2011 Paper 1.2 Answers METHOD 1 METHOD 2IB Questionbank Mathematics Higher Level 3rd edition 12 (iii) area = ∫3 − π 0 (sin2x sinx ) d x M1 Note: Award M1 for an integral that

IB Questionbank Mathematics Higher Level 3rd edition 1

2011 Paper 1.2 Answers

1. (a) METHOD 1

f′(x) = q – 2x = 0 M1 f′(3) = q – 6 = 0 q = 6 A1 f(3) = p + 18 – 9 = 5 M1 p = –4 A1

METHOD 2

f(x) = –(x – 3)2 + 5 M1A1 = –x2 + 6x – 4 q = 6, p = –4 A1A1

(b) g(x) = –4 + 6(x – 3) – (x – 3)2 (= –31 + 12x – x2) M1A1

Note: Accept any alternative form that is correct. Award M1A0 for a substitution of (x + 3).

[6]

2. (a) A2 =

+−

−

1222

aa

a (M1)A1

(b) METHOD 1

det A2 = 4a2 + 2a – 2a = 4a2 M1 a = ±2 A1A1 N2

METHOD 2

det A = –2a M1 det A = ±4 a = ±2 A1A1 N2

[5]


3. (a)

A1

Note: Award A1 for intercepts of 0 and 2 and a concave down curve in the given domain

Note: Award A0 if the cubic graph is extended outside the domain [0, 2].

(b) ∫ =−+2

01d)2)(1( xxxkx (M1)

Note: The correct limits and = 1 must be seen but may be seen later.

∫ =++−2

0

23 1d)2( xxxxk A1 2

0

234

31

41

++− xxxk = 1 M1

1438

4 =

++−k (A1)

k = 38

A1

[6]

4. (a) AB = 22 )32(1 −+ M1

= 3488− A1

= 322 − A1


(b) METHOD 1

3π

arg,4π

z arg 2 1 −=−= z A1A1

Note: Allow 3π

and4π

.

Note: Allow degrees at this stage.

4π

3π

BOA −=

= )12π

(accept 12π

− A1

Note: Allow FT for final A1.

METHOD 2

attempt to use scalar product or cosine rule M1

22

31BOAcos

+= A1

12π

BOA = A1

[6]

5. (a)

A3

Note: Award A1 for each correct branch with position of asymptotes clearly indicated. If x = 2 is not indicated, only penalise once.


(b)

A3

Note: Award A1 for behaviour at x = 0, A1 for intercept at x = 2, A1 for behaviour for large │x│.

[6]

6. (a) CB = b – c, AC = b + c A1A1

Note: Condone absence of vector notation in (a).

(b) CBAC• = (b + c) • (b – c) M1 = │b│2 – │c│2 A1 = 0 since │b│=│c│ R1

Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.

so AC is perpendicular to CB i.e. BCA is a right angle AG [5]

7. (a) area of AOP = 21r2 sin θ A1

(b) TP = r tan θ (M1)

area of POT = 21r(r tan θ)

= 21r2 tan θ A1

(c) area of sector OAP = 21r2 θ A1

area of triangle OAP < area of sector OAP < area of triangle POT R1


θθθ tan21

21

sin21 222 rrr <<

sin θ < θ < tan θ AG [5]

8. x = 2ey ye

1− M1

Note: The M1 is for switching the variables and may be awarded at any stage in the process and is awarded independently. Further marks do not rely on this mark being gained.

xey = 2e2y – 1 2e2y – xey – 1 = 0 A1

48

e2 +±

=xxy M1A1

y = ln

+±

482xx

therefore h–1(x) = ln

++

482xx

A1

since ln is undefined for the second solution R1

Note: Accept y = ln

++

482xx

.

Note: The R1 may be gained by an appropriate comment earlier. [6]


9. (a) METHOD 1

P(3 defective in first 8) = 87

98

109

1110

1211

132

143

154

38

××××××××

M1A1A1

Note: Award M1 for multiplication of probabilities with decreasing denominators. Award A1 for multiplication of correct eight probabilities.

Award A1 for multiplying by

38

.

= 19556

A1

METHOD 2

P(3 defective DVD players from 8) =

815

511

34

M1A1

Note: Award M1 for an expression of this form containing three combinations.

=

!7!8!15

!6!5!11

!1!3!4×

M1

= 19556

A1

(b) P(9th selected is 4th defective player│3 defective in first 8) = 71

(A1)

P(9th selected is 4th defective player) = 71

19556

× M1

= 1958

A1

[7]


10. (a) let the first three terms of the geometric sequence be given by u1, u1 r, u1r2

∴ u1 = a + 2d, u1r = a + 3d and u1r2 = a + 6d (M1)

da

da

da

da

23

36

+

+=

+

+ A1

a2 + 8ad + 12d2 = a2 + 6ad + 9d2 A1

2a + 3d = 0

a = –23d AG

(b) u1 =

==29

,23

,2

211

dru

dru

d M1

r = 3 A1

geometric 4th term u1r3 =

227d

A1

arithmetic 16th term a + 15d = dd 1523

+− M1

= 2

27d A1

Note: Accept alternative methods. [8]

11. (a) 3

21

2dd

xxx

y−= A1

− 2

21

2 xx = 0

x = 0, ±2

−

=825

,2,825

,2,89

0,at 0ddx

y A1A1A1

Note: Award A2 for all three x-values correct with errors/omissions in y-values.


(b) at x = 1, gradient of tangent = 23

(A1)

Note: In the following, allow FT on incorrect gradient.

equation of tangent is y – 2 = 23

(x – 1)

+=21

23xy (A1)

meets x-axis when y = 0, –2 = 23

(x – 1) (M1)

x = 31

−

coordinates of T are

− 0,31

A1

(c) gradient of normal = –32

(A1)

equation of normal is y – 2 =

+−=−−38

32

)1(32

xyx (M1)

at x = 0, y = 38

A1

Note: In the following, allow FT on incorrect coordinates of T and N.

lengths of PN = 952

PT ,913

= A1A1

area of triangle PTN = 952

913

21

×× M1

= 913

(or equivalent e.g. 18676

) A1

[15]

12. (a) using the factor theorem z + 1 is a factor (M1) z3 + 1 = (z + 1)(z2 – z + 1) A1


(b) (i) METHOD 1

z3 = – 1 ⇒ z3 + 1 = (z + 1)(z2 – z + 1) = 0 (M1)

solving z2 – z + 1 = 0 M1

z = 2

3i12

411 ±=

−± A1

therefore one cube root of – 1 is γ AG

METHOD 2

γ2 =

23i1

23i1

2+−

=

+ M1A1

γ3 = 4

312

3i12

3i1 −−=

+−×

+− A1

= –1 AG

METHOD 3

γ = 3π

i

23i1

e=+−

A1

γ3 = eiπ – 1 A1

(ii) METHOD 1

as γ is a root of z2 – z + 1 = 0 then γ2 – γ + 1 = 0 M1R1 ∴ γ2 = γ – 1 AG

Note: Award M1 for the use of z2 – z + 1 = 0 in any way. Award R1 for a correct reasoned approach.

METHOD 2

γ2 = 2

3i1+− M1

γ – 1 = 2

3i11

23i1 +−

=−+

A1


(iii) METHOD 1

(1 – γ)6 = (–γ2)6 (M1) = (γ)12 A1

= (γ3)4 (M1) = (–1)4 = 1 A1

METHOD 2

(1 – γ)6

= 1 – 6γ + 15γ2 – 20γ3 + 15γ4 – 6γ5 + γ6 M1A1

Note: Award M1 for attempt at binomial expansion.

use of any previous result e.g. = 1 – 6γ + 15γ 2 + 20 – 15γ + 6γ 2 + 1M1 = 1 A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.

(c) METHOD 1

A2 =

+

=

2

2

10

1

10

11

0

1

γ

γγγ

γ

γ

γ

γ A1

A2 – A + I =

+−

−++−

111

0

11

1

2

2

γγ

γγγγ

M1

from part (b) γ2 – γ + 1 = 0

)1(1

11 2 +−=−+ γγ

γγγ = 0 A1

)1(1

111 2

22+−=++ γγ

γγγ = 0 A1

hence A2 – A + I = 0 AG

METHOD 2

A2 =

−−

+−

23i1

0

12

3i1

A1A1A1

Note: Award 1 mark for each of the non-zero elements expressed in this form.

verifying A2 – A + I = 0

(d) (i) A2 = A – I


⇒ A3 = A2 – A M1A1

= A – I – A A1 = –I AG

Note: Allow other valid methods.

(ii) I = A – A2 A

–1 = A–1A – A–1

A2 M1A1

⇒ A–1 = I – A AG

Note: Allow other valid methods. [20]

13. (a) (i)

A2

Note: Award A1 for correct sin x, A1 for correct sin 2x.

Note: Award A1A0 for two correct shapes with 2π

and/or 1 missing.

Note: Condone graph outside the domain.

(ii) sin 2x = sin x, 0 ≤ x ≤ 2π

2 sin x cos x – sin x = 0 M1 sin x (2 cos x – 1) = 0

x = 0, 3π

A1A1 N1N1


(iii) area = ∫ −3π

0d )sin(sin2 xxx M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with sin x and sin 2x subtracted in either order.

= 3π

0

cos2cos21

+− xx A1

=

+−−

+− 0cos0cos21

3π

cos3π2

cos21

(M1)

= 21

43−

= 41

A1

(b) ∫∫ −=

−6π

0 2

21

0 sin44

sin4d

4 θ

θx

x

x × 8 sin θ cos θ dθ M1A1A1

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of dθ, first A1 for correct limits, second A1 for correct substitution for dx.

θθdsin86π

0

2∫ A1

θθd2cos446π

0∫ − M1

= [ ]6π

02sin24 θθ − A1

= 03π

sin23π2

−

− (M1)

= 33π2− A1


(c) (i)

M1 from the diagram above

the shaded area = ∫∫ −−=ba

yyfabxxf0

1

0d)(d)( R1

= ab – ∫ −b

xxf0

1 d)( AG

(ii) f(x) = arcsin ⇒4x

f–1(x) = 4 sin x A1

∫∫ −=

6π

0

2

0dsin4

3π

d4

arcsin xxxx

M1A1A1

Note: Award A1 for the limit 6π

seen anywhere, A1 for all else correct.

= [ ]60cos4

3π

π

x−− A1

= 3243π

+− A1

Note: Award no marks for methods using integration by parts. [25]



1. area of triangle POQ = 21

82 sin 59° M1

= 27.43 (A1)

area of sector = π82

36059

M1

= 32.95 (A1) area between arc and chord = 32.95 – 27.43 = 5.52 (cm2) A1

[5]

2. u4 = u1 + 3d = 7, u9 = u1 + 8d = 22 A1A1

Note: 5d = 15 gains both above marks

u1 = –2, d = 3 A1

Sn = 2n

(–4 + (n – 1)3) > 10000 M1

n = 83 A1 [5]

3. (a) a = 10e–0.2t (M1)(A1) at t = 10, a = 1.35 (m s–2) (accept 10e–2) A1

(b) METHOD 1

d = ∫ −−10

0

2.0 d)e1(50 tt (M1)

= 283.83... A1 so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1

METHOD 2

s = ∫50(1 – e–0.2t)dt = 50t + 250e–0.2t (+ c) M1 Taking s = 0 when t = 0 gives c = –250 M1 So when t = 10, s = 283.3... so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1

[6]

4. (a) det A = cos 2θ cos θ + sin 2θ sin θ M1A1 = cos (2θ – θ) A1

Note: Allow use of double angle formulae if they lead to the correct answer

= cos θ AG


(b) cos2 θ = sin θ A1 θ = 0.666, 2.48 A1A1

[6]

5.

M1A1A1A1A1A1A1

Note: Award A1 for both vertical asymptotes correct, M1 for recognizing that there are two turning points near the origin, A1 for both turning points near the origin correct, (only this A mark is dependent on the M mark) A1 for the other pair of turning points correct, A1 for correct positioning of the oblique asymptote, A1 for correct equation of the oblique asymptote, A1 for correct asymptotic behaviour in all sections.

[7]

6. (a) P(x < 1.4) = 0.691 (accept 0.692) A1

(b) METHOD 1

Y ~ B(6, 0.3085...) (M1) P(Y ≥ 4) = 1 – P(Y ≤ 3) (M1) = 0.0775 (accept 0.0778 if 3 s.f. approximation from (a) used) A1


METHOD 2

X ~ B(6, 0.6914...) (M1) P(X ≤ 2) (M1) = 0.0775 (accept 0.0778 if 3 s.f. approximation from (a) used) A1

(c) P(x < 1 | x < 1.4) = )4.1(P

)1P(<

<

x

x M1

= ...6914.0...06680.0

= 0.0966 (accept 0.0967) A1 [6]

7. (a) x3 + 1 = 1

13 +x

(–1.26, –1) (= ( 3 2− , –1)) A1

(b) f′(–1.259...) = 4.762... (3 × 32

2 ) A1

g′(–1.259...) = –4.762... (–3 × 32

2 ) A1

required angle = 2arctan

...762.4

1 M1

= 0.414 (accept 23.7°) A1

Note: Accept alternative methods including finding the obtuse angle first. [5]

8. let the length of one side of the triangle be x consider the triangle consisting of a side of the triangle and two radii

EITHER

x2 = r2 + r2 – 2r2 cos 120° M1

= 3r2

OR

x = 2r cos 30° M1


THEN

x = 3r A1

so perimeter = 33 r A1 now consider the area of the triangle

area = 3 × 21r2sin 120° M1

= 3 × 2

43r A1

2

433

33

r

r

A

P=

= r

4 A1

Note: Accept alternative methods [6]

9. let x = distance from observer to rocket let h = the height of the rocket above the ground

METHOD 1

t

h

dd

= 300 when h = 800 A1

x = 21

22 )360000(360000 +=+ hh M1

360000dd

2 +=

h

h

h

x A1

when h = 800

t

h

h

x

t

x

dd

dd

dd

×= M1

= 360000

3002 +h

h A1

= 240 (m s–1) A1


METHOD 2

h2 + 6002 = x2 M1

2h = 2xh

x

dd

A1

x

h

h

x=

dd

=

=54

1000800

A1

t

h

dd

= 300 A1

t

h

h

x

t

x

dd

dd

dd

×= M1

= 54

× 300

= 240 (m s–1) A1

METHOD 3

x2 = 6002 + h2 M1

2xt

hh

t

x

dd

2dd

= A1A1

when h = 800, x = 1000

t

h

t

x

dd

1000800

dd

×= M1A1

= 240 m s–1 A1

METHOD 4

Distance between the observer and the rocket = 21

22 )800600( + = 1000 M1A1 Component of the velocity in the line of sight = sin θ × 300 (where θ = angle of elevation) M1A1

sin θ = 1000800

A1

component = 240 (m s–1) A1 [6]


10. 21

21

21

ayx =+

0dd

21

21 2

121

=+−−

x

yyx M1

x

y

y

x

x

y−=−=

2

12

1

dd

A1

Note: Accept 21

21

1dd

x

a

x

y−= from making y the subject of the equation,

and all correct subsequent working

therefore the gradient at the point P is given by

p

q

x

y−=

dd

A1

equation of tangent is y – q = )( pxp

q−− M1

(y = pqqxp

q++− )

x-intercept: y = 0, n = ppqpq

pq+=+ A1

y-intercept: x = 0, m = qpq + A1

n + m = ppqppq +++ M1

= qppq ++2

= 2)( qp + A1 = a AG

[8]

11. (a)

−

−

−

=

−

−

=

z

y

x

150

SR,311

PQ (M1)

point S = (1, 6, –2) A1


(b)

=

−

−

=

142

PS311

PQ A1

−

−

=×

2713

PSPQ

m = –2 A1

(c) area of parallelogram PQRS = 222 )2(7)13(PSPQ −++−=× M1

= 222 = 14.9 A1

(d) equation of plane is –13x + 7y – 2z = d M1A1 substituting any of the points given gives d = 33 –13x + 7y – 2z = 33 A1

(e) equation of line is r =

−

−

+

2713

000

λ A1

Note: To get the A1 must have r = or equivalent.

(f) 169λ + 49λ + 4λ = 33 M1

λ = 22233

(= 0.149...) A1

closest point is

−−3711

,7477

,74143

(= (–1.93, 1.04, –0.297)) A1

(g) angle between planes is the same as the angle between the normals (R1)

cos θ = 6222

1227113

×

×−−×+×− M1A1

θ = 143° (accept θ = 37.4° or 2.49 radians or 0.652 radians) A1 [17]

12. (a) P(x = 0) = 0.607 A1


(b) EITHER

Using X ~ Po(3) (M1)

OR

Using (0.6065...)6 (M1)

THEN

P(X = 0) = 0.0498 A1

(c) X ~ Po(0.5t) (M1) P(x ≥ 1) = 1 – P(x = 0) (M1) P(x = 0) < 0.01 A1 e–0.5t < 0.01 A1 –0.5t < ln (0.01) (M1) t > 9.21 months therefore 10 months A1N4

Note: Full marks can be awarded for answers obtained directly from GDC if a systematic method is used and clearly shown.

(d) (i) P(1 or 2 accidents) = 0.37908… A1 E(B) = 1000 × 0.60653... + 500 × 0.37908... M1A1 = $796 (accept $797 or $796.07) A1

(ii) P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000) + P(0, 1000, 1000) + P(1000, 500, 500) + P(500, 1000, 500) + P(500, 500, 1000) (M1)(A1)

Note: Award M1 for noting that 2000 can be written both as 2 × 1000 + 1 × 0 and 2 × 500 + 1 × 1000.

= 3(0.6065...)2(0.01437...) + 3(0.3790...)2(0.6065...) M1A1 = 0.277 (accept 0.278) A1

[18]


13. prove that 1 + 21

132

22

421

...21

421

321

−

−+

−=

++

+

+

n

nn

n

for n = 1

LHS = 1, RHS = 4 – 02

21+ = 4 – 3 = 1

so true for n = 1 R1 assume true for n = k M1

so 1 + 21

132

22

421

...21

421

321

−

−+

−=

++

+

+

k

kk

k

now for n = k + 1

LHS: 1 + 2kk

kk

++

++

+

+

−

21

)1(21

...21

421

321 132

A1

= 4 – k

kk

k

++

+− 2

1)1(

221

M1A1

= 4 – kk

kk

21

2)2(2 ++

+ (or equivalent) A1

= 4 – 1)1(2

2)1(−+

++k

k (accept 4 –

k

k

2

3+) A1

Therefore if it is true for n = k it is true for n = k + 1. It has been shown to be true for n = 1 so it is true for all n (∈ +). R1

Note: To obtain the final R mark, a reasonable attempt at induction must have been made.

[8]

14. (a) METHOD 1

∫e2x sin x dx = –cos xe2x + ∫2e2x cos x dx M1A1A1

= –cos xe2x + 2e2x sin x – ∫4e2x sin x dx A1A1 5∫e2x sin x dx = –cos xe2x + 2e2x sin x M1

∫e2x sin x dx = 51

e2x(2 sin x – cos x) + C AG

METHOD 2

∫sin xe2x dx = xxx xx

d2

ecos

2esin 22

∫− M1A1A1

= xxxx xxx

d4

esin

4e

cos2esin 222

∫−− A1A1

4ecos

2sine

dsine45 22

2xx

x xxxx −=∫ M1

Cxxxx xx +−=∫ )cossin2(e51

dsine 22 AG

(b) xxy

y x dsine1

d 2

2 ∫∫ =−

M1A1


arcsin y = 51

e2x (2 sin x – cos x)(+ C) A1

when x = 0, y = 0 51

=⇒C M1

y = sin

+−51

)cossin2(e51 2 xxx A1

(c) (i)

A1 P is (1.16, 0) A1

Note: Award A1 for 1.16 seen anywhere, A1 for complete sketch.

Note: Allow FT on their answer from (b)

(ii) V = xy dπ....162.1

0

2∫ M1A1

= 1.05 A2

Note: Allow FT on their answers from (b) and (c)(i). [25]



1. (a) (i) P(A∪ B) = P(A) + P(B) = 0.7 A1

(ii) P(A∪ B) = P(A) + P(B) – P(A∩ B) (M1) = P(A) + P(B) – P(A)P(B) (M1) = 0.3 + 0.4 – 0.12 = 0.58 A1

(b) P(A∩ B) = P(A) + P(B) – P(A∪ B) = 0.3 + 0.4 – 0.6 = 0.1 A1

P(A│B) = )P(

)P(B

BA∩ (M1)

= 4.01.0

= 0.25 A1

[7]

2. METHOD 1

z = (2 – i)(z + 2) M1 = 2z + 4 – iz – 2i z(1 – i) = – 4 + 2i

z = i1

i24−

+− A1

z = i1i1

i1i24

++

×−+−

M1

= – 3 – i A1

METHOD 2

let z = a + ib

2ii++

+ba

ba = 2 – i M1

a + ib = (2 – i)((a + 2) + ib) a + ib = 2(a + 2) + 2bi – i(a + 2) + b a + ib = 2a + b + 4 + (2b – a – 2)i attempt to equate real and imaginary parts M1 a = 2a + b + 4(⇒ a + b + 4 = 0) and b = 2b – a – 2(⇒ – a + b – 2 = 0) A1

Note: Award A1 for two correct equations.

b = –1;a = –3 z = – 3 – i A1

[4]


3. (a) u1 = 27

r−=

127

281

M1

r = 31

A1

(b) v2 = 9 v4 = 1 2d = – 8 ⇒ d = –4 (A1) v1 = 13 (A1)

2N

(2 × 13 – 4(N – 1)) > 0 (accept equality) M1

2N

(30 – 4N) > 0

N(15 – 2N) > 0 N < 7.5 (M1) N = 7 A1

Note: 13 + 9 + 5 + 1 – 3 – 7 – 11 > 0 ⇒ N = 7 or equivalent receives full marks.

[7]

4. (a) AB = b – a A1

CB = a + b A1

(b) CBAB• = (b – a)•(b + a) M1 = │b│2–│a│2 A1 = 0 since │b│=│a│ R1

Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.

so AB is perpendicular to CB i.e. CBA is a right angle AG [5]


5. (a) 1cos21

cossin22cos1

2sin2 −+

=+ θ

θθθ

θ M1

Note: Award M1 for use of double angle formulae.

= θ

θθ2cos2

cossin2 A1

= θθ

cossin

= tan θ AG

(b)

4π

cos1

4π

sin

8π

tan+

= (M1)

4π

sin

4π

cos1

8π

cot+

= M1

=

2222

1+

= 1 + 2 A1 [5]

6. R is ‘rabbit with the disease’ P is ‘rabbit testing positive for the disease’


(a) P(P) = P(R∩ P) + P(R′∩ P) = 0.01 × 0.99 + 0.99 × 0.001 M1 = 0.01089 (= 0.0109) A1

Note: Award M1 for a correct tree diagram with correct probability values shown.

(b) P(R′|P) =

=×+×

×

01089.000099.0

99.001.099.0001.099.0001.0

M1A1

01.0001.0

01089.000099.0

< = 10 % (or other valid argument) R1

[5]

7. METHOD 1

area = ∫3

0darctan xx A1

attempting to integrate by parts M1

= xx

xxx d1

1]arctan[

2

3

0

30

+− ∫ A1A1

= 3

0

230 )1ln(

21

]arctan[

+− xxx A1

Note: Award A1 even if limits are absent.

= 4ln21

3

π− A1

−= 2ln

33π

METHOD 2

area = ∫− 3π

0dtan

33π

yy M1A1A1

= 3π

0]cos[ln33π

y+ M1A1

=

−=+ 2ln

33π

21

ln33π

A1

[6]

8. (a) (i) (g ○ f)(x) = 23

,32

1−≠

+x

x (or equivalent) A1

(ii) (f ○ g)(x) = x

2 + 3, x ≠ 0 (or equivalent) A1


(b) EITHER

f(x) = (g–1 ○ f ○ g)(x)⇒ (g ○ f)(x) = (f ○ g)(x) (M1)

32

321

+=+ xx

A1

OR

(g–1 ○ f ○ g)(x) = 3

21

+x

A1

2x + 3 = 3

21

+x

M1

THEN

6x2 + 12x + 6 = 0 (or equivalent) A1 x = –1, y = 1 (coordinates are (–1, 1)) A1

[6]

9. Attempt at implicit differentiation M1

e(x+y)

+−=

+ y

x

yxxy

x

y

dd

)sin(dd

1 A1A1

let x = 0, y = 0 M1

e0

+

x

y

dd

1 = 0

x

y

dd

= –1 A1

let x = π2,π2 −=y

e0

+−−=

+ y

x

yx

x

y

dd

π)2sin(dd

1 = 0

so x

y

dd

= –1 A1

since both points lie on the line y = –x this is a common tangent R1

Note: y = –x must be seen for the final R1. It is not sufficient to note that the gradients are equal.

[7]


10. (a) f (x – a) ≠ b (M1) x ≠ 0 and x ≠ 2a (or equivalent) A1

(b) vertical asymptotes x = 0, x = 2a A1 horizontal asymptote y = 0 A1

Note: Equations must be seen to award these marks.

maximum

−b

a1

, A1A1

Note: Award A1 for correct x-coordinate and A1 for correct y-coordinate.

one branch correct shape A1 other 2 branches correct shape A1

[8]

11. (a)

−=

−

−

=

13

4AC ,

314

AB A1A1

Note: Accept row vectors.


(b)

=

−

−−=×

16168

134314ACABkji

M1A1

normal n =

221

so r •

•

=

221

121

221

(M1)

x + 2y + 2z = 7 A1

Note: If attempt to solve by a system of equations: Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer.

(c) r =

+

221

735

λ (or equivalent) A1

1(5 + λ) + 2(3 + 2λ) + 2(7 + 2λ) = 7 M1 9λ = –18 λ = –2 A1

Note: λ = –

16168

if41

is used.

distance = 222 2212 ++ (M1) = 6 A1

(d) (i) area = 222 1616821

ACAB21

++=× (M1)

= 12 (accept 57621

) A1

(ii) EITHER

volume = 31

× area × height (M1)

= 31

× 12 × 6 = 24 A1

OR

volume = ( ))ACAB(AD61

×• M1

= 24 A1

(e) 222 16168ACAB ++=×


614134ADAC −=×

kji

M1

= │–19i – 20j + 16k│ A1

EITHER

222222 1616821

16201921

++>++ M1

therefore since area of ACD bigger than area ABC implies that B is closer to opposite face than D R1

OR

correct calculation of second distance as 222 162019

144

++ A1

which is smaller than 6 R1

Note: Only award final R1 in each case if the calculations are correct. [19]

12. (a) (i) f′(x) = 2

ln1

x

xx

x − M1A1

= 2

ln1

x

x−

so f′(x) = 0 when ln x = 1, i.e. x = e A1

(ii) f′(x) > 0 when x < e and f′(x) < 0 when x > e R1 hence local maximum AG

Note: Accept argument using correct second derivative.

(iii) y ≤ e1

A1


(b) f′″(x) = 4

2 2)ln1(1

x

xxx

x −−−

M1

= 4

ln22

x

xxxx +−−

= 3

ln23

x

x+− A1

Note: May be seen in part (a).

f″(x) = 0 –3 + 2 ln x = 0 M1

x = 23

e

since f″(x) < 0 when x < 23

e and f″(x) > 0 when x > 23

e R1

then point of inflexion

23

23

2e

3,e A1

(c)

A1A1A1

Note: Award A1 for the maximum and intercept, A1 for a vertical asymptote and A1 for shape (including turning concave up).


(d) (i)

A1A1

Note: Award A1 for each correct branch.

(ii) all real values A1

(iii)

(M1)(A1)

Note: Award (M1)(A1) for sketching the graph of h, ignoring any graph of g.

–e2 < x < –1 (accept x < –1) A1 [19]

13. (a) (cos θ + i sin θ)3 = cos3 θ + 3cos2 θ(i sin θ) + 3 cos θ(isin θ)2 + (isin θ)3 (M1) = cos3 θ – 3 cos θ sin2 θ + i(3 cos2 θ sin θ – sin3 θ) A1


(b) from De Moivre’s theorem (cos θ + i sin θ)3 = cos 3θ + i sin 3θ (M1) cos 3θ + i sin 3θ = (cos3 θ – 3 cos θ sin2

θ) + i(3 cos2 θ sin θ – sin3 θ)

equating real parts cos 3θ = cos3 θ – 3 cos θ sin2 θ M1 = cos3 θ – 3 cos θ (1 – cos2 θ) A1 = cos3 θ – 3 cos θ + 3 cos3 θ = 4 cos3 θ – 3 cos θ AG

Note: Do not award marks if part (a) is not used.

(c) (cos θ + i sin θ)5 = cos5 θ + 5 cos4 θ (i sin θ) + 10 cos3 θ(i sin θ)2 + 10 cos2 θ(i sin θ)3 + 5cos θ (i sin θ)4 + (i sin θ)5 (A1)

from De Moivre’s theorem cos 5θ = cos5 θ – 10 cos3 θ sin2 θ + 5 cos θ sin4 θ M1 = cos5 θ – 10 cos3 θ (1 – cos2 θ) + 5cos θ(1 – cos2 θ)2 A1 = cos5 θ – 10 cos3 θ + 10 cos5 θ + 5 cos θ – 10 cos3 θ + 5 cos5 θ ∴ cos 5θ = 16 cos5 θ – 20 cos3

θ + 5 cos θ AG

Note: If compound angles used in (b) and (c), then marks can be allocated in (c) only.

(d) cos 5θ + cos 3θ + cos θ = (16 cos5 θ – 20 cos3 θ + 5 cos θ) + (4 cos3 θ – 3 cos θ) + cos θ = 0 M1 16 cos5 θ – 16 cos3 θ + 3 cosθ = 0 A1 cos θ (16 cos4 θ – 16 cos2 θ + 3) = 0 cos θ (4 cos2 θ – 3)(4 cos2 θ – 1) = 0 A1

21

;23

;0cos ±±=∴ θ A1

2π

;3π

;6π

±±±=∴θ A2


(e) cos 5θ = 0

5θ = ... ...;27π

;25π

;23π

;2π

(M1)

θ = ... ...;107π

;105π

;103π

;10π

(M1)

Note: These marks can be awarded for verifications later in the question.

now consider 16 cos5 θ – 20 cos3 θ + 5 cos θ = 0 M1 cos θ (16 cos4 θ – 20 cos2 θ + 5) = 0

cos2θ = 32

)5)(16(440020 −±; cos θ = 0 A1

cos θ = 32

)5)(16(440020 −±±

32

)5)(16(440020

10π

cos−+

±= since max value of cosine ⇒ angle

closest to zero R1

855

8.4

)5(42545.4

10π

cos+

=−+

= A1

855

107π

cos−

−= A1A1

[22]



1. (a) (i) median = 104 grams A1

Note: Accept 105.

(ii) 30th percentile = 90 grams A1

(b) 80 – 49 (M1) = 31 A1

Note: Accept answers 30 to 32. [4]

2. (a) f ′(x) = 3x2 – 6x – 9 (= 0) (M1) (x + 1)(x – 3) = 0 x = –1; x = 3 (max)(–1, 15); (min)(3, –17) A1A1

Note: The coordinates need not be explicitly stated but the values need to be seen.

y = –8x + 7 A1 N2

(b) f″(x) = 6x – 6 = 0 ⇒ inflexion (1, –1) A1 which lies on y = – 8x + 7 R1AG

[6]

3. METHOD 1

540sin

7sin

=C

M1(A1)

DCB = 64.14...° A1 CD = 2 × 5cos 64.14... M1

Note: A1so allow use of sine or cosine rule.

CD = 4.36 A1


METHOD 2

let AC = x cosine rule 52 = 72 + x2 – 2 × 7 × x cos 40 M1A1 x2 – 10.7... x + 24 = 0

x = 2

244...)7.10(...7.10 2 ×−± (M1)

x = 7.54; 3.18 (A1) CD is the difference in these two values = 4.36 A1

Note: Other methods may be seen. [5]

4. (a) f(a) = 4a3 + 2a2 – 7a = –10 M1

4a3 + 2a2 – 7a + 10 = 0 (a + 2) (4a2 – 6a + 5) = 0 or sketch or GDC (M1) a = –2 A1

(b) substituting a = –2 into f(x) f(x) = 4x3 – 4x + 14 = 0 A1

EITHER

graph showing unique solution which is indicated (must include max and min) R1

OR

convincing argument that only one of the solutions is real (–1.74, 0.868±1.12i) R1

[5]

5. (a) 2x2 + x – 3 = (2x + 3)(x – 1) A1

Note: Accept 2

+23

x (x – 1).

Note: Either of these may be seen in (b) and if so A1 should be awarded.


(b) EITHER

(2x2 + x – 3)8 = (2x + 3)8(x – 1)8 M1 = (38 + 8(37)(2x) + ...)((–1)8 + 8(–1)7(x) + ...) (A1) coefficient of x = 38 × 8 × (–1)7 + 37 × 8 × 2 × (–1)8 M1 = –17 496 A1

Note: Under FT, final A1 can only be achieved for an integer answer.

OR

(2x2 + x – 3)8 = (3 – (x – 2x2))8 M1 = 38 + 8(–(x – 2x2)(37) + ...) (A1) coefficient of x = 8 × ( – 1) × 37 M1 = –17 496 A1

Note: Under FT, final A1 can only be achieved for an integer answer. [5]


6.

α = 2 arcsin ⇒

(75.4

α = 1.396... = 80.010° ...) M1(A1)

β = 2 arcsin ⇒

(55.4

β = 2.239... = 128.31°...) (A1)

Note: Allow use of cosine rule.

area P = 21

× 72 × (α – sin α) = 10.08... M1(A1)

area Q = 21

× 52 × (β – sin β) = 18.18... (A1)

Note: The M1 is for an attempt at area of sector minus area of triangle.

Note: The use of degrees correctly converted is acceptable.

area = 28.3(cm2) A1 [7]

7. (a) 1dd)2( 34

0

0

2

2 =++ ∫∫− xkxxk M1

134

38

=+kk

k = 41

A1

Note: Only FT on positive values of k.


(b) (i) E(X) = ∫∫ ++−

34

0

0

2

2 d41

d)2(41

xxxxx M1

= 92

34

41

+−

×

= 91

− (= –0.111) A1

(ii) median given by a such that P (X < a) = 0.5

5.0d)2(41

2

2 =+∫− xxa

M1

a

x

2

3

3)2(

−

+ = 2 (A1)

(a + 2)3 – 0 = 6 a = 3 6 – 2 (= –0.183) A1

[7]

8. (a) equation of line in graph a = t6025

− + 15 A1

+−= 15125ta

(b) 15125

dd

+−= tt

v (M1)

v = 2

245t− + 15t + c (A1)

when t = 0, v = 125 m s–1

v = 2

245t− + 15t + 125 A1

from graph or by finding time when a = 0 maximum = 395 m s–1 A1


(c) EITHER

graph drawn and intersection with v = 295 m s–1 (M1)(A1) t = 57.91 – 14.09 = 43.8 A1

OR

295 = 2

245t− + 15t + 125 ⇒ t = 57.91...; 14.09... (M1)(A1)

t = 57.91… – 14.09… = 43.8 (8 30 ) A1 [8]

9. logx+1 y = 2

logy+1 x = 41

so (x + 1)2 = y A1

41

)1( +y = x A1

EITHER

x4 – 1 = (x + 1)2 M1 x = –1, not possible R1 x = 1.70, y = 7.27 A1A1


OR

41

2 )22( ++ xx – x = 0 M1 attempt to solve or graph of LHS M1 x = 1.70, y = 7.27 A1A1

[6]

10. METHOD 1

equation of journey of ship S1

r1 =

2010

t

equation of journey of speedboat S2 ,setting off k minutes later

r2 =

−−+

3060

)(3070

kt M1A1A1

Note: Award M1 for perpendicular direction, A1 for speed, A1 for change in parameter (e.g. by using t – k or T, k being the time difference between the departure of the ships).

solve

−−+

=

3060

)(3070

2010

ktt (M1)

Note: M mark is for equating their two expressions.

10t = 70 – 60t + 60k 20t = 30 + 30t – 30k M1

Note: M mark is for obtaining two equations involving two different parameters.

7t – 6k = 7 –t + 3k = 3

k = 1528

A1

latest time is 11:52 A1


METHOD 2

x

y

time takent - k

timetaken t

S (26,52)22 5

26 5

10 58

(A)O

B (70,30)

SB = 22 5 M1A1 (by perpendicular distance) SA = 26 5 M1A1 (by Pythagoras or coordinates)

t = 510

526 A1

t – k = 530

522 A1

k = 1528

leading to latest time 11:52 A1

[7]

11. (a)

=

−

−

kz

y

x

13

212311120

212311120

−

− = 0 – 2( – 2 + 6) + ( – 1 + 2) = –7 M1A1

since determinant ≠ 0 ⇒ unique solution to the system planes R1 intersect in a point AG

Note: For any method, including row reduction, leading to the explicit

solution

−+−

721

,7

10,

756 kkk

, award M1 for an attempt at

a correct method, A1 for two correct coordinates and A1 for a third correct coordinate.


(b) 212

31112

+−

+−

a

a

a

= a((a + 1)(a + 2) – 3) – 2(–1(a + 2) + 6) + (–1 + 2(a + 1)) M1(A1) planes not meeting in a point ⇒ no unique solution i.e. determinant = 0(M1) a(a2 + 3a – 1) + (2a – 8) + (2a + 1) = 0 a3 + 3a2 + 3a – 7 = 0 A1 a = 1 A1

[5]

(c) 21

31244403121 rr

k

+

−

M1

312

655044403121 rr

k

+

+

(A1)

23 54

4400044403121 rr

k

−

+

(A1)

for an infinite number of solutions to exist, 4 + 4k = 0 ⇒ k = –1 A1

x + 2y + z = 3 y + z = 1 M1

−+

=

11

1

011

λz

y

x

A1

Note: Accept methods involving elimination.

Note: Accept any equivalent form e.g.

−+

−

=

−+

=

11

1

120

or 11

1

102

λλz

y

x

z

y

x

.

Award A0 if =

z

y

x

or r = is absent.

[14]


12. (a) P(X < 30) = 0.4 P(X < 55) = 0.9 or relevant sketch (M1)

given Z = σµ−X

P(Z < z) = 0.4 σµ−

⇒30

= –0.253... (A1)

P(Z < z) = 0.9 σµ−

⇒55

= 1.28... (A1)

µ = 30 + (0.253...) × σ = 55 – (1.28...) × σ M1 σ = 16.3, µ = 34.1 A1

Note: Accept 16 and 34.

Note: Working with 830 and 855 will only gain the two M marks.

(b) X ~ N(34.12..., 16.28...2) late to school ⇒ X > 60 P(X > 60) = 0.056... (A1) number of students late = 0.0560... × 1200 (M1) = 67 (to nearest integer) A1

Note: Accept 62 for use of 34 and 16.

(c) P(X > 60 | X > 30) = )30P()60P(

>

>

X

X M1

= 0.0935 (accept anything between 0.093 and 0.094) A1

Note: If 34 and 16 are used 0.0870 is obtained. This should be accepted.

(d) let L be the random variable of the number of students who leave school in a 30 minute interval since 24 × 30 = 720 A1 L ~ Po(720) P(L ≥ 700) = 1 – P(L ≤ 699) (M1) = 0.777 A1

Note: Award M1A0 for P (L > 700) = 1 – P(L ≤ 700) (this leads to 0.765).

(e) (i) Y ~ B(200, 0.7767…) (M1) E(Y) = 200 × 0.7767… = 155 A1

Note: On FT, use of 0.765 will lead to 153.


(ii) P(Y > 150) = 1 – P(Y ≤ 150) (M1) = 0.797 A1

Note: Accept 0.799 from using rounded answer.

Note: On FT, use of 0.765 will lead to 0.666. [17]

13. (a) A2 =

−

− θθθθ

θθθθ

cossinsincos

cossinsincos

=

+−−−

+−

θθθθθθθθθθθθ

22

22

cossinsincoscossincossinsincossincos

M1(A1)

=

−−

−

θθθθθθθθ22

22

sincoscossin2cossin2sincos

A1

=

− θθθθ

2cos2sin2sin2cos

AG


(b) let P(n) be the proposition that

−=

− θθθθ

θθθθ

nn

nnn

cossinsincos

cossinsincos

for all n ∈ + P(1) is true A1

−=

− θθθθ

θθθθ

cossinsincos

cossinsincos 1

assume P(k) to be true A1

Note: Must see the word ‘true’ or equivalent, that makes clear an assumption is being made that P(k) is true.

−=

− θθθθ

θθθθ

kk

kkk

cossinsincos

cossinsincos

consider P(k + 1)

−

−=

−

+

θθθθ

θθθθ

θθθθ

cossinsincos

cossinsincos

cossinsincos 1 kk

(M1)

−

−=

θθθθ

θθθθ

cossinsincos

cossinsincos

kk

kk A1

+−−−

+−=

θθθθθθθθθθθθθθθθ

coscossinsinsincoscossincossinsincossinsincoscos

kkkk

kkkkA1

++−

++=

θθθθ)1cos()1sin()1sin()1cos(

kk

kk A1

if P(k) is true then P(k + 1) is true and since P(1) is true then P(n) is true for all n ∈ + R1

Note: The final R1 can only be gained if the M1 has been gained.


(c) EITHER

A–1 =

−−−

−−

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

θθθθ

from formula

=

−

θθθθ

cossinsincos

A1

A–1A = AA–1 =

−

−=

−

−

θθθθ

θθθθ

θθθθ

θθθθ

cossinsincos

cossinsincos

cossinsincos

cossinsincos

M1

Note: Accept either just A–1A or just AA–1.

=

1001

A1

∴A–1 is inverse of A

OR

A–1 =

−

+ θθθθ

θθ cossinsincos

sincos1

22 M1

A–1 =

−

θθθθ

cossinsincos

A1

putting n = –1 in formula gives inverse A1 [13]

14. (a) volume = ∫h

yx0

2dπ (M1)

∫h

yy0

dπ M1

= 2

π2

π2

0

2hy

h

=

A1


(b) t

V

dd

= –3 × surface area A1

surface area = πx2 (M1) = πh A1

since V = π

22

π 2 Vh

h=⇒ M1A1

π2

π3dd V

t

V−= A1

Vt

Vπ23

dd

−= AG

Note: Assuming that t

h

dd

= –3 without justification gains no marks.

[6]

(c) V0 = 5000π (= 15700 cm3) A1

Vt

Vπ23

dd

−=

attempting to separate variables M1

EITHER

∫∫ −= tV

Vdπ23

d A1

ctV +−= π232 A1

c = π50002 A1 V = 0 M1

3133

2π5000π

32

==⇒ t hours A1

OR

∫∫ −=T

tV

V

0

0

π5000dπ23

d M1A1A1

Note: Award M1 for attempt to use definite integrals, A1 for correct limits and A1 for correct integrands.

[ ] TV π2320

π5000 −= A1

T = 3133

2π5000π

32

= hours A1

[16]

Documents

2011 Paper 1.2 Answers METHOD 1 METHOD 2IB Questionbank Mathematics Higher Level 3rd edition 12 (iii) area = ∫3 − π 0 (sin2x sinx ) d x M1 Note: Award M1 for an integral that