Northern Beaches Secondary College Manly Selective Campus 04 HSC Trial Examination Mathematics Extension General Instructions Total marks 70 Reading time 5 minutes. Working time hours. Write using blue or black pen. Board-approved calculators and templates may be used. All necessary working should be shown in every question. Multiple choice questions are to be completed on the special answer page. Each free response questions Attempt Questions -4
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Multiple choice section Answer each of the following ten (0) questions on the special answer sheet provided. tan Q. is equivalent to? tan cos sin tan cot Q. The remainder theorem when P(x) = x 3 x 4x + 7 is divided by x + 3 is: 4 4 8 8-6 Q3. The point P(, ) divides the interval joining A (-, -4) and B (x, y ) in the ratio :. The coordinates of B are? (4, 5) (6, 8) (0, -) (0, 4) Q4. The term independent of x in the expansion 3 6 8 54 4 3 x x is? Page of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Q5. A particle is moving along a straight line. The displacement of the particle from a fixed point O is given by x. The graphs below show acceleration against displacement. Which of the graphs below best represents a particle moving in simple harmonic motion? (A) acceleration (B) acceleration 4 4-3 - - 3 x - - x - - -4-4 (C) acceleration (D) acceleration 4 4 - - x - - x - - -4-4 Q6. The definite integral the substitutionu log e x. The value of the integral is? e dx is evaluated using e x log x e e e e e Page 3 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Q7. The function shown in the diagram below has the equation y = Asin - Bx. Which of the following is true? y.5 0.5 - - x -0.5 - -.5 (A) A, B (B) A, B (C) A, B (D) A, B Q8. Three English books, four Mathematics books and five Science books are randomly placed along a bookshelf. What is the probability that the Mathematics books are all next to each other? 3!5! 4!9!! 4!3!5!! 4! 9! Page 4 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Q9. A, B, C and D are points on a circle. The line l is tangent to the circle at C. ACE =. A B l D C E What is ADC in terms of? A) 90 B) 80 C) 80 D) x dx? Q0. What is the indefinite integral for cos x sec A) x sin x tan x c 4 B) x sin x tan x c 4 C) D) x sin x tan x c 4 x sin x tan x c 4 End of Multiple Choice Page 5 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Free response questions answer each question in a separate Booklet Question : Start a new Booklet 5 Marks a) Solve x + 3 x 0 b) A container ship brings a total of 00 cars into Australia. Of these cars, three hundred have defective brakes. A total of five hundred cars are unloaded at Sydney. Find an expression for the probability that one hundred of the cars unloaded at Sydney have defective brakes. (NOTE: You don t need to simplify or evaluate the expression). c) How many times must a die be rolled so that the probability of at least one six exceeds 0.5? d) Line A is defined by the equation y = x +. Line B is defined by the equation y = mx + b If the acute angle between the two is 45 degrees, what are the possible values of m? Question : continued on next page Page 6 of 5
Question: continued. Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination e) In the diagram above the point R is the fourth vertex of the rectangle POQR. The points P and Q move such that POQ = 90. (i) Show that st = 4 (ii) Find the locus of the point R f) Find the general solution of the equation: sin cos = g) The diagram shown below shows a circle and AB is a chord of length 6 cm. The tangent to the circle at P meets AB produced at X. PX = 4 cm. Find the length of BX. Page 7 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Question : Start a new Booklet 5marks a) (i) Show that a zero of the function f(x) = log e x x lies between x = and x =. (ii) Use one application of Newton s Method with an initial approximation of x = 5 to obtain an improved estimate to the solution of the equation: log e x x = 0. (State your answer to one decimal place.) b) The rate of change of the temperature T of a cool item placed in a hot environment is determined by the equation. dt dt = k(s T) where k is a constant and T is the temperature of the object, and S is the temperature of the environment. (i) Show that T = S Ae kt is a solution to the differential equation: dt dt = k(s T). Jamie is cooking a large roast in an oven set to 60 o C. The roast will be cooked when the thermometer shows that the temperature of the centre of the roast is 50 o C. When Jamie started cooking, the temperature of the centre of the roast was 4 o C and 30 minutes later it was 60 o C. (ii) How long will it take for the roast to be cooked? 3 Question continued on next page Page 8 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Question continued n!!! c) (i) Show that n n n (ii) Use mathematical induction to show that, for all integers n, 3 n.... 3! 3! 4!!! n n d) Given the function f(x) = x x 3 8 (i) State the vertical asymptote. (ii) Sketch the graph of the function. Clearly show on your diagram the x intercepts. 3 (Your diagram should be one third of a page in size) Page 9 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Question 3: Start a new Booklet 5 marks a) Find x + e e x dx using the substitution u = e x 3 b) A particle is moving in a straight line. At time t seconds it has displacement x metres from a fixed point O on the line. The velocity is given by. x = 8 x3 The acceleration of the particle is given by.. x. The particle is initially metres to the right of O. (i) Show that.. x = 3 64 x5 (ii) Find an expression for x in terms of t 3 c) The letters of the word INTEGRAL are arranged in a row. Calculate the probability that there are three letters between the letters N and T. 3 Question 3 continued on next page Page 0 of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Question 3 continued d) During the medieval wars, the enemy wanted to attack a fortress with a 5 metre opening along a front wall. The strategy was to stand at the point P, on a line 8 metres from the opening and perpendicular to the wall, as per the diagram. The archer stands x metres away from the wall, thus giving an angle of vision, α, through which to fire arrows from a cross-bow. (i) Show that the angle of vision α is given by = tan - 3 x tan - 8 x (ii) Determine the distance x which gives the maximum angle of vision α. 3 Page of 5
Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination Question 4: Start a new Booklet 5 marks a) Let,, be the roots of P(x) = x 3 5x + 3x 5. Find the value of + + 3 0 b) Let (3 + x) 0 =r = 0 a r x r (i) Write down an expression for ar (ii) Show that a r + a r (iii) = 40 r 3r + 3 Find the value of r which produces the greatest coefficient in the expansion of (3 + x) 0 c) A particle is moving in a straight line in simple harmonic motion. At time t it has displacement x metres from a fixed point O on the line where x = (cost + sint) At time t, the velocity of the particle is ẋ ms- and the acceleration is.. x ms - (i) Show that.. x = -4(x ) (ii) Find the extreme positions of the particle during its motion. Question 4 continued on next page. Page of 5
Question 4continued Manly Selective Campus 04 HSC Mathematics Extension - Trial Examination d) (i) Show that: ( x) n + x n = x x n (ii) By considering the expansion of ( x) n + x n or otherwise, express n n 0 n 3 n +... + (-) n n n n n in simplest form. 3 END OF EXAMINATION PAPER Page 3 of 5
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04 MSC HSC X Trial Solutions Q Let t tan tan t then tan t sin B Q x = 3 3 3 3 4 3 + 7 = 4 8 B Q3 Q4 ( ) x x 4 ( 4) y y 5 T k 4 4k 3 k Ckx C x 3 4 4k k k k C 3 x 4 k 4k k independent of x 4 k 0 k the term is C 3 54 x x 4 A D Q5 For simple harmonic motion A x n x Page of 5
04 MSC HSC X Trial Solutions Q6 e e e dx dx x x x x log e log e du u u du u u e D Q7 A Domain is D : Bx x B B B B C Q8 4!9!! Q9 80 B B Q0 x sin x tan x c C 4 Page of 5
04 MSC HSC X Trial Solutions Q x + 3 x 0 nb x ± x + 3 x x 0 x (x + 3) x 0 (x + 3)(x )(x + ) 0 Q -a marks: correct solution mark: x or and (x + 3) x 0 therefore x 3 or < x < Q -b (p + q) 500 P fective = p = 4 P non-defective = q = 3 4 00 defective = 500 C 400 p 00 q 400 = 500 C 400 00 4 3 400 4 marks: correct solution mark: p = 4, q = 3 and considering (p+q)500 4 Page 3 of 5
04 MSC HSC X Trial Solutions p 6 = 6 q not 6 = 5 6 p no 6's = 5 6 n Q-c Q-d 5 6 n > 0 5 0 5 > 5 6 n ln(0 5) > nln 5 6 ln(0 5) ln 5 6 < n 3 8 < x n = 4 Therefore 4 throws of dice required. y = x + m = m = m tan = m m + m m = m + m - = m + m = m + m -m = m + m = m marks: correct solution mark: 5 6 n > 0.5 Or bald correct answer Note: many students failed to recognise that ln 5 6 is negative requiring inequality sign to be reversed marks: correct solution mark: correct formula m = -3 m = 3 Page 4 of 5
04 MSC HSC X Trial Solutions Q-e-i st = 4 m OQ = as as = s m OP = at at = t m OQ m OP = - s t = - st = -4 mark: correct demonstration Q as, as P at, at As shape is a rectangle R = as + at, as + at Q-e-ii x = a(s + t) x = (s + t) a (s + t) = s + st + t (s + t) st = s + t x 4a (-4) = s + t y = a s + t marks: correct demonstration mark: applying (s + t) = s + st + t Note: too many students were unable to write down coords of R, but wasted time deriving them y = a x 4a + 8 y = x 4a + 8a 4ay = x + 3a x = 4a(y 8a) sin cos = Q-f sincos = sin = = + n marks :correct solution mark: correct base value = 4 + n Page 5 of 5
04 MSC HSC X Trial Solutions Q-g XB. AX = XP (6 + x)x = 6 x + 6x 6 = 0 (x + 8)(x ) = 0 x = -8 or x = distance marks: correct solution mark: correct equation (6 + x) x = 6 Q Q-a-i f(x) = log e x x f() = log e = 0 = neg f() = log e = 0 9 = pos mark correct explanation based on correct values calculated. Therefore at least one root must lie between and as graph moves from negative to positive. Q-a-i f(x) = log e x x f (x) = x + x = x + x a = a f(x) f (x) = 5 = 735 log e 5 5 5 + ( 5) marks correct solution mark correct substitution into correct formula Qb-i Version T = S Ae kt dt dt = kae kt = ks ks + kae kt = k S + S Ae kt = k( S + T) = k(s T) Verision T = S Ae kt dt dt Ae kt = S T = kae kt = k(s T) from mark fully explained in explanation of substitution. Page 6 of 5
04 MSC HSC X Trial Solutions T = S Ae kt at t = 0 T = 4 4 = 60 Ae 0 A = 56 qb-ii Q-c at t = 30 T = 60 60 = 60 56e -30k e -30k = 00 56-30k = ln 00 56 k = ln 00 56 50 = 60 56e kt e kt = 0 56 t = ln 0 56 k 30 = 0 0483 = ln 0 56 ln 00 56 30 = 85min 0sec (n + )! n + (n + )! = = = = (n + )! n + (n + )(n + )! n + (n + )(n + )! n + (n + )! (n + ) (n + ) (n + )! (n + )! 3 marks correct solution marks correct value for k mark correct to line () mark correct solution Page 7 of 5
04 MSC HSC X Trial Solutions! + 3! + 3 4! +... n (n + )! = (n + )! Qc-ii Qc-ii For n = LHS = RHS =! + 3! + 3 4! +... n (n + )! = (n + )! Assume true for n = k ie.! + 3! + 3 4! +... k (k + )! = (k + )! RTP true for n = k + ( + )! = LHS =! + 3! + 3 4! +... = (k + )! + k + (k + )! ( + )! = 3 marks correct solution clearly = (k + )! k + (k + )! = (k + )! = RHS k (k + )! + k + ((k + ) + )! showing use of assumption and or substitution or matching algebra marks solution partially showing use of assumption and or substitution or matching algebra mark correct for n = Therefore proved true for all n by process of mathematical induction. Qd-i f(x) = x x 3 8 x = (x ) x + x + 4 x = is vertical asymptote mark correct solution x not accepted. Qd-ii 3 marks - correct shape - correct asymptotes - correct x intercepts Page 8 of 5
04 MSC HSC X Trial Solutions Q3 Q3-a Q3b-i x + e e x dx u = e x = e x e ex dx = e u du = e u + C = e ex + C. x = 8 x3 = v v = 8 x6 du = e x dx d.. x = v = 6 dx 8 x5 = 3 64 x5 3 marks correct solution marks for correct expression u u e du = e + C mark for e x + ex = e x. e ex marks for correct solution. mark for v = 8 x6 Page 9 of 5
04 MSC HSC X Trial Solutions. x = 8 x3 dx dt = 8 x3 dt dx = 8 x 3 = -8x-3 Q3b-ii t = 8x -3 dx = -8x- - t = 4 x + C at t = 0 x = 0 = 4 4 + C C = - t = 4 x 3 marks for correct solution marks - t = 4 + C and C = - x - t = 4 + C and correct x primitive from incorrect value for C mark - t = 4 x t + = 4 x x = 4 t + x = 4 t + Note; Positive answer only to agree original conditions Q3c INTEGRAL N N N N T Total Number of arrangements= 8! Number of restricted arrangements= 4! 6! T T T Prob = 4! 6! 8! = 7 3 marks correct solution marks for n(s)=8! and any two correct of 4 or! or 6! mark - 8! -any two of 4 or! or 6! Page 0 of 5
04 MSC HSC X Trial Solutions Q3-d Q3-d-ii tan(qpr) = 8 x = tan - 3 x tan - 8 x = QPS QPR tan(qps) = 3 x QPR = tan - 8 x QPS = 3 tan- x = tan - 3 x tan - 8 x = tan - 3 x tan - 8 x d dx = 3 8 x + 3 + x x + 8 x at d dx = 0 3 8 x + 3 = x x + 8 x 8x x + 64 = 3x x + 69 8 x + 69 = 3 x + 64 x = 50 5 x = 04 Test for maximum x = 0 d dx = 3 00 + 3 + 8 0 00 + 8 = 0 0004 0 x = 0 d dx = -0 000004 Therefore change in gradient positive, zero to negative therefore maximum. mark for correct solution 3 marks correct solution marks - x = 04 mark - correct for d dx - value of x correctly obtained from incorrect d dx Page of 5
04 MSC HSC X Trial Solutions + + = b a = 5 + + = c a = 3 = d a = 5 Q4-a + + = + + ( + + ) = + + + + + = + + + ( + + ) 3 = + + + 5 5 9 4 5 = + + 3 marks: correct solution marks : partial correct with a least values of ratios of coefficients correct with correct required expansion mark: at least two ratios correct only Or mark: correct required expansion only + + = 4 4 5 4 = 4 5 (3 + x) 0 x r term = 0 C r 3 0 r. (x) r a r = 0 C r 3 0 r. () r mark: correct solution Page of 5
04 MSC HSC X Trial Solutions a r = 0 C r 3 0 r. () r a r + = 0 C r + 3 9 r. () r + a r + = 0 C r + 3 9 r r +. () a 0 r C r 3 0 r. () r = 0! 39 r. () r + (9 r)!(r + )! = = (0 r) 3(r + ) 40 r 3r + 3 40 r 3r + 3 40 r 3r + 3 37 5r 7 4 r r = 8 (0 r)!r! 0! 3 0 r. () r marks: correct solution mark: partial correct with both terms correct mark: correct solution Test r = 8 0 C 8 3 8 = 74 0 3 r = 7 0 C 7 3 3 7 = 58 0 3 r = 9 0 C 9 3 9 = 53 0 3 4c-i x = (cost + sint) = cos t + sintcost + sin t x = + sin t. x = cos t.. x = -4sin t = -4( + sin t ) = -4(x ) marks: corrcet solution mark: correct expression for velocity 4c-ii x = + sin t - sin t 0 + sin t 0 x therefore extremes of position are x = 0 and x = marks: correct solution mark: only one correct extreme Page 3 of 5
04 MSC HSC X Trial Solutions 4 d -i ( x) n + x n = ( x) + x n = + x x n = x x n = x n x mark: correct solution Page 4 of 5
04 MSC HSC X Trial Solutions Determine the coefficient of x on both sides of the equation. RHS = ( x) n + x n = n n k = 0 (-) k n C k x k k = 0 n C k x k coefficient of x n n 0 n 3 n +.... + (-) n n n n n 4d-ii LHS = x n x = x x n n = k = 0 n = k = 0 (-) k n k x (n k) x k (-) k n k x k n coefficient of x at k n = k = + n If n is odd, then k is not an integer therefore cannot exist. therefore n n 0 n 3 n +... + (-) n n n n n = 0 If n is even then (-) therefore + n n + n is coefficient of x n n 0 n 3 n +... + (-) n n n n n = (-) + n n + n 3 marks : correct solution marks: partial correct with significant progress to equating coefficients of squared term mark: one expression for the squared term correct Page 5 of 5