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M16/5/MATHL/HP1/ENG/TZ2/XX/M

18 pages

Markscheme

May 2016

Mathematics

Higher level

Paper 1

This markscheme is the property of the International

Baccalaureate and must not be reproduced or distributed

to any other person without the authorization of the IB

Assessment Centre.

• Once a correct answer to a question or part-question is seen, ignore further correct working.

However, if further working indicates a lack of mathematical understanding do not award the final A1. An exception to this may be in numerical answers, where a correct exact value is followed by

an incorrect decimal. However, if the incorrect decimal is carried through to a subsequent part,

and correct FT working shown, award FT marks as appropriate but do not award the final A1 in that part.

Examples

Correct answer seen Further working seen Action

1. 8 2

(incorrect decimal value)

Award the final A

(ignore the further working)

2. 1 sin 4 4

x (^) sin x Do not award the final A

3. log^ a^ −^ log b log (^ a^ −^ b ) Do not award the final A

3 N marks

Award N marks for correct answers where there is no working.

• Do not award a mixture of N and other marks.
• There may be fewer N marks available than the total of M , A and R marks; this is deliberate as it penalizes candidates for not following the instruction to show their working.

4 Implied marks

Implied marks appear in brackets eg (M1) , and can only be awarded if correct work is seen or if

implied in subsequent working.

• Normally the correct work is seen or implied in the next line.
• Marks without brackets can only be awarded for work that is seen.

5 Follow through marks

Follow through ( FT ) marks are awarded where an incorrect answer from one part of a question is

used correctly in subsequent part(s). To award FT marks, there must be working present and

not just a final answer based on an incorrect answer to a previous part.

• If the question becomes much simpler because of an error then use discretion to award fewer FT marks.
• If the error leads to an inappropriate value ( eg ), do not award the mark(s) for the final

answer(s).

• Within a question part, once an error is made, no further dependent A marks can be awarded,

but M marks may be awarded if appropriate.

• Exceptions to this rule will be explicitly noted on the markscheme.

sin θ =1.

6 Misread

If a candidate incorrectly copies information from the question, this is a misread ( MR ). A candidate should be penalized only once for a particular misread. Use the MR stamp to indicate

that this has been a misread. Then deduct the first of the marks to be awarded, even if this is an

M mark, but award all others so that the candidate only loses one mark.

• If the question becomes much simpler because of the MR , then use discretion to award fewer marks.
• If the MR leads to an inappropriate value ( eg ), do not award the mark(s) for the

final answer(s).

7 Discretionary marks (d)

An examiner uses discretion to award a mark on the rare occasions when the markscheme does

not cover the work seen. In such cases the annotation DM should be used and a brief note written next to the mark explaining this decision.

8 Alternative methods

Candidates will sometimes use methods other than those in the markscheme. Unless the question specifies a method, other correct methods should be marked in line with the markscheme.

If in doubt, contact your team leader for advice.

• Alternative methods for complete questions are indicated by METHOD 1 , METHOD 2 , etc.
• Alternative solutions for part-questions are indicated by EITHER... OR.
• Where possible, alignment will also be used to assist examiners in identifying where these alternatives start and finish.

9 Alternative forms

Unless the question specifies otherwise, accept equivalent forms.

• As this is an international examination, accept all alternative forms of notation.
• In the markscheme, equivalent numerical and algebraic forms will generally be written in brackets immediately following the answer.
• In the markscheme, simplified answers, (which candidates often do not write in examinations), will generally appear in brackets. Marks should be awarded for either the form preceding the

bracket or the form in brackets (if it is seen).

Example : for differentiating f ( x ) = 2 sin (5 x − 3), the markscheme gives

f ′^ ( ) x = (^) ( 2cos(5 x −3) 5 (^) ) A

Award A1 for ( 2cos(5 x^ −3) 5^ ) , even if 10 cos (5 x − 3)is not seen.

10 Accuracy of Answers

Candidates should NO LONGER be penalized for an accuracy error (AP).

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

the required accuracy. When this is not specified in the question, all numerical answers should be

given exactly or correct to three significant figures. Please check work carefully for FT.

sin θ =1.

( = 10cos(5 x −3))

Section A

1. EITHER

eliminating a variable, x , for example to obtain y + 3 z = − 16 and − 5 y − 3 z = 8 M1A

attempting to find the value of one variable M

point of intersection is ( 1, 2,− − 6) A1A1A

OR

attempting row reduction of relevant matrix, eg.

M

correct matrix with two zeroes in a column, eg.

A

further attempt at reduction M

point of intersection is ( 1, 2,− − 6) A1A1A

Note: Allow solution expressed as x = − 1 , y = 2 , z = − 6 for final A marks.

[6 marks]

2. x = − 1

A1A1A1A1A

Note: Award A1 for correct shape, A1 for x = − 1 clearly stated and asymptote shown,

A1 for y = 3 clearly stated and asymptote shown, A1 for

and A1 for (0, 2).

[5 marks]

3. (a) EITHER

use of a diagram and trig ratios

eg ,

tan cot

O A

A O

from diagram, tan 2

A

O

 −^ =

R

OR

use of

π sin π 2 cos tan 2 π sin cos 2

 −^  =^ =

  ^ 

R

THEN

π cot tan 2

AG

[1 mark]

(b) [ ]

cot cot 2 tan tan

d arctan 1

x x x

α α α α

 (A1)

Note: Limits (or absence of such) may be ignored at this stage.

= arctan(cot α) − arctan(tan α) (M1)

π

= − α − α (A1)

π 2 2

= − α A

[4 marks]

Total [5 marks]

6. (a)

n n n n n nx x x

A1A

Note: Award A1 for the first two terms and A1 for the next two terms.

Note: Accept

n

r

notation.

Note: Allow the terms seen in the context of an arithmetic sum.

Note: Allow unsimplified terms, eg , those including powers of 1 if seen.

[2 marks]

(b) (i) EITHER

using u 3 (^) − u 2 (^) = u 4 (^) − u 3 (M1)

n n n n n n n n

− = − A

attempting to remove denominators and expanding (or vice versa) M 2 3 2 3 n − 9 n = n − 6 n + 5 n (or equivalent, eg ,

2 3 2 6 n − 12 n = n − 3 n + 2 n ) A

OR

using u 2 (^) + u 4 (^) = 2 u 3 (M1)

( 1)( 2) ( 1) 6

n n n n n n

− −

• = − (A1)

attempting to remove denominators and expanding (or vice versa) M 3 2 2

6 n + n − 3 n + 2 n = 6 n − 6 n (or equivalent) (A1)

THEN

3 2 n − 9 n + 14 n = 0 AG

(ii) n n ( − 2)( n − 7) = 0 or ( n − 2)( n − 7) = 0 (A1)

n = 7 only (as n ≥ 3 ) A

[6 marks]

Total [8 marks]

7. (a) P( A ∪ B ) = P( ) A + P( B ) − P( A ∩ B )

= P( ) A + P( B ) − P( )P( A B ) (M1)

2 = p + p − p A

2 = 2 p − p AG

[2 marks]

(b)

P (^) ( ( )) P( | ) P( )

A A B

A A B

A B

(M1)

Note: Allow (^) P ( A ∩ A ∪ B )if seen on the numerator.

P( )

P( )

A

A B

(A1)

2 2

p

p p

A

2 p

A

[4 marks]

Total [6 marks]

8. let P( ) n be the proposition that (^) ( )

2 n n + 5 is divisible by 6 for n

∈

consider P(1) :

when n = 1 , (^) ( ) ( )

2 2 n n + 5 = 1 × 1 + 5 = 6 and so P(1) is true R

assume P( ) k is true ie , (^) ( )

2 k k + 5 = 6 m where k , m

∈ M

Note: Do not award M1 for statements such as “let n = k ”.

consider P( k + 1):

( )

2 ( k + 1) ( k + 1) + 5 M

( )

2 = ( k + 1) k + 2 k + 6

3 2 = k + 3 k + 8 k + 6 (A1)

( ) ( )

3 2 = k + 5 k + 3 k + 3 k + 6 A

( )

2 = k k + 5 + 3 ( k k + 1) + 6 A

k k ( + 1)is even hence all three terms are divisible by 6 R

P( k + 1)is true whenever P( ) k is true and P(1) is true, so P( ) n is true for n

∈ R

Note: To obtain the final R1 , four of the previous marks must have been awarded.

[8 marks]

Section B

10. (a) EITHER

n and 2

 p    =      

d A1A

and n ≠ k d R

OR

p

p

× = −

n d M1A

the vector product is non-zero for p ∈ R

THEN

L is not perpendicular to Π AG

[3 marks]

(b) METHOD 1

(2 + p λ) + ( q + 2 λ) + 3(1 + λ) = 9 M

( q + 5) + ( p + 5) λ= 9 (A1)

p = − 5 and q = 4 A1A

METHOD 2

direction vector of line is perpendicular to plane, so

^ p        ⋅ =            

M

p = − 5 A

(2, q , 1)is common to both L and Π

either

q

or by substituting into x + y + 3 z = 9 M

q = 4 A

[4 marks]

continued…

Question 10 continued

(c) (i) METHOD 1

α is the acute angle between^ n and^ L

if

sin 11

θ = then

cos 11

α = (M1)(A1)

attempting to use cos α

n d

n d

or sin θ

n d

n d

M

2

p

p

× +

A1A

2 2 ( p + 5) = p + 5 M

10 p = − 20 (or equivalent) A

p = − 2 AG

METHOD 2

α is the angle between n and L

if

sin

θ = (^) then

sin 11

α = (M1)A

attempting to use sin α

×

n d

n d

M

2 2 2

2

p p

p

× +

A1A

2 2 p − p + 3 = p + 5 M

− p + 3 = 5 (or equivalent) A

p = − 2 AG

(ii)

x y q p z

and (A1)

x = 6 and y = q − 4 (A1)

this satisfies Π so 6 + q − 4 − 3 = 9 M

q = 10 A

[11 marks]

Total [18 marks]

Question 11 continued

(ii)

π sin 2 0 0, , π 2

h =  h = A

Note : Award A1 for

π sin 2 0 0, , π 2

h =  h = from an incorrect

2

2

d

d

h

t

.

(iii) METHOD 1

d

d

h

t

is a minimum at h = 0, πand the container is widest at these values

R

d

d

h

t

is a maximum at

π

h = and the container is narrowest at this value

R [7 marks]

Total [19 marks]

12. (a) EITHER

7 7 2 π^2 π cos i sin 7 7

w

(M1)

= cos 2π + i sin 2π A

= 1 A

so w is a root AG

OR

7 z = 1 = cos(2π k ) + i sin(2π k ) (M1)

2 π 2 π cos sin 7 7

k k z i

A

2 π 2 π 1 cos sin 7 7

k z i

A

so w is a root AG

[3 marks]

(b) (i) (^) ( )

2 3 4 5 6 ( w − 1) 1+ w + w + w + w + w + w

2 3 4 5 6 7 2 3 4 5 6 = w + w + w + w + w + w + w − 1 − w − w − w − w − w − w M

7 = w − 1 ( = 0) A

(ii)

7 w − 1 = 0 and w − 1 ≠ 0 R

so

2 3 4 5 6 1 + w + w + w + w + w + w = 0 AG

[3 marks]

continued…

Question 12 continued

(c) the roots are

2 3 4 5 6 1, w , w , w , w , w and w A

7 points equidistant from the origin A

approximately correct angular positions for

2 3 4 5 6 1, w , w , w , w , w and w A

Note: Condone use of cis notation for the final two A marks.

Note: For the final A mark there should be one root in the first quadrant, two in the

second, two in the third, one in the fourth, and one on the real axis.

[3 marks]

(d) (i) (^) ( )

• 2 4

α w w w

∗ = + +

( ) ( )

2 4 w w w

∗ ∗ ∗ = + + A

since

6 w w

∗ = , (^) ( )

2 5 w w

∗ = and (^) ( )

4 3 w w

∗ = R

• 6 5 3  α = w + w + w AG

continued…