NCUK - Sogangmaths.sogang.ac.kr/jlee/Syllabus-n/17-01/FM/14-15.v4.pdf · 2018. 11. 30. · NCUK. THE NCUK INTERNATIONAL FOUNDATION YEAR. IFYFMOOl Further Mathematics Examination

IFYFMOOl Further Mathematics

NCUK THE NCUK INTERNATIONAL FOUNDATION YEAR

IFYFMOOl Further Mathematics Examination

Examination Session Time Allowed Semester Two 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 4 questions ONLY. This section carries 60% of the exam marks.

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

• Answers must not be written during the first 10 minutes.

• A formula booklet and graph paper will be provided.

• An approved calculator may be used in the examination.

• Show ALL workings in your answer booklet.

• Examination materials must not be removed from the examination room.

DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED BY THE INVIGILATOR

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IFYFMOOl Further Mathematics

Section A Answer ALL questions. This section carries 40 marks.

Question A1

The complex number z = 3 - 4i.

a) Evaluate 2z - 3z·. [ 1 ]

b) Calculate the value of arg(z), giving your answer in radians. [ 1 ]

c) Find the value of 11 __1_1. [ 3 ] z-l

Question A2

Let A and B be matrices with A = (; ~) and B = (-;1 ~1)' [5 ]Solve for the matrix C, the equation A(C -I)(B - 21) = I.

Question A3

1m 1.5 m ) )< E <

A I I B c ! 0

30 N

The diagram above shows a horizontal non-uniform rod AB of mass 10 kg and length 6 m. The rod is in equilibrium and rests on two supports at C and D where AC =1 m and BD =1.5 m. The reaction force at the support C is 50 N and a weight of 30 N rests on the rod at E where AE =3m.

a) Find the magnitude of the reaction force at D. [ 2 ]

b) Find the distance of the centre of mass of the rod from A, giving your answer [ 3 ] to 3 significant figures.

In this question, 1 mark will be awarded for the correct use ofsignificant figures.


IFYFM001Further Mathematics

Question A4

The quadratic equation X Z + 2x + 6 = 0 has roots a and fJ.

a) Write down the values of a + fJ and afJ. [ 1 ]

b) Find a quadratic equation with integer coefficients that has roots [4]

3 - ~ and 3 - f!...P a

Question A5

a) Show that cosh-lx = In(x ± 'l!xz - 1 ), x ~ 1. [ 3]

b) Find all solutions of the equation [ 2 ]

3coshzx-Bcoshx = 3.

Give your answers in terms of natural logarithms.

Question A6

A particle of mass 0.6 kg moves along the positive x-axis and is acted upon by a

single force of magnitude -.!.. newtons, where x is the displacement of the particle Sx

from the origin O. The force is directed towards O. The particle passes through the point A where OA = 1 m, with velocity 1.5 ms'.

a) Find the velocity of the particle, v in terms of x. [4]

b) Find the velocity of the particle when it passes through the point B, where [ 1 ] OB = 4m.

Question A7

a) The plane nl passes through the points (1, 2, -2), (3, -2, 1) and (5, 1, -4).

Find the cartesian equation of nl . [3 ]

b) The plane nz has equation r. (-4i + j + 2k) = 1.

Show that nl and nz intersect at right angles. [ 2 ]

Section A continues on the following page


IFYFMOO1 Further Mathematics

Question A8

1 1a) Show that 2 = ----- [ 1 ] (2r-l)(2r+l) 2r-l 2r+l

b) Hence evaluate by the method of differences [ 3 ]

r=n

L (2r - 1)1(2r + 1)'r=l

You must show all of your working.

c) Hence find as an exact fraction, the value of [ 1 ]

r=10

L (2r - 1)1(2r + 1)'r=6


IFYFMOO1 Further Mathematics

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

Question B1

a)

20 N

~ -----~-==-

A box of mass 5 kg is moving on a rough horizontal plane. A constant force of 20 N acts on the box at an angle of 30° above the horizontal and the coefficient of friction between the mass and the plane is 0.25.

i. Find the magnitude of the reaction force on the box due to the plane. [ 2 ]

ii. Find the acceleration of the box. [ 3 ]

b) A particle P is projected with speed 12 rns' up a rough plane which is

inclined to the horizontal at an angle of sin-1 ~ degrees. The coefficient of 5

friction between the particle and the plane is 0.6.

i. Find the acceleration of P. [ 3 ]

ii. Find the distance travelled by P before it comes to instantaneous rest [2] on the plane.

c) A sphere A of mass 3 kg is travelling at speed 5 ms' when it collides with a stationary sphere B of mass 8 kg. After the collision the direction of motion of A is reversed and its speed is 2 ms'. All motion is in a straight line.

i. Find the speed of B after the collision. [ 2]

ii. Find the coefficient of restitution between the spheres. [ 1 ]

iii. After the collision B collides with a wall which is perpendicular to its path. The coefficient of restitution between B and the wall is e.

Find the inequality satisfied bye if A and B are to collide again. [ 2 ]

Section B continues on the following page


IFYFM001Further Mathematics

Question 82

a) A particle of weight 20 N is suspended in equilibrium by two strings in a

vertical plane. The strings make angles of 35° and 50° with the horizontal and have tensions T1 and Tz respectively.

[ 5]Calculate the values of T1 and Tz .

b) c B

o A

The diagram shows a uniform rectangular lamina OABC from which a square of side 2m has been removed. OA=CB=5m and AB=OC=8m. The centre of the square is 2m from OC and 3m from OA. The mass of the lamina with the square removed is M.

i. Show that the mass of the removed square is ~ M. [ 1 ] 9

ii. Find the distance of the centre of mass of the lamina from OC. [ 2 ]

iii. Find the distance of the centre of mass of the lamina from OA. [ 2 ]

iv. Masses of 5M, 4M and 2M are now placed at C, B and A respectively and the lamina is suspended freely under gravity from point C.

[5]Find the angle that CB makes with the vertical.


IFYFM001 Further Mathematics

Question 83

a) i. Find [(D), ['(D) and ["(D) where [3]

[(x) = In(1 + em).

ii. Using the first three terms of the Maclaurin series for [(x), find an [ 2]

approximate value for In(1 + :e). iii. Using the first three terms of the Maclaurin series for [(x), find an [3]

approximation for

LO.S

[(x)dx.

b) Evaluate the following, giving numerical answers:

i. (4 dx [3]

J2 ..Jx 2 - 4x + 8

ii. [4]

Question 84

a) The variables x and yare related by the differential equation

d2y dy dx 2 + 6 dx + 13y = 26x - 1.

i. Solve the auxiliary equation and hence write down the complementary [ 2 ] function.

ii. Find the particular integral. [ 3 ]

iii. Calculate the particular solution given that y = 2 and dy = 1 when x = o. [ 3 ] dx

b) The curve C has equation

y=

i. Write down the equation of the vertical asymptote. [ 1 ]

ii. Determine the equation of the other asymptote. [2]

iii. Find the points where the curve crosses the axes. [2]

iv. Sketch the curve in your answer booklet. [ 2 ]

Section 8 continues on the following page



Question 85

a) Let v be a complex number which satisfies the equation

Iv-Z+3il = Iv-4il·

[ 2 ] Describe the locus of v geometrically.

b) The complex number w satisfies the equation

[w - Z+ 3il = Zlw - 4il·

Show that the locus of w is a circle and find the centre and radius of the circle. [ 6 ]

c) Let z be a complex number such that z = cose + isint),

i. Given that [ 3 ]

z" + z?' = ZcosnO (n a positive integer), show that

1Z8cos70 = Zcos70 + 14cos50 + 4Zcos30 + 70cosO. ii. Hence find the exact value of [4]

Question 86

a) Show that the equation of the tangent to the ellipse x: + y: = 1 at the point [4]a b (h, k) can be written as x~ + y~ = 1.

a b

b) An ellipse E has its directrices at x = ± 16~ and its foci at (±4~, 0).

Find the eccentricity of E. [ 2]

c) Find the equation of E in the form x: + y: = 1, where the values of aand b [ 2]a b should be found.

d) A vertical line is drawn through the focus with positive x coordinate which intersects E at the point P, where P has positive y coordinate.

Find the y coordinate of P. [ 1 ]

e) Hence show that the equation of the tangent to E at P is x + Zy = 16~. [ 2]

f) The tangent to E at P intersects the y-axis at Q and the normal to E at P intersects the y-axis at R.

Find the area of triangle PQR. [4]

THIS IS THE END OF THE EXAMINATION

V4 1415 © 2015 Northern Consortium UK Ltd Page B of B



THE NCUK INTERNATIONAL FOUNDATION YEAR

IFYFM001 Further Maths Examination 2014-15

Mark Scheme



Notice to Markers This mark scheme should be used in conjunction with the NCUK Centre Marking and Recording results policy, available from the secure area of the NCUK website (http://www.ncuk.ac.uk). Contact your Principal/ Academic Manager if you do not have login details. Significant Figures: All correct answers should be rewarded regardless of the number of significant figures used, with the exception of question A3 b). For this question, 1 discretionary mark is available which will only be awarded to students who correctly give their answer to the number of significant figures explicitly requested. Error Carried Forward: Method marks should be awarded for correct application of mathematical techniques,

procedures and methods. The result does not have to be correct (i.e. to match that given in the marking scheme) for the mark to be awarded. An incorrect result may be due to using results from an earlier part of the question that were incorrect (the error is carried forward) and transcription or minor arithmetic errors.

Accuracy marks can only be awarded when the answer matches that given in the marking scheme. This applies even where the method applied is correct and the error is due to the student using an earlier, incorrectly derived result in his/her workings. Exceptionally, the marking scheme may state that ‘follow-through’ is allowed. In such cases accuracy marks may be awarded provided that the erroneous answer is solely due to the use of a result of an earlier part of the question being in error – i.e. the error has been carried forward.

When this happens, write ECF next to the ticks or FT for follow-through as stated in the mark scheme.

M=Method A=Answer B=Correct answer independent of method



Section A Answer ALL questions. This section carries 40 marks.

Question A1 [ A ] a) 6 – 8i -3(3 + 4i) = -3 - 20i [ A1 ]

b) - tan-1 (4/3) = -0.927 [ A1 ]

c) 1 – 1/(z-1) = 1 – 1/(2-4i) = 1 – (2+4i)/20 = (18 - 4i)/20 = 9/10 – i/5

|9/10 – i/5| = √85/10

[ M1 ]

[ A1 ]

[ A1 ]

Question A2 [ B ] Pre-multiplying by A-1 (C – I)(B-2I) = A-1 Post-multiplying by (B-2I) and adding I C = A-1(B-2I)-1 + I A-1 = -1/2 5 43 2

(B-2I)-1 = 3 42 3 = 3 42 3

So C = -1/2 5 43 2

3 42 3

1 00 1

= -1/2 7 85 6 +

1 00 1

= 9/2 45/2 2

[ M1 ]

[ M1 ]

[ B1 ]

[ A1 ]

[ A1 ]



Question A3 [ F ] a) Resolving vertically 50 + RD = 30 + 10g

So RD = 78 N.

[ M1 ]

[ A1 ]

b) Taking moments about A, if x is the distance of the CoM from A 1(50) + 4.5(78) = x(10g) + 3(30) So x = 3.173 x = 3.17 to 3sf* *This mark can be given as a FT mark if the candidate obtains the wrong answer but rounds to 3sf correctly.

[ M1 ]

[ A1 ]

[ A1 ]

Question A4 [ E ] a) = -2, = 6

[ A1 ]

b) Sum of roots of new equation is

6 – ( / = 6 +4/3 = 22/3 Product of roots is 10 - 3( / = 14 Therefore the new equation is x2 -22x/3 + 14 = 0* *Allow FT mark for incorrect values in part a) With integer coefficients 3x2 - 22x + 42 = 0 (or equivalent) Correct answer only.

[ M1 ]

[ M1 ]

[ A1 ]

[ A1 ]



Question A5 [ G ] a) Let x = coshy

So x =( ey + e-y) / 2 Rearranging and multiplying by ey gives e2y -2xey + 1 = 0 Treating as a quadratic in ey gives ey =( 2x √4 4)/2 All above working (or equivalent) must be seen for 3 marks So y = cosh-1 = ln( √ 1)

[ M1 ]

[ M1 ]

[ M1 ]

b) (3coshx +1)(coshx – 3) = 0 So x = cosh-1 (3) = ln (3 + √8 and ln(3 - √8

[ A1 ]

[ A1 ]

Question A6 [ S ] a) Using F = ma

-1/5x = mv dv/dx so -dx/5x = 0.6 v dv integrating both sides -ln(x)/5 + C = 0.3v2 x = 1 when v = 1.5 So C = 27/40 so v2 = (27 - 8lnx)/12

v=

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

b) When x = 4 v = 27 8 4 /12 = 1.151 ms-1

[ A1 ]



Question A7 [ Q ] a) Using the first two points a direction in the plane is 2i + 4j - 3k

Using the first and third points gives 4i + j + 2k (any two directions) Hence a normal vector is (-2i+4j-3k)x(-4i+j+2k) = 11i+16j +14k So equation is (using any point) 11x + 16y + 14z = 15

[ M1 ]

[ A1 ]

[ M1 ]

b) Planes are perpendicular if the normals are perpendicular Using (-4i +j+2k).(11i+16j+14k) (use of dot product) = -44 + 16 +28 =0 So they are perpendicular

[ M1 ]

[ A1 ]

Question A8 [ D ] a) Must be seen

[ B1 ]

b) Sum now is ½ { 1/1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + …. + 1/(2n-3) – 1/(2n-1) + 1/(2n-1) – 1/(2n+1) } At least first 4 terms M1 and last 4 terms for other M1 = ½ { 1 – } or

[ M2 ]

[ A1 ]

c) = 10/21 – 5/11 = 5/231

[ A1 ]



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

Question B1 [ F ] a) i. Resolving vertically, if R is the reaction force

R + 20sin30 = 5g So R = 39 N

[ M1 ]

[ M1 ]

ii. Applying F = ma in the direction of motion, Fr is the frictional force. 20cos30 - Fr = 5a But Fr = μR So a = [20cos30 –(0.25)(39)]/5 = 1.51ms-2

[ M1 ]

[ M1 ]

[ A1 ]

b) i. Resolving perpendicular to the plane,R is the reaction R = mg(4/5) Applying F = ma up the plane -mg(3/5) – μmg(4/5) = ma a = -3g/5 – 12g/25 = -27g/25 = -10.584 ms-2

[ M1 ]

[ M1 ]

[ A1 ]

ii. Using v2 = u2 + 2as s = 122/(2x10.584) = 6.80 m

[ M1 ]

[ A1 ]

c) i. Using conservation of linear momentum, vB is the speed of B afterwards 3(5) + 0 = 3(-2) + So vB = 21/8 = 2.625 ms-1

[ M1 ]

[ A1 ]

ii. e = (2+ 2.625)/5 = 0.925* *Allow FT mark on the value of e if VB is incorrect.

[ A1 ]

iii. There will be another collision if vB > 2

So > 2/2.625 or > 0.762 Correct answer only.

[ M1 ]

[ A1 ]



Question B2 [ F/L ] a) i. Resolving horizontally T1cos35 = T2cos50

Resolving vertically T1sin35 + T2sin50 = 20 Substituting for T2 from the first equation in to the second T1sin35 + T1cos35sin50/cos50 = 20 So T1 = 12.9 N And T2 = 16.4 N

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

[ A1 ]

b) i. (8x5 – 2x2)ρ = M, so ρ = M/36

Hence mass of square is M/9. Method must be seen.

[ M1 ]

ii. Taking moments about OC M ̅ = (M + M/9)(2.5) – (M/9)(2) So ̅ = 23/9

[ M1 ]

[ A1 ]

iii. Taking moments about OA M = (M+ M/9)(4) – (M/9)(3) So = 37/9

[ M1 ]

[ A1 ]

iv. Total mass is now 12M Taking moments about OC 12M ̅ = 23M/9 +5M(0) + 4M(5) + 2M(5) So ̅ = 2.71 Taking moments about OA 12M = 37M/9 + 5M(8) + 4M(8) + 2M(0) So = 6.34 If the angle is θ tanθ = (10 -6.34)/2.71 θ = 53.5°

[ M1 ]

[ A1 ]

[ M1 ]

[ A1 ]

[ A1 ]



Question B3 [ J/N ] a) i. f(0) = ln(1+1) = ln2

f’(x) =

so f’(0) =

f’’(x) =

so f’’(0) = ( 2/9 – 1/9)/4 =

[ A1 ]

[ A1 ]

[ A1 ]

ii. Series is f(x) = ln2 +x/6 + x2/72 Putting x = -3/2 gives ln(1+1/√ ≅ 0.474

[ M1 ]

[ A1 ]

iii. ≅. 2.

2 . = 0.368

[ M1 ]

[ M1 ]

[ A1 ]

b) i. = = = 0.881

[ M1 ]

[ M1 ]

[ A1 ]

ii. = 3 - √ dx

= 3 √9 1 = 3.289

[ M2 ]

[ M1 ]

[ A1 ]



Question B4 [ C/P ] a) i. Auxiliary equation is m2+6m+13 =0

This has roots -3 2i Hence the CF is 2 2

[ A1 ]

[ A1 ]

ii. Using the trial fuction y = px + q and substitutiing 0 + 6p + 13(px + q) = 26x – 1 13p = 26 so p=2 6p + 13q = -1 so q = -1 So PI is y = 2x-1

[ M1 ]

[ A1 ]

[ A1 ]

iii. General solution is 2 2 2 1 Using y=2 when x = 0 2 = A -1 so A = 3 dy/dx = 2 2 2 2 3 2 2 2 So 1 = 2B – 3A +2 so B = 4 Hence PS is 3 2 4 2 2 1

[ M1 ]

[ M1 ]

[ A1 ]



b) i. x = 3/2 [ B1 ]

ii. By long division or other method

(remainder not required) So the equation is y = x/2 + 3/4

[ M1 ]

[ A1 ]

iii. When x = 0 y = 2/3 so (0, 2/3) When y = 0 x = √2. So ( √2, 0

[ B1 ]

[ B1 ]

iv. Shape and position x > 3/2 Shape and position x < 3/2

[ B1 ]

[ B1 ]



Question B5 [ A/M ] a) The perpendicular bisector of

The line segment joining (2, -3) to (0, 4) (both coordinates)

[ B1 ]

[ B1 ]

b) Putting = x + iy gives |(x-2) + i(y+3)| = 2|x + i(y-4)| Squaring gives (x-2)2 + (y+3)2 = 4{ x2 + (y-4)2 } 3x2 + 3y2 - 38y + 4x + 51 = 0 x2 + y2 - (38/3)y + 4x/3 + 17 = 0. (x + 2/3)2 – 4/9 + (y - 19/3)2 – 361/9 + 17 = 0 (x+ 2/3)2 +(y+ 19/3)2 = 212/9 Which is the equation of a circle Centre (-2/3, 19/3) Radius √

[ M1 ]

[ M1 ]

[ M1 ]

[ M1 ]

[ A1 ]

[ A1 ]

c i. We have, using n = 1, z + 1/z = 2cos So 128cos7 = (z + 1/z)7 = z7 + 7z5 + 21z3 + 35z + 35/z + 21/z3 + 7/z5 +1/z7 = (z7+1/z7) + 7(z5+ 1/z5) + 21(z3 + 1/z3) + 35(z+1/z) = 2cos7 + 14cos5 + 42cos3 + 70cosθ

[ M1 ]

[ M1 ]

[ M1 ]

ii. Integral is equal to

2 7 14 5 42 3 70 * = [ sin7θ sin5θ 14sin3θ 70sinθ = { ( sin + sin + 14sin + 70sin ) – (0) } = *If the factor is missing throughout award M0 A1 M1 A1 if all other working correct.

[ M1 ]

[ A1 ]

[ M1 ]

[ A1 ]



Question B6 [ I ] a) Differentiating gives 2x/a2 + 2y/b2 (dy/dx) = 0. (differentiation)

So dy/dx = -xb2/(ya2) At (h, k) dy/dx = -hb2/(ka2) (derivative at (h. k) ) Therefore equation of the tangent is (y-k) = - (x-h) (tangent at (h, k) ) i.e. ka2y + xhb2 = h2b2 + a2k2 OR yk/b2 + xh/a2 = h2/a2 + k2/b2 = 1 since (h, k) is on the ellipse. (or by parametric method)

[ M1 ]

[ A1 ]

[ M1 ]

[ M1 ]

b) a/e = 16√3 and ae =4 Dividing gives e2 = ¼ so e = ½

[ M1 ]

[ A1 ]

c) From b) above a = 8√3 Using b2 = a2(1 – e2) , b = 12. So we have √ 1

[ A1 ]

[ A1 ]

d) If x = 4√3 , y2 = 144(1 – ¼ ) so y = 6√3

[ A1 ]

e) From part a) tangent is √ √ 1 (use of part a) ) Which gives x/16 + y/8 = √3 or x+2y = 16√3 (multiplication by constant)

[ M1 ]

[ A1 ]

f) Tangent intersects y axis at Q (0, 8√3 Equation of normal is (y- 6√3) = 2(x - 4√3 Therefore R is the point (0, -2√3) Using QR as the base of the triangle (or any other valid method) Area = ½ (10√3)(4√3) = 60 square units

[ A1 ]

[ M1 ]

[ A1 ]

[ A1 ]

14-15.V4FINAL_Further_Maths_V4_MS_141514-15.V4

Documents

NCUK - Sogangmaths.sogang.ac.kr/jlee/Syllabus-n/17-01/FM/14-15.v4.pdf · 2018. 11. 30. · NCUK. THE NCUK INTERNATIONAL FOUNDATION YEAR. IFYFMOOl Further Mathematics Examination