2007 Paper 1 Solutions

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

∴x < –3 or –2 < x < –1 or x > 7

2(i) Rg = [0, ∞) ⊄ Df

Hence fg does not exist.

Rf = IR \{0} ⊆ Dg Hence gf exists.

gf(x) = g( 1

x – 3 ) = 1

(x – 3)2 ∴gf : x → 1

(x – 3)2 , x ∈IR, x ≠ 3

+ – + – + –3 –2 –1 7

(ii) Let y = 1

x – 3 , x ≠ 3

x – 3 = 1y

x = 1y + 3

∴f –1 : x → 1x + 3, x ∈IR , x ≠ 0

3(a) (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

–2

3

13

(1) Check that the circle passes through origin. When z = 0, |2‐3i| = sq rt 13. Drawing accurately using the same scale in both axes also help realize this. (2) Label centre and radius. 

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 . The current tends towards 2/3.

Alternatively, you can solve for 4 ln 43

c = − and then obtain

| 2 – 3I | = 344

te−

which can be simplified to either

2 – 3I = 344

te−

or

−( 2 – 3I ) = 344

te−

. The latter is correct as t = 0 , I = 2 satisfy the equation.

5. y = 2x + 7x + 2 =

2(x + 2) + 3x + 2 = 2 +

3x + 2 ∴A = 2, B = 3

y = 1x

↓ translate 2 units in the negative x-direction

y = 1

x + 2

↓ scale // y–axis by factor of 3

y = 3

x + 2 + 2

↓ translate 2 units in the positive y-direction

5. y = 2 + 3

x + 2 = 2x + 7x + 2

x = –2

y = 2 (0,

72 )

(– 72 , 0)

Your description of the transformations should be formal rather than left right up down.  

6(i) →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

(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

Check your cross product by performing a dot product with the OA and OC vectors and obtaining zero will ensure that the answer is right.

7(i) 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 (iii) z = 2e–5iπ/6, 2e–iπ/2, 2e–iπ/6, 2eiπ/6, 2eiπ/2, 2e5iπ/6

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)

Argument of −64 is π and modulus is 1. k can also be from 0 to 5. Both acceptable though here it is better to have the angles in + and − relationship.

Question says non-trigo from

means evaluate cos π6 .

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 – 3λ

2 + 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) cos θ =⎪⎪⎪⎪

⎪⎪⎪⎪

⎝⎜⎜⎛

⎠⎟⎟⎞–3

1–3

• ⎝⎜⎜⎛

⎠⎟⎟⎞3

–12

32+12+32 32+12+22 = ⎪⎪⎪

⎪⎪⎪–9 – 1 – 6

19 14=

16 19 14

θ = 11.180° Answer = 90° − 11.180 = 78.8° (to 1 decimal place)

(iii) One point, C on plane p is (1, 0, 7).

1 1 02 0 24 7 3

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

Distance = 0 3

1n 2 19 1 4 3 2

CA∧

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

i i = 814

or 2.14 (to 3 s.f.)

9(i) From GC α = 0.619, β = 1.512 (to 3 decimal places) (ii) Let the sequence converge to a value x.

i.e. xn → x and xn+1 → x as n → ∞

Then x satisfies x = 13 ex ⇒ 3x = ex ⇒ ex − 3x = 0.

Since α and β are the roots to ex − 3x = 0, x is 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 exn – xn =

13 (exn – 3xn)

From the graph, this is 13 (exn – 3xn) < 0 if α < xn < β

But 13 (exn – 3xn) > 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 ∞.

Key into GC and change u(nMin) = 0, 1, 2 to see the respective behaviours in the table.

In GC, when x1 = 2, the terms gets very large by x6 and error after that because it is too big for the GC to display.

Treat xn as x – they are just dummy variables.

This part is quite tough.

10. a + 3d = ar ⇒ d =

13 (ar – a)

a + 5d = ar2 ⇒ d = 15 (ar2 – a)

Make d the subject so that we can eliminate it to get an equation in r.

∴15 (ar2 – a) =

13 (ar – a) [a is non zero, so can cancel away on both sides]

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

(ii) (3r – 2)(r – 1) = 0

r = 23 , 1 (r = 1 is rejected since d =

13 (ar – a) = 0 but Q states d 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 [a can be cancelled away without affecting the inequality sign as a>0]

19n – n2 > 72 n2 – 19n + 72 < 0. Show graph below.

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

Here you have to give a reason why r = 1 is rejected as the question ask you to show convergent so it is obvious r =1 has to be rejected.

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

Gradient at the point is – 32 sin θ.

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 ⇒ y = sin3 θ +

32 sin θ cos2 θ

1

1 Sketched for the required domain for t. Go windows to change. It is a curve, not a straight line. Label intercepts.

1

1

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

When x = 0, t = π2

When x = 1, t = 0

= ∫ 0π/2 sin3 t (–2 cos t sin t dt)

= 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

When x = 0, cos 0t = so t corresponds to 2π .

When x = 1, cos 1t = so t corresponds to 0.

2007 Paper 2 Solutions

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

⎣⎢⎢⎡

⎦⎥⎥⎤1.15 0.6 0.55 8.28

1.2 0.45 0.3 6.842.15 0.9 0.65 13.05

⇒ By GC, 3.5, 2.6, 4.9x y z= = =

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

2(i) Let Pn be the statement: un = 1n2 , 1n ≥

When n = 1: LHS = u1 = 1 RHS = 112 = 1 = LHS ∴P1 is true.

Assume that Pk is true for some k 1≥ 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 [Recurrence relation given by Q]

= 1k2 –

2k + 1k2(k + 1)2 [By assumption]

= (k + 1)2 – (2k + 1)

k2(k + 1)2 = 2

22 1 2 1( 1)

k k kk

+ + − −+

= 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 Mathematical Induction, Pn is true for all 1n ≥ .

(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

(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 − 0 = 1

(iv) ∑n=2

N

2n – 1n2(n – 1)2 =

32212 +

53222 +...+

2N – 1N2(N – 1)2

Compare to

2 21

2 1( 1)

N

n

nn n=

++∑ =

32212 +

53222 +...+

2N – 1N2(N – 1)2 + 2 2

2 1( 1)N

N N++

Therefore, ∑n=2

N

2n – 1n2(n – 1)2 = ∑

n=1

N–1

2n + 1n2(n + 1)2 = 1 – 2

1( 1 1)N − +

= 1− 1

N2

[Replace N by N−1 in the sum in (ii)]

Remember it is N for the last term not n.

For sum to be convergent, it means when N goes to infinity, the sum must tends towards a finite number. The sum to infinity is equal to 1, NOT approximately 1. |r| < 1 is only a condition for sum to be convergent when it is a GP SUM. Not applicable in (iii) as it is not a GP sum!

3(i) 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 +...)

= 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 ⇒ 4 4x− < < and 22 1x < ⇒ 2 12

x < ⇒ 1 12 2

x− < <

Therefore, – 12

< x < 12

[ you can also give answer as 12

x < .]

You should also know how to derive the expansions of those in the formula list by 1st principle like ex, sin x, ln(1+x), cos x.

You can double check your answer with the expansion in the formula list!

4(i) ∫5π/30 sin2 x dx = ∫5π/3

0 1 – cos 2x

2 dx

= 12 [ x –

sin 2x2 ]5π/3

0 = 12 [

5π3 +

34 ]

= 5π6 +

38

∫5π/30 cos2 x dx = ∫5π/3

0 1 – sin2 x dx

= [ x ]5π/30 – [

5π6 +

38 ] =

5π3 –

5π6 –

38 =

5π6 –

38

(ii)(a) Area = ∫π/20 x2 sin x dx u = x2 v’ = sinx

u' = 2x, v = −cosx

= [–x2 cos x ]π/20 – ∫π/2

0 –2x cos x dx

= 0 + 2 ∫π/20 x cos x dx u = x v’ = cos x

u' = 1, v = sin x

= 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 (3 decimal places)

Give your answer to 3 decimal places hints that GC can be used to evaluate the answer!

Exact value means you have to perform integration and also evaluate

sin 3π as 3

2, etc.

You should know you can check your answer using GC!

5(i) To survey, say a sample of 50 shoppers who visit a particular shopping mall to find out whether how many times they visit the mall in a month. 50 shoppers can be 25 females and 25 males.

Quota sampling is appropriate in this situation because the sampling frame is not available (do not have information on the total population of shoppers visiting the mall) so random sampling methods cannot apply. A disadvantage of quota sampling is that the sample is biased as it is up to the surveyor to choose who he prefers to survey to meet the quota.

(ii) Impossible since we do not have a list of all the shoppers.

State the population – which is the shopper who visits the mall State the purpose of survey – which is how many times they visit the mall in a month State the mutually exclusive subgroups – females and males.

Your answer should show appreciation that sampling frame is not available so you cannot do simple random and systematic because you need the list of all the people and number them. You cannot do stratified because you do not have proportion of each subgroup in the population!

Do state possible or not. Possible as answers is not acceptable.

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) A~N(240, 182.4) since n large, np=240>5, n(1−p)= 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) B ~ Po(3) since n is large, np = 3 < 5 P(2 ≤ B < 5) = P(2 B≤ ≤ 4) = ( 4) ( 1)P B P B≤ − ≤ = 0.616

Q says using approximation, so approximation must be used. Remember to state the conditions including n is large which is often left out. Remember to do continuity correction for (i) !!

7) x = 4626150 = 30.84 (exact) = unbiased estimate for population mean

s2 = 1

149 (147 691 – 46262

150 ) = 33.7259 = 33.7 (3 s.f.)

= unbiased estimate for population variance H0 : μ = 30 H1 : μ > 30

2

30.84 3033.7359 150

XZs n

μ− −= = = 1.772

p-value = 0.038 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 because by Central Limit Theorem, X is normally distributed approximately.

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 event in (iii) includes the case in (ii). So the probability in (iii) is higher.

9(i)(a) 12! = 479 001 600 (b) 6! (2!)6 = 46 080 (ii)(a) (12 – 1)! = 39 916 800

(b) (6 – 1)! 6! = 86 400

(c) (6 – 1)! 2 = 240

It is only × 2, not ×26. It is because once the wife of the first man choose to his left, the rest of the wives has no choices. So either left or right – 2 choices.

(i) P(SSS) = 18

14

12 =

164

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

(ii) P(SSF) + P(SFS) + P(FSS) + 164

= 18

14

12 +

18

34

14 +

78

18

14 +

164 =

21256

(iii) P(A|B) = ( )( )

P A BP B∩ =

P(SFS) + P(FSS)17/256 =

3128 +

7256

17/256 = 1317

11. The regression line is x = –0.2597t + 66.194 x = −0.260 t + 66.2 (ans to 3 s.f.)

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

Here x is dependent and t is independent. Read the Q. So x on t.

(i) Correlation coefficient r = –0.994 which is closer to –1. This indicates that t and y = ln x

have a stronger linear correlation than t and x. (ii) Regression line is ln x = –0.012343t + 4.62061

ln 15 = –0.0123t + 4.62 ⇒ t = 155 min.

Here x is dependent and t is independent. Read the Q. So line of ln x on t is to be used even if you are estimating t given x = 15.