higher extended unit tests 2010-11© pegasys 2010 dingwall academy 2. 4. mathematics higher grade...

© Pegasys 2010

Contents

3 Unit Tests (at levels A, B and C)

Detailed marking schemes

Detailed marking schemes

Higher Still

Higher Mathematics

Extended Unit Tests 2010-2011

(more demanding tests covering all levels)

H I G

H

E

R

S T I

L L

Pegasys Educational Publishing

M

A

T

H

E

M

A

T

I

C

S

© Pegasys 2010

DINGWALL ACADEMY

MATHEMATICS

Higher Grade Extended Unit Test - UNIT 1

Time allowed - 50 minutes

Read Carefully 1. Full credit will be given only where the solution contains appropriate working. 2. Calculators may be used. 3. Answers obtained by readings from scale drawings will not receive any credit.

4. This Unit Test contains questions graded at all levels.

© Pegasys 2010

1. A sequence is defined by u 1n = 2u n + 3 with u 1 = 2.

The value of u 3 is

A. 17

B. 13

C. 9

D. 7

2. Here are two statements about the line ST where S is the point (3, 2) and T the point (7, – 1).

(1) The length of ST is 5 units.

(2) The gradient of ST is4

3.

Which of the following is true?

A. Neither statement is correct

B. Only statement (1) is correct

C. Only statement (2) is correct

D. Both statements are correct

3. The gradient of the tangent to the curve with equation 32xy – 4 x at the point (2, 8) is

A. 20

B. 16

C. 4

D. –2

Section A

In this section the correct answer to each question is given by one of the alternatives A, B, C or D.

Indicate the correct answer by writing A, B, C or D opposite the number of the question.

Rough working may be done on the paper provided. 2 marks will be given for each correct answer.

© Pegasys 2010

4. Two functions, f and g, are defined on suitable domains as 12)( xxf and 1)( 3 xxg .

The value of ))((21fg is

A. 9

B. 7

C. – 7

D. – 9

5. The point A(–2, 4) lies on the graph with equation ).(xfy The graph of the related function

3)2( xfy is drawn.

The coordinates of the image of point A are

A. (–4, –7)

B. (–4, 1)

C. (0, –7)

D. (0, 1)

End of Section A

© Pegasys 2010

6. (a) The line L 1 passes through the point (–5, 2) and makes an angle of 45º with the

positive direction the x - axis.

Find the equation of L 1 3

(b) The line L 2 is perpendicular to L 1 and

passes through the point with coordinates

(–6, –3).

Find the coordinates of P, the point of

intersection of the lines L 1 and L 2 . 4

7. The graph shows part of the graph of y = f(x). It crosses the x and y axes at (–2, 0), (3, 0)

and (0, –4) as shown.

Sketch the graph of the related function

2)1( xfy

3

8. A function is defined as )3()( 2 xxxg .

(a) Find the points where the graph of g(x) cuts the x and y axes. 3

(b) Find the stationary points of this function g and determine the nature of each. 6

Section B

ALL QUESTIONS SHOULD BE ATTEMPTED

In this section credit will be given for all correct working.

–2 x

y

O 3

–4

x

y

0

L1

(–5, 2)

(–6, –3) L2

P

© Pegasys 2010

9. Given that x

xf1

)( ( 22 x ) , evaluate f ′(4) 4

10. Two functions are defined on suitable domains as

x

xxf1

)( and 2)( 2 xxg .

Find an expression, in its simplest form, for ))(( xfg . 3

11. For what value(s) of x is the function

3 + 75x – x3 decreasing ? 4

12. A farmer intends to use 120 metres of fencing to form the perimeter a rectangular enclosure.

The width of the rectangle will be x metres.

(a) Show that the area that can be enclosed is given by

)60()( xxxA 1

(b) Calculate the maximum area that can be enclosed, justifying your answer. 4

END OF QUESTION PAPER

© Pegasys 2010

Higher Grade Unit Tests 2010/2011 Marking Scheme - UNIT 1

1 A

2 B

3 A

4 C

5 D

6(a) ans: 7 xy (3 marks)

●1 substitutes values ●

1 1m , (a,b) = (– 5, 2)

●2 forms equation ●

2 )5(2 xy

●3 simplifies equation ●

3 7 xy

6(b) ans: P(8, –1) (4 marks)

●1 finds L 2 ●

1 9 xy

●2 equates lines ●

2 79 xx

●3 finds x coordinate ●

3 x = – 8

●4 finds y coordinate ●

4 y = –1

7 ans: graph drawn (3 marks)

●1 correct shape of graph ●

1

●2 graph through (–3, –2) and (2, –2) ●

2

●3 graph through (–1, 2) ●

3

8(a) ans: (0, 0), (3, 0) (3 marks)

●1

makes g(x) = 0 ●1 0)3(2 xx

●2 finds values of x ●

2 02 x , 03 x so (0, 0),(3, 0)

●3 substitutes x = 0

●

3 (0, 0)

8(b) ans: max at (0,0)

min at (2, –4) (6 marks)

●1 knows g'(x) = 0 ●

1 g'(x) = 0 at SP

●2 finds derivative ●

2 3x

2– 6x

●3 solves for x

●

3 3x(x – 2) = 0, x = 0, 2

●

4 finds corresponding y values

●

4 0, –4

●

5 sets up table of values (or 2

nd derivative)

●

5 table of values or takes 2

nd derivative

●6 states nature of SPs ●

6 max at (0,0);

min at (2, –4)

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

Award 2 marks for each

correct answer

10 marks

(–1,2)

x

y

O (2, – 2) (–3, – 2)

© Pegasys 2010

9 ans: 8

13 (4 marks)

●1 prepares to differentiate ●

1 2

1

2

3

2

xx

●2 differentiates ●

2 2

1

2

3x ..........

●3 differentiates a negative power ●

3 ..... 2

3

x

●4 substitutes for x ●

4

2

3×2 +

8

1

10 ans: 2

2 1

xx (3 marks)

●1 substitutes ●

1 2)

1( 2

xx

●2 removes brackets ●

2 2

12

2

2 x

x

●3 states answer ●

3

2

2 1

xx

11 ans: x < – 5 or x > 5 (4 marks)

●1 knows to differentiate ●

1 75 – 3x

2

●2 knows derivative < 0 ●

2 75 – 3x

2 < 0

●3 attempts to solve for x ●

3 graph drawn (or any acceptable method)

●4 answer ●

4 x < – 5 or x > 5


Total: 40 marks

© Pegasys 2010

12(a) ans: proof (1 mark)

●1 finds area ●

1 proves expression for area

12(b) ans: 900m2

(4 marks)

●1 differentiates ●

1 xxA 260)('

●2 solves for x ●

2 x = 30

●3 justifies answer ●

3 sets up table of values or uses 2

nd deriv.

●4 finds maximum area ●

4 maximum = 30 × 30 = 900m 2


Total: 45 marks

© Pegasys 2010

Higher Still - 2010/2011

MATHEMATICS



Read Carefully 1. Full credit will be given only where the solution contains appropriate working. 2. Calculators may be used.

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


© Pegasys 2010

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

Trigonometric formulae:

© Pegasys 2010

1. The roots of a quadratic equation can be described as:

I Real II Equal III Distinct IV Non-real

Which of the above can be used to describe the roots of the equation 0543 2 xx ?

A. I and II

B. I and III

C. II and IV

D. IV only

2. If sinα = 13

5 , then the value of sin2α is:

A. 13

10

B. 13

12

C. 169

60

D. 169

120

3. The circle with centre (3, –2) and radius 5 units has equation:

A. 5)2()3( 22 yx

B. 5)2()3( 22 yx

C. 25)2()3( 22 yx

D. 25)2()3( 22 yx

Section A




© Pegasys 2010

4. When 782 2 xx is expressed in the form qpx 2)(2 .

The value of q is

A. −1

B. 1

C. −2

D. 2

5. A curve for which 22 xdx

dypasses through the point (2, 3).

The equation of the curve is

A. 322 xxy

B. 122 2 xxy

C. y 522 xx

D. 322 2 xxy

End of Section A

© Pegasys 2010

6. Find the equation of the tangent to the circle with equation 0924422 yxyx

at the point (4, –6). 4

7. Solve the equation 3 = 2cos2xº + 4sinxº in the interval 3600 x . 4

8. If 6523 xkxx is exactly divisible by (x – 2), find k and hence fully factorise the

expression. 4

9. Two curves with equations 2)2( 2 xy and 262 xxy meet at A and B

as shown in the diagram.

(a) Calculate the coordinates of A and B. 3

(b) Find the area between the two curves.

i.e the shaded area in the diagram. 4

10. Find k , such that the line with equation kxy is a tangent to the curve xxy 52 5

and find the coordinates of the point of contact.

Section B



A

B

262 xxy

2)2( 2 xy

© Pegasys 2010

11. A circle has equation 25)4()2( 22 yx .

The straight line with equation 5 xy

cuts the circle at A and B.

Find the coordinates of A and B. 6

.



x

y

A

B

0

© Pegasys 2010

1 D

2 D

3 D

4 A

5 C

6 ans: 03643 yx (4 marks)

●1 centre of circle ●

1 centre is (–2, 2)

●2 finds gradient of radius

●

2 mrad =

6

8

●3 finds gradient of tangent ●

3 mtan =

4

3

●4 substitutes in equation of straight line ●

4 )4(

4

36 xy

7 ans: 30o, 150

o (4 marks)

●1 substitutes for x2cos ●

1 xx sin4)sin21(23 2

●2 multiplies out and simplifies

●

2 01sin4sin4 2 xx

●3 solves for sinxº ●

3 sinxº =

2

1


4 x = 30

o, 150

o

8 ans: (x – 2)(x +1)(x + 3) (4 marks)

●1 knows to use syn.div. ●

1 evidence

●2 uses x = 2 in division ●

2

0841221

24242

6512

kkk

kk

k

●3 finds k ●

3 k = 2

●4 completes factorising ●

4 (x – 2)(x +1)(x + 3)

Give 1 mark for each


Illustration for awarding each mark


correct answer

10 marks

© Pegasys 2010

9(a) ans: A(3, 7), B(5, 7) (3 marks)

●1 equates lines ●

1 2462 22 xxxx


2 x = 0 or 5

●3 finds coordinates of A and B

●

3 A(0, 2) , B(5, 7)

9(b) ans 3

241 sq units (4 marks)

●1 sets up integration and simplifies ●

1 dxxx

5

0

2 )210(

●2 integrates ●

2

5

0

3

2

3

25

xx

●3 substitutes ●

3 ]0[]125

3

2255[

4 evaluates

4

10 ans: k = −9 , (3, −6) (5 marks)

●1 equates equations to get ●

1 062 kxx

●2 finds 042 acb ●

2 36 + 4k = 0

●3 finds k ●

3 k = −9

4 uses k

4 0962 xx so x = 3

●5 finds y y = −6

11 ans: A(−2,−7) , B(5, 0) (6 marks)

●1 substitutes for y ●

1 25]4)5[()2( 22 xx

●2 expands ●

2 251244 22 xxxx 0

●3 simplifies ●

3 01032 xx

●4 factorises ●

4 0)5)(2( xx

●5 finds x coords ●

5 x = −2 or 5

●6 y coords, states A and B ●

6 y = −7 or 0 so A(−2,−7) B(5, 0)

3

241

5

Total: 40 marks

© Pegasys 2010

Higher Still - 2010/2011

MATHEMATICS



Read Carefully 1. Full credit will be given only where the solution contains appropriate working. 2. Calculators may be used.

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


© Pegasys 2010

FORMULAE LIST

Scalar Product: .andbetweenangletheisθwhere,cosθ. bababa

or

ba .

3

2

1

3

2

1

332211

b

b

b

and

a

a

a

where babababa

A

A

AAA

AAA

BABABA

BABABA

2

2

22

sin21

1cos2

sincos2cos

cossin22sin

sinsincoscoscos

sincoscossinsin

axaax

axaax

xfxf

sincos

cossin

)()(

f x f x dx

axa

ax C

axa

ax C

( ) ( )

sin cos

cos sin

1

1

Table of standard derivatives:

Table of standard integrals:

Trigonometric formulae:

© Pegasys 2010

1. The point (7, 3) lies on the graph of )1(log xy a .

The value of a is

A. 2

B. 3

C. 4

D. 5

2. A(−2, −3, 1) , P(0, 3, 6) , and B(4, 15, 16) are three collinear points.

P divides AB in the ratio

A. 1:1

B. 1:2

C. 1:3

D. 1:4

3. dxx )12cos(4 equals

A. 2 cx )12sin(

B. cx )12sin(4

C. cx )12sin(8

D. cx )12sin(2

Section A




© Pegasys 2010

4. 4)25( xdx

d equals

A. 3)25(4 x

B. 3)25(8 x

C. 3)25(6 x

D. 5)25(10

1x

5. The vectors with components

k

k

1 and

4

4

k

are perpendicular.

The value of k is:

A. 1

B. -1

C. 2

D. -2

End of Section A

© Pegasys 2010

6. R and S are the points with coordinates (2, 5, −1) and (6, −3, 11) respectively.

(a) State the coordinates of the mid point of RS. 1

(b) T divides RS in the ratio 3:1. Find the coordinates of T 3

7. Triangle PNQ has coordinates P(5, 10, – 3) N(2, – 6, 0) and Q(8, 6, 3).

Find the size of angle PNQ. 5

.

8. Evaluate dxx 2)13( 3

9. Express cos2xº + 3 sin2xº in the form kcos(2x – α)º 0 ≤ x ≤ 180 and hence

solve the equation cos2xº + 3 sin2xº = 1. 6

10. (a) Simplify: log a 12 + log a 4 1

(b) Simplify: 5log 9 3 – log 9 27 3

11. Find )sin3cos2( 2 xxdx

d , expressing your answer in terms of 3x and 2x 3

12. The number of germs present in a laboratory sample after t hours is given by the formula

tetG 6150)( .

State how many germs were present in the sample initially, and find how many minutes

the germs will take to double in number. 5



Section B



1

-⅔

© Pegasys 2010

1 A

2 B

3 A

4 B

5 C

6(a) ans: (4, 1, 5) (1 mark)

1 answer 1 (4, 1, 5)

6(b) ans: T(5, −1, 8) (3 marks)

●1 Finds RS ●

1 RS=

12

8

4

●2 Finds RT ●

2 RT=

12

8

4

4

3

●3 answer ●

3 T(5, −1, 8)

7 ans: 28º (5 marks)

●1 finds NP and NQ ●

1 NP=

3

16

3

; NQ=

3

12

6

●2 finds scalar product ●

2 18 + 0 + 9 = 201

●3 finds magnitudes of NP and NQ

●

3 √274 and √189

●4 subs into formula ●

4 PNQcos

189274

201

●5 finds angle ●

5 28º



correct answer

10 marks

© Pegasys 2010

8 ans: 9

83 (3 marks)

●1 integrates ●

1

3

1

3

)13( 3

x

●2 substitutes values ●

2

9

)12(

9

)13( 33

●

3 evaluates answer ●

3

9

83

9

27

9

8

9 ans: 60o, 180

o (6 marks)

●1 expands )2cos( xk

o ●

1 sin2sincos2cos xkxk

●2 calculates k = 2 ●

2 132 k : k =2

●2 calculates α = 60

o ●

3

1

3tan : 60

●

4 uses answer ●

4 )602cos(2 x º = 1

●5 evaluates (2x – 60) ●

5 60

o, 300

o

6 evaluates x ●

6 60

o, 180

o

10(a) ans: log a 48 (1 mark)

●1 answer ●

1 lo48 3a

10(b) ans: 1 (3 marks)

●1 changes 3log5 9

1 27log3log 9

5

9

●

2 uses laws of logs ●

2 )

27

243(log 9

●3 answer ●

3 19log9

11 ans: xx 2sin3sin6 (3 marks)

●1 differentiates 1

st term ●

1 xx 3sin623sin3

●2 differentiates 2

nd term ●

2 xxcossin2

●3 answer ●

3 xx 2sin3sin6

12 ans: 50, 26 minutes (5 marks)

1 50

1 t = 0, G(0) = 50

2 equates values

2 te 6150100 or 261 te

3 takes logs

3 1∙6tlne = ln2

4 finds t hours

4 t = 0∙433 hours

5 answer

5 0∙433 hours = 26 minutes


Total: 40 marks

higher extended unit tests 2010-11© pegasys 2010 dingwall academy 2. 4. mathematics higher grade...

Documents