 (ar2 – a) =

13 (ar – a)

3(r2 – 1) = 5(r – 1) 3r2 – 5r + 2 = 0 Shown (ii) (3r – 2)(r – 1) = 0

r = 23 , 1 (rejected since d =

13 (ar – a) is non–zero)

Since | r | = 23 < 1, the geometric

series is convergent.

S∞ = a

1 – 23

= 3a

(iii) d = 13 (

23 a – a) = –

a9

S = n2 [2a – (n – 1)

a9 ] > 4a

n2 (

199 –

n9 ) > 4

19n – n2 > 72 n2 – 19n + 72 < 0

From GC, 5.2 < n < 13.8 Hence set of possible values of n = {6, 7, 8, ..., 13}. 11i)

(ii)

dxdt = 2 cos t (– sin t)

dydt = 3 sin2 t cos t

dydx =

3 sin2 t cos t– 2 cos t sin t = –

32 sin t

Equation of tangent is

y – sin3 θ = – 32 sin θ (x – cos2 θ)

When y = 0: 32 sin θ (x – cos2 θ) = sin3 θ

x – cos2 θ = 23 sin2 θ

x = cos2 θ + 23 sin2 θ

When x = 0:

y – sin3 θ = 32 sin θ cos2 θ

1

1

3

y = sin3 θ + 32 sin θ cos2 θ

Area of triangle OQR

= 12 OQ × OR

= 12 (cos2 θ +

23 sin2 θ)(sin3 θ +

32

sin θ cos2 θ)

= 112 sin θ (3 cos2 θ + 2 sin2 θ)(2

sin2 θ + 3 cos2 θ)

= 112 sin θ (3 cos2 θ + 2 sin2 θ)2

(iii) Area = ∫10 y dx

= ∫0π/2 sin3 t (–2 cos t sin t) dt

When x = 0, t = π2

When x = 1, t = 0

= 2 ∫π/20 cos t sin4 t dt Shown

Let u = sin t

dudt = cos t

When t = 0, u = 0

When t = π2 , u = 1

= 2 ∫10 u4 du

= 2 [ u5

5 ]10

= 25

GCE ‘A’ Level H2 Maths Nov 2007 Paper 2 1) Let x, y, z = price of 1 kg of pineapples, mangoes, lychees respectively 1.15x + 0.6y + 0.55z = 8.28 1.2x + 0.45y + 0.3z = 6.84 2.15x + 0.9y + 0.65z = 13.05 Augmented matrix =

⎣⎢⎡

⎦⎥⎤1.15 0.6 0.55 8.28

1.2 0.45 0.3 6.842.15 0.9 0.65 13.05

Rref =

⎣⎢⎡

⎦⎥⎤1 0 0 3.5

0 1 0 2.60 0 1 4.9

∴Lee Lian paid 1.3×3.5 + 0.25×2.6 + 0.5×4.9 = $7.65 2i)

Let Pn be the statement: un = 1n2

When n = 1: LHS = u1 = 1

RHS = 112 = 1 = LHS

∴P1 is true. Assume that Pk is true for a k ∈

Z+ i.e. uk = 1k2

Prove that Pk+1 is also true i.e.

uk+1 = 1

(k + 1)2

LHS = uk+1

= uk – 2k + 1

k2(k + 1)2

= 1k2 –

2k + 1k2(k + 1)2

= (k + 1)2 – (2k + 1)

k2(k + 1)2

= k2

k2(k + 1)2

= 1

(k + 1)2 = RHS

∴Pk true ⇒ Pk+1 true also. Since P1 is true, and Pk true ⇒ Pk+1 true, by Math Induction, Pn is true for all n ∈ Z+

(ii) ∑n=1

N

2n + 1n2(n + 1)2

= ∑n=1

N un – un+1

= u1 – u2 + u2 – u3 + u3 – u4 : + uN – uN+1 = u1 – uN+1

= 1 – 1

(N + 1)2

= (N + 1)2 – 1

(N + 1)2

= N2 + 2N(N + 1)2 =

N(N + 2)(N + 1)2

(iii) As N → ∞, 1

(N + 1)2 → 0

Hence ∑n=1

N

2n + 1n2(n + 1)2

= 1 – 1

(N + 1)2 is convergent.

Sum to infinity = 1.

(iv) ∑n=2

N

2n – 1n2(n – 1)2

= 3

2212 + 5

3222 +...+ 2N – 1

N2(N – 1)2

= ∑n=1

N–1

2n + 1n2(n + 1)2

= 1 – 1

N2

3i) Let y = (1 + x)n

dydx = n(1 + x)n–1

d2ydx2 = n(n – 1)(1 + x)n–2

d3ydx3 = n(n – 1)(n – 2)(1 + x)n–3

When x = 0, y = 1, dydx = n,

d2ydx2 =

n(n – 1), d3ydx3 = n(n – 1)(n – 2).

∴y = 1 + nx + n(n – 1)

2! x2 +

n(n – 1)(n – 2)3! x3 +...

(ii) (4 – x)3/2(1 + 2x2)3/2

= 43/2(1 – x4 )3/2(1 +

32 2x2 +...)

4

= 8(1 + 32 (–

x4 ) +

32

12

2! (– x4 )2 +

32

12⎝⎛

⎠⎞–

12

3! (– x4 )3 +...)(1 + 3x2 +...)

= 8(1 – 3x8 +

3x2

128 + x3

1024 +...) (1

+ 3x2 +...)

= 8(1– 3x8 +

387x2

128 – 1151x3

1024 +...)

= 8 – 3x + 387x2

16 – 1151x3

128 +...

(iii) The expansion is valid

for | x4 | < 1 and | 2x2 | < 1

| x | < 4 and | x | < 12

i.e. – 12 < x <

12

4i) ∫5π/30 sin2 x dx

= ∫5π/30

1 – cos 2x2 dx

= 12 [ x –

sin 2x2 ]5π/3

0

= 12 [

5π3 +

34 ]

= 5π6 +

38

∫5π/30 cos2 x dx

= ∫5π/30 1 – sin2 x dx

= [ x ]5π/30 – [

5π6 +

38 ]

= 5π3 –

5π6 –

38 =

5π6 –

38

(iia) Area

= ∫π/20 x2 sin x dx

= [–x2 cos x ]π/20 – ∫π/2

0 –2x cos

x dx

= 0 + 2 ∫π/20 x cos x dx

= 2[ x sin x]π/20 – 2 ∫π/2

0 sin x dx

= 2[ π2 ]– 2 [ – cos x ]π/2

0

= π + 2 [0 – 1] = π – 2

(b) Volume

= π∫π/20 ( x2 sin x )2 dx

= 5.391 5i) To survey shoppers in a shopping centre to find out whether they like a certain product, we can pick 10 shoppers from each age group. Quota sampling is appropriate in this situation because the age groups are well defined. A disadvantage of quota sampling is that the sample is biased. (ii) Impossible since we do not have a list of all the shoppers. 6) Let X = no. of people out of 10 with gene A. X ~ B(10, 0.24) P(X ≤ 4) = 0.933 (i) Let A = no. of people out of 1000 with gene A. A ~ B(1000, 0.24) ≈ N(240, 182.4) since n large, np = 240 > 5, nq = 760 > 5 P(230 ≤ A ≤ 260) = P(229.5 ≤ A ≤ 260.5) = 0.717 (ii) Let B = no. of people out of 1000 with gene B. B ~ B(1000, 0.003) ≈ Po(3) since n > 50, np = 3 < 5 P(2 ≤ B < 5) = 0.616

7) x = 4626150 = 30.84

s2 = 1

149 (147 691 – 46262

150 )

= 33.7259 H0 : μ = 30 H1 : μ > 30

Since p–value = 0.038 < 0.05, we reject H0 . There is sufficient evidence at the 5% level to conclude that the population mean time for a student to complete the project exceeds 30 hours. Since n = 150 is large, there is no need to assume that the population is normal. 8) Let C, T = mass of a chicken, turkey respectively. C ~ N(2.2, 0.52) T ~ N(10.5, 2.12) (i) P(3C > 7)

= P(C > 73 )

= 0.39486 = 0.395 (ii) P(3C > 7) P(5T > 55) = 0.39486 P(T > 11) = 0.160 (iii) 3C + 5T ~ N(3×2.2 + 5×10.5, 32 0.52 + 52 2.12) = N(59.1, 112.5) P(3C + 5T > 62) = 0.392 (iv) The probability in part (iii) includes the cases when e.g. a chicken costs $6 and a turkey costs $57 so that the total price is > $62. 9ia) 12! = 479 001 600 (b) 6! (2!)6 = 46 080 (iia) (12 – 1)! = 39 916 800 (b) (6 – 1)! 6! = 86 400 (c) (6 – 1)! 2 = 240

5

10)

1/2 S 1/4 S 1/2 F

1/8 S 1/4 S 3/4 F 3/4 F 1/4 S 1/8 S 3/4 F

7/8 F 1/8 S 7/8 F 7/8 F

(i) P(SSS) = 18

14

12 =

164

(ii) P(SSF) + P(SFS) +

P(FSS) + 164

= 18

14

12 +

18

34

14 +

78

18

14 +

164

= 164 +

3128 +

7256 +

164

= 17

256 + 1

64

= 21

256

(iii) P(SFS) + P(FSS)

17/256

=

3128 +

7256

17/256 = 1317

11)

The regression line is x = –

0.2597t + 66.194 When t = 300, x = –11.7 which is negative. But the concentration cannot be negative. Hence the linear model is not suitable.

(i) Correlation coefficient r = –0.994 which is closer to –1. This indicates that t and y = ln x have a high negative correlation. (ii) Regression line is ln x = –0.0123t + 4.62 ln 15 = –0.0123t + 4.62 ⇒ t = 155 min.