NATIONAL Mathematics QUALIFICATIONS Higher Prelim ... · PDF fileSample Question A line has...

Pegasys 2014 Current Higher Units 1 and 2 + Vectors

Read carefully

Calculators may NOT be used in this paper.

Section A - Questions 1 - 20 (40 marks)

Instructions for the completion of Section A are given on the next page.

For this section of the examination you should use an HB pencil.

Section B (30 marks)

1. Full credit will be given only where the solution contains appropriate working.

2. Answers obtained by readings from scale drawings will not receive any credit.

Mathematics

Higher Prelim Examination 2014/2015

Paper 1

Assessing Units 1 & 2 + Vectors

Time allowed - 1 hour 30 minutes

NATIONAL

QUALIFICATIONS


Sample Question

A line has equation .14 xy

If the point )7,(k lies on this line, the value of k is

A 2

B 27

C 15

D 2

The correct answer is A 2. The answer A should then be clearly marked in pencil with a

horizontal line (see below).

Changing an answer

If you decide to change an answer, carefully erase your first answer and using your pencil, fill in the

answer you want. The answer below has been changed to D.

Read carefully

1 Check that the answer sheet provided is for Mathematics Higher Prelim 2014/2015 (Section A).

2 For this section of the examination you must use an HB pencil and, where necessary, an eraser.

3 Make sure you write your name, class and teacher on the answer sheet provided.

4 The answer to each question is either A, B, C or D. Decide what your answer is, then, using

your pencil, put a horizontal line in the space below your chosen letter (see the sample question below).

5 There is only one correct answer to each question.

6 Rough working should not be done on your answer sheet.

7 Make sure at the end of the exam that you hand in your answer sheet for Section A with the rest

of your written answers.


FORMULAE LIST

Circle:

The equation 02222 cfygxyx represents a circle centre ),( fg and radius cfg 22 .

The equation ( ) ( )x a y b r 2 2 2 represents a circle centre ( a , b ) and radius r.

A

A

AAA

AAA

BABABA

BABABA

2

2

22

sin21

1cos2

sincos2cos

cossin22sin

sinsincoscoscos

sincoscossinsin

Scalar Product: .andbetweenangletheisθwhere,cosθ. bababa

or

ba .

3

2

1

3

2

1

332211

b

b

b

and

a

a

a

where babababa

Trigonometric formulae:

Table of standard derivatives:

Table of standard integrals:

)(xf )(xf

axaax cossin

axaax sincos

)(xf )(xf dx

axsin Caxa

cos1

axcos Caxa

sin1


1. The remainder when 12 23 xx is divided by )2( x is

A 19

B 7

C 13

D 11

2. A sequence is defined by the recurrence relation baUU nn 1 with .40 U

An expression in terms of a and b for 2U is

A aba 24

B ba 516

C 4 ba

D baba 24

3. Any line perpendicular to the line with equation 563 xy , has as its gradient

A 2

1

B 6

1

C 5

3

D 2

4. The rate of change of the function 2

1)(

xxf when 1x is

A 1

B 1

C 2

D 2

SECTION A

ALL questions should be attempted


5. Part of the graph of the function )(xfy is

shown opposite.

Which of the following graphs represents the

related function 4)( xfy ?

A B

C D

6. Given that 32sin A , the exact value of A2cos is

A 97

B 31

C 1

D 92

(3,4)

y

o 3

)(xfy

x

-4

y

x o

(3, 4)

y

x

o

y

o x (3, 4)

(3,4)

y

x o

8


( )

7. The minimum turning value of the function xxxf 12)( 2 is

A 6

B 0

C 36

D 12

8. A vector v is given by

What is the length, in units, of 3v ?

A 7

B 21

C 15

D 49

9. dxxx )1(4 2 is

A 1

B 1

C 8

D 0

10. For the equation 033 2 kxx to have real roots, k must take the values

A 6

B 66 k

C 6k or 6k

D 6k or 6k

0

1

-3

2

6


11. Part of the graph of the curve )(xfy is shown below.

Which of the following is/are true for this function f ?

1. 0)0( f

2. 0)1( f

3. 0)3( f

4. 0)4( f

A 3 only

B 1, 3 and 4 only

C 1 and 3 only

D 3 and 4 only

12. The equation of a circle , centre ( 3 , 2 ), with the x-axis as a tangent is

A 9)2()3( 22 yx

B 4)2()3( 22 yx

C 4)2()3( 22 yx

D 9)2()3( 22 yx

13. If 240sin)cos( x for 1800 x then x has the value

A 30

B 60

C 120

D 150

x

y

O

(3, 4)


14. A recurrence relationship is defined as bUU nn 401 .

If the limit of the recurrence relationship is 50, the value of b is

A 20

B 30

C 3133

D 125

15. Given that the points S(-4, 5, 1), T(-16, -4, 16) and U(-24, -10, 26) are collinear,

calculate the ratio in which T divides SU.

A 2 : 3

B 3 : 2

C 2 : 5

D 3 : 5

16. The function f is defined as 1,1

4)(

x

x

xxf . The value of ))1(( ff equals

A 3

8

B 8

C 2

D 2

17. 1)( 2 axxxf has a stationary value when 2x . The value of a is

A 4

B 1

C 0

D 5


18. O is the centre of a circle of radius 4cm.

Angle POQ is 50 radians.

The area of the shaded sector POQ in square centimetres is

A 1

B 2

C 4

D 8

19. If 1)( xxf where 0x , then the composite function ))(( xff is equal to

A 2x

B 1x

C 0x

D 1x

20. The diagram opposite is made from 4 semi-circular arcs

each of radius 1 centimetre. The resulting shape has two

axes of symmetry as shown by the dotted lines.

The area of the shaded shape in square centimetres is

A 2

B 4

C 2

D 4

O P

Q

4cm

[ END OF SECTION A ]


21. Triangle ABC has B(8,11) as one of its vertices.

The line through A which meets side BC at D has as its equation 213 yx .

(a) If side BC has a gradient of 2 , find the equation of BC. 2

(b) Establish the coordinates of D. 3

(c) If D is in fact the mid-point of side BC, write down the coordinates of C. 1

(d) Show clearly that this triangle is right-angled at B. 3

(e) Hence find the equation of the circle passing through the points A, B and C. 3

22. Angle x is acute and is such that 42tan x .

(a) Show clearly that the exact value of xsin is

31 . 3

(b) Hence show that 22sin94x . 4

SECTION B


x

y

O

B(8,11)

A

C

D


23. A sequence is defined by the recurrence relation qpUU nn 1 , where

p and q are constants.

(a) Given that 200 U , 121 U and 7q , find the value of p. 2

(b) Given that 3213 UUUS , calculate the value of 3S . 3

24. The small box below is in the shape of a cuboid.

The box has dimensions x12 , kx and 2 as shown.

All the lengths are in centimetres.

The volume of the box is 60 cubic centimetres.

(a) By forming an equation for the volume of the box, and simplifying it, show

that the following equation can be formed

0)3012(2 2 kxkx . 2

(b) Hence find the value of k, given that the above equation has equal roots. 4

kx x12kx

x12

2

[ END OF SECTION B ]

[ END OF QUESTION PAPER 1 ]


Read carefully

1. Calculators may be used in this paper.

2. Full credit will be given only where the solution contains appropriate working.

3. Answers obtained from readings from scale drawings will not receive any credit.

Mathematics


Paper 2

Assessing Units 1 & 2 + Vectors

Time allowed - 1 hour 10 minutes

NATIONAL

QUALIFICATIONS


FORMULAE LIST

Circle:

The equation 02222 cfygxyx represents a circle centre ),( fg and radius cfg 22 .

The equation ( ) ( )x a y b r 2 2 2 represents a circle centre ( a , b ) and radius r.

A

A

AAA

AAA

BABABA

BABABA

2

2

22

sin21

1cos2

sincos2cos

cossin22sin

sinsincoscoscos

sincoscossinsin

Scalar Product: .andbetweenangletheisθwhere,cosθ. bababa

or

ba .

3

2

1

3

2

1

332211

b

b

b

and

a

a

a

where babababa

Trigonometric formulae:

Table of standard derivatives:

Table of standard integrals:

)(xf )(xf

axaax cossin

axaax sincos

)(xf )(xf dx

axsin Caxa

cos1

axcos Caxa

sin1


1. The diagram below, which is not drawn to scale, shows part of the graph of the

curve 326 xxy .

(a) Find the coordinates of the point A, the maximum turning point

of this curve. 4

(b) The line through A, with gradient 5, intersects the curve at a further two

points, one of which is B.

Find algebraically the coordinates of B.

Your answer must be accompanied with the appropriate working. 5

2. Solve algebraically the equation

01sin32cos xx , for 3600 x . 5


A

x

y

O

326 xxy

B


3. A curve has as its derivative 2

92

xx

dx

dy .

(a) Given that the point ( 1 , 6) lies on this curve, express y in terms of x. 4

(b) Hence find p if the point T(1, p) also lies on this curve. 1

(c) Find the equation of the tangent to this curve at T. 3

4. The circle below has as its equation 046822 yxyx .

QR is a tangent to the circle.

(a) Prove that the point P(2,-8) lies on the circumference of the circle. 1

(b) Hence find the equation of the tangent to the circle at the point Q, where PQ is

a diameter of the circle. . 5

(c) Find the length of the line QR.

Give your answer correct to 1- decimal place. 3

P(2,-8)

R

Q

x

y

o

046822 yxyx


5. A is the point (3, -3, 0), B is (2, -3, 1) and C is (4, k, 0).

(a) (i) Express BA and BC in component form.

(ii) Show that cos ABC = )146(2

3

2 kk 7

(b) If angle ABC = 30 ͦ, find the possible values of k. 5

6. The diagram below shows a sketch of part of the graphs of 28163 xxy and

38124 23 xxxy .

The curves intersect at the point (0,3) and at A.

The dotted line shown is parallel to the x-axis and passes through (0,3) and A.

(a) Establish the coordinates of the point A. 2

(b) Hence calculate the area enclosed between the two curves. 5

x

y

o

A 3


7. Two functions in x are defined on suitable domains as

)(3)( axxf and 1)( axxg , where a is a constant.

Given that 9))1(( agf , find the value of a where 0a . 4

8. (a) Given that 8

2

4

dxx show clearly that the exact value of k is 4 . 4

(b) Hence show that kk cossin can be written in the form p and state the

value of p. 2

0

k

[ END OF QUESTION PAPER 2 ]


Indicate your choice of answer with a

single mark as in this example

Mathematics


Paper 1 - Section A - Answer Sheet

NATIONAL

QUALIFICATIONS

NAME :

TEACHER :

CLASS :

You should use an HB pencil.

Erase all incorrect answers thoroughly.

Section A

Section B

40

30

Total (P1)

70

Total (P2)

60

Overall Total

130

%

Please make sure you have filled in all your details above before handing in this answer sheet.

A B C D

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

A B C D


Higher Grade - Paper 1 2014/2015 ANSWERS - Section A

1 C

2 D

3 A

4 D

5 B

6 A

7 C

8 B

9 B

10 D

11 D

12 C

13 A

14 B

15 B

16 A

17 A

18 C

19 D

20 B

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

A B C D


Higher Prelim Paper 1 2014/2015 Marking Scheme

21(a) ans: y + 2x = 27 (2 marks)

●1 substitutes info into equation ●

1 y – 11 = – 2(x – 8)

●2 rearranges ●

2 y + 2x = 27

(b) ans: D(12, 3) (3 marks)

●1 knows to use system of equations ●

1 evidence

●2 solves for one variable ●

2 y = 3

●3 solves for other variable and states point D ●

3 x = 12; D(12, 3)

(c) ans: C(16, –5) (1 mark)

●1 establishes coordinates of C ●

1 C(16, –5)

(d) ans: proof (3 marks)

●1

establishes coordinates of A ●1 when x = 0; y = 7

●2 finds gradients of AB and BC ●

2 2

168

511;

2

1

08

711

BCAB mm

●3 reason ●

3 since 1 BCAB mm ; AB is perp to BC so

triangle is right angled at B

(e) ans: (x – 8)² + (y – 1)² = 10 (3 marks)

●1 knows AC is diameter and finds midpoint ●

1 midpoint AC = (8, 1) [centre]

●2 finds radius ●

2 (8, 1) to (0, 7) = √(8² + 6²) = 10

●3 subs into equation of circle ●

3 (x – 8)² + (y – 1)² = 100

22(a) ans: proof (3 marks)

●1 interprets information ●

1 triangle drawn or any acceptable method

●2 finds hypotenuse ●

2 23184)2( 22

●3 finds sinx and simplifies ●

3

3

1

23

2sin x

(b) ans: proof (4 marks)

●1 finds cosx ●

1

23

4cos x

●2 substitutes values in expansion of sin2x ●

2

23

4

3

122sin x

●3 starts to simplify ●

3

29

82sin x

●4 rationalises denominator to answer ●

4 2

9

4

18

28

2

2

29

8

Give 1 mark for each Illustration(s) for awarding each mark


23(a) ans: p = 0·25 (2 marks)

●1 substitutes values ●

1 72012 p

●2 solves for p ●

2 250 p

(b) ans: 31·5 (3 marks)

●1 finds U2 ●

1 U2 = 10

●2 finds U3 ●

2 U3 = 9·5

●3 totals ●

3 31·5


●1 subs values ●

1 )12(260 xkx

●2 rearranges to answer

●

2 2kx

2 – 24kx + 60 = 0 …….

(b) ans: 6

5k (4 marks)

●1 knows condition for equal roots ●

1 042 acb for equal roors

●2 identifies a, b and c ●

2 30;12; ckbka

●3 substitutes values and simplifies ●

3 0120144304144 22 kkkk

●4 solves and discards ●

4

6

5;0)56(24 kkk


Total: 70 marks


Higher Prelim Paper 2 2014/2015 Marking Scheme

1(a) ans: A(4, 32) (4 marks)

●1 finds derivative and equates to zero ●

1 0312 2 xx

dx

dy

●2 solves for x ●

2 4;0;0)4(3 xxx

●3 chooses appropriate value and subs ●

3 when x = 4, y = 6(4²) – 4³ = 32

●4 states coordinates of A ●

4 A(4, 32)

(b) ans: (– 1, 7) (5 marks)

●1 establishes equation of AB ●

1 y – 32 = 5(x – 4); y = 5x + 12

●2 equates equations of line and curve ●

2 5x + 12 = 6x² – x³ ; x³ – 6x² + 5x + 12 =

●3 knows to use synthetic division ●

3

0321

1284

125614

●4 solves and chooses solution ●

4 (x – 4)(x – 3)(x + 1) = 0; x = – 1

●5 subs and states coordinates of B ●

5 y = 5(–1) + 12 = 7

2 ans: 210o; 330

o (5 marks)

●1 subs for cos 2x

o ●

1 01sin3)sin21( 2 xx

●2 simplifies and arranges to quadratic ●

2 02sin3sin2 2 xx

●3 factorises ●

3 0)2sin)(1sin2( xx

●4 knows to discard and solve ●

4

2

1sin x

●5 finds angles ●

5 210

o; 330

o



3(a) ans: 1492 x

xy (4 marks)

●1 knows to integrate function ●

1

292 xxy

●2 integrates including C ●

2 C

xxy

92

●3 subs and solves for C ●

3 14;916 CC

●4 states y in terms of x ●

4 14

92 x

xy

(b) ans: p = 24 (1 mark)

●1 subs values ●

1 2414

1

912 p

(c) ans: y + 7x = 31 (3 marks)

●1 subs x = 1 into

dx

dy ●

1 792 m

●2 subs into equation ●

2 )1(724 xy

●3 rearranges ●

3 317 xy

4(a) ans: proof (1 mark)

●1 subs point to make true statement ●

1 2² + (– 8)² – 8(2) + 6(–8) – 4 = 0

and conclusion so point lies on circle

(b) ans: 5y + 2x = 22 (5 marks)

●1 uses acceptable method to find Q ●

1 stepping out method or any acceptable

●2 establishes coordinates of Q ●

2 Q(6, 2)

●3 finds gradient of PQ ●

3

2

5

26

82

PQm

●4 finds perpendicular gradient ●

4 mperp =

5

2

●5 subs in straight line equation and simplifies ●

5 );6(

5

22 xy 5y + 2x = 22

(c) ans: 6·5 units (3 marks)

●1 establishes coordinates of R ●

1 5y + 0 = 22; y = 4·4; R(0, 4·4)

●2 uses distance formula ●

2 QR² = 6² + 2·4²

●3 finds QR correctly rounded ●

3 QR = 6·5 units




(b) ans: k = -2 , k = -4 (5 marks)



6(a) ans: A(– 2, 3) (2 marks)

●1 equates equations ●

1 3 – 16x – 8x² = 3; –8x(2 + x) = 0

●2 solves and states coordinates ●

2 x = 0 or – 2; A(– 2, 3)

(b) ans: 10⅔ units² (5 marks)

●1 sets up limits ●

1

0

2.......

●2 upper – lower ●

2 dxxxxxx

0

2

232 )38124(8163

dxxxx 24204 20

2

3

●3 integrates ●

3

0

2

234

2

24

3

20

4

4

xxx=

0

2

23

4 123

20

x

xx

●4 substitutes values ●

4 )]])2(12

3

)2(20)2[()0[( 2

34

●5 evaluates ●

5

3

210)

3

210()]48

3

16016[)0[(

7 ans: A(–2, 9) (4 marks)

●1 finds g(a + 1) 2 ●

1 g(a + 1) = a(a + 1) – 1 = a² + a – 1

●2 finds f(g(a + 1) ●

2 f(g(a + 1) = 3(a² + a – 1 – a) = 3(a² – 1)

●3 equates f(g(a + 1) to 9 ●

3 3(a² – 1) = 9

●4 solves for a ●

4 a = 2

8 (a) ans: proof (4 marks)

●1 integrates ●

1 k

x 0

22

●2 subs values and equates to

8

2 ●

2

82

22 k

●3 starts to simplify ●

3 2216 k

●4 solves for k ●

4

4;4

kk

(b) ans: p = 2 (2 marks)

●1 uses exact values ●

1

2

2

2

1

2

1

●2 rationalises denominator and states p ●

2 2

2

22

2

2

2

2 ; p = 2


Total: 60

marks

NATIONAL Mathematics QUALIFICATIONS Higher Prelim ... · PDF fileSample Question A line has...

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Transcript of NATIONAL Mathematics QUALIFICATIONS Higher Prelim ... · PDF fileSample Question A line has...