2014 baulk ham hills trial

7/23/2019 2014 Baulk Ham Hills Trial

http://slidepdf.com/reader/full/2014-baulk-ham-hills-trial 1/21

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2

Section I

10 marks

Attempt questions 1-10

Allow about 15 minutes for this section

Use the multiple choice answer sheet for Questions 1-10.

1. z = a + ib , where a

0 and b

0 , which of the following statements is false

(A) z – z = 2bi

(B) | z|

2= | z|| z|

(C) | z| + | z| = | z + z|

(D)

arg( z) + arg( z) = 0

2. The number of points of intersection of the graphs y = | x| and y = x

2 – 4

is

(A) 0

(B) 1

(C) 2

(D) 4

3. The equation 48 x3 – 64 x

2+ 25 x – 3 = 0 has roots , , .

If = , a possible value of is

(A) – 12

(B) 14

(C) 12

(D) 38



3

4. The substitution t = tan 2

is used to find d

cos .

Which of the following gives the correct expression for the required integral?

(A)

dt

2 1 – t

2

(B) 2tdt

1 – t

2

(C) 2dt

1 – t

2

(D) 4tdt

1 + t

2 2

5. The cross section perpendicular to the x axis between two curves y = x and

y = 2 x is a circle. If the curves are drawn between x = 0 and x = 4, the volume of

the resulting horn is given by

(A)

0

4

x dx

(B) 0

4

x dx

(C)

0

4

x2

dx

(D)

0

4

x4

dx

6. An ellipse has the equation x2

100

+ y2

36

= 1 . The distance between the foci is:

(A) 8

(B) 16

(C) 20

(D) 25



4

7. The graph of y = f ( x) is shown below.

The graph of y = f (2 – x) is:

(A) (B)

(C) (D)

8. If a particle moves in a straight line so that its velocity at any particular time is given

by v = sin-1 x , then the acceleration is given by

(A) – cos-1 x

(B) cos-1 x

(C) – sin

-1 x

1 – x2

(D)

sin-1 x

1 – x2



5

9. Given that dydx

= y2

+ 1 and at x = 0, y =1, then

(A) y = y2 x + x + 1

(B) y = tan x +

4

(C)

y = tan x –

4

(D) y = loge y

2+ 12

10. From a set of n objects of which two are white and the rest are black, four objects are

to be chosen at random without replacement.

The probability that both white objects will be chosen is twice the probability that

neither white will be chosen.

The total number of objects is:

(A) 5

(B) 7

(C) 8

(D) 9

End of Section I



6

Section II

90 marks

Attempt questions 11-16

Allow about 2hours and 45 minutes for this section

Answer each question on the appropriate pages of your answer booklet. Extra pages are

available.

In questions 11-16, your responses should include relevant mathematical reasoning and/or

calculations.

Question 11 (15 marks) Answer on the appropriate page in your answer booklet

(a) Let z = 1 – 3i and w = 2 + i

(i)

Express zw in the form a + ib 1

(ii) Express zw in modulus- argument form 2

(iii) Hence find x if 2 (cos x – isin x)2 + i

= 1 – 3i5

2

(b) Evaluate

0

2

sin x

1 + cos2 x

dx 3

(c) Sketch the region on the Argand diagram defined by: 3

(d) (i) Find the constants A,B and C such that 2

x

2+ x + 1

( x + 2)

x

2

+ 2 x + 3

= A

x + 2 +

Bx + C

x

2

+ 2 x + 3

(ii) Hence find x

2+ x + 1

( x + 2) x

2+ 2 x + 3

2

End of Question 11



7


(a) Find

x( x + 1)10

dx 3

(b) Find

e x

+ e2 x

1 + e2 x

dx 3

(c) Find the equations of the tangents to the curve x2

+ y2

= xy + 3 when x=1. 4

(d) A stone monument has a rectangular base measuring 16 metres by 12 metres. The

cross section formed by any slice perpendicular to the base is a trapezium with top

edge 8 metres and bottom edge 16 metres. It is 5 metres high at the front and

10 metres high at the back.

(i)

Show that the area of a trapezium at a distance of x

metres from the front 3 of the stone monument is .

(ii) Find the volume of the stone monument. 2

End of Question 12



8


(a)

Sketch the following curves on separate diagrams. There is no need to use calculus.

(i) y = x2( x + 3) 2

(ii) y = x

2( x + 3)

2

(iii) y =1

x2( x + 3)

2

(iv) y = x2( x + 3) 2

(v) 2

(b) Consider the polynomial P( z) = z3

+ az2

+ bz + c where a,b and c are all real.

If P( i) = 0 where is real and non zero:

(i) Explain why P( – i) = 0 1

(ii) Show that P( z) has only one real zero. 1

(iii)

Hence show that c=ab where b>0 3

End of Question 13



9


(a) If is a complex cube root of unity, prove that: 2

(b) Factorise P( x) = x6 – 3 x

2+ 2 over the field of complex numbers given that it has 3

two roots of multiplicity of at least two.

(c) P, Q and R represent the complex numbers , . If 2

sketch a diagram and discuss the geometric properties of , giving reasons

for your answer.

(d) (i) Prove that x

2

x

2+ 1

n + 1 =1

x

2+ 1

n –

1

x

2+ 1

n + 1 1

(ii) Given I n =

0

1

1 x

2+ 1

n dx , prove that 2nI n + 1 = 2 – n + (2n – 1) I n 2

(iii) Hence evaluate

0

11

x

2+ 1

3 dx 2

(e) Prove by mathematical induction for 3

End of Question 14



10


(a) The locus of a point is defined by the equation | z – 2| = 2Re z – 1

2 .

(i)

If z = x + iy explain why x 12. 1

(ii) Show that the locus is a branch of the hyperbola 3 x2 – y

2= 3. 2

(iii) Sketch the locus showing its asymptotes and vertex. 2

(b) (i) Prove that the equation of the normal to the curve xy = c2 at the point 2

P cp, c

p is given by p

3 x – py = c

p

4 – 1

.

(ii)

The normal at P meets the x axis at M , and the tangent at P meets the y axis 3

at N. Prove that the locus of the midpoint of MN as P varies is given by

2c2 xy = c

4 – y

4.

(c) AB is a fixed chord of a circle and C is a variable point on the major arc AB. The angle

bisectors of

CAB and

ABC meet the circle again at P and Q respectively.

Let

CAB = 2 ,

ABC = 2 , and

BCA = 2 .

(i) Show that

PCQ = + + 2 1

(ii) Hence explain why the length of PQ is constant. 2

(iii) Use the sine rule to show that ABPQ

= 2sin 2

End of Question 15



11


(a) A food parcel is dropped vertically from a helicopter which is hovering 2000 metres

above a group of stranded bushwalkers. After 10 seconds a parachute opens

automatically. Air resistance is neglected for the first 10 seconds but then the effect of the

open parachute is to supply a resistance of 2 Mv newtons where M kg is the mass of the

parcel plus parachute and V ms-1 is the velocity after t seconds ( t 10 seconds).

Take the position of the helicopter to be the origin, the downwards direction as positive

and the value of g, the acceleration due to gravity, as 10 ms-2.

(i) Use calculus to find the equations of motion in terms of t for the parcel before 2

the parachute opens and prove that the velocity at the end of 10 seconds is

100 ms-1

and the distance fallen at the end of 10 seconds is 500 metres.

(ii) Show that the velocity of the parcel after the parachute opens is given by 3

v = 5 + 9 5e-2(t – 10)

for t

10

(iii) Find x, the distance fallen as a function of t and calculate the height of the 2

parcel above the bushwalkers 2 minutes after it leaves the helicopter.

(iv) Calculate the terminal velocity of the parcel. 1

(v)

Draw a neat sketch of velocity against time from the instant the parcel leaves 2

the helicopter.

(b) Find given that j = 1

10

means the product of terms from 2

j=1 to j =10.

(c) Form a polynomial of the least possible positive degree with integer coefficients 3

and one root of which is 3 12 + 3 18 .

END OF PAPER



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