2013 HSC Mathematics Extension 2 Marking Guidelines

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1 3 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer B A 3 D 4 A 5 B 6 D 7 C 8 C 9 B A

2 3 HSC Mathematics Extension Marking Guidelines Section II Question (a) (i) Correct answer z + w ( i 3)+ ( i 3) 3 3i Question (a) (ii) Correct solution Finds modulus or argument w + 3 arg w tan 3 π 3 Hence w cos π 3 + isin π 3 in modulus argument form. Question (a) (iii) Correct solution Attempts to use de Moivre s theorem By de Moivre s theorem and part (ii) w 4 4 cos 4 π 3 + isin 4 π 3 4 cos8π + isin8π 4

3 3 HSC Mathematics Extension Marking Guidelines Question (b) Correct solution Finds value of one of A, B or C x + 8x + ( x 3) x + A x 3 + Bx + C x + ( Ax + ) + ( Bx + C )( x 3 ) ( x 3) ( x + ) We need: x + 8x + Ax ( + ) + ( Bx + C )( x 3 ) Make particular choices for x and substitute: x 3: 44 A A 4 x : A 3C 8 3C C x : 3A B C B + B 3 Hence A 4, B 3 and C Question (c) Correct factorisation Attempts to use formula to solve a quadratic equation, or equivalent merit Find the zeros by using the quadratic formula: ( ( 4i ) 4.5.) z 4i ± 4i ± 36 i ± 3i The zeros are 5i and i and therefore z + 4iz + 5 z + 5i ( z i) 3

4 3 HSC Mathematics Extension Marking Guidelines Question (d) Correct solution 3 Correct primitive, or equivalent merit Uses a suitable substitution to get an indefinite integral, or equivalent merit Use the substitution x sinθ dx dθ cosθ When x, sinθ, then θ When x, sinθ, then θ π Hence π x 3 x dx sin 3 θ sin θ. cosθ dθ π sin 3 θ cos θ dθ π sinθ dθ cos θ cos θ π cos θ sinθ cos 4 θ sinθ dθ cos3 θ + cos5 θ π Alternative: Use the substitution u x du dx x Where x, u When x, u Hence x 3 + x dx ( u) u du u u 3 du 3 3 u 5 5 u 5 4

5 3 HSC Mathematics Extension Marking Guidelines Question (e) Correct sketch 3 Sketches the correct boundary Obtains x y 4, or equivalent merit z x + iy ( x, y real) z + z ( x + iy) + ( x iy) x + ixy y + x ixy y x y ( x y ) Hence ( x y ) 8, so x 4 y 4. y x 5

6 3 HSC Mathematics Extension Marking Guidelines Question (a) Correct solution 4 Makes significant progress towards finding a primitive in terms of t, or 3 equivalent merit Makes complete substitution including the limits, or equivalent merit Correctly substitutes for cos x and attempts to write dx in terms of t, or equivalent merit First use the t-substition, t tan x t Then dx dt and cos x + t + t When x, t tan and when x π, t tan π 4 π Hence 4 + 5cosx dx t + t 4+ t + t dt + 5 ( t ) dt 9 t dt 3 t ( 3 + t ) dt Next, do partial fraction: ( 3 t) ( 3 + t ) A 3 t + B 3 + t A ( 3 + t ) + B ( 3 t ) ( 3 t) ( 3 + t ) A( 3 + t)+ B 3 t Choose t 3, so 6A and A 3 Choose t 3, so 6B and B 3 Hence π 4 + 5cosx dx 3 3 t t dt [ + ln ( 3 + t )] 3 ln 3 t 3 ln 3 + t 3 t 3 ln 3 ln 3 ln 6

7 3 HSC Mathematics Extension Marking Guidelines Question (b) Correct solution Correct implicit differentiation, or equivalent merit Differentiate both sides of the given identity implicitly with respect to x: y dy dx + y dy dx 5 y + y dy dx 5 y + y y y dy dx 5 dy dx 5 y y 5 y y 7

8 3 HSC Mathematics Extension Marking Guidelines Question (c) Correct solution 4 Makes substantial progress towards evaluating the integral 3 Correct integral for volume, or equivalent merit Attempts to use cylindrical shells, or equivalent merit e x x Volume of thin cylindrical shell: π ( 4 x) e x δ x. Hence the volume V of the solid formed is V 3 π ( 4 x) e x dx 3 π ( 4e x xe x )dx π 4e x 3 π 8π e 3 e 3 xe x dx π xe x dx 3 Using integration by parts: 3 xe x dx xe x 3 [ xe x e x 3 ] 3 e x dx ( 3e 3 e 3 ) ( e e ) 3e 3 e 3 e 3 Hence V 8π ( e 3 e) π e 3 8π ( e 3 e) 4πe 3 4πe 3 8πe Volume is 4πe( e )units 3 8

9 3 HSC Mathematics Extension Marking Guidelines Question (d) (i) Correct solution Finds correct gradient, or equivalent merit xy c, so y c x Now dy dx c x Hence the slope of the tangent at cp, c p is c c p p The equation of the tangent is y c p p p y cp x + cp x + p y cp ( x cp) 9

10 3 HSC Mathematics Extension Marking Guidelines Question (d) (ii) Correct solution Finds the coordinates of A and B, or equivalent merit Find coordinates of A and B using part (i). A : y, so x cp Acp, B : x, so p y cp y c p B, c p Now find distances from P: PO : ( cp ) + c p c p + p PA : PB : ( cp cp ) + c p ( cp ) + c p c p c p + p c p + p All distances are equal, so A, B, O are on a circle with centre O.

11 3 HSC Mathematics Extension Marking Guidelines Question (d) (iii) Correct proof The slope of PQ is m c q c p cq cp q p q p p q pq q p pq Using part (i), the tangent to the hyperbola at Q is x + q y cq. At C, y and x cq ie C (cq, ). Hence the slope of BC is c p m cq c cpq pq As m m, the lines PQ and BC are parallel.

12 3 HSC Mathematics Extension Marking Guidelines Question 3 (a) (i) Correct solution 3 n Obtains I n n x x dx, or equivalent unit Attempts to use integration by parts, or equivalent merit Use integration by parts with u ( x ) n, v u nx( x ) n, v x Hence I n x x n n x n x + n x x n dx ( x ) n ( x ) n n dx + n ( x ) n ni n + ni n dx dx Solving for I n ( + n) In ni n I n n n + I n

13 3 HSC Mathematics Extension Marking Guidelines Question 3 (a) (ii) Correct solution Finds I, or equivalent merit I x dx area of quarter circle with radius π 4 From the recursion relation in part (i), I I I I I π 4 5π 3 3

14 3 HSC Mathematics Extension Marking Guidelines Question 3 (b) (i) Correct sketch Sketches y ƒ ( x), or equivalent merit 4

15 3 HSC Mathematics Extension Marking Guidelines Question 3 (b) (ii) Correct sketch 3 Sketches graph showing three correct vertical asymptotes, or equivalent merit Sketches graph of y ƒ ( x), or equivalent merit First sketch ƒ x y x Now sketch ƒ x 5

16 3 HSC Mathematics Extension Marking Guidelines Question 3 (c) (i) Correct solution Finds ABC α + β, or equivalent merit ADC 8 α + β ABC 8 ADC α + β ACB 9 As ACB is right-angled, sin α + β AC AC (angle sum in triangle ACD) r r sin α + β (opposite angles of cyclic quadrilateral) (angle in semicircle, AB is diameter) 6

17 3 HSC Mathematics Extension Marking Guidelines Question 3 (c) (ii) Correct solution Observes that ADB is a right angle or ABD β, or equivalent merit Join BD. ADB 9 ABD ACD β As ABD is right-angled, sinβ AD AB AD r AD r sinβ (angle in semicircle, AB is diameter) (angles standing on the same arc AD) As AED is right-angled, cosα AE AD AE ADcosα r cosα sinβ 7

18 3 HSC Mathematics Extension Marking Guidelines Question 3 (c) (iii) Correct solution Similarly, CE r cosβ sinα Now AC AE + CE r cosα sinβ + r cosβ sinα r ( cosα sinβ + cosβ sinα) From part (i) AC r sin( α + β), so sin α + β cosα sinβ + cosβ sinα. 8

19 3 HSC Mathematics Extension Marking Guidelines Question 4 (a) Correct solution 3 Correctly finds both relevant areas, or equivalent merit Correctly finds one relevant area, or equivalent merit The area of the triangle with corners (, ), (t, ) and (t, ln t) is given by ( t ) lnt. The area under the graph y ln x between x and x t is given by t ln xdx lnxdx t x ln x t x x dx t t (integration by parts) t lnt x t t lnt ( t ). As the graph y ln x is concave down, the area under the graph is greater than the area of the triangle. ie t lnt ( t ) > ( t )lnt t lnt ( t )lnt > t lnt t ( t ) > t lnt ( t + ) > t lnt > t t +,fort > 9

20 3 HSC Mathematics Extension Marking Guidelines Question 4 (b) Correct solution 3 Uses the induction hypothesis, or equivalent merit Establishes initial case Show true for n z + ie true for n Assume true that z k k, for some k Show true for n k + ie z k + k + LHS z k + z k + i z k z k + i z k k + i k k + k k + RHS By the principle of Induction, z n n for all integers n.

21 3 HSC Mathematics Extension Marking Guidelines Question 4 (c) (i) Correct solution As sec θ + tan θ we apply the binomial theorem sec n θ ( sec θ) n ( + tan θ) n n k n k tank θ Question 4 (c) (ii) Correct solution Attempts to use part (c) (i), or equivalent merit sec 8 θ dθ sec 6 θ sec θ dθ 3 k 3 3 k 3 k tank θ sec θ dθ 3 k tan k θ sec θ dθ 3 k k + tank + θ + C since d dθ tanθ sec θ k

22 3 HSC Mathematics Extension Marking Guidelines Question 4 (d) (i) Correct solution 5 B D 3 A 3 E 4 NOT TO SCALE C DAE is common AB AE and 4 AC AD So ABC and AED have pairs of sides in the same ratio, and the included angles equal. Hence ABC and AED are similar. Question 4 (d) (ii) Correct solution ADE BCE (corresponding angles in similar triangles ABC and AED) BDE 8 ADE 8 BCE Hence BCED is cyclic (opposite angles are supplementary).

23 3 HSC Mathematics Extension Marking Guidelines Question 4 (d) (iii) Correct solution Uses cosine rule in a suitable triangle, or equivalent merit 5 B D 3 A 6 3 E 4 NOT TO SCALE C AED is isosceles, so by similarity with ABC, ABC is also isosceles with BC AC and ABC BAC α Applying the cosine rule in ADC CD cosα Applying the cosine rule in BDC CD cosα Eliminate cosα from the two equations: CD CD cosα cosα 6 6 cosα 45 CD CD 45 CD 36 3 CD 4 CD CD 3

24 3 HSC Mathematics Extension Marking Guidelines Question 4 (d) (iv) Correct solution Makes some progress Let O be the centre of the circle passing through B, C, E and D (BC ED is a cyclic quadrilateral). ABC α, then COD α (angle at the centre is twice the angle on the circumference standing on the same arc). 45 CD From part (iii) cosα 36 Now cosα cos α From the cosine rule in CDO CD r + r r cosα r + 9 r 9 r Hence r 9 and r 3 5 4

25 3 HSC Mathematics Extension Marking Guidelines Question 5 (a) Correct solution 3 Shows that zw wz i w z ( sinθ cosφ cosθ sinφ), or equivalent merit Shows the fact that z z ( cosθ + isinθ), or equivalent merit Area absinc ie A z w sin( θ φ) A z w sin( θ φ) zw wz z ( cosθ + isinθ) w ( cosφ isinφ ) w cosφ + isinφ w z i cosφ sinθ i cosθ sinφ i w z sinθ cosφ cosθ sinφ i w z sin θ φ i A 4i A z cosθ isinθ 5

26 3 HSC Mathematics Extension Marking Guidelines Question 5 (b) (i) Correct solution Correctly finds the value of b, or equivalent merit P() a + b + c + e 3 (remainder theorem) P( ) a b + c + e (as x is a root) Subtracting b 3, so b 3 Hence 4ax 3 + 3bx + cx 4a + 3b c (as x is a double root) P x P P 4a + c 3b Question 5 (b) (ii) Correct solution Slope of tangent at x is P () 4a + 3b + c + 3b 4a + c

27 3 HSC Mathematics Extension Marking Guidelines Question 5 (c) (i) Correct solution The probability that a car completes all 4 days is Question 5 (c) (ii) Correct solution Demonstrates some understanding of binomial probability using n 8, or equivalent merit The probability that a car does not complete all 4 days is (.7) 4.76 Probability that no car completes all 4 days: Probability that precisely one completes all 4 days: (.4) Probability that precisely two complete all 4 days: (.4) Therefore the probability that, or complete all 4 days: (.76) 6.76 ( + 8(.76) (.4 ) + 8 (.4 ) ).7 Hence the probability that at least 3 complete all 4 days is.7.3 7

28 3 HSC Mathematics Extension Marking Guidelines Question 5 (d) (i) Correct solution Terminal velocity occurs when acceleration is zero ( v ), ie mg kv v T mg k v T ± ie v T mg k mg k 8

29 3 HSC Mathematics Extension Marking Guidelines Question 5 (d) (ii) Correct solution 3 Finds an expression for x in terms of v and an undetermined constant, or equivalent merit Obtains equation of motion, or equivalent merit Integrating we get x v T g ln ( v + v T )+ C When t, x and v u. Hence v T g ln ( v + u T )+ C Equation of motion for the ball going up: m v mg kv mg + k mg v mg + v v T v g v T ( v T + v ) We know v dv dx g v T dv dx, so ( v + v T ) so C v T g ln v + u T x v T g ln ( v + u T ) ln v T + v v T g ln v + u T v T + v When the ball reaches its maximum height, H v and x H. Hence v T g ln v + u T v T g v T u ln + v T or dx dv v T g v T + v 9

30 3 HSC Mathematics Extension Marking Guidelines Question 5 (d) (iii) Correct solution Makes some progress H O kv mg x, v ie x v T g ln v T v T v When the ball hits the ground, x H, v w H v T g ln v T v T w x xh, vw Equation of motion for the ball coming down: m v mg kv mg k mg v v T v g ( v v T ) v T mg v We know v dv dx, so dx dv v T g v T v Hence x v T g ln ( v T v )+ C When x, v, but v T g H v T g u ln + v T u ln + v T g ln v T v T w v T + u v T v T + u v T v T v T w v T v T w ( v T + )( u v T ) w v T v T v T v T v T w + u v T u w v T v T ( dividing by u w v T ) ie u + w + v T ( from part (ii)) w u + v T C v T g ln v T x v T g ln v T ln v T v 3

31 3 HSC Mathematics Extension Marking Guidelines Question 6 (a) (i) Correct solution Solves P ( x), or equivalent merit P(x) x 3 5x + 4x + 6 P (x) 6x 3x + 4 (for max/mim) ie x 5x + 4 ( x 4) ( x ) x 4 or P (x) x 3 P (4) 48 3 > min P () 3 < max P(4) min at (4,) Test boundary: x, P() 6 minimum value of P(x) is when x 4 Question 6 (a) (ii) Correct solution ( ) 5x x + x + 8x + 6 ( x + ) x + x + 4 LHS x + x + 8x + 6x + x + 8x + 6 x 3 + x + 4x + 6 Since P(x), ie x 3 5x + 4x + 6 (from part (i)) then x 3 + x + 4x + 6 5x x + ( x + 4) ie x + 5x. 3

32 3 HSC Mathematics Extension Marking Guidelines Question 6 (a) (iii) Correct solution Shows that ( m + n) + ( m + n + 4) 5 ( m + n ) ( m + n) + ( m + n + 4) mn m + n + From part (ii) ( x + ) ( x + ( x + 4) ) 5x Since x then x + ( x + 4) 5x x + Let x m + n ( m + n) + ( m + n + 4) 5 ( m + n ) Now ( m n) m mn + n m + n mn m + n + and m + mn + n mn + mn ( m + n) 4mn ( m + n) + ( m + n + 4) 5 4mn m + n + mn m + n +. m + n +, or equivalent merit 3

33 3 HSC Mathematics Extension Marking Guidelines Question 6 (b) (i) Correct answer The string has length a, so PS + PS a. This characterises an ellipse with focal length a and focal points S and S. Also, for an ellipse, SS ae, which is given. Question 6 (b) (ii) Correct solution For an ellipse, the distance from the centre O to the directrix is a e. Hence the distance from P to the directrix is a e OQ a e acosθ. By the focus-directrix definition of an ellipse, SP epm (where M is on the directrix) SP e a e acosθ So, SP a aecosθ 33

34 3 HSC Mathematics Extension Marking Guidelines Question 6 (b) (iii) Correct solution Finds the length of S P, or equivalent merit sinβ Q S P S ae + OQ a SP ae + acosθ a a( ecosθ) ae+ cosθ a a + aecosθ ae+ ( cosθ) a + ecosθ e + cosθ + ecosθ. 34

35 3 HSC Mathematics Extension Marking Guidelines Question 6 (b) (iv) Correct solution Finds the component of one tension force in the vertical direction, or equivalent merit The vertical component of the force along SP is Tsinβ T e + cosθ + ecosθ Now sin( QPS) QS PS ae OQ a( ecosθ) ae acosθ a( ecosθ) e cosθ ecosθ The vertical component of the force along SP is therefore T sin( QPS) T e cosθ ecosθ Therefore mg T e + cosθ + ecosθ e cosθ ecosθ ( ecosθ ) ( e cosθ )( + ecosθ ) T e + cosθ e cos θ T e cos θ e + cosθ e cosθ ecos θ ( e cosθ + e cosθ ecos θ) T [ e cos θ cosθ e cosθ] cosθ e cos θ. T e 35

36 3 HSC Mathematics Extension Marking Guidelines Question 6 (b) (v) Correct solution 3 Makes substantial progress Finds QP, or equivalent merit As P lies on an ellipse of eccentricity e r b QP a e sinθ Since ellipse has b a e cos( QPS) r SP cos( β) r SP a e sinθ a ( ecosθ) e sinθ ecosθ a e sinθ a ( + ecosθ) e sinθ + ecosθ Hence the horizontal force is mrw T e sinθ + T e sinθ ecosθ + ecosθ T e sinθ e. cosθ 36

37 3 HSC Mathematics Extension Marking Guidelines Question 6 (b) (vi) Correct solution mrw mg T e sinθ cosθ e cos θ T e e cos θ e sinθ e cosθ rw g tanθ e tanθ rw g e. 37

38 3 HSC Mathematics Extension Marking Guidelines Mathematics Extension 3 HSC Examination Mapping Grid Section I Question Content Syllabus outcomes 8. H5, E8 3. E4 3. E E4 5.5 E E E E7 9.5 E6 8. HE3, E Section II Question Content Syllabus outcomes (a) (i). E3 (a) (ii). E3 (a) (iii).4 E3 (b) 7.6 E4 (c). E3 (d) 3 4. E8 (e) 3.5 E3 (a) 4 4. E8 (b).8 E4, E6 (c) 4 4., 5. E7, E8 (d) (i) 3.3 E4 (d) (ii) 3.3, 8. PE3, E3, E4 (d) (iii) 3.3 E3, E4 3 (a) (i) 3 4. E8 3 (a) (ii) 4. E8 3 (b) (i).7 E6 3 (b) (ii) 3.8 E6 3 (c) (i) 8., 8. E 3 (c) (ii) 8., 8. E 3 (c) (iii) 8., 8. E 4 (a) 3 4., 8.3 PE3, E 4 (b) 3.,., 8. HE, E. E3 38

39 3 HSC Mathematics Extension Marking Guidelines Question Content Syllabus outcomes 4 (c) (i) 8. HE3, E 4 (c) (ii) 4. E8 4 (d) (i) 8. P4, H5 4 (d) (ii) 8. PE3, E 4 (d) (iii) 8. H5, E 4 (d) (iv) 8. PE3, E 5 (a) 3. E3 5 (b) (i) 7., 8. PE3,E4 5 (b) (ii) 7., 8. PE3,E4 5 (c) (i) 8. HE3, E 5 (c) (ii) 8. HE3, E 5 (d) (i) 6. E5 5 (d) (ii) 3 6. E5 5 (d) (iii) 6. E5 6 (a) (i) 8. H6, E6 6 (a) (ii) 8.3 PE3, E4 6 (a) (iii) 8.3 PE3, E4 6 (b) (i) 3. E3, E4 6 (b) (ii) 3. E3, E4 6 (b) (iii) 8. H5 6 (b) (iv) 6.3 E5 6 (b) (v) E5 6 (b) (vi) 6.3 E5 39

2017 HSC Mathematics Extension 2 Marking Guidelines

2017 HSC Mathematics Extension 2 Marking Guidelines 07 HSC Mathematics Etension Marking Guidelines Section I Multiple-choice Answer Key Question Answer C B 3 D 4 C 5 B 6 A 7 A 8 B 9 C 0 B NESA 07 HSC Mathematics Etension Sample Answers Section II Question