IFYFM00 Further Maths THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY) Further Mathematics Examination Session Summer 009 Time Allowed hours 0 minutes (Including 0 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 0 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 significant figures. Full marks will only be given for full and detailed answers. Students will receive a formula book. V 0809 Page of 7
IFYFM00 Further Maths Section A Answer ALL questions. This section carries 40 marks. Question A Let z 4 i and w i. (i) Find the value of z w. [ ] Express z. w w + z* in the form a + bi, where z * is the complex conjugate of [ 4 ] Question A Let the matrix (i) Find B. 6 6 B 4 6. 6 8 Hence show that B B (iii) Hence find B. [ ] is a multiple of the identity matrix, I. [ ] [ ] Question A A uniform rod, AB, of length 80 cm and mass 6 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 0 cm from A as shown in Figure. The string makes an angle of 4 with the horizontal. C T A F 4 P B 6 g Figure V 0809 Page of 7
IFYFM00 Further Maths (i) Find the tension in the string. [ ] Show that the magnitude of the reaction force, F, at the hinge A is 6 7 g. [ ] Question A4 Solve the differential equation d y dy + 7 + 0y 0 dt dt d y completely, given that y 9 and dt when t 0. [ ] Question A The lines l and l have equations l : r, 4,6 + s,, l : r 4,,0 + t,, ( ) ( ) ( ) ( 4) (i) Show that l and l intersect and find the coordinates of the point of intersection. [ ] Find the Cartesian equation of the plane in which l and l lie. [ ] Question A6 A particle of mass. of length. 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. s to complete one revolution. Let the acceleration due to gravity, g, be 9. 8 ms, and take π as. 4. kg is attached to the end B of a light inelastic string AB (i) Calculate the tension in the string. [ ] Find the radius of the circular path of B. [ ] V 0809 Page of 7
IFYFM00 Further Maths Question A7 (i) Express ( r + )( r + ) in the form Hence find the exact value of A B + r + r + ( r + )( r ) r +. [ ]. [ ] Question A8 The curve, C, has equation y 7 cosh x + sinh x. (i) Find the exact value of the x -coordinate of the turning point on C in terms of a natural logarithm. [ ] Hence find the exact value of y. [ ] (iii) Determine the nature of the turning point. [ ] V 0809 Page 4 of 7
IFYFM00 Further Maths Section B Answer 4 questions. This section carries 60 marks. Question B B A 4 Figure Figure shows two particles A and B, of mass kg and 4 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 4 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 -. (i) Draw a diagram to show the forces acting on the bodies A and B. [ ] If the coefficient of friction, μ, is 0. 4, calculate the acceleration of body B. [ ] (iii) Find in Newtons, the tension in the string. [ ] (iv) After travelling from rest for. s, the string breaks. Calculate the time taken for B to come to rest given that it does not reach the pulley. [ 4 ] (v) What is the total distance that B travels? [ ] Question B (i) Use integration to find the centre of mass of a uniform semicircular lamina of radius cm. You may use the formula for the area of a circle. [ 6 ] A cm B cm F C E D V 0809 Figure Page of 7
IFYFM00 Further Maths (iii) 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. If the semicircles have radii 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. μ (iv) A particle of mass m, where m is the mass of the lamina, is added at E. Find the value of μ so that AB makes an angle of 4 with the vertical. [ ] [ ] [ 4 ] Question B (i) Show that the point P ( 4cost,sint) lies on the ellipse x + y 6. [ ] Find the equation of the tangent at P. [ 4 ] (iii) Find the equation of the normal at P. [ ] (iv) 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. [ ] (v) x y [ ] Find the equation of the locus of V as t varies in the form +. b a (vi) Find the eccentricity and foci of the locus of V. [ ] Question B4 (a) (i) In an Argand diagram, the point P represents the complex number z, where z 8 λ i z, where z x + iy. Given that ( ) λ is a real parameter, show that as λ varies the locus of P is a circle, and find its centre and radius. If in (i) above z μ( 4 + i), where μ is real, prove that there is only one possible position for the point P and find its coordinates. (b) Use De Moivre s theorem to solve ( z ) 8 for z. [ 7 ] [ 4 ] [ 4 ] V 0809 Page 6 of 7
IFYFM00 Further Maths Question B (a) The roots of the cubic x + bx + cx + d are α, β and γ. (i) Show that α + β + γ b + bc d. [ ] Given the cubic x + x 4x + 8 evaluate α β β γ γ α γ α β (b) Differentiate f ( x) e cos x. a suitable number of times and hence find the first three non-zero terms of its Maclaurin series. [ 4 ] [ 6 ] Question B6 (a) (i) Express 4 + x + (b) Hence find x in the form ( px + q) + r / 8 d x / 4x + x +. [ ] in terms of a natural logarithm. Find the surface area of a parabolic mirror obtained by rotating the parabola x 4t, y 8t from ( 0,0) to ( 4,8) about the x -axis, giving your answer correct to one decimal place. [ 8 ] [ ] V 0809 Page 7 of 7
IFYFM00 Further Maths THE NCUK INTERNATIONAL FOUNDATION YEAR (IFY) Further Mathematics Mark Scheme V 0809 Page of 0
IFYFM00 Further Maths 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 A(i) z 4 i, w i, z w i. w i w + z* i + 4 + i i 7 + 4i ( i)( 7 4i) M ( 7 + 4i)( 7 4i) i 7i 4 49 + 6 7 9i 7 9 i A 6 6 6 A(i) 6 6 6 6 6 6 B 4 6 4 6 6 6 8 6 8 6 0 6 6 6 6 0 0 B B 6 4 6 0 0 I 6 0 6 8 0 0 B B I, B I so B M / /. / 7 / A (iii) ( ) I V 0809 Page of 0
IFYFM00 Further Maths A(i) Let a AB, and b AP a. mga T mga 6 g 40 4 g So T ( a + b) sinθ 0 Resolving forces vertically T sin θ + FV 6g so Taking moments about A, ( a + b) sin θ 0 MA 4 g 6 F V 6 g g Resolving forces horizontally So 4 g 4 F H T cos θ g 6 4 6 F FV + FH g + g 7g A4 The auxiliary equation is n + 7n + 0 0. This factorises as ( n + )( n + ) 0 with roots n,. t t The complementary function is Ae + Be. For a particular integral try y C. Then 0 C 0, so C. t t The general solution is y Ae + Be +. When t 0, A + B + 9, A + B 4 d t t Differentiating: y Ae Be dt When t 0, A B Solving, A 7, B t t So y 7e e + A(i) l : r (, 4,6) + s(,, ) l : r ( 4,,0 ) + t(,, 4) If the lines intersect, then at the point of intersection (, 4,6) + s (,,) ( 4,,0) + t(,, 4) giving the equations: s 4 + t s + t 4 + s t s + t 6 + s 4t s 4t 6 Solving the first two equations, s, t, and these values satisfy the third equation also, so the lines intersect. There r (, 4,6) (,,) (,, 4) r,, 4 + s,, + t,, 4 The normal is (,,) (,, 4) ( 9,,) so the equation of the plane is 9 x + y + z 6 The vector equation of the plane is ( ) ( ) ( ) V 0809 Page of 0
IFYFM00 Further Maths A6(i) Resolving horizontally T sinθ mrω m( l sinθ ) ω π 0 so T mlω.. π 6.7 6. 7 N.. Resolving vertically T cos θ mg mg. 9.8 cos θ 0.6 T 6.7 Then r l sinθ.sinθ. cos. 0.6.47.44 m θ A7(i) A + ( r + )( r + ) r + r + A ( r + ) + B( r ) + B r gives A, A r gives B B So + ( r + )( r + ) r + r + ( )( ) r r + r + r r + r r + + + + K + K 6 7 8 7 8 4 6 4 7 4 4 4 A8(i) y 7 cosh x + sinh x dy 7 sinh x + cosh x 0 at a turning point. dx 7 tanh x + 0, tanh x 7 7 7 x tanh ln ln ln ln 4 7 + 7 7 + 6 7 ln 4 ln 4 ln 4 ln 4 y 7cosh( ln 4) + sinh( - ln4) ( e + e ) + ( e e ) 7 7 + 7 + 40 + 4 + 4 8 4 4 8 d y (iii) 7 cosh x + sinh x 8 > 0 dx So this is a local minimum. V 0809 Page 4 of 0
IFYFM00 Further Maths Section B B(i) R B T B F R A A T 4 g 4 g Diagram (Resolving forces on A perpendicular to the slope: R A g cos 4.) Resolving forces on A down the slope: a g sin 4 T, where a is the initial acceleration Resolving forces on B vertically: R B 4g Resolving forces on B horizontally: 4a T F T μ RB T 0.4 4g Adding the second and fourth equations: a + 4a g sin 4 0.4 4g 9.8 sin 4 0.4 4 9.8 So a.07. ms + 4 (iii) T 4a + μ4g 4.07 + 0.4 4 9.8 4. 4. N (iv) The velocity after t s when the string breaks is u at.07..6ms Now 4a μ4g, so a μg. 9 ms u.6 The time to rest is t 0.80647 0. 806 s a.9 (v) The distance travelled before the string breaks is s at. 7m The distance travelled after the string breaks is s at. 7 m The total distance travelled is s + s.646. 6 m V 0809 Page of 0
IFYFM00 Further Maths B(i) 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. Divide the lamina up with vertical strips of width δ x. The mass of the strip at distance x from the origin is ρ δx where ρ is the density. The mass of the lamina is π ρ πρ Taking moments about the origin we see that, x,0 (iii) letting the centre of gravity be ( ) πρ x ρ x 0 ( ) d x / ρ x 0 0 ρ MA 0ρ 0 So x πρ π 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 0 ρ 0ρ. Taking moments about the origin, 0 0 0 ρx 0ρ 6 πρ + πρ + ( 70 + 0π )ρ π π So x 6 + π. MA 4 If the side of the lamina hangs at angle θ to the vertical, then 0 tan θ 0.07 4 + π 6 + π 4 So θ 6.7 6. 7 X, Y Taking moments: 4 + π ( m + μm) X m 6 + π so X 4 4 ( + μ) μ ( m + μ m) Y μm( ) so Y + μ (iv) After the particle is added let the new centre of gravity be ( ) If the system now hangs at 4 to the vertical then X Y. 4 + π μ So 4( + μ) + μ 4 + π 0( + μ) + 0μ 0 + 40μ 4 + π μ 0.497 0.49 MA 40 V 0809 Page 6 of 0
IFYFM00 Further Maths B(i) 6cos t sin t + cos t + sin t 6 Differentiating implicitly x y dy + 0 6 dx dy x 4cost cost So dx 6y 6 sin t 4sin t cost Therefore the equation of the tangent is y + c 4sin t where, since the tangent must pass through P cost sin t + cos t c sin t + 4cost 4sin t sin t sin t cost So the equation is y + 4sin t sin t or 4 ( sin t ) y + ( cost) x 0 (iii) 4sin t The gradient of the normal at P is cost 4sin t So the equation of the normal is y x + c cos t 4sin t 6 9 where c sin t 4cost sin t sin t cost 4sin t 9 Therefore the equation is y x + sin t cost or ( cost) y 4( sin t) x 9sin t cost (iv) 9 At Q, y 0, so x cost 4 9 At R, x 0, so y sin t 9 9 So the coordinates of V are cos t, sin t 4 (v) x y At V, cost, sin t 9 / 4 9 / x y so the equation of the locus of V is + ( ) 9 / 4 ( 9 / ) x y (vi) For the ellipse + b a Here b, so e a, b a ( e ) ( 9 / ) 6 9 ( 9 / 4) e 9 7 ± which is ±,0 ±, 0 4 0 ±.) The foci are at ( ae,0) (Accept (., 0) V 0809 Page 7 of 0
IFYFM00 Further Maths B4a(i) Equating real and imaginary parts of 8 i( z ) z λ, x 8 λ y x 8 λ y y λ ( x ) y λ x x 8 y So, eliminating λ, we get y x Then ( x 8)( x ) y x + y 0x + 6 0 x + y 9 This can be written as ( ) This is a circle centre (,0) radius. If z μ( 4 + i), x 4μ y μ So 6μ + 9μ 40μ + 6 0 μ 40μ + 6 0 ( μ 4) 0 4 μ 6 Then P, b ( z ) 8 8( cos0 + isin 0) So, by de Moivre s theorem, 0 + nπ 0 + nπ z 8 cos + isin for n 0,, π π 4π 4π ( cos0 + isin 0), cos + isin, cos + isin, + i, i So z, i, i V 0809 Page 8 of 0
IFYFM00 Further Maths Ba(i) For the cubic x + bx + cx + d with roots α, β, γ we have α + β + γ b αβ + αγ + βγ c αβγ d So b ( α + β + γ ) ( α + β + γ )( α + β + γ + αβ + αγ + βγ ) α + β + γ + αβ + α β + αγ + α γ + βγ + β γ + 6αβγ bc ( α + β + γ )( αβ + αγ + βγ ) α β α γ αβ β γ αγ βγ 9αβγ d αβγ Adding these α β γ β γ α b + bc d α + β + γ γ γ α β α β γ α α β + γ α β γ β γ α β αβγ α β γ d + b bc + d b bc 4 8 + 4 ( )( ) b f ( x) e cos x so ( 0 ) f ( x) e cos x e sin x so ( 0) f ( x) e cos x + e sin x + e sin x e cos x e sin x so ( 0 ) 0 f ( x) e sin x + e cos x so ( 0 ) f Then ( x) f ( 0) + xf ( 0) + f ( 0) + f ( 0) + K f f f x x f!! x x + +K V 0809 Page 9 of 0
IFYFM00 Further Maths B6a(i) 4 + x + ( x + ) + 4 / 8 x MA 4x d x + x + ( x + ) / / + / 8 d x d u Let u x +. Then. dx When x, u 8 4 When x, u 0 / 4 du The integral 0 u + / 4 4 u sinh 0 sinh 8 ln + + 8 64 7 ln + ln 4 ln 8 8 dx dy b S π y + dt dt dt 0 ( 8t ) π 8t + 8 dt 0 64π t t + dt 0 ( + ) / 64π t 0 8π ( ) 4. 08 4. to decimal place. V 0809 Page 0 of 0