Prelim Examination 2010 / 2011 (Assessing Units 1 & 2) MATHEMATICS. Advanced Higher Grade. Time allowed - 2 hours

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1 Prelim Examination 00 / 0 (Assessing Units & ) MATHEMATICS Advanced Higher Grade Time allowed - hours Read Carefully. Calculators may be used in this paper.. Candidates should answer all questions. Full credit will only be given where the solution contains appropriate working

2 All questions should be attempted. Use Gaussian elimination to solve the following system of equations x + y + z = x + y + 6z = 0 x + y - z = 0. Find the solution corresponding to y = -. sin xx ( ). Given that y =, 0 x <, show that dy dx = x + x x sin ( x) ( x).. Two complex numbers, z and z, are given by where k is a real number. z = i and z = 6 + ki, (a) Given that z z is a purely real number, find the value of k. (b) Express z z in the form x + iy, where x and y are real numbers.. The term independent of x in the expansion of 6 t x x is 60. Given that t > 0, find the value of t.. (a) The function f(x) is given by ( x) Express f(x) in the form ( x) x + x + 7 f =, 0 x < π. ( x + )( x + 9) p qx + r f = + where p, q and r are integers. x + ( x 9), + π (b) Hence show that f ( x) dx = + ln [Turn over Pegasys 00

3 6. For each of the following statements, decide whether it is true or false and prove your conclusion. A For all natural numbers n, if n is divisible by then n + 6n 9 is divisible by 9. B The sum of any prime number p and any even integer q is always odd. dy sec y 7. Find the solution of the differential equation = given that t dt e π y = when t = A curve is defined by the parametric equations x = t, y = + t t t for all t. (a) Find the coordinates of the stationary point of this curve. d y (b) Use dx to determine the nature of this stationary point. 9. The first two terms of a series are + and +. (a) (b) If the series is arithmetic, show that the common difference is, and that the sum of the first six terms is 6 6. If the series is geometric, show that it has a sum to infinity, and that this sum is ( + ). 0. Use the substitution ( ) x + = θ + to show that 0 dx x + 0 = ln. 6 [Turn over Pegasys 00

4 x + 8. Part of the graph of the function f(x) =, x k, x k is shown below. The graph has a vertical asymptote with equation x =, a non-vertical asymptote L and crosses the y-axis at the point P. The points (a, b) and (c, d) are the stationary points of the graph. y x = (c,d) L (a,b) O P x (a) Write down the value of k. (b) Determine algebraically the equation of the non-vertical asymptote, L. (c) (d) Write down the coordinates of the point P where the graph of f crosses the y-axis. The graph has stationary points with coordinates (a, b) and (c, d) as shown in the diagram. Use calculus to determine the values of a, b, c and d and justify the nature of the stationary points. (d) State the coordinates of the stationary points of the graph with equation g( x) = f ( x). [END OF QUESTION PAPER] Pegasys 00

5 Marking Scheme Advanced Higher Grade 00 / 0 Prelim (Assessing Units & ) ans: x = 7z, y = z +,' z = z' (or equivalent) marks 6 0 correct augmented matrix 0 correct first modified system 0 0 correct second modified system 0 0 correct third modified system x = 7z, y = z +,' z = z' (or equivalent) ans: x = 7, (y = -), z = - mark ans: Proof marks attempts to substitute into quotient rule differentiates correctly differentiates correctly and correct denominator simplifies correctly x = 7, (y = -), z = - (or equivalent) f g fg substitutes into " " g x x or ( x) ( x) ( x) ( ) sin sin ( ) x ( x) ( ) or x x and x ( ) ( x) dy x + x sin ( x) =. dx x ( x) Pegasys 00

6 (a) ans: k = marks starts correctly continues correctly i or 8 ki + i( k) (or equivalent) k = (b) ans: 0 + i (or equivalent) marks correct complex conjugate continues correctly ans: t = marks finds correct general term simplifies to find correct expression for power of x solves for r correctly finds correct term 6 + i 6 + i + i i + i 0 + i (or equivalent) 6 6 ( x ) r r x r = t = r t x r (a) ans: ( x) f = + marks x + x + 9 correct method one value correct other two values correct x + x + 7 p( x + 9) + ( qx + r)( x + ) p =, q = 0 or r = f ( x) = + x + x + 9 (b) ans: Proof marks integrates correctly integrates correctly substitutes correctly completes proof correctly ln x + tan x ln + tan ln+ tan 0 π ln Pegasys 00

7 6 ans: A True & Proof; B False & Proof marks 7 A - correct conclusion A starts proof A completes proof B correct conclusion B completes proof ans: y 7 = sin + t e marks rearrange correctly correct strategy; i.e. rewrite and integrate integrate correctly substitute correctly True ( k ) + 6( k) 9 9( + k ) k which is divisible by 9 False p = and (e.g.) q = gives p + q = 6 which is even dy dt = t sec y e t cos ydy = e dt t e sin y = + C ( 0) sin π = e + C 6 7 y = sin + t e 7 (accept sin y = + t e ) Pegasys 00

8 8(a) ans:, marks attempts to substitute into correct formula differentiates correctly (twice) equates with zero and solves for t correctly correct coordinates ( ) ( ) " substitutes into " dy y t = dx x t 8t + t + t t =, 8(b) ans: Maximum (turning point) marks finds second derivative correctly substitutes correctly correct sign and conclusion 8 6 t + + t t t t t t + t + ( + ) d y <0 MaximumT.P. dx Pegasys 00

9 9(a) ans: Proof marks correct method + ( + ) correct value = correct method 6 ( ) + + correct value =... = 6 6 9(b) ans: Proof marks correct method correct conclusion correct method correct value < < S exists + ( + )( + ) ( + ) ( + ) =... = ( + ) conjugate surd used to form answer 0 ans: Proof 6 marks starts substitution correctly continues substitution correctly changes limits correctly integrates correctly substitutes correctly completes proof correctly ( θ + ) dθ θ x = θ =, x = 0 θ = [ θ + lnθ ] { + ln + ln } 0 + ln ln =... = ln ( ) ( ) ( ( ) Pegasys 00

10 (a) ans: k = mark correct value k = (b) ans: y = x + marks correct method continues correctly x x y = x + + x y = x + (c) ans: (, 8) 0 mark correct coordinates (, 8) 0 (accept -8) (d) a =, b =, c = & d = 8 ans: (, ) (,8) MinimumT. P. MaximumT. P. & differentiates correctly solves f ( x) = 0 correctly correct y-coordinates correct formula for second derivative correct nature of both points marks f ( x) = 9 ( x ) x =, a =, b =, c = & d = 8and/or (, ) & (,8) 8 f ( x) = ( x ) f ( ) < 0 MaximumT.P. & f () > 0 MinimumT.P. (d) ans: (, )(,, 6) marks correct point correct point (, ) (, 6) TOTAL MARKS = 70 Pegasys 00