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MEI Mathematics in Education and Industry MEI STRUCTURED MATHEMATICS METHODS FOR ADVANCED MATHEMATICS, C3 Practice Paper C3-D Additional materials: Answer booklet/paper Graph paper List of formulae (MF2) TIME 1 hour 30 minutes INSTRUCTIONS • • • Write your Name on each sheet of paper used or the front of the booklet used. Answer all the questions. You are permitted to use a graphical calculator in this paper. INFORMATION • • • • • The number of marks is given in brackets [] at the end of each question or part-question. You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used. Final answers should be given to a degree of accuracy appropriate to the context. The total number of marks for this paper is 72. You are reminded of the need for clear presentation in your answers. MEI February 2005 Section A (36 marks) 1 You are given that y2 = 4x + 7. (i) Use implicit differentiation to find (ii) Make x the subject of the equation. Find dy in terms of y. dx [2] dx dx 1 and hence show that in this case = . dy dy dy dx [3] (i) Expand ( e x + e − x ) . (ii) Hence find (i) Sketch the graph of y = 3 x − 6 . [2] (ii) Solve the equation 3 x − 6 = x + 4 and illustrate your answer on your graph. [4] 4 Find ∫ x sin 3x dx . [4] 5 Make x the subject of t = ln 6 The function f(x) is defined as f(x) = 2 3 2 ∫ (e x [1] + e − x ) dx . 2 [3] 5 . ( x − 3) P [4] lnx . The graph of the function is shown in Fig. 6. x Q Fig. 6 (i) Give the coordinates of the point, P, where the curve crosses the x-axis. [1] (ii) Use calculus to find the coordinates of the stationary point, Q, and show that it is a maximum. [6] © MEI February, 2005 MEI Structured Mathematics Practice Paper C3-D Page 2 7 An oil slick is circular with radius r km and area A km2. The radius increases with time at a dr rate given by = 0.5 , in kilometres per hour. dt dA (i) Show that [4] = πr . dt (ii) Find the rate of increase of the area of the slick at a time when the radius is 6 km. [2] Section B (36 marks) 8 Fig. 8 shows the graph of y = x 1 + x . The point P on the curve is on the x-axis. P Fig. 8 (i) Write down the coordinates of P. (ii) Show that [1] dy 3x + 2 = . dx 2 1 + x [4] (iii) Hence find the coordinates of the turning point on the curve. What can you say about the gradient of the curve at P? [4] 1 1 32 2 + = − x 1 x d x u u du . ∫−1 ∫0 Evaluate this integral, giving your answer in an exact form. 0 (iv) By using a suitable substitution, show that (v) What does this value represent? [7] Use your answer to part (ii) to differentiate y = x 1 + x sin2 x with respect to x. (You need not simplify your result.) [2] © MEI February, 2005 MEI Structured Mathematics Practice Paper C3-D Page 3 9 The functions f(x) and g(x) are defined by f(x) = x2 , g(x) = 2x − 1, for all real values of x. (i) (ii) State the ranges of f(x) and g(x). Explain why f(x) has no inverse. [3] Find an expression for the inverse function g-1(x) in terms of x. Sketch the graphs of y = g(x) and y = g -1(x) on the same axes. [4] (iii) Find expressions for gf(x) and fg(x). [2] (iv) Solve the equation gf(x) = fg(x). Sketch the graphs of y = gf(x) and y = fg(x) on the same axes to illustrate your answer. (v) Show that the equation f(x + a) = g 2(x) has no solution if a > © MEI February, 2005 MEI Structured Mathematics [4] 1 . 4 Practice Paper C3-D [5] Page 4 Qu Answer Section A 1 (i) dy y 2 = 4 x + 7 ⇒ 2 y. = 4 dx dy 2 ⇒ = dx y (ii) 2 x= 3 ( A1 2 ) (i) e +2+e (ii) = ∫ (e + 2 + e B1 M1 A1 3 B1 −2 x 2x Comment M1 1 2 dx 1 y 1 y −7 ⇒ = .2 y = = 4 dy 4 2 2 y 2x = Mark 1 −2 x )dx B1 B1 B1 1 2x 1 e + 2 x − e −2 x + c 2 2 (i) B1 B1 One for each exponential term, one for both 2x and 3 constant. One for two half lines; one for correct orientation and meeting at (2, 0) . 2 At P −3 x + 6 = x + 4 1 ⇒ x= 2 At Q 3x − 6 = x + 4 (ii) ⇒ x=5 The solution is x = 12 , 5 . As shown on graph 4 ∫ x sin 3x dx; u= x⇒ du dv 1 = 1, = sin 3 x ⇒ v = − cos x dx dx 3 1 1 = − x cos 3x + ∫ cos 3x dx 3 3 1 1 = − x cos 3x + sin 3x + c 3 9 M1 A1 A1 E1 4 Choice of u M1 1 − cos 3 x 3 A1 M1 A1 Correct form c must be seen 4 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 2 5 6 t = ln (i) 5 ( x − 3) 1 ( x − 3) = − ln 2 5 ( x − 3) ⇒ −2t = ln 5 x − ( 3) ⇒ e −2t = ⇒ x = 5e −2t + 3 5 P(1,0) Rules of logs Change to exponentials M1 M1 A1 A1 4 B1 1 (ii) 1 x. − 1.ln x dy = x 2 x dx 1 − ln x = x2 At Q, gradient is zero, so x = e . Q is (e, 1e ) . 1 x 2 (− ) − (1 − ln x).2 x x x4 −3 + 2 ln x = x3 When x = e , this is –ve, so Q is a maximum. d2 y = dx 2 7 (i) (ii) dA = 2π r dr dA dA dr 1 ⇒ = . = × 2π r = π r dt dr dt 2 A = π r2 ⇒ So when r = 6 , dA = 6π (= 15.707...) . dt The area increases at 15.7 km2h-1, to 3sf. quotient rule M1 A1 M1 A1 =0 M1 Or equivalent methods A1 6 For 1/e. M1 A1 M1 A1 4 Substituting M1 A1 2 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 3 Section B 8 (i) P(−1,0) (ii) B1 1 y = x 1+ x 1 = x(1 + x) 2 1 dy 1 −1 = 1.(1 + x) 2 + x. (1 + x) 2 dx 2 2(1 + x) + x = 1 2(1 + x) 2 3x + 2 = 2 1+ x (iii) At a turning point, gradient is zero. x = − 23 there. Then ⇒ y=− 2 1 3 3 M1 Product rule A1 Any correct expression Combining fractions M1 E1 4 M1 A1 2 3 9 These are the coordinates of the TP. A1 At P the gradient is undefined. B1 =− (iv) ∫ 0 −1 Accept a simplified form with √3 in the bottom. Accept reference to infinity or to 4 vertical. x 1 + xdx 1 = ∫ (u − 1)u 2 du, u = 1 + x 0 du ⇒ =1 dx 1 Integral in u 1 M1 E1 Change of limits M1A1 Integrating = ∫ (u − u )du 3 2 1 2 0 1 2 5 2 3 = u2 − u2 3 0 5 4 =− 15 The magnitude represents the area bounded by the curve and axis between P and O. It is negative because the curve is below the axis (except at the end points). (v) y = x 1 + x sin 2 x dy 3 x + 2 sin 2 x + x 1 + x .2 cos 2 x ⇒ = dx 2 1 + x A1 B1 B1 One for geometry 7 and one for sign. M1A1 2 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 4 9 (i) B1 Range of f is [0, ∞) , B1 of g (−∞, ∞) . f has no inverse because (say) for any value of f > 0 there are E1 2 corresponding values of x 3 y = 2x −1 (ii) 1 1 y+ 2 2 1 1 ⇒ g −1 ( x) = x + 2 2 ⇒x= One mark for one line, and one mark for second correctly related M1 A1 B1 B1 4 (iii) B1 B1 gf ( x) = 2 x 2 − 1 fg( x) = (2 x − 1) 2 2 (iv) M1 A1 2 x 2 − 1 = (2 x − 1) 2 ⇒ 0 = 2x2 − 4x + 2 ⇒ 0 = 2( x − 1) 2 ⇒ x =1 B1 y = (2x − 1)2 B1 y = (2x − 1)2 y = (2x − 1)2 4 (v) f ( x + a) = ( x + a) 2 g 2 ( x) = 2(2 x − 1) − 1 = 4 x − 3 B1 f ( x + a) = g 2 ( x) ⇒ ( x + a)2 = 4 x − 3 M1 ⇒ x + (2a − 4) x + a + 3 = 0 ⇒ There are two roots to this equation if M1 (2a − 4) 2 > 4(a 2 + 3) A1 2 Both fns correct Equating 2 i.e. 4a − 16a + 16 > 4a + 12 1 ⇒ 16a < 4 ⇒ a < 4 2 Using b2 − 4ac Correct inequality 2 A1 5 Result MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 5 MEI, February 2005 MEI Structured Mathematics Practice paper C3-D Mark Scheme Page 6