MARKSCHEME

May 2004

MATHEMATICS

Higher Level

Paper 1

14 pages

M04/511/H(1)M+

INTERNATIONAL

BACCALAUREATE

BACCALAURÉAT

INTERNATIONAL

BACHILLERATO

INTERNACIONAL

c

– 2 –

M04/511/H(1)M+

This markscheme is confidential and for the exclusive use of
examiners in this examination session.

It is the property of the International Baccalaureate and must
not be reproduced or distributed to any other person without
the authorization of IBCA.

Paper 1 Markscheme

Instructions to Examiners

Note: Where there are 2 marks (e.g. M2, A2) for an answer do NOT split the marks unless

otherwise instructed.

1

Method of Marking

(a)

All marking must be done using a red pen.

(b)

In this paper, the maximum mark is awarded for a correct answer, irrespective of the method
used. Thus, if the correct answer appears in the answer box, award the maximum mark and
move onto the next question; in this case there is no need to check the method.

(c)

If an answer is wrong, then marks should be awarded for the method according to the
markscheme. Examiners should record these marks using the abbreviations shown on the
markscheme. (A correct answer incorrectly transferred to the answer box is awarded the
maximum mark.)

2

Abbreviations

The markscheme may make use of the following abbreviations:

(C)

Marks awarded for Correct answers (irrespective of working shown)

(M) Marks awarded for Method

(A)

Marks awarded for an Answer or for Accuracy

(R)

Marks awarded for clear Reasoning

Note: In general, it is not possible to award (M0)(A1).

Examiners should use (d) to indicate where discretion has been used. It should only be used for
decisions on follow through and alternative methods. It must be accompanied by a brief note to
explain the decision made.

Follow through (ft) marks should be awarded where a correct method has been attempted but error(s)
are made in subsequent working which is essentially correct.

y Penalize the error when it first occurs
y Accept the incorrect result as the appropriate quantity in all relevant subsequent working
y If the question becomes much simpler then use discretion to award fewer marks
y Use (d) to indicate where discretion has been used. It should only be used for decisions on

follow through and alternative methods. It must be accompanied by a brief note to explain the
decision made.

3

Using the Markscheme

(a)

This markscheme presents a particular way in which each question may be worked and how it
should be marked. Alternative methods have not always been included. Thus, if an answer is
wrong then the working must be carefully analysed in order that marks are awarded for a
different method in a manner which is consistent with the markscheme. Indicate the awarding
of these marks by (d).

– 3 –

M04/511/H(1)M+

Where alternative methods for complete questions are included, they are indicated by
METHOD 1, METHOD 2, etc. Other alternative (part) solutions, are indicated by
EITHER….OR. Where possible, alignment will also be used to assist examiners to identify
where these alternatives start and finish.

It should be noted that G marks have been removed, and gdc solutions will not be indicated
using the OR notation as on previous markschemes.

(b)

Unless the question specifies otherwise, accept equivalent forms. For example:

.

sin

for tan

cos

θ

θ

θ

On the markscheme, these equivalent numerical or algebraic forms will be written in brackets
after the required answer. Paper setters will indicate the required answer, by allocating full
marks at that point. Further working should be ignored, even if it is incorrect. For example: if
candidates are asked to factorize a quadratic expression, and they do so correctly, they are
awarded full marks. If they then continue and find the roots of the corresponding equation, do
not penalize, even if those roots are incorrect, i.e. once the correct answer is seen, ignore
further working, unless it contradicts the answer. This includes more than the required number
of solutions, unless otherwise specified in the markscheme.

(c)

As this is an international examination, all alternative forms of notation should be accepted. For
example: 1.7,

, 1,7; different forms of vector notation such as , , u ;

for arctan x.

1 7

⋅

G

u u

tan

−1

x

4

Accuracy of Answers

If the level of accuracy is specified in the question, a mark will be allocated for giving the answer to
the required accuracy.

There are two types of accuracy error. Candidates should be penalized once only IN THE PAPER
for an accuracy error (AP). Award the marks as usual then write –1(AP) against the answer and also
on the front cover.

Rounding errors: only applies to final answers not to intermediate steps.

Level of accuracy: when this is not specified in the question the general rule unless otherwise stated
in the question all numerical answers must be given exactly or to three significant figures applies.

y If a final correct answer is incorrectly rounded, apply the AP

OR

y If the level of accuracy is not specified in the question, apply the AP for answers not given to 3

significant figures. (Please note that this has changed from May 2003).

Incorrect answers are wrong, and the accuracy penalty should not be applied to incorrect answers.

Note: If there is no working shown, and answers are given to the correct two significant figures,
apply the AP. However, do not accept answers to one significant figure without working.

5

Graphic Display Calculators

Many candidates will be obtaining solutions directly from their calculators, often without showing
any working. They have been advised that they must use mathematical notation, not calculator
commands when explaining what they are doing. Incorrect answers without working will receive no
marks. However, if there is written evidence of using a graphic display calculator correctly, method
marks may be awarded. Where possible, examples will be provided to guide examiners in awarding
these method marks.

– 4 –

M04/511/H(1)M+

Calculator penalties

Candidates are instructed to write the make and model of their calculator on the front cover. Please
apply the following penalties where appropriate.

(a)

Illegal calculators

If candidates note that they are using an illegal calculator, please report this on a PRF, and deduct
10 % of their overall mark. Note this on the front cover. The most common examples are:

Texas Instruments: TI-89 (plus); TI-92 (plus); TI-Voyage 200
Casio: fx 9970; fx2.0 algebra; classpad
HP: 38 -95 series

(b)

Calculator box not filled in.

Please apply a calculator penalty (CP) of 1 mark if this information is not provided. Note this on the
front cover.

Examples

1.

Accuracy

A question leads to the answer 4.6789….

y 4.68 is the correct 3 s.f. answer.
y 4.7, 4.679 are to the wrong level of accuracy : both should be penalized the first time this type

of error occurs.

y 4.67 is incorrectly rounded - penalize on the first occurrence.

Note: All these “incorrect” answers may be assumed to come from 4.6789…, even if that value is not
seen, but previous correct working is shown. However, 4.60 is wrong, as is 4.5, 4.8, and these should
be penalized as being incorrect answers, not as examples of accuracy errors.

2.

Alternative solutions

The polynomial

is a factor of

.

2

4

3

x

x

−

+

3

2

(

4)

(3 4 )

3

x

a

x

a x

+

−

+ −

+

Calculate the value of the constant a.

METHOD 1

Using the information given it follows that

(M1)(A1)

3

2

2

(

4)

(3 4 )

3 (

4

3)(

1)

x

a

x

a x

x

x

x

+

−

+ −

+ ≡

−

+

+

Comparing coefficients of

(M1)

2

(or )

x

x

(A1)(A1)

4

3 (or 3 4

1)

a

a

− = −

−

= −

giving a

= 1

(A1)

(C6)

METHOD 2

(M1)(A1)

2

4

3 (

3)(

1)

x

x

x

x

−

+ =

−

−

EITHER

(M1)(A1)

1 (

4) (3 4 ) 3 0

a

a

+

− + −

+ =

Solving, a

= 1

(M1)(A1)

OR

(M1)(A1)

27 9(

4) 3(3 4 ) 3 0

a

a

+

− +

−

+ =

Solving, a

= 1

(M1)(A1)

(C6)

Note that the first line of METHOD 2 applies to both EITHER and OR alternatives.

– 5 –

M04/511/H(1)M+

3.

Follow through

Question
Calculate the acute angle between the lines with equations

and

.

4

4

1

3

s

 

 

=

+

 

 

−

 

 

r

2

1

4

1

t

 

 

=

+

 

 

−

 

 

r

Markscheme

Angle between lines

= angle between direction vectors (May be implied)

(A1)

Direction vectors are

and

(May be implied)

(A1)

4
3

 

 

 

1

1

 

 

−

 

(M1)

4

1

4

1

cos

3

1

3

1

θ

       

=

       

−

−

       

i

(A1)

( )

(

)

( )

(

)

2

2

2

2

4 1 3

1

4

3

1

1

cos

θ

× + × − =

+

+ −

(= 0.1414….)

(A1)

1

cos

5 2

θ

=

(1.43 radians)

(A1)

(C6)

81.9

θ

=

D

Examples of solutions and marking

Solutions

Marks allocated

1.

4

1

4

1

cos

3

1

3

1

θ

       

=

       

−

−

       

i

(A0)(A1)

7

cos

5 2

θ

=

(A1)ft

8.13

θ

=

D

Total 5 marks

2.

4

2

1

4

cos

17 20

θ

   

   

−

   

=

i

(A1)ft

0.2169

=

(A1)ft

77.5

θ

=

D

Total 4 marks

END OF EXAMPLES

– 6 –

M04/511/H(1)M+

(A1)(A1) implied
(M1)

(A0)(A0) wrong vectors implied
(M1) for correct method, (A1)ft

QUESTION 1

The two equations for a and b are

(M1)(A1)

1

0

a b

− + + =

(M1)(A1)

3

8

a b

− − + =

Solution is

(A1)(A1)

(C6)

5,

6

a

b

= −

=

QUESTION 2

(A1)(A1)(A1)

2

cos

y x

x C

=

+

+

(M1)

2 0 cos0 C

= +

+

(A1)

1

C

=

(A1)

(C6)

2

cos

1

y x

x

=

+

+

QUESTION 3

(A1)(A1)

(C2)

0.524,

0.238

x

y

=

=

(A1)(A1)

(C2)

1.69,

0.346

x

y

=

= −

(A1)(A1)

(C2)

4.63,

1.81

x

y

=

= −

Note:

Do not penalize any additional solutions.

QUESTION 4

(A1)(A1)

1

1

(1

) 6 ,

8

1

u

u

r

r

+ =

=

−

Dividing

(M1)

2

3

1

4

r

−

=

(A1)

2

1
4

r

⇒

=

(A1)(A1)

(C6)

1

1

,

12

2

r

u

= −

=

Note:

If no working shown, award (C2) for one correct answer.

– 7 –

M04/511/H(1)M+

QUESTION 5

Clockwise rotation matrix

(M1)(A1)

cos( 45)

sin ( 45)

sin ( 45)

cos( 45)

−

−

−





= 



−

−





1

1

2

2

1

1

2

2

























=









−

















Reflection matrix

(M1)(A1)

cos( 90)

sin ( 90)

sin ( 90)

cos( 90)

−

−





= 



−

−

−





0

1

1 0





−





=









−









(M1)

1

1

0

1

2

2

1 0

1

1

2

2









−







= 



−







 −









T

(A1)

(C6)

1

1

2

2

1

1

2

2





−









=





−

−









QUESTION 6

We require

(accept diagram showing this)

(M1)(A1)

P(

) 0.8

X

x

≤

=

(A1)

1

(0.8) 0.84 (16)

−

Φ

=

(M1)(A1)

25

0.84

3

x

−

=

(A1)

(C6)

27.5

x

=

Note:

Award (M0)(A0)(A0) if

.

25

0.8,

27.4

3

x

x

−

=

=

QUESTION 7

(a)

(M1)(A1)

(C2)

2

2

15

3

p

p

p

− =

⇒ =

Note:

Do not penalize if

also appears.

5
2

p

= −

(b)

(A1)(A1)

2

3

d

d

4

3

0

d

d

y

y

y

x y x

x

x

−

−

=

Note:

Award (A1) for

and (A1) for

.

2

3

d

3

d

y

x y x

x

−

−

d

4

0

d

y

y

x

=

(A1)

d

d

dy

12

9

0

11

9

d

d

d

y

y

x

x

x

− −

= ⇒

=

(A1)

(C4)

9

Gradient

11

=

( 0.818)

=

– 8 –

M04/511/H(1)M+

QUESTION 8

(A1)

5

− =

−

a b

j k

(M1)(A1)

1

3

2

2

1

2









× =

−









−





i

j

k

b c

(A1)

4

6

7

=

+

+

i

j

k

(M1)

(

) (

) 6 5 7 1

− ⋅ × = × − ×

a b

b c

(A1)

(C6)

23

=

QUESTION 9

(a)

Range is [0.5, 1].

(A1)(A1)

(C2)

Note:

Penalize one mark for an open interval.

(b)

Let

(M1)

2

1

1

y

x

=

+

2

1

y yx

+

=

(A1)

2

1

1

or

1

y

x

y

y





−

=

−









(A1)

1

1

or

1

y

x

y

y









−





= −

−

−

















(A1)

(C4)

1

1

1

1

( )

or

1

accept

x

x

f

x

y

x

x

x

−



 



−

−





= −

−

−

= −



 





 









 







Note:

Award (C3) for

with correct working, (C2) for

with correct working.

1 x

x

−

1 x

x

−

±

QUESTION 10

(M1)(A1)

1

,

2

1,

3

x

y

z

µ

µ

µ

= +

=

−

=

(M1)(A1)

(1

) 2(2

1) 3

1

µ

µ

µ

+

+

− −

=

(A1)

1

µ

=

P is (2, 1, 3) (Accept any form of notation, including vectors.)

(A1)

(C6)

– 9 –

M04/511/H(1)M+

QUESTION 11

(a)

(M1)(A1)

2

0

0

d

1

arctan

4

2

2

m

m

x

x

x





 

=

 





+

 





∫

(A1)

(C3)

1

arctan

2

2

m

 

=

 

 

(b)

1

1

arctan

2

2

3

m

  =

 

 

(M1)

2

arctan

2

3

m

  =

 

 

(A1)

2

tan

2

3

m

 

=

 

 

(A1)

(C3)

2

2 tan

3

m

 

=

 

 

( 1.57)

=

Note:

If

is replaced by m, do not award the final (A1) for ft.

2

m

QUESTION 12

(a)

Mean

(A1)

10 0.4 4

= ×

=

Standard deviation

(M1)(A1)

(C3)

10 0.4 0.6 1.55

=

×

×

=

(b)

(M1)

P(

2) 1 P(

0) P(

1)

X

X

X

≥

= −

= −

=

(A1)

10

9

1 0.6

10 0.6

0.4

= −

− ×

×

(A1)

(C3)

0.954

=

QUESTION 13

(a)

(M1)(A1)

1

1 2 1 1

P( )

1

3

3 3 3 3

G

= × + × + ×

(A1)

(C3)

2
3

=

(b)

(M1)

3

P(

)

P( )

D

G

G

∩

(A1)

1 2
9 3

= ÷

(A1)

(C3)

1
6

=

– 10 –

M04/511/H(1)M+

1
3

2
3

1
3

1
3

1
3

G

G

G

S

S

D

1

D

2

D

3

1

QUESTION 14

(M1)

2

2

d

d (e )

e d

d

d

d

d

x

x

v

x

x

u

x

x

x

x

x

=

=

∫

∫

∫

(A1)

2

e

2 e d

x

x

x

x

x

=

−

∫

Note:

Award (M1)(A1) if u and v are reversed, leading to

.

3

3

e

e

d

3

3

x

x

x

x

x

−

∫

(A1)

2

d (e )

e

2

d

d

x

x

x

x

x

x

=

−

∫

(A1)

2

e

2 e

2 e d

x

x

x

x

x

x

=

−

+

∫

(A2)

(C6)

2

e

2 e

2e

x

x

x

x

x

C

=

−

+

+

Note:

Award only (A1) if the C is missing.

QUESTION 15

(a)

Median

= 1.04

(M1)(A1)

(C2)

Note:

Award the (M1) for the appropriate line drawn at 30.

(b)

The frequency table based on mid-values is

3

1.35

6

1.25

12

1.15

15

1.05

15

0.95

9

0.85

Frequency

Mid-value

Mean

(M1)

(9 0.85 15 0.95

3 1.35)

60

×

+ ×

+ … + ×

=

(A1)

(C4)

1.05

=

– 11 –

M04/511/H(1)M+

(A1)(A1)

QUESTION 16

METHOD 1

Note:

If no working shown or if working is incorrect, award (C3) for one correct interval.

The critical values occur when

(M1)(A1)

12

3

6, 24

12

x

x

x

+

= ± ⇒ =

−

Consider

value of function at 0 is 1 which is

.

(A1)

]

, 6]:

− ∞

3

≤

Consider

value of function at 10 is 11 which is not

.

(A1)

[6, 24]:

3

≤

Note:

The discontinuity at x

= 12 does not cause any problems since the

value of the function is very large in its vicinity.

Consider

value of function at 30 is which is

.

(A1)

[24, [ :

∞

7
3

3

≤

The required solution set is therefore

.

(A1)

(C6)

]

, 6] [24, [

− ∞

∪

∞

Notes: Penalize one mark for open end at 6 and/or 24.

Award the final (A1) for the symbol

∪ or the word “or”.

METHOD 2

(M1)(A1)(A1)

(A1)(A1)(A1)

(C6)

]

, 6] [24, [

− ∞

∪

∞

Notes: Award (A1) for each interval, and (A1) for the symbol

∪ or the word “or”.

Penalize one mark for open end at 6 or 24.

– 12 –

M04/511/H(1)M+

QUESTION 17

METHOD 1

Let

(M1)

2

u x

=

(A1)

d

2

d

u

x

x

=

Therefore,

(M1)

d

( ) 3 ln 3

d

u

u

f x

x

′

=

×

(A1)

2

3 ln 3 2

x

x

=

×

( ) 2

f x

′

=

solution

= 0.607.

(A2)

(C6)

METHOD 2

2

3

x

y

=

(M1)

2

ln

ln 3

y x

=

(M1)(A1)

1 d

2 ln 3

d

y

x

y x

=

d

2 ln 3

d

y

xy

x

=

(A1)

2

( ) 2

3 ln 3

x

f x

x

′

=

×

( ) 2

f x

′

=

solution

= 0.607.

(A2)

(C6)

QUESTION 18

(A1)

2

1
2

A

r

θ

=

(M1)(A1)

2

d

d

1

d

d

d

2

d

A

r

r

r

t

t

t

θ

θ

=

+

(A1)

π

1

3

2

9 0.1

4

2

= × × + × ×

(A1)(A1)

(C6)

2

5.16 cm per sec

=

2

3π

9

cm per sec

2

20





=

+









Note:

Award (A1) for the correct value, and (A1) for the correct units.

– 13 –

M04/511/H(1)M+

QUESTION 19

(a)

(M1)(A1)

(C2)

2

3

( ) 3 cos

sin

f x

x

x x

x

′

=

−

(b)

(A2)

(C2)

1.19

x

=

(c)

(A2)

(C2)

0.823

x

=

Notes: In parts (b) and (c), do not penalize any additional solutions.

In part (c), award (M1)(A0) if candidates have

, without x

= 0.823.

( ) 0

0

f x

x

′′

= ⇒ =

QUESTION 20

Let AD

= h so that

(A1)

AB

cot 30 and AC

cot 20

h

h

=

=

D

D

Using the cosine rule in

,

ABC

∆

(M1)(A1)

2

2

2

2

cot 20

cot 30

2500 100 cot 30 cos 45

h

h

h

=

+

−

D

D

D

D

(A1)

2

2

2

(cot 20

cot 30 ) 100 cot 30 cos 45

2500 0

h

h

−

+

−

=

D

D

D

D

(M1)

2

2

2

2

2

100cot 30 cos 45

(100cot 30 cos 45 )

10000(cot 20

cot 30 )

2(cot 20

cot 30 )

h

−

+

+

−

=

−

D

D

D

D

D

D

D

D

(no units required)

(A1)

(C6)

13.6

=

– 14 –

M04/511/H(1)M+