UVI FURTHER MATHS VACATION WORK...= 19.35 as a first approximation to α, apply the Newton–Raphson procedure once to h(t) to find a second approximation to α, giving your answer

UVI Further Maths Vacation Work

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UVI FURTHER MATHS VACATION WORK

The topics you will study next term rely on a good understanding of the material you have studied this year. You need to complete the following vacation work, and mark your work with the answers provided. Use your notes and textbook to help you. Your work is due in on the first lesson back, and you will be tested on this in your start up test.

Please also complete the UVI MATHS VACATION WORK



1. Show that can be written in the form

Find the values of the constants A and B.

(5 marks) ___________________________________________________________________________________

2. Use proof by contradiction to show that there exist no integers a and b for which 25a + 15b = 1.

(4 marks) ___________________________________________________________________________________

3. A curve has parametric equations x = cos 2t, y = sin t, t−π π .

(a) Find an expression for ddyx

in terms of t.

Leave your answer as a single trigonometric ratio. (3 marks)

(b) Find an equation of the normal to the curve at the point A where 56

t π= − .

(5 marks) ___________________________________________________________________________________

4. Showing all steps, find cot 3 dx x∫ .

(3 marks) ___________________________________________________________________________________

5. A triangle has vertices A(−2, 0, −4), B(−2, 4, −6) and C(3, 4, 4). By considering the side lengths of the triangle, show that the triangle is a right-angled triangle.

(6 marks) ___________________________________________________________________________________

6. The functions p and q are defined by 2p : x x→ and q : 5 2x x→ − .

(a) Given that pq(x) = qp(x), show that 23 10 10 0x x− + = (4 marks)

(b) Explain why 23 10 10 0x x− + = has no real solutions.

(2 marks) ___________________________________________________________________________________

7. Prove by contradiction that there are infinitely many prime numbers.

(6 marks) ___________________________________________________________________________________



8. In a rainforest, the area covered by trees, F, has been measured every year since 1990. It was found that the

rate of loss of trees is proportional to the remaining area covered by trees. Write down a differential equation relating F to t, where t is the numbers of years since 1990.

(2 marks) ___________________________________________________________________________________

9. At the beginning of each month Kath places £100 into a bank account to save for a family holiday. Each subsequent

month she increases her payments by 5%. Assuming the bank account does not pay interest, find (a) the amount of money in the account after 9 months.

(3 marks) Month n is the first month in which there is more than £6000 in the account.

(b) Show that log 4log1.05

n >

(4 marks) Maggie begins saving at the same time as Kath. She initially places £50 into the same account and plans to increase her payments by a constant amount each month. (c) Given that she would like to reach a total of £6000 in 29 months, by how much should Maggie

increase her payments each month? (2 marks)

___________________________________________________________________________________

10. Find 2cos 6 dx x∫ .

(5 marks) ___________________________________________________________________________________

11. (a) Prove that .

(3 marks)

(b) Hence solve, in the interval 0 2x π , the equation .

(3 marks) ___________________________________________________________________________________



12. A large arch is planned for a football stadium. The parametric equations of the arch are 8( 10)x t= + ,

2100y t= − ,–19 ≤ t ≤ 10 where x and y are distances in metres. Find (a) the cartesian equation of the arch,

(3 marks)

(b) the width of the arch, (2 marks)

(c) the greatest possible height of the arch.

(2 marks) ___________________________________________________________________________________

13.

Find the values of the constants A, B, C and D.

(5 marks) ___________________________________________________________________________________

14. The volume of a sphere V cm3 is related to its radius r cm by the formula 343

V r= π . The surface area of the

sphere is also related to the radius by the formula 24S r= π . Given that the rate of decrease in surface area,

in cm2 s–1, is d 12dSt= − ,

find the rate of decrease of volume ddVt

(4 marks) ___________________________________________________________________________________

15. Find

3sin dx x∫ .

(4 marks) ___________________________________________________________________________________

16. 21h( ) 40ln( 1) 40sin , 0 5 4tt t t t = + + −

.

The graph y = h(t) models the height of a rocket t seconds after launch. (a) Show that the rocket returns to the ground between 19.3 and 19.4 seconds after launch.

(2 marks) (b) Using t0 = 19.35 as a first approximation to α, apply the Newton–Raphson procedure once to h(t) to

find a second approximation to α, giving your answer to 3 decimal places. (5 marks)

(c) By considering the change of sign of h(t) over an appropriate interval, determine if your answer to part

(b) is correct to 3 decimal places. (3 marks)

___________________________________________________________________________________



17. (a) Show that in KLM∆ with 3 0 6KL = + −i j k

and 2 5 4LM = + +i j k

, 66.4KLM∠ = ° to one decimal place.

(7 marks) (b) Hence find LKM∠ and LMK∠ .

(3 marks)

___________________________________________________________________________________

TOTAL FOR PAPER IS 93 MARKS



SECTION B: MECHANICS

Answer ALL questions.

9. At time t seconds, a 2 kg particle experiences a force F N, where 28 64 12

t t= +−

F

(a) Find the acceleration of the particle at time t seconds.

(3) The particle is initially at rest at the origin. (b) Find the position of the particle at time t seconds.

(6) (c) Find the particle’s velocity when t = 1.

(3)

(Total 12 marks) ____________________________________________________________________________

10. An archer shoots an arrow at 10 m s−1 from the origin and hits a target at (10, −5) m. The initial velocity of the arrow is at an angle θ above the horizontal. The arrow is modelled as a particle moving freely under gravity.

(In this question, take g = 10 m s−2.)

(a) Show that (tan θ − 1)2 = 1.

(11) (b) Find the possible values of .

(3)

(Total 14 marks) _____________________________________________________________________________



11. Figure 3 shows a 5000 kg bus hanging 12 m over the edge of a cliff with 1000 kg of gold at the front. The gold sits on a wheeled cart. A group of n people, each weighing 70 kg, stands at the other end. The bus is 20 m long.

Figure 3 (a) Write down the total clockwise moment about the cliff edge in terms of n.

(7) (b) Find the smallest number of people needed to stop the bus falling over the cliff.

(2) (c) One person needs to walk to the end of the bus to retrieve the gold. Find the smallest number

of people needed to stop the bus falling over the cliff in this situation, including the one retrieving the gold.

(4)

(Total 13 marks) _____________________________________________________________________________

12. A car travels along a long, straight road for one hour, starting from rest. After t hours, its acceleration is a km h−2, where a = 180 − 360t. (a) Find the speed of the car, in km h−1 in terms of t.

(2) The speed limit is 40 km h−1. (b) Find the range of times during which the car is breaking the speed limit. Give your answer in

minutes. (4)

(c) Find the average speed of the car over the whole journey.

(5)

(Total 11 marks) _____________________________________________________________________________

TOTAL FOR PAPER IS 100





1 Scheme Marks AOs

Pearson Progression Step

and Progress descriptor

States that:

M1 2.2a 5th

Decompose algebraic

fractions into partial fractions − two linear factors.

Equates the various terms.

Equating the coefficients of x:

Equating constant terms:

M1* 2.2a

Multiplies both of the equations in an effort to equate one of the two variables.

M1* 1.1b

Finds A = 8 A1 1.1b

Find B = −2 A1 1.1b

(5 marks)

Notes

Alternative method

Uses the substitution method, having first obtained this equation:

Substitutes to obtain 272

− B = 27 (M1)

Substitutes to obtain 275

A = 43.2 (M1)



2 Scheme Marks AOs



Begins the proof by assuming the opposite is true.

‘Assumption: there do exist integers a and b such that ’

B1 3.1 7th

Complete proofs using proof by contradiction.

Understands that

‘As both 25 and 15 are multiples of 5, divide both sides by 5 to

leave ’

M1 2.2a

Understands that if a and b are integers, then 5a is an integer, 3b is an integer and 5a + 3b is also an integer.

M1 1.1b

Recognises that this contradicts the statement that ,

as 15

is not an integer. Therefore there do not exist integers a and

b such that ’

B1 2.4

(4 marks)

Notes



3 Scheme Marks AOs



(a) Finds d 2sin 2

dx tt= − and d cos

dy tt=

M1 1.1b 6th

Differentiate simple functions

defined parametrically

including application to tangents and

normals.

Writes −2sin 2t = − 4sin t cos t M1 2.2a

Calculates d cos 1 cosecd 4sin cos 4y t tx t t= = −−

A1 1.1b

(3)

(b) Evaluates d

dyx

at 56

t π= −

A1 ft 1.1b 6th

Differentiate simple functions

defined parametrically

including application to tangents and

normals. Understands that the gradient of the tangent is 12

, and then the

gradient of the normal is −2.

M1 ft 1.1b

Finds the values of x and y at 56

t π= −

5 1cos 26 2

x π = ×− =

and 5 1sin6 2

y π = − = −

M1 ft 1.1b

Attempts to substitute values into 1 1( )y y m x x− = −

For example, 1 122 2

y x + = − −

is seen.

M1 ft 2.2a

Shows logical progression to simplify algebra, arriving at: 122

y x= − + or 4 2 1 0x y+ − =

A1 2.4

(5)

(8 marks)

Notes

(b) Award ft marks for a correct answer using an incorrect answer from part a.



4 Scheme Marks AOs



States that

cos3cot 3sin3

xxx

= M1 2.2a 6th

Integrate using trigonometric

identities. Makes an attempt to find

cos3 dsin 3

x xx

∫

Writing ( )( ) ( )

f 'd ln f

fx

x xx

= ∫ or writing ln (sin x) constitutes an

attempt.

M1 2.2a

States a fully correct answer1 ln sin33

x C+ A1 1.1b

(3 marks)

Notes



5 Scheme Marks AOs



Demonstrates an attempt to find the vectors AB

, AC

and BC

M1 2.2a 5th

Find the magnitude of a

vector in 3 dimensions.

Finds ( )0,4, 2AB = −

, ( )5,4,8AC =

and ( )5,0,10BC =

A1 1.1b

Demonstrates an attempt to find | |AB

, | |AC

and| |BC

M1 2.2a

Finds ( ) ( ) ( )2 2 2| | 0 4 2 20AB = + + − =

Finds ( ) ( ) ( )2 2 2| | 5 4 8 105AC = + + =

Finds ( ) ( ) ( )2 2 2| | 5 0 10 125BC = + + =

A1 1.1b

States or implies in a right-angled triangle 2 2 2c a b= + M1 2.2a

States that 2 2 2| | | | | |AB AC BC+ =

B1 2.1

(6 marks)

Notes



6 Scheme Marks AOs



(a) States or implies that ( )2pq( ) 5 2x x= − M1 2.2a 5th

Find composite functions. States or implies that 2qp( ) 5 2x x= − M1 2.2a

Makes an attempt to solve ( )2 25 2 5 2x x− = − . For example, 2 225 20 4 5 2x x x− + = − or 26 20 20 0x x− + = is seen.

M1 1.1b

States that 23 10 10 0x x− + = . Must show all steps and a logical progression.

A1 1.1b

(4)

(b) ( )( )2 4 100 4 3 10 20 0b ac− = − = − < M1* 2.2a 5th

Find the domain and range of composite functions.

States that as 2 4 0b ac− < there are no real solutions to the equation.

B1* 3.2b

(2)

(6 marks)

Notes

(b) Alternative Method

M1: Uses the method of completing the square to show that 25 653 0

3 9x − + =

or

25 6533 9

x − = −

B1: Concludes that this equation will have no real solutions.



7 Scheme Marks AOs



Begins the proof by assuming the opposite is true.

‘Assumption: there is a finite amount of prime numbers.’

B1 3.1 7th

Complete proofs using proof by contradiction. Considers what having a finite amount of prime numbers means

by making an attempt to list them:

Let all the prime numbers exist be

M1 2.2a

Consider a new number that is one greater than the product of all the existing prime numbers:

Let

M1 1.1b

Understands the implication of this new number is that division by any of the existing prime numbers will leave a remainder of 1. So none of the existing prime numbers is a factor of N.

M1 1.1b

Concludes that either N is prime or N has a prime factor that is not currently listed.

B1 2.4

Recognises that either way this leads to a contradiction, and therefore there is an infinite number of prime numbers.

B1 2.4

(6 marks)

Notes

If N is prime, it is a new prime number separate to the finite list of prime numbers, .

If N is divisible by a previously unknown prime number, that prime number is also separate to the finite list of prime numbers.



8 Scheme Marks AOs



Attempts to write a differential equation.

For example, ddF Ft∝ or is seen.

M1 3.1a 7th

Construct simple differential equations.

States A1 3.1a

(2 marks)

Notes



9 Scheme Marks AOs



(a) Recognises that it is a geometric series with a first term 100=a and common ratio 1.05=r

M1 3.1a 6th

Use geometric sequences and

series in context. Attempts to use the sum of a geometric series. For example,

( )9

9

100 1 1.05

1 1.05S

−=

− or

( )9

9

100 1.05 1

1.05 1S

−=

− is seen.

M1* 2.2a

Finds 9 £1102.66S = A1 1.1b

(3)

(b) States

( )100 1.05 16000

1.05 1

n −>

− or

( )100 1 1.056000

1 1.05

n−>

−

M1 3.1a 5th

Use arithmetic sequences and

series in context. Begins to simplify. 1.05 4n > or 1.05 4n− < − M1 1.1b

Applies law of logarithms correctly log1.05 log 4n > or log1.05 log 4n− < −

M1 2.2a

States log 4log1.05

n > A1 1.1b

(4)

(c) Uses the sum of an arithmetic series to

state ( )29 100 28 60002

d + = M1 3.1a 5th

Use arithmetic sequences and

series in context. Solves for d. d = £11.21 A1 1.1b

(2)

(9 marks)

Notes M1 Award mark if attempt to calculate the amount of money after 1, 2, 3,….,8 and 9 months is seen.



10 Scheme Marks AOs



Selects 2cos 2 2cos 1x x≡ − as the appropriate trigonometric identity.

M1 2.2a 6th


identities. Manipulates the identity to the question: 2cos12 2cos 6 1x x≡ − M1 1.1b

States that ( ) ( ) 2 1cos 6 d 1 cos12 d

2x x x x= +∫ ∫

M1 1.1b

Makes an attempt to integrate the expression, x and sin x are seen.

M1 1.1b

Correctly states 1 1 sin122 12

x x C + +

A1 1.1b

(5 marks)

Notes

Student does not need to state ‘+C’ to be awarded the third method mark. Must be stated in the final answer.



11 Scheme Marks AOs



(a) Writes tanx and secx in terms of sinx and cosx. For example,

sin 1tan sec cos cos

1 sin1 sin1

xx x x x

xx

− − =−−

M1 2.1 5th

Understand the functions sec, cosec and cot.

Manipulates the expression to find sin 1 1cos 1 sin

xx x− × −

M1 1.1b

Simplifies to find A1 1.1b

(3)

(b) States that or B1 2.2a 6th

Use the functions sec, cosec and cot

to solve simple trigonometric

problems.

Writes that or M1 1.1b

Finds A1 1.1b

(3)

(6 marks)

Notes



12 Scheme Marks AOs



(a) Rearranges ( )8 10x t= + to obtain 80

8xt −

= M1 1.1b 8th

Use parametric equations in

modelling in a variety of contexts.

Substitutes 808

xt −= into 2100y t= −

For example,280100

8xy − = −

is seen.

M1 1.1b

Finds 21 564 2

y x x= − + A1 1.1b

(3)

(b) Deduces that the width of the arch can be found by substituting 10t = ± into ( )8 10x t= +

M1 3.4 8th



Finds x = 0 and x = 160 and deduces the width of the arch is 160 m.

A1 3.2a

(2)

(c) Deduces that the greatest height occurs when d 0 2 0 0dy t tt= ⇒ − = ⇒ =

M1 3.4 8th



Deduces that the height is 100 m. A1 3.2a

(2)

(7 marks)

Notes



13 Scheme Marks AOs



Makes an attempt to set up a long division.

For example: is seen.

M1 2.2a 5th

Divide polynomials by

linear expressions with a remainder.

Award 1 accuracy mark for each of the following: 2x seen, 2x seen, −21 seen.

For the final accuracy mark either D = 138 or 138

6x + or the remainder

is 138 must be seen. 2

3 2

3 2

2

2

2 216 8 9 12

6

2 9

2 12

21 12

21 126

138

x xx x x x

x x

x x

x x

x

x

+ −

+ + − +

+

−

+

− +

− −

A4 1.1b

(5 marks)

Notes

This question can be solved by first writing and then solving for A, B, C and D. Award 1 mark for the setting up the problem as described. Then award 1 mark for each correct coefficient found. For example:

Equating the coefficients of x3: A = 1

Equating the coefficients of x2: 6 + B = 8, so B = 2

Equating the coefficients of x: 12 + C = −9, so C = −21

Equating the constant terms: −126 + D = 12, so D = 138.



14 Scheme Marks AOs



Recognises the need to use the chain rule to find d

dVt

For example d d d dd d d dV V r St r S t= × × is seen.

M1 3.1a 8th

Construct differential

equations in a range of contexts.

Finds 2d 4dV rr= π and d 8

dS rr= π

M1 2.2a

Makes an attempt to substitute known values. For

example,2d 4 1 12

d 1 8 1V rt r

π −= × ×

π

M1 1.1b

Simplifies and states d 6dV rt= −

A1 1.1b

(4 marks)

Notes



Scheme Marks AOs



15 Recognises the need to write ( )3 2sin sin sinx x x≡ M1 2.2a 6th


identities. Selects the correct trigonometric identity to write

( ) ( )2 2sin sin sin 1 cosx x x x≡ − . Could also write 2sin sin cosx x x−

M1 2.2a

Makes an attempt to find ( ) 2sin sin cos dx x x x−∫ M1 1.1b

Correctly states answer 31cos cos3

x x C− + + A1 1.1b

(4 marks)

Notes



16 Scheme Marks AOs



(a) Finds and M1 3.1a 7th

Use numerical methods to solve

problems in context.

Change of sign and continuous function in the interval ⇒ root

A1 2.4

(2)

(b) Makes an attempt to differentiate h(t) M1 2.2a 7th



Correctly finds A1 1.1b

Finds and M1 1.1b

Attempts to find 1x

01 0 1

0

h( ) 0.2903...19.35h ( ) 13.6792...

xx x xx

= − ⇒ = −′ −

M1 1.1b

Finds A1 1.1b

(5)

(c) Demonstrates an understanding that x = 19.3705 and x = 19.3715 are the two values to be calculated.

M1 2.2a 7th



Finds and M1 1.1b

Change of sign and continuous function in the interval ⇒ root

A1 2.4

(3)

(10 marks)

Notes

(a) Minimum required is that answer states there is a sign change in the interval and that this implies a root in the given interval.



17 Scheme Marks AOs



(a) Demonstrates an attempt to find the vectors KL

, LM

and KM

M1 2.2a 6th

Solve geometric problems using

vectors in 3 dimensions.

Finds ( )3,0, 6KL = −

, ( )2,5,4LM =

and ( )5,5, 2KM = −

A1 1.1b

Demonstrates an attempt to find | |KL

, | |LM

and| |KM

M1 2.2a

Finds ( ) ( ) ( )2 2 2| | 3 0 6 45KL = + + − =

Finds ( ) ( ) ( )2 2 2| | 2 5 4 45LM = + + =

Finds ( ) ( ) ( )2 2 2| | 5 5 2 54KM = + + − =

A1 1.1b

Demonstrates an understanding of the need to use the Law of Cosines. Either 2 2 2 2 cosc a b ab C= + − × (or variation) is seen, or attempt to substitute into formula is made

( ) ( ) ( ) ( )( )2 2 254 45 45 2 45 45 cosθ= + −

M1 ft 2.2a

Makes an attempt to simplify the above equation. For example, 36 90cosθ− = − is seen.

M1 ft 1.1b

Shows a logical progression to state 66.4θ = ° B1 2.4

(7)

(b) States or implies that KLM∆ is isosceles. M1 2.2a 6th

Solve geometric problems using

vectors in 3 dimensions.

Makes an attempt to find the missing

angles 180 66.421...2

LKM LMK −∠ = ∠ =

M1 1.1b

States 56.789...LKM LMK∠ =∠ = ° . Accept awrt 56.8° A1 1.1b

(3)

(10 marks)

Notes

(b) Award ft marks for a correct answer to part a using their incorrect answer from earlier in part a.



H9 Scheme Marks AOs



a Use of Newton’s second law. M1 3.1b 8th

Understand general

kinematics problems with

vectors.

2=

Fa M1 1.1b

24 32 6

t t = + −

(m s−2) A1 1.1b

(3)

b Integrate a M1 1.1a 8th

Solve general kinematics

problems using calculus of

vectors.

2 32 11 2

t t = + + −

v c (m s−1) A1 1.1b

0=c because initially at rest. A1 2.4

Integrate v M1 1.1a

3 4

2 13 41 13 2

t t

= + + −

r c (m)

A1 1.1b

c = 0 because initially at origin. A1 2.4

(6)

c Subsititute t = 1 M1 1.1a 6th

Understand general

kinematics problems with

vectors.

2 11 2

= + − v

M1 1.1b

31

= −

(m s−1) A1 1.1b

(3)

(12 marks)

Notes



a Moment from bus = 5000 × 2 × g M1 3.1a 5th

Find resultant moments by considering direction.

= 10 000g (N m) A1 1.1b

Moment from gold = 1000 × 12 × g M1 3.1b

= 12 000g (N m) A1 1.1b

Moment from people = 70 × 8 × n × g M1 3.1a

= 560ng (N m) A1 1.1b

Total moment = (22 000 − 560n)g (N m) A1 1.1b

(7)

b Forming an equation or inequality for n and solving to find (n = 39.28…)

M1 1.1b 5th

Solve equilibrium problems involving

horizontal bars.

Need 40 people. A1 3.2a

(2)

c New moment from gold and extra person is 1070 × 12 × g (N) M1 3.1a 5th

Solve equilibrium problems involving

horizontal bars.

New total moment = (22840 − 560n)g (N m) M1 1.1b

n = 40.78… A1 3.2a

42 people (including the extra) A1 2.4

(4)

(13 marks)

H11 Scheme Marks AOs








a Use of suvat equations M1 1.1a 8th

Derive formulae for projectile

motion.

10 cosx t θ= A1 1.1b

2110 sin2

y t gtθ= − M1 1.1b

210 sin 5t tθ= − A1 1.1b

Substitute x = 10 and y = −5 M1 1.1a

Solve x equation for t M1 1.1b

1cos

tθ

= A1 1.1b

Substitute into y equation M1 1.1a

25 10 tan 5secθ θ− = − A1 2.1

Use of 2 2sec 1 tanθ θ= + M1 2.1

( )2tan 1 1θ − = legitimately obtained A1 2.1

(11)

b Solve for tan θ M1 1.1a 8th

Solve problems in unfamiliar

contexts using the concepts of friction and

motion.

tan θ = 0 or tan θ = 2 A1 1.1b

θ = 0 or 63.43…(°) (accept awrt 63) A1 1.1b

(3)

(14 marks)

Notes






a Integrate a w.r.t. t M1 1.1a 5th

Use integration to determine

functions for velocity and/or displacement.

2180 180a t t= − A1 1.1b

(2)

b 2180 180 40t t− > M1 3.1a 7th

Solve general kinematics

problems in less familiar contexts.

( )( )20 3 2 3 1 0t t− − < A1 1.1b

1 23 3

t< < A1 2.4

Breaking the speed limit between 20 and 40 minutes. A1 3.2a

(4)

c Integrate v w.r.t. t M1 1.1a 5th

Use integration to determine

functions for velocity and/or displacement.

( )2 390 60x t t C= − + A1 1.1b

When 1, 30t x= = A1 3.1b

Average speed =distance

time M1 1.1b

30 km h−1 A1 1.1b

(5)

(11 marks)

Notes





Documents

UVI FURTHER MATHS VACATION WORK...= 19.35 as a first approximation to α, apply the Newton–Raphson procedure once to h(t) to find a second approximation to α, giving your answer