advanced higher - extended unit tests a

Pegasys 2005

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T T H

M A M A

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Higher Still

Advanced Higher Mathematics

Extended Unit Tests A

(more demanding tests covering all levels)

Contents

3 Extended Unit Tests Detailed marking schemes

Pegasys Educational Publishing

Pegasys 2005

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.

Time allowed - 50 minutes

MATHEMATICS

Advanced Higher Grade

Extended Unit Tests A - UNIT 1

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All questions should be attempted 1. Differentiate the following with respect to x :

yxx

=+−

ln

11

(3)

2. In the expansion of (1 + px)(1 + qx)5, where ,0and, ≠≠∉ qpRqp

the coefficient of x2 is zero and the coefficient of x3 is –270. Find the values of p and q. (4)

3. The function f is defined by f(x) = exsin x, where 20 ≤≤ x .

Find the coordinates of the stationary points of f and determine their nature. (5)

4. Use the substitution to evaluate the definite integral tx sin3=

dxx

x2−

9∫

23

(5) 0

5. Use Gaussian elimination to solve the following system of equations

5353553251

⋅=−⋅=+−⋅=++

cacbacba

(5)

6. A particle is moving along a straight line so that at a time t its displacement x from a fixed point on the line is given by

x t= +2 3 16cos( ). Prove that v2 – ax is always constant ,where v is the velocity and a is the acceleration of the particle at time t. (4)

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7. The function f is defined by

f xx xx

x( ) =+−

≠ ±3

2

91

1 ,

(a) (i) Write down the equations of the vertical asymptotes (1)

(ii) Show that has a non-vertical asymptote and obtain its equation. (2) )(xfy =

(iii) Find the point(s) of intersection with the two axes. (1) (b) Find the coordinates and nature of the stationary points of f(x) . (5) (c) Sketch the graph of . )(xfy =

(You must show all of the above results in your sketch ) (3)

[ END OF QUESTION PAPER ]

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Advanced Higher Grade - Extended Unit Tests A Marking Scheme – UNIT 1

Give 1 mark for each •

21 2− x

34

7 4674

172 64π π

, . , .

−

Max, Min

1.

ans:

3 marks • 1 knowing to use chain rule • 2 knowing to use quotient rule • 3 completing simplification

2. ans: p = -6, q = 3 4 marks

• 1 using Binomial Expansion • 2 product of brackets • 3 creating a system of equations • 4 solving equations

3. ans:

5 marks • 1 differentiating using product rule • 2 equating to zero • 3 solving for x • 4 evaluating y coordinates • 5 justifying nature

Illustration(s) for awarding each mark

• 1

• 2

• 3 answer

• 1

11

11

−+

×+−

xx

ddx

xx

( )11

21 2

−+

×−

xx x

1 5 10 102 3+ + + +( ) ( ) ( ) ..qx qx qx

1 5 10 105 10

2 2 3 3

2 2 3

+ + + +

+ + + +

qx q x q xpx pqx pq x

......

2 010 10 2702 2

q pq pq+ =

+ = −

( )e xx sin cos+sin cosx x+ = 0

x =34

74

π π,

y = −7 46 172 64. , .from or nature table′′f x( )

.

• 2

• 3

• 4 answer

• 1 • 2

• 3

• 4

x

• 5

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Give 1 mark for each • 4. ans: 0.1438 5 marks

• 1 dealing with substitution • 2 finding dx and new limits • 3 simplifying expression • 4 integrating correctly • 5 evaluating correctly

5. ans: a b c= = − =2 1, , 05. 5 marks

• 1 using augmented matrix • 2 first modified system • 3 second modified system • 4 finding one value • 5 finding other 2 values

6. ans: proof 4 marks

• 1 knowing how to calculate v • 2 knowing how to calculate a • 3 substituting into statement • 4 completing the proof


• 1 9 92 2− =x xcos

• 2 dx t dt= 3 06

cos , , limits = π

• 3 tan tdt06π

∫

• 4 [ ]− ln(cos )t 06π

• 5 answer

• 1 113

1 1 1.5 - 2 3 5.5 0 -5 3.5

• 2

−−−−

18304230

5.1111

• 3

−−−

510004230

5.1111

• 4 c = 0.5 • 5 b = -1 and a = 2

• 1 vdxdt

t= = − +6 3 16sin( )

• 2 advdt

t= = − +18 3 16cos( )

• 3 v ax

t t

2

2 236 3 16 36 3 16

− =

+ + +sin ( ) cos ( )

• 4 36

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Give 1 mark for each • 7. (a) i) ans: x = ±1 1 mark

• 1 stating equations

ii) ans: y x= 2 marks • 1 restating the function • 2 stating equation

iii) ans: (0,0) 1 mark

(b) ans: (-3.565, -6.610) Max ( 3.565, 6.610) Min 5 marks • 1 knowing to differentiate using

quotient rule • 2 knowing to solve f’(x)=0 • 3 solving f’(x) = 0 • 4 finding y coordinates • 5 justifying nature

(c) ans: sketch 3 marks

• 1 sketch showing all relevant pointsand turning points

• 2 showing how curve approaches asymptotes

• 3 completing curve


• 1 answer

• 1 y xx

x= +

−10

12

• 2 y = x • 1 for answer (0,0)

• 1 22

24

)1(912)(

−−−

=′x

xxxf

• 2 x x4 212 9 0− − =

• 3 x x2 12 1802

3565=+

= ±, .

• 4 y = ±6 610.

• 5 ′′ − <′′ >

ff

( . )( . )

356 0356 0

so Max so Min

Total : 38 marks

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Time allowed - 50 minutes

MATHEMATICS

Advanced Higher Grade

Extended Unit Tests A - UNIT 2

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 2005

All questions should be attempted 1. Differentiate the following with respect to x :

y x= −sin ( )1

(3) 2. A curve has parametric equations and . 42 2 −= tx 34 43 tty −= Find the equation of the tangent to the curve when 1−=t . (4) 3. (a) Verify that z = 3 is a solution of the equation . (1) 09922 23 =−−− zzz (b) Express 2 z as a product of a linear factor and a 992 23 −−− zz quadratic factor with real coefficients.

Hence find all the solutions of (3) 09922 23 =−−− zzz 4. The first, fourth and eighth terms of an arithmetic sequence are in geometric progression.

Find : (i) the relationship, in its simplest form, between a, the first term, and d, the common difference; (3) (ii) the value of r the common ratio. (1)

5. (a) Find partial fractions for (3) 2 1x +

( )( )2 12x x− +

(b) Hence show that

=

+−+

1345ln

21

)1)(2(122 dx

xxx (5) ∫

3

5

6. Prove by induction that for all positive integers , n, (5) r

∑=

n

11

11)1(

1+

−=+ nrr

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7. (a) Find the stationary point lying between the lines 2=x and 3=x of the

curve given by the equation

(4) .0,0,633 >>=+ yxxyyx

(b) By considering 2

2

xdyd , determine the nature of this stationary point. (5)


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( )1

2 1x x−

( )ddx

sin−1

iz 21 21±−=

( )1

12

− x

( )( )1

12

−×

x

ddx

x

11

12−

×x x

dxdt

tdydt

t t= = −4 12 123 2,

dydx

t t m= − ⇒ =3 3 62

( )( )z z z− + +3 2 4 32

z =− ± −4 8

4

zi

=− ±2 2

2iz 21 2

1±−=

Give 1 mark for each •

1. (a) ans:

3 marks • 1 knowing to use the chain rule

• 2 knowing

• 3 completing the simplification 2. ans: y = 6x + 19

4 marks • 1 finding coordinates of point • 2 differentiating w.r.t. x • 3 finding gradient of tangent • 4 finding the equation of line

3. (a) ans : Proof 1 mark

• 1 knowing to sub z = 3 into the polynomial

(b) ans : (or equiv.) 3 marks • 1 writing the expression as a linear

and quadratic factor • 2 using quadratic formula to solve

the quadratic • 3 finding the complex roots


• 1

• 2

• 3

• 1 (-2, 7)

• 2

• 3

• 4 answer • 1 3 2 -2 -9 -9

6 12 9

2 4 3 0 • 1

• 2

• 3

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4. (i) ans: a = 9d 3 marks

• 1 knowing how to find u u u1 4, , 8

• 2 using uu

uu

2

1

3

2= in the geometric

sequence • 3 solving equation

(ii) ans: r =43

1 mark

5. (a) ans : 1

2 12xx

x−−

+

3 marks • 1 knowing to express fraction as a

sum • 2 knowing to find A, B, C • 3 calculating A, B, C (b) ans: proof

5 marks • 1 knowing to express the integral in

PF’s

• 2 integrating 1

2x −

• 3 integrating x

x 2 1+

• 4 evaluating integral • 5 completing proof


• 1 knowing to try for one value of n • 2 assume true for n=k • 3 attempt to prove true for n=k+1 • 4 simplifying • 5 concluding statement

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

• 1 u a u a d u a1 4 83 7d= = + = +, ,

• 2 a d

aa da d

+=

++

3 73

• 3 answer • answer

• 1 ( )( )

2 12 1 2 12 2

xx x

Ax

Bx Cx

+− +

=−

+++

• 2 2 1 1 22x A x Bx C x+ = + + + −( ) ( )( )• 3 A B C= = − =1 1, , 0

• 1 1

1 123

5

xx

xdx

−−

+

∫

• 2 ( )ln x − 2

• 3 ( )12

12ln x +

• 4 ln ln ln ln312

26 112

10−

− −

• 5 12

4513

ln

• 1 n LHS RHS t= ⇒ = = ⇒112

12

, rue

• 2 1

11

111 r r kr

k

( )+= −

+=∑

• 3

( )( )

11

11

11 2

1

1

1

r r

r r k k

r

k

r

k

( )

( )

+=

++

+ +

=

+

=

∑

∑

• 4 11

2−

+k

• 5 By induction, true ∀n

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7. (a) ans: 2 243

53,

4 marks • 1 differentiating w.r.t. x

• 2 solving dy

= 0dx

• 3 substituting into original equation • 4 solving for x and y

(b) ans: Maximum 5 marks

• 1 differentiating w.r.t. x • 2 “ “ • 3 rearranging equation • 4 substituting for x and y • 5 conclusion

• 1 3 3 6 62 2x ydydx

y xdydx

+ = +

• 2 6 33 6

02

2

2

2y xy x

yx−

−= ⇒ =

• 3 xx

xx3

2 3 2

26

2+

=

• 4 ( )x x x y3 343

5316 0 2 2− = ⇒ = =,

• 1 6 6 32

22

2x ydydx

yd ydx

+

+

• 2 12 62

2

dydx

xd ydx

+

• 3 d ydx

dydx

ydydx

x

y x

2

2 2

6 2 6

3 6=

−

−

−

• 4 d ydx

2

2

43

53

2 43

6 2

3 2 6 20=

−

−

<

• 5 Answer

Total : 37 marks


Pegasys 2005

MATHEMATICS Advanced Higher Grade

Extended Unit Tests A - UNIT 3 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 2005

All questions should be attempted 1. (a) Use the Euclidean Algorithm to find integers x and y such that

123x + 337y = 1. (4) (b) Express 2132 in base 7. (3) 4

2. (a) Find the first four terms in the Maclaurin series for ( )x3

71ln − (4) (b) Hence show l (1) 3 18 81 243

( )n ln ....3 7 37 49 343 24012 3 4− = − − − − +x x x x x

3. The n x n matrices A and B satisfy the equation IAAB 47 +=

Where I is the n x n identity matrix. If A and B are both invertible, show that )7(4

11 IBA −=− (3)

4 . The position, s(t) metres, from the origin at a time t seconds, of a particle satisfies

the differential equation

xsdtds

tdsd sin1704113 2

2

=−+

If the particle starts from rest at the origin, find s(t). (9) 5. One face of an irregular tetrahedron has two of its edges defined by the following equations

27

15

12

31

22

11 −

=−

=−+−

=−−

=− zyxandzyx

(a) Show that these lines intersect and find the point of intersection. (5) (b) Calculate the size of the acute angle between these two edges. (3) (c) Find the equation of the face defined by these two edges. (3)


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Give 1 mark for each • 1. (a) ans: x y= = −137 50,

4 marks • 1 knowing to find the gcd of 337 & 123 • 2 finding the gcd • 3 knowing to rearrange the algorithm • 4 correctly rearranging the algorithm (b) ans: 3147

3 marks • 1 converting to base 10 • 2 repeated division by 7 • 3 recording generated remainders

2. (a) ans : − − − −73

4918

34381

2401324

2 3x x x 4x

4 marks • 1&2 finding f i iv− ( )0• 3 substituting above into

Maclaurins expansion • 4 simplifying expression (b) ans: proof

1 mark • 1 applying rules of logarithms


• 1&2

337 2 123 91123 1 91 3291 2 32 2732 1 27 527 5 5 25 2 2 1

= += +

= += += +

= +

( )( )

( )( )( )

( )• 3&4 123(137) + 337(-50) = 1

• 1 2 1 3 4 1 4 2 4 1582 3( ) ( ) ( ) ( )+ + + =

• 2 158 7 22 422 7 3 13 7 0 3

÷ =÷ =

÷ =

rr

r

• 3 answer

• 1&2

′ = −

′′ = −

′′′ = −

= −

f

f

f

f

( )

( )

( )

( )

073

0499

068627

014406

814

• 3

− −×

−×

−×

73

499 2

68627 3

1440681 4

2 3x x x! !

4x!

• 4 answer

• 1 ( )ln ln

ln ln

3 7 3 173

3 173

− = −

= + −

x x

x

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• 1 making I the subject of the formula

• 2 using AA-1=I • 3 identifying A-1

4. ans: ttee tt cos11sin741310

13153 3

1

−−−= −S 9 marks

• 1 creating and solving auxiliary equation• 2 stating the complementary function • • 3, 4 & 5 finding the particular integral • • 6 stating the general solution

• 7 finding ds

dt

• 8 evaluating constants using initial conditions

• 9 stating particular solution 5. (a) ans: (1, 2, 1) 5 marks

• 1 creating parametric equations • 2 equating corresponding coordinates • 3 solving two from three equations for

parameters • 4 showing parameters satisfy third

equation • 5 finding coordinates

• 1 4 7

74

I AB A

IA B I

= −

=−( )

• 2 ( )I A B I= −

14

7

• 3 answer • 1 4&,04113 3

12 −===−+ mmmm

• 2 tt BeAeS 431

−+= • 3 Let S tDtC cossin +=

• 4 tDtCS

tDtCScossin

sincos−−=′′

−=′

• 5 11,7 −=−= DC

• 6 ttBeAeS tt cos11sin7431

−−+= −

• 7 ttBeAeS tt sin11cos74 431 3

1

+−−=′ − • 8 13

1013153 , −== BA

• 9 answer

• 1 x t y t z tx t y t z t

= + = − + = += − − = + = +

1 1

2 2

1 2 2 32 5 2

, ,, ,

1

2

17

• 2

t tt t

t t

1 2

1 2

1 2

32 3

3 2 6

+ = −− − =

− =

• 3

t tt t t t1 2

1 21 2

32 3 0 3

+ = −− − =

⇒ = = −,

• 4 ( ) ( )3 0 3 3− − =

• 5 answer


}

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5. (b) ans : 70.9 o

3 marks • 1 identifying the direction vectors • 2 using dot product • 3 calculating angle

(c) ans: 7 5 18x y z+ + = 3 marks

• 1 finding the normal to the plane • 2 calculating constant • 3 stating equation

• 1 12

3

112

−

−

and

• 2 cosθ =

−

•−

×

12

3

112

14 6

• 3 answer

• 1 ( ) ( )i j k i j k

i j k− + × − + +

= − − −

2 3 27 5

• 2 − − − = −7 10 1 18 • 3 answer


Total : 35 marks

advanced higher - extended unit tests a

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