national mathematics qualifications higher prelim ... · higher prelim examination 2013/2014 paper...

Pegasys 2013

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 2013/2014

Paper 1

Assessing Units 1 & 2

Time allowed - 1 hour 30 minutes

NATIONAL

QUALIFICATIONS

Pegasys 2013

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 2013/2014 (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.

Pegasys 2013

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

Pegasys 2013

1. The line 013 yax is parallel to the line with equation 032 yx .

The value of a is

A 31

B 6

C 2

D 2

2. The function f is defined as 0,10

)(2

xx

xxf . The value of ))2(( ff equals

A 739

B 31

C 3

D 31

3. If x

xf1

)( , where 0x , then )2(f equals

A 1

B 21

C 41

D 41

4. A sequence is defined by the recurrence relation 21 nn kUU with 120 U .

An expression in terms of k for 2U is

A 212 k

B kk 212 2

C 2212 2 kk

D 16 2 kk

SECTION A

ALL questions should be attempted

Pegasys 2013

5. )cos(32 A is equal to

A )sin3(cos21 AA

B )sin3(cos21 AA

C )sincos3(21 AA

D )sincos3(21 AA

6. The diagram shows the area bounded by the curves )(xfy and )(xgy

Which of the following gives the value of the shaded area?

A 1

3)()( dxxfxg

B

3

1)()( dxxgxf

C 1

3)()( dxxfxg

D 1

3)()( dxxgxf

x

y

O 1 3

)(xfy

)(xgy

Pegasys 2013

7. If 0dx

dy for all values of x then

A 0y for all x

B y cannot have a maximun value

C y is always decreasing

D 0y for all x

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

shown opposite.

Which of the following graphs could represent the

related function 3)( xfy ?

A B

C D

9. Which of the following statements is false for the circle 01022 yyx ?

A The circle passes through the origin

B The circle has radius 5 units

C The circle passes through the point )10,10(

D The circle has its centre on the y-axis

y

o

3

(5, 3)

)(xfy

x

y

x o

)6,5(

y

x o

)6,5(

y

o x 5

6

5

y

x o

6

Pegasys 2013

10.

dxx 2

1

equals

A Cx 2

1

B Cx 2

1

2

C Cx 21

21

D Cx

23

21

11. The quadratic equation 0)1(32 2 kxx has no real roots.

Which one of the following describes the possible values of k ?

A 81k

B 81k

C 81k

D 81k

12. When the polynomial 1573 23 xx is divided by 3x the remainder is

A 9

B 129

C 33

D 33

13. P is the point )0,4( , Q is )3,0( and R is )1,1( .

The fourth vertex S of the parallelogram PQRS has coordinates

A )2,3(

B )2,5(

C )2,5(

D )2,3(

Pegasys 2013

14. If 210sincos x for 1800 x , then x has the value

A 120

B 150

C 30

D 60

15. An arrow is fired into the air reaching a height of 'h' metres after 't' seconds where

23182)( ttth . The time taken, in seconds, to reach its maximum height is

A 29

B 3

C 2

D 9

16. The line mxy passes through the point of intersection of the lines ax and by , if

and only if

A b

am

B b

am

C a

bm

D a

bm

17. The solution to the inequality 0)2)(3( xx is

A 23 x

B 3x and 2x

C 2x and 3x

D 32 x

Pegasys 2013

18. If 12 xdx

dy and 4y when 1x , then y is equal to

A xx 2

B 22 xx

C 42 xx

D 62 xx

19. The diagram below shows part of a quadratic function cbxaxxf 2)( .

Which one of the following statements is true

A 0a

B 0c

C 042 acb

D 042 acb

20. If 4

3

4

1

2

1

8 yx , then y equals

A 2

3

2

x

B 2

3

2

1

8 y

C 3

2

2x

D 3

2

2

x

o x

f(x)

[ END OF SECTION A ]

Pegasys 2013

21. The diagram below, which is not drawn to scale, shows a circle, centre C, with

equation 0105141822 yxyx .

PQ is a diameter of this circle.

(a) Write down the coordinates of C. 1

(b) Hence find the coordinates of P. 2

(c) Find the equation of the circle, centre P and passing through C. 3

22. Two functions are defined on suitable domains as 12)( xxf and 2)2()( xxg .

Given that the function h is such that ))(()( xfgxh , express h in the form

cbxaxh 2)()( , where a, b and c are integers,

and hence write down the minimum value of h and the corresponding replacement for x. 6

23. Solve the equation 0cos622sin12 AA , for 2

0 A . 5


P

C

Q( 13 , 10 )

x

y

O

0105141822 yxyx

Pegasys 2013

24. A recurrence relation is given as 60801 nn uu .

(a) Given that 2201 u , find the initial value, 0u , of this sequence. 2

(b) State why this sequence has a limit and hence find the difference between the initial

value and the limit of the sequence. 4

25. An equation is given as 023 bxaxx , where a and b are constants.

(a) It is known that 1x and 2x are two roots of this equation.

Use the above roots to establish the values of the constants a and b. 4

(b) Hence find the third root of this equation. 3

[ END OF SECTION B ]

[ END OF QUESTION PAPER 1 ]

Pegasys 2013

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

Time allowed - 1 hour 10 minutes

NATIONAL

QUALIFICATIONS

Pegasys 2013

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

Pegasys 2013

1. In the diagram triangle ABC has vertices A( 8 , 4 ), B( 20 , p ) and C( 2 , 17 ) as shown.

AD is perpendicular to side BC. [Diagram not drawn to scale]

(a) Given that the gradient of side BC is21 , find the value of p. 3

(b) Find the equation of the altitude AD in the form Ax + By + C = 0. 3

(c) By considering the gradients of side AB and the altitude AD, calculate the size

of the shaded angle DAB. 3

2. In the diagram below the triangle and the rectangle have equal areas.

(a) Show clearly that the following quadratic equation, in x, can be constructed from

the information given.

0)22( 22 pxpx 4

(b) For what value of p does the above quadratic equation have equal roots? 4


x

y

O

A( 8 , 4 )

B( 20 , p )

C( 2 , 17 )

D

2p

2x + p x2

x

Pegasys 2013

3. The diagram below shows parts of the graphs of 432 xxy and 3234 xxy .

The curves intersect on the y-axis and at the point with x-coordinate a.

(a) Find, algebraically, the value of a. 4

(b) Hence calculate the area enclosed between the two curves. 4

4. Part of the graph of nxky sin is shown in the diagram.

(a) Write down the value of n. 1

The broken line has equation 3y .

(b) Given that the x-coordinate of the point P is 26o, calculate the value of k correct

to 1 decimal place. 3

x

y

O

3234 xxy

432 xxy

a

180o

x

y

P 3y

o 26o

Pegasys 2013

5. A function is defined on a suitable domain as )(xf xxx

21.

(a) Find the derivative of the function, )(xf , expressing your answer with

positive indices. 4

(b) Find the equation of the tangent to the curve xxx

y 21, at the point

where 1x . 3

6. A circle has a radius of 10 units and the point )2,3( as its centre.

(a) Write down the equation of this circle. 1

(b) Given that the point P( 5 , k ) lies on this circle, find k where 0k . 4

(c) Find the equation of the tangent to this circle at the point P. 4

(d) Show clearly that this tangent passes through the centre of the circle with

equation 04316422 yxyx . 2

7. The diagram below shows part of the curve with equation .3452 23 xxxy

(a) Find the coordinates of the stationary point marked A. 4

(b) Find the coordinates of B, one of the points where the curve crosses the x-axis. 2

.3452 23 xxxy

A

x

y

O

B

Pegasys 2013

8. Rectangle ABCD measures 6 units by 1 unit as shown. The diagram is not drawn to scale.

Angle BAC p radians. Triangle BCE is isosceles with BEBC .

(a) Show clearly that pCBE 2 . 3

(b) Hence calculate the exact value of EBC ˆcos . 4

A B

C D

E

p

1

6

[ END OF QUESTION PAPER ]

Pegasys 2013

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

Pegasys 2013

Higher Grade - Paper 1 2013/2014 ANSWERS - Section A

1 B

2 D

3 C

4 C

5 A

6 D

7 B

8 C

9 C

10 B

11 B

12 D

13 B

14 D

15 B

16 C

17 D

18 B

19 D

20 A

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

A B C D

Pegasys 2013

Higher Grade Paper 1 2013/2014 Marking Scheme

21(a) ans: C(9, 7) (1 mark)

●1 states centre of circle ●

1 C(9, 7)

(b) ans: P(5, 4) (2 marks)

●1 uses a stepping out method or midpoint ●

1 evidence

●2 finds coordinates of P ●

2 P(5, 4)

(c) ans: (x – 5)² + (y – 4)² = 25 (3 marks)

●1

finds radius of circle ●1 r = 5 units

●2 starts to write equation of circle ●

2 (x – 5)² + (y – 4)²

●3 completes equation ●

3 (x – 5)² + (y – 4)² = 25

22 ans: 0; 2

1x (6 marks)

●1 substitutes f(x) into g(x) ●

1 2)212())(( xxfg

●2 simplifies and multiplies bracket ●

2 144)12())(( 22 xxxxfg

●3 start to complete square ●

3 1]

4

1)

2

1[(41)(4 22 xxx

●4 completes square ●

4 22 )

2

1(411)

2

1(4 xx =

●5 states minimum value of function ●

5 minimum value = 0

●6 states corresponding value for x ●

6

2

1x

23 ans: 4

A or

2

(5 marks)

●1 replaces sin 2A ●

1 0cos62)cossin2(12 AAA

●2 multiplies and takes common factor ●

2 0)6sin12(cos2 AA

●3 makes each bracket equal to zero and ●

3 0cos2 A or 0)6sin12( A

finds first solution 2

A

●4 simplifies second bracket ●

4

2

1

2

1

12

6

12

6sin A

●5 finds solution ●

5

4

A

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

Pegasys 2013

24(a) ans: u0 = 200 (2 marks)

●1 substitutes into RR ●

1 0·8u0 + 60 = 220

●2 finds u0

●

2 u0 = 200

(b) ans: 100 (4 marks)

●1 states condition for limit ●

1 limit exists since – 1 < 0·8 < 1

●2 knows how to find limit ●

2

20

60

801

60

L

●3 finds limit ●

3 300

●4 states difference ●

4 100

25(a) ans: a = 2; b = –2 (4 marks)

●1 knows to set up system of equations ●

1 evidence

●2 uses synthetic division with first root

●

2

011

11

111

baaa

aa

ba

●3 uses synthetic division with second root ●

3

642321

64242

112

baaa

aa

ba

●4 solves for a and b ●

4 a = 2; b = –2

(b) ans: x = –1 (3 marks)

●1 substitutes for a and b and uses syth.div. ●

1

0231

231

21211

●2 factorises ●

2 0)1)(2)(1( xxx

●3 states third factor ●

3 x = –1


Total: 70 marks

Pegasys 2013

Higher Grade Paper 2 2013/2014 Marking Scheme

1(a) ans: p = 8 (3 marks)

●1 finds expressions for gradient of BC ●

1

18

17

220

17

ppmBC

●2 equates to – ½ ●

2

2

1

18

17

p

●3 solves for p ●

3 p = 8

(b) ans: y – 2x + 12 = 0 (3 marks)

●1 states gradient of AD ●

1 mAD = 2

●2 substitutes into y – b = m(x – a) ●

2 y – 4 = 2(x – 8)

●3 rearranges into required form ●

3 y – 2x + 12 = 0

(c) ans: 45o (3 marks)

●1 finds gradient of AB ●

1

3

1

820

48

ABm

●2 takes tan

–1 of both gradients ●

2 463)2(tan;418)

3

1(tan 11

●3 subtracts to find angle DAB ●

3 45

o

2(a) ans: proof (4 marks)

●1 finds expressions for both areas ●

1 p(2x + p) and x(2 – x)

●2 equates expressions and starts to simplify ●

2 2xp + p² = 2x – x²

●3 brings all terms to LHS ●

3 2xp + p² – 2x + x² = x² – 2x + 2xp + p²

●4 rearranges to required form ●

4 x² + (2p – 2) x + p² = 0

(b) ans: 2

1p (4 marks)

●1 knows to make discriminant equal to 0 ●

1 b² – 4ac = 0 [stated or implied]

●2 lists values of a, b and c ●

2 a = 1; b = 2p – 2; c = p²

●3 substitutes and simplifies ●

3 (2p – 2)² – 4 . 1 . p² = 0; 4p² – 8p + 4 – 4p² = 0

●4 solves for p ●

4

2

1p


Pegasys 2013

3(a) ans: a = 3 (4 marks)

●1 knows to equate the two functions ●

1 a

2 – 3a + 4 = 4 + 3a

2 – a³

●2 brings all terms to LHS and simplify ●

2 032 23 aaa

●3 factorises ●

3 0)1)(3(;0)32( 2 aaaaaa

●4 solves for a and chooses correct value of a ●

4 a = 0, 3, –1; a = 3

(b) ans: 11·25 units² (4 marks)

●1 sets up integral ●

1

3

0

232 )43(34 dxxxxx

3

0

32 32 dxxxx

●2 integrates ●

2

3

0

243

2

3

43

2

xxx

●3 substitutes limits ●

3

3

0

243

2

)3(3

4

)3(

3

)3(2

●4 evaluates ●

4 11·25 units²

4(a) ans: n = 6 (1 mark)

●1 states value of n ●

1 n = 6

(b) ans: k = 7·4 (3 marks)

●1 substitutes and equates ●

1 3)266sin( k

●2 finds expression for k ●

2

156sin

3k

●3 value of k correctly rounded ●

3 k = 7·4

5(a) ans: 23

11

x2 (4 marks)

●1 starts to prepare to differentiate ●

1 )( 2

121 xxx

●2 completes preparation ●

2 2

1 xx

●3 differentiates both terms ●

3 2

3

2

11

x

●4 writes with positive indices ●

4

2

3

2

11

x

(b) ans: 2y = 3x – 3 (3 marks)

●1 substitutes into function for point

●

1 (1, 0)

●2 substitutes into derivative for gradient ●

2 m = 3/2

●3 states equations of tangent ●

3 y – 0 = 3/2(x – 1) ; 2y = 3x – 3


Pegasys 2013

6(a) ans: (x + 3)² + (y – 2)² = 100 (1 mark)

●1 subs into equation of circle ●

1 (x + 3)² + (y – 2)² = 100

(b) ans: k = –4 (4 marks)

●1 substitutes point into circle equation ●

1 (5 + 3)² + (k – 2)² = 100

●2 simplifies ●

2 64 + (k – 2)² = 100; (k – 2)² = 36

●3 solves for k ●

3 k – 2 = ±6; k = 8 or k = –4

●4 chooses appropriate value for k ●

4 k = –4

(c) ans: 3y – 4x + 32 = 0 (4 marks)

●1 finds gradient of CP ●

1

4

3CPm

●2 finds perpendicular gradient ●

2

3

4tan m

●3 substitutes into equation ●

3 )5(

3

44 xy

●4 simplifies ●

4 3y – 4x + 32 = 0 [or equivalent]

(d) ans: proof (2 marks)

●1 finds centre of circle ●

1 centre of circle is (2, –8)

●2 substitutes into equation of tangent ●

2 3(–8) – 4(2) + 32 = 0 which is correct

tangent passes through centre of circle

7(a) ans: A(–2, 9) (4 marks)

●1 knows 0

dx

dy ●

1 evidence stated or implied

●2 differentiates and equates to zero ●

2 04106 2 xx

dx

dy at SP

●3 solves for x and chooses required value ●

3 2(3x – 1)(x + 2) = 0; x = –2 or 1/3

●4 finds coordinates of point A ●

4 A(–2, 9)

(b) ans: B(1, 0) (2 marks)

●1 makes y = 0 and uses synthetic division ●

1

0372

372

34521

●2 factorises, solves and chooses approp. value ●

2 (x – 1)(2x + 1)(x + 3) = 0

states coordinates of B B(1, 0)


Pegasys 2013

8(a) ans: proof (3 marks)

●1 finds expression for ACB ●

1 ACB = 90 – p

●2 knows CEB = ACB ●

2 CEB = 90 – p

●3 finds expression for CBE ●

3 CBE = 180 – 2(90 – p) = 2p

(b) ans: 7

5 (4 marks)

●1 uses expansion for cos2p ●

1 ppp 22 sincos2cos

●2 finds value of cos

p and sin

p ●

2

7

1sin;

7

6cos pp

●3 subs into expansion ●

3 22 )

7

1()

7

6(2cos p

●4 simplifies to answer ●

4

7

5

7

1

7

6

Total: 60 marks


national mathematics qualifications higher prelim ... · higher prelim examination 2013/2014 paper...

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