maths.sogang.ac.krmaths.sogang.ac.kr/jlee/Syllabus/09-02/FM/Final.pdfIFYFMOOl Further Maths o NCUK THE NORTHERN CONSORTIUM THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY) Further Mathematics

IFYFMOOl Further Maths

o NCUK

THE NORTHERN CONSORTIUM

THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY)

Further Mathematics

Examination Session Time Allowed Summer 2009 3 hours 10 minutes

(Including 10 minutes reading time)

INSTRUCTIONS TO STUDENTS

SECTION A Answer ALL questions. This section carries 40% of the exam marks.

SECTION B Answer FOUR questions. This section carries 60% of the exam marks.

The marks for each part of the question are indicated in square brackets [ ].

• No answers must be written during the first 10 minutes. • Write your Candidate Number clearly on the Answer Book in the space

provided. • Write your answers in the Answer Book provided. Additional sheets will

be provided on request. • Clearly write the number and parts of questions attempted at the start

of each answer. • No written material is allowed in the examination room. • No mobile phones are allowed in the examination room. • An approved calculator may be used in the examination. • State the units where necessary. • Where appropriate, working should be carried out to 4 significant

figures and answers given to 3 significant figures. • Full marks will only be given for full and detailed answers. • Students will receive a formula book.

Page 1 of 7 V30809


Section A Answer ALL questions.

This section carries 40 marks.

Question A1

Let z = 5 - 3i and w =-2 + 4i .

(i) Find the value of 3z - w. [ 1 ]

w (ii) Express in the form a + bi, where z* is the complex conjugate [4 ]

w+z* of z.

Question A2

-12 -10 -10J

Let the matrix B = 5 3 5 . (

10 10 8

(i) Find B 2 • [ 2 ]

(ii) Hence show that B 2 - B is a multiple of the identity matrix, I. [ 1 ]

(iii) Hence find B-1 • [ 2 ]

Question A3

A uniform rod, AB, of length 70 cm and mass 5 kg, is freely hinged at a point A on a vertical wall. The rod is held in equilibrium horizontally by a light inextensible string attached to the wall at C and to a point P on the rod 45 cm from A as shown in Figure 1. The string makes an angle of 30° with the horizontal.

C

PA ~=---'----------,-'--~"------ B

5g

Figure 1

Page 2 of 7 V30809


(i) Find the tension in the string. [ 2 ]

(i i) Show that the magnitude of the reaction force, F, at the hinge A is [ 3 ]

5Ji5l --g.

9

Question A4

Solve the differential equation

d 2 y dy--9--lOy =60 dt 2 dt

.completely, given that y =2 and

dy-d1

= 14 when 1 =O.

[ S ]

Question AS

The lines I[

II : 12 :

and 12 have equations

r =(5,-4,6)+ s(2,- 5,4) r =(4,10, - 4)+ 1(3,4, - 2)

(i)

(i i)

Show that II and 12 intersect and find the coordinates of the point of

intersection.

Find the Cartesian equation of the plane in which II and 12 lie.

[ 3 ]

[ 2 ]

Question A6

A particle of mass 1.6 kg is attached to the end B of a light inelastic string AB of length 1.5 m, the other end of which is fixed at A. The particle moves in a horizontal circle whose centre is vertically below A with a constant angular speed. The particle takes 1.2 s to complete one revolution.

Let the acceleration due to gravity, g, be 9.8 ms -2, and take 1i as 3.14.

(i)

(i i)

Calculate the tension in the string.

Find the radius of the circular path of B.

[ 2 ]

[ 3 ]

Page 3 of 7 V30809


Question A7

(i)

(i i)

2 A B Express in the form - +- .

(r+1Xr+3) r+1 r+3

13 2 Hence find the exact value of I ( X ).

r=3 r + 1 r + 3

[ 2 ]

[ 3 ]

Question AS

The curve, C, has equation y = 13 cosh x + 12 sinh x .

(i)

(ii)

( iii)

Find the exact value of the x -coordinate of the turning point on terms of a natural logarithm.

Hence find the exact value of y.

Determine the nature of the turning point.

C in [ 3 ]

[ 1 ]

[ 1 ]

Page 4 of 7 V30809


Section B Answer 4 questions. This section carries 60 marks.

Question 81

B

Figure 2

Figure 2 shows two particles A and B, of mass 5 kg and 6 kg respectively. They are connected by a light inextensible string which passes over a light smooth pulley P. Particle A rests on a smooth plane inclined at an angle of 30° to the horizontal. Particle B rests on a rough horizontal plane. The string is parallel to the line of greatest slope of the inclined plane. Let the acceleration due to gravity, g be 9.8 ms-2

•

(i) Draw a diagram to show the forces acting on the bodies A and B. [ 2 ]

(ii) If the coefficient of friction, j.i, is 0.15, calculate the acceleration of [ 5 ]

body B.

(iii) Find in Newtons, the tension in the string. [ 1 ]

(iv) After travelling from rest for 1s, the string breaks. Calculate the time [ 4 ] taken for B to come to rest given that it does not reach the pulley.

(v) What is the total distance that B travels? [ 3 ]

Question 82

(i) Use integration to find the centre of mass of a uniform semicircular [ 6 ] lamina of radius 4 cm. You may use the formula for the area of a circle.

lOcm BA r-o::::::::----------.....-_ I I I I

I 4cm: ,

I I : c

---01--------- ,------------------~--------- _.I I ,,I ,I I II , ,I

II

E D

Figure 3

Page 5 of 7 V30809

IFYFMOOI Further Maths

(ii)

(iii)

(iv)

A semicircular section AFE is removed from a uniform rectangular lamina ABDE and placed at BCD to form the uniform lamina ABCDEF as shown in Figure 3. If the semicircles have radii 4 cm find the position of the centre of mass of the lamina.

the lamina ABCDEF is freely suspended from A. Find the angle AB makes with the vertical.

A particle of mass pm, where m is the mass of the lamina, is added at

E. Find the value of J1 so that AB makes an angle of 45° with the

vertical.

[ 3 ]

[ 2 ]

[ 4 ]

Question 83

(i)

(ii)

(iii)

(iv)

(v)

x2 y2 Show that the point P = (4cost, 7sint) lies on the ellipse -+ = 1.

16 49

Find the equation of the tangent at P.

Find the equation of the normal at P.

The normal meets the axes at the points Q and R. Lines are drawn

parallel to the axes through the points Q and R;. these lines meet at

the point V . Find the coordinates of V .

x2 y2 Find the equation of the locus of V as t varies in the form -2 +-2 = 1.

a b

[ 1 ]

[ 4 ]

[ 3 ]

[ 2 ]

[ 3 ]

(vi) Find the eccentricity and foci of the locus of V . [ 2 ]

Question 84

(i)

(ii)

(a)

(b)

In an Argand diagram, the point P represents the complex number z, where z =x + iy. Given that z 2 =Ai(z 8), where

A is a real parameter, show that as A varies the locus of P is a circle, and find its centre and radius.

If in (i) above z=J1(4-3i), where J1 is real, prove that there is

only one possible position for the point P and find its coordinates.

Use De Moivre's theorem to solve (z - 2Y=-8 for z .

[ 7 ]

[ 4 ]

[ 4 ]

Page 6 of 7 V30809


Question 85

(a) The roots of the cubic x 3 + bx2 + ex + d are a, f3 and r.

( i) Show that a 3 + /33 +r3 =_b 3 + 3bc -3d. [ 5 ] 3(ii) Given the cubic x - 4x2 + X + 5 evaluate [ 4 ]

a f3 ~ /3 r a.

r a

2x(b) Differentiate j(x) =e cosx a suitable number of times and hence find [ 6 ] the first four terms of its Maclaurin series.

Question 86

(a) ( i) Express 4x2 -12x + 25 in the form (px + qY+ r . [ 2 ]

( ii) Hence find 3f.J

dx in terms of a natural logarithm. [ 8 ]

3/2 4x 2 -12x + 25

(b) Find the surface area of a parabolic mirror obtained by rotating the [ 5 ] (2 (

parabola X=-, y=- from (0,0) to (4,2) about the x-axis, giving4 2

your answer correct to one decimal place.

Page 7 of 7 V30809

IFYFM001 Further Maths

NCUK THE NORTHERN CONSORTIUM

THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY)

Further Mathematics

Mark Scheme

V30809 Page 1 of 10


Level of accuracy: If a question specifies how many decimal places or significant figures are required, there is a mark for this. Otherwise accept any reasonable level of accuracy and alternative form.

Error carried forward: Where numerical errors have been made, students lose a mark/marks at that stage but may be awarded marks for using correct methods subsequently if the student demonstrates basic understanding.

Section A

A1(i) z = 5 - 3i, W = -2 + 4i, 3z - W = 17 -13i . I

W - 2 + 4i (ii) --=----- I

W + z* - 2 + 4i + 5 + 3i - 2 + 4i

= I 3 + 7i

_ (-2+4iX3-7i) MI

- (3+7iX3-7i)

- 6 + 14i + 12i + 28

9+49 22 + 26i 11 13.

= =-+-1 Al 58 29 29

-10 -10 -10

B2 [-12 -~or~2 -10J [-6 -~O~A2(i) = 5 3 3 5 = 5 9 2

10 10 8 10 10 8 10 10 14

B2 [-6 -10 -IOJ [-12 -10 -10j [6(ii) - B = 5 95 - 535 =0 6 I

0

~j =61 10 10 14 10 10 8 0 0

(iii) B(B - I) = 61 ,

B- 1 B-1 so =-- MI 6

13 5 5 - - -

6 3 3

= 5 -

1 -

5 - Al

6 3 6 5 5 7 - - -3 3 6

V30809 Page 2 of 10


1A3(i) Let a = - AB , and b = AP - a .

2 Taking moments about A, mga - T(a + b)sin e= 0

So T = mga = 5 x g x 35 = 70 g MIAI (a + b)sin e 45 x 1 9

2

(ii) Resolving forces vertically T sin e+ Fv = 5g so Fv = 5g - 70 g x .!- = -.!..Q g I 9 2 9

Resolving forces horizontally FH

= T cose = 70 g x .J3 = 35 .J3g I 929

.-------,------

1 J (10)2 (35 J2 5SO IFI=~F;-+FH- = gg + 9.J3g =9Ji51g I

A4 The auxiliary equation is n" - 9n -10 = 0 . This factorises as (n + 1Xn -10) = 0 with roots n = -1, 10.

+ Be lOIThe complementary function is Ae-I • I

For a particular integral try y = C . Then - 10C = 60 , so C = -6.

+ Be lOIThe general solution is y = Ae -I - 6. I

When 1 = 0, A + B - 6 = 2, A + B = 8 I

Differentiating: dy = -Ae-' + 10Be lol

dl Whenl=0,-A+IOB=14 I

Solving, A = 6, B = 2

So y = 6e-1 + 2e lOI - 6 I

A5(i) II: r=(5,-4,6)+s(2,-5,4)

12 : r=(4,10,-4)+1(3,4,-2) If the lines intersect, then at the point of intersection (5, - 4,6) + s(2, - 5,4) = (4,10, - 4) + 1(3,4, - 2) giving the equations:

5 + 2s = 4 + 31 2s - 31 = -I - 4 - 5s = 10+ 41 5s + 41 = -14 6 + 4s = -4 - 21 4s + 21 = -10 I Solving the first two equations, s = -2, I = -1, and these values satisfy the third equation also, so the lines intersect. I There r = (5,-4,6)-2(2,-5,4)= (1,6,-2) I

(ii) The vector equation of the plane is r = (1,6, - 2) + s(2, - 5,4) + 1(3, 4, - 2)

The normal is (2, - 5,4)x (3,4, - 2) = (- 6,16,23) I

so the equation of the plane is - 6x + 16y + 23z = 44 I

V30809 Page 3 of 10


2A6(i) Resolving horizontally TsinO = mrm = m(lsinO)m2 1

2)2 8soT=mlm2 =1.6x1.5x ~ =-Tr2=65.73~65.7N 1( 1.2 1.2 (ii) Resolving vertically T cos 0 = mg 1

cosO = mg = 1.6 x 9.8 = 0.2386 1 T 65.73

Then r = l sin 0 = 1.5 sin 0 = 1.5.Jl- cos 20

= 1.5.Jl- 0.2386 2 = 1.457 ~ 1.46 m 1

2 A BA7(i) ---;------,--,-----------,- = -- +-

(r+lXr+3) r+l r+3

2 = A(r + 3) + B(r + 1)

r = -1 gives 2 = 2A, A = 1 1 r=-3 gives 2=-2B B=-1 1

2 1-1So =--+-

(r+lXr+3) r+l r+3 13 2 13 1 1.1 1

(ii) I =I--I,=3 (r + 1Xr + 3) ,=3 r + 1 ,=3 r + 3 111 111 111

=-+-+-+ ... +------ ... ------ 1 4 5 6 14 6 7 14 15 16 1 1 1 1

=-+----- 1 4 5 15 16 60 + 48 - 16 - 15 77

= = 1 240 240

A8(i) y = 13coshx+ 12sinhx

dy = 13 sinh x + 12 cosh x = 0 at a turning point. 1 dx

1213 tanh x + 12 = 0, tanh x = -- 1

13

-I( 12) 1 (l- tt J 1 (13-12) 1 (1) 1x=tanh -13 =2"ln 1++t =2"ln 13+12 =2"ln 25 =-ln5

(ii) y = 13cosh(-ln5)+ 12sinh(-ln5) = .!.2(e-1n5 + e1n5 )+ g(e-1n5 _ e1n5

)

2 2

= ~ (.!. + 5) + 6(.!. _ 5) = 13 + 325 + 12 - 300 = 5 1 2 5 5 10

d2

(iii) --? = 13 cosh x + 12 sinh x = 5 > 0 dx So this is a local minimum. 1

V30809 Page 4 of 10


Section B

B1(i)

B F

6g

5g

Diagram 2

(ii) (Resolving forces on A perpendicular to the slope: RA = 5gcos30°.)

Resolving forces on A down the slope: Sal = 5g sin 30° - T , where a l is the initial acceleration 1 Resolving forces on B vertically: Rs = 6g 1

Resolving forces on B horizontally: 6a) = T - F = T - f.1R s = T - 0.15 x 6g 1

Adding the second and fourth equations: Sa) + 6a] = 5g sin 30° - 0.15 x 6g 1

So a) =5x9.8xsin300-0.15x6 x 9.8=1.425::::::1.43ms-2 1 5+6

(iii) T=6al +f.16g=6xl.425+0.15x6x9.8=17.37::::::17.4N 1

(iv) The velocity after l] s when the string breaks is

u2 = all] = 1.425 xl = 1.425 ms- 1 1

Now 6a2 = -f.16g, so a2 = -f.1g = -1.47ms-2 1

The time to rest is l2 = _!!.l- = 1.425 = 0.9697 :::::: 0.970 s 2 a2 1.47

(v) The distance travelled before the string breaks is Sf = ~a)lj2 = 0.7127m 1 2

The distance travelled after the string breaks is S2 = -~a2l/ = 0.6911 m 1

The total distance travelled is Sj + S2 = 1.4038 :::::: 1.40 m 1

V30809 Page 5 of 10

B2(i)

(ii)

(iii)

(iv)


Place the semicircular lamina to the right of the y -axis, with its centre at the

origin. By symmetry the centre of gravity will lie on the x -axis. I Divide the lamina up with vertical strips of width &. The mass of the strip at

distance x from the origin is 2p .J16 - x 2 & where p is the density. I

The mass of the lamina is -.!.. Jl"4 2 P = 8np I 2

Taking moments about the origin we see that, letting the centre of gravity be (x, 0)

8np X = 4J2 p x.J16 - x 2 dx o

=Hp(16-X')'" I 128

=-p MIAI 3

So x = 128p = 16 I 3 x 8np 3n

Once again, by symmetry, the centre of gravity lies on the x -axis. Call this point (x, 0). (not the same x) The mass of the lamina is 8 xl 0P =80 p. I Taking moments about the origin,

80pi = 80p x 5 - 8np x ~ + 8np x (10 + ~J = (400 + 80n)p3n 3n

So x = 5 + n. MIAI If the side of the lamina hangs at angle {} to the vertical, then

4tan{) =-- ~ 0.4913 I

5+n So {} = 26.17° ~ 26.2° I

After the particle is added let the new centre of gravity be (X, Y) Taking moments:

- ( ) - 5 + n(m + j.1Jn )X = m 5 + n so X =-- I 1+ ,u

(m+ j.1Jn)Y = j.1Jn(-4) so Y = -4,u I 1+ ,u

If the system now hangs at 45° to the vertical then X = 4 - Y. SO 5 + n =4 _ - 4,u

1+,u 1+,u

5 + n = 4(1 + ,u)+ 4,u = 4 + 8,u 1+ n

,u =-- = 0.5177 ~ 0.518 MIAI 8

V30809 Page 6 of 10

B3(i)

(ii)

(iii)

(iv)

(v)

(vi)


2 216cos t 49sin t • 2 17 + =cos-t+sm t= 1

16 49 Differentiating implicitly

2x + 2y dy = 0 16 49 dx

So dy = _ 49x = _ 49x 4cost = _ 7cost 1 dx 16y 16x7sint 4sint

... h . f' h . 7 cos tTherelore t e equatIOn 0 t e tangent IS y = - -,- x + c 1 4smt

where, since the tangent must pass through P

. 7 cos t 4 7 sin 2 t + 7 cos 2t 7 c = 7 sm t +-- cos t = = 1

4 sin t sin t sin t .. 7cost 7

So t he equatIOn IS Y = --,-x + -. 4smt smt

or 4(sint)y+7(cost)x=28 1

f hlP' 4 sin tTh d· t at - e gra lent 0 e norma IS 1 7 cost

. f h 1 . 4 sin tSo t he equatIOn 0 t e norma IS Y = --x + c 7cost

. 4sint 4 49-16. 33.h -- sm t = - sm t 1were c = 7 sm t - cos t =

7 cost 7 7 ... h .. 4sint 33.

Therelore t e equatIOn IS Y =-- x + - sm t 7 cost 7

or 7(cost)Y - 4(sint)x = 33 sin t cost 1

33 At Q, Y = 0, so x = --cost 1

4

At R, x = 0, so y = 33 sin t 7

So the coordinates of V are [ - 3: cos t, 3; sin t) 1

At V cost = __x_ sint = -y- 2 , 33/4' 33/7

x2 y2 so the equation of the locus of V is + = 1 1

(33/4)" (33/7)222 g2For the ellipse ':;-+~=1, b 2 =a 2 (1-e 2 ),so e= 1--

2 a b a

Here e= 1- (33/7)2 =Jl_16 = {33 = J33 1 (33/4)2 49 V49 7

The foci are at (±ae,O) which is [ ±3: x ~ ,oJ ~ [± 33: ,oJ (Accept (±6.77,O).) 1

V3 0809 Page 7 of 10


B4a(i) Equating real and imaginary parts of z - 2 = ,1i(z - 8),

x-2 x-2=-~ ,1=---- 1

y

y = ,1(x - 8) ,1=--L 1 x-8

I,· '~ x - 2 yS0, e lmmatmg /t" we get - --- =--- 1

y x-8

Then -(x-2Xx-8)=y2 1

x 2+ y2 -10x+16=0 1

32This can be written as (x - 5y + y2 = 9 =This is a circle centre (5,0) radius 3. 2

(ii) If z = ,u(4-3i),

x = 4,u

y = -3,u 1

So 16,u 2 + 9,u 2 - 40,u + 16 = 0

25,u2 -40,u + 16 = 0 1

(5,u-4)2=0

4 ,u=- 1

5

Then P = (~ -gJ 1 5' 5

b (z - 2Y= -8 = 8(cos Jr + i sin Jr) So, by de Moivre's theorem,

,r;:;( [Jr + 2nJr J .. (Jr + 2nJr Jj'z - 2 = ?v 8 cos 3 + Ism 3 for n = 0, 1, 2 1

Jr, . Jr J 2[ 3Jr .. 3Jr J 2[ 5Jr .. 5Jr J= 2 Cos3"+lsm3"' Cos 3 +1sm 3 , Cos3 +1sm 3 1[

= 1+ i.J3, - 2, 1- i.J3 1

So z=3+i.J3,0,3-i.J3 1

V30809 Page 8 of 10


B5a(i) For the cubic x 3 +bx 2+cx+d with roots a,fJ,r we have

a + fJ + r =-b afJ + ar + fJr = c

afJr = -d 1 3So _b =(a+fJ+rY =(a+fJ+rXa 2+fJ2 +r 2+2afJ+2ar+2fJr)

=a 3 +fJ3 +r 3 +3afJ2 +3a2fJ+3ar2 +3a2r+3fJr2 +3fJ2r+6afJr 1

3bc = -3(a + fJ + r XafJ + ar + fJr)

=- 3a 2 fJ - 3a 2r - 3afJ 2 - 3fJ 2r - 3ar 2 - 3fJr 2 - 9afJr 1

- 3d = 3afJr 1

Adding these - b3+ 3bc - 3d = a 3+ fJ3 + r 3 1

a fJ r r a fJ a fJ r (ii) fJ r a=a -fJ +r

a fJ r fJ r a r a fJ 3 fJ3 3=ar- a - -r 1

= -3d + b3 - 3bc + 3d =b3 - 3bc 1

=(-4y-3(-4)1=-64+12=-52 1

3 f3

b /(x) = e 2x cosx so /(0)= 1 1

j'(x) = 2e 2x cosx-e 2x sinx so 1'(0)= 2 1

/"(x) = 4e 2x cosx - 2e 2x sinx - 2e 2x sin x - e 2x cos x = 3e 2x cosx - 4e 2x sinx

so /"(0)=3 1

/'"(x) = 6e 2x cos x - 3e 2x sin x - 8e 2x sin x - 4e 2x cos x

=2e 2 'cosx-Ile2xsinx so /"'(0)=2 1 2 3x x

Then /(x) = /(0)+ xj'(O) + 2I .!"(O) + 3! /"'(0)+ ... 1

3x 2 x3

= 1+2x+-+-+,.. 1 2 3

V30809 Page 9 of 10

1


B6a(i) 4x 2-12x+ 25 = (2x - 3)2 + 16 MIAI

(ii) f d X = f d X

3/2~4x2 -12x+25 3/2J(2x-3Y +16

du Let u =2x - 3 . Then - = 2 . 2

dx When X :::: 3, U :::: 3

3 When x = - U = 0 I2'

1 3 d The integral = - f~ 0 U ) I

2 0 u"+4

1 • _I U 3

2smh I:::: [ 4"] a

1 . h- I 3 =-sm - I 2 4

I=~ln(l+ ~J2 4 ~16-r1

1 (3 5) 12ln 2 I:::: 2lnl4" + 4" =

b S~2+ (~r + (t)'dl 2~ 2ffH(~r +(H dl

4

:::: Jr f2t~dt 4 a

= Jr [~ (t 2+ 1t 2]4 2 4 3 a

:::: Jr (17m -1)= 36.18 6

= 36.2 to 1 decimal place. I

V30809 Page 10 of 10

maths.sogang.ac.krmaths.sogang.ac.kr/jlee/Syllabus/09-02/FM/Final.pdfIFYFMOOl Further Maths o NCUK THE NORTHERN CONSORTIUM THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY) Further Mathematics

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