2017 HSC Mathematics Extension 2 Marking Guidelines

Transcription

1 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

2 NESA 07 HSC Mathematics Etension Sample Answers Section II Question (a) (i) Provides correct value z = i 3 π argz = 3 Question (a) (ii) Provides correct solution Finds argument of w, or equivalent merit w = + i π arg w = 4 arg z = arg z arg w w = π 3 π 4 = = 4π 3π 7π

3 NESA 07 HSC Mathematics Etension Sample Answers Question (b) Provides correct solution Finds the slope of the asymptote, or equivalent merit y = 4 a = a = 3 b = 4 b = b tanα = = = a 3 3 π α = 6 Question (c) Provides correct sketch Sketches one correct region, or equivalent merit π argz 0 4 and z ( i) 3

4 NESA 07 HSC Mathematics Etension Sample Answers Question (d) Provides correct solution 3 Obtains correct primitive, or equivalent merit Correctly deals with dθ, or equivalent merit θ t = tan π 3 3 dt + t dθ = + cos θ + t t t + t 3 dt = 0 3 = dt 0 3 = t 0 = 3 0 = 3 4

5 NESA 07 HSC Mathematics Etension Sample Answers Question (e) Provides correct solution Finds the height of the shells in terms of y, or equivalent merit r = y h = y = (3 y) 3 = 3 y V = π 0 rh dy = π 0 y 3 3 y dy 5

6 NESA 07 HSC Mathematics Etension Sample Answers Question (f) Provides correct solution 3 Attempts to use a double angle result, or equivalent merit Obtains correct integrand in terms of θ, or equivalent merit = sin θ d = sinθ cosθ dθ π 4 sin θ d = sinθ cosθ dθ sin θ 0 0 π 4 sinθ = sinθ cosθ dθ cos θ 0 π 4 = sin θ dθ 0 π 4 = ( cosθ )dθ 0 π 4 = θ sin θ 0 π π = sin ( 0) 4 π = 4 6

7 NESA 07 HSC Mathematics Etension Sample Answers Question (a) (i) Provides correct solution e u = e ƒ = ( ) e + u = e v = e + v = e ƒ ( ) = = = (e + )e (e )e (e + ) + + e e e e (e + ) e (e + ) ƒ ( ) > 0 for all as e > 0 and (e + ) > 0 ƒ ( ) is increasing for all. 7

8 NESA 07 HSC Mathematics Etension Sample Answers Question (a) (ii) Provides correct solution e ƒ ( ) = e + = e + e ƒ ( ) e e + e e = e + e = ( ) + e = e + (e ) = e + = [ ƒ ( )] = ƒ ( ) the function is odd. Question (a) (iii) Provides correct description e lim = e + ie, ƒ ( ) (from below) 8

9 NESA 07 HSC Mathematics Etension Sample Answers Question (a) (iv) Provides correct sketch ƒ ( 0 ) = e 0 e 0 + = + = 0 (0, 0) lies on the curve. 9

10 NESA 07 HSC Mathematics Etension Sample Answers Question (a) (v) Provides correct sketch e + y = e y = ƒ ( ) 0

11 NESA 07 HSC Mathematics Etension Sample Answers Question (b) Provides correct solution 3 Correctly evaluates the discriminant, or equivalent merit Uses quadratic formula, or equivalent merit z + ( + 3i)z + ( + 3i) = 0 a = b = + 3i c = + 3i z = = ( + 3i) ± ( + 3i) 4()( + 3i) () ( + 3i) ± 4 + i + 9i 4 i ( + 3i) ± 9i = ( + 3i) ± 3i = 3i + 3i 3i 3i z = and z = 6i z = and z = z = z = 3i

12 NESA 07 HSC Mathematics Etension Sample Answers Question (c) Provides correct primitive 3 Correctly uses parts, or equivalent merit Attempts to use parts, or equivalent merit I = tan d Using integration by parts, u = tan dv = d du = d v = + I = tan d + = tan d + + = tan d + = tan d + = tan tan + c tan = tan + + c = ( tan + tan ) + c

13 NESA 07 HSC Mathematics Etension Sample Answers Question (d) (i) Provides correct solution Recognises that P( )= ( α ) q( ), or equivalent merit Let P( ) = ( α ) Q( ) P ( ) = ( α ) Q( )+ ( α ) Q ( ) = ( α )[ Q( )+ ( α ) Q ( )] P( α ) = (α α ) Q( α ) = 0 P ( α ) = (α α )[ Q( α ) + (α α ) Q ( α )] = 0 Question (d) (ii) Provides correct solution Attempt to solve P ( )= 0, or equivalent merit P( ) = P ( ) = = (4 9 + ) = (4 )( ) Roots of P ( ) are = 0,, 4 P( 0 ) = 4 P 4 P( ) = 0 = ( ) is a factor of P() α = 3

14 NESA 07 HSC Mathematics Etension Sample Answers Question 3 (a) Provides correct solution As ( r s ) 0 r rs + s 0 r + s rs Question 3 (b) (i) Provides correct proof Attempts to use sum of the threewise products of roots, or equivalent merit a = sum of roots = α + + β + α β c = sum of products of 3 roots OR = α β + α + αβ + β α α β β α β = β + + α + β α = a a = c The equation whose roots are reciprocals of the roots is found by replacing by. + a b c d ie + a + b + c 3 + d 4 Since the reciprocals give the same set of roots we can match the coefficients so that d = and c = a. 4

15 NESA 07 HSC Mathematics Etension Sample Answers Question 3 (b) (ii) Provides correct solution Uses sum of the roots in pairs, or equivalent merit b = sum of pairs of roots = α + αβ + α + β + + β α β α α β β α β = + αβ β α αβ α β = αβ + β β α α α β α + β α β Now + = from 3 part (a) β α αβ αβ α β + and similarly αβ + β α αβ b + + = 6 5

16 NESA 07 HSC Mathematics Etension Sample Answers Question 3 (c) Provides correct solution 4 Obtains correct epression for in terms of v, k and g, or equivalent merit 3 Separates the variables, or equivalent merit Rewrites acceleration as v dv, or equivalent merit d dv = v = g kv d dv g + kv = d v d v = dv g + kv v = dv g + kv = log (g + kv + c k e ) g g when = 0 v = so that v = k 4k hence 0 = log g kg e + + c k 4k c = log 5g e k 4 5g = log e log g + kv k Maimum height occurs when v = 0 k 4 e ( ) 5g ( ) H= log log g k e 4 e 5 = log e k 4 6

17 NESA 07 HSC Mathematics Etension Sample Answers Question 3 (d) Provides correct solution 4 Obtains an integral for the volume, or equivalent merit 3 Finds the area of the cross-section in terms of, or equivalent merit Finds the height of PQR in terms of, or equivalent merit 3 The line BC has equation y = y = + QP = + = + To find height of QRP d = 4 height = + d = + 4 d By similar triangles = 4 7

18 Area QRP = QP + 4 = = Volume = d = = = units 3 3 8

19 NESA 07 HSC Mathematics Etension Marking Guidelines Question 3 (e) Provides correct solution Obtains c d = i( d a), or equivalent merit c d = i(d a) π (rotation by ) c = d + id ia = ( + i)d ia 9

20 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (a) (i) Provides correct values Multiply both sides by = ( A + ) ( + ) + ( B ) ( + + ) at = 0 6 = A + B A + B = 8 at = 6 = (A + ) + (B )5 6 = A + + 5B 0 A + 5B = 4 Subtract 4B = 6 B = 4 A = 4 0

21 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (a) (ii) Provides correct solution Obtains correct primitive, or equivalent merit m m d = + d m m + = + d + d m m m ( ) = m d + ln d ln( + ) + ( + ) + 0 ( ) m m = tan ( + ) + ln( + + ) log + tan m m ( ) ln(m m + ) + log 0 0 m π m + m + π = tan (m + ) + ln + tan ( m ) + m m + m + m + = ln tan m tan m m m + + ( + ) + ( )

22 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (a) (iii) Provides correct answer m m d =ln + tan (m + ) + tan (m ) m m m 0 as m π π ln ( ) + + =π

23 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (b) (i) Provides correct solution Construct DB EAD = EBD, as they stand on the same arc, ED of C. Question 4 (b) (ii) Provides correct solution Construct AB EDA = EBA angles standing on arc EA of C. EBA = CBA = CFA angles standing on arc AC in other circle C. so EDA = EBA = CFA = AFC 3

24 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (b) (iii) Provides correct solution 3 Makes substantial progress, or equivalent merit Attempts to find an eterior angle, or equivalent merit Let AFC = EDA = α EAD = EBD = β In EAD, AED = π α β In EFG, EGF = π (α + (π α β )) = β so EGF = EBD =β Eterior angle (EGF) equals opposite interior angle (CBD) and so BCGD is a cyclic quadrilateral. 4

25 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (c) (i) Provides correct solution Obtains one correct component of R, or equivalent merit R sinθ = mg R cosθ = mrw Dividing g tanθ = rw r g = h rw gh w = r 5

26 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (c) (ii) Provides correct solution Correctly resolves N, or equivalent merit Horizontal: T sinθ N cosθ = mrw T sinθ cosθ N cos θ = mrw cosθ Vertical: T cosθ + N sinθ = mg T sinθ cosθ + N sin θ = mg sinθ N (sin θ + cos θ ) = mg sinθ mrw cosθ 6

27 NESA 07 HSC Mathematics Etension Marking Guidelines Question 4 (c) (iii) Provides correct solution Observes N 0, or equivalent merit As particle moves in a circle of radius r it must have N 0 h mg sinθ cosθ 0 r sinθ cotθ cosθ 0 sinθ cotθ cosθ tanθ cotθ In first quad so π θ 4 7

28 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (a) (i) Provides correct answer I = d 0 = d 0 = ( ) 3 0 = (0 ) 3 = 3 3 8

29 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (a) (ii) Provides correct solution 3 Correctly uses integration by parts, simplifying where possible, or equivalent merit Attempts to use integration by parts, or equivalent merit I n = n d 0 Let u = n dv = d du = (n ) n d v = ( 3 ) 3 3 n I = ( ) + ( ) (n n n ) d ( n ) d 3 0 = n ( ) ( n ) = n n d n d n n I = I I 3 3 n n In + = I n n I n = I 3 3 n n n n I n = I n + n n n 9

30 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (a) (iii) Provides correct answer 4 I 5 = I = I = from part (i) = 05 30

31 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (b) (i) Provides correct solution Differentiates with respect to, or equivalent merit + y = a (a constant) Differentiating implicitly: dy + y = 0 d dy = d y y = y = At P, gradient of tangent = and equation of tangent is d c d y d = ( c) c c y c d = d + d c y c + d = d c + c d 3

32 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (b) (ii) Provides correct solution 3 Obtains an epression for OA + OB in terms of c and d, or equivalent merit Finds an epression for OA, or equivalent merit At A, y = 0 d = d c + c d = OA = At B, = 0 d d c + c d d c + c d d y c = d c + c d OB = d c + c d c d c + c d d c + c d OA + OB = + d c = d c + c + d + c d = c ( d + c ) + d ( d + c ) = ( c + d )( d + c ) But P lies on + y = a c + d = a OA + OB = a a = a 3

33 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (c) (i) Provides correct solution Attempts to solve the simultaneous equations, or equivalent merit y y At P, + = and = a b c d y y + = a b c d y = a c d + b c a b + d y a c = b d b + d a c = y b d c a a c ( b + d ) = ( a c ) b d 33

34 NESA 07 HSC Mathematics Etension Marking Guidelines Question 5 (c) (ii) Provides correct solution 3 Finds product of the two slopes, or equivalent merit Finds slope of tangent to ellipse, or equivalent merit Ellipse: y + = a b y dy + = 0 a b d dy b = d a y b Gradient of tangent at P = a y y Hyperbola: = c d d Similarly gradient of tangent at P = c y b d Product of gradients = a y c y b d = y a c a c (b + d ) b d from part (i) a c b d a c = ( ) = (b + d a c But, conics have some foci so ae = ce and b = a ( e ) and d = c (E ) b = a a e d = c E c b + d = a a e + c E c = a a e + a e c = a c ) 34

35 NESA 07 HSC Mathematics Etension Marking Guidelines Product of gradients = = (b + d = a c a c ) ( ) a c 35

36 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (a) (i) Provides correct solution α = cosθ + isinθ By de Moivre s theorem, α k = cos(kθ ) + isin(kθ ) α k = cos ( kθ ) + isin ( kθ ) = cos(kθ ) isin(kθ ) α k + α k = cos kθ Question 6 (a) (ii) Provides correct solution 3 Sums the series, multiplies top and bottom by epand, or equivalent merit ( α ) and attempts to Recognises that there are n + terms, or equivalent merit There are (n + ) terms in the geometric series with first term α n and ratio α α n ( α n+ ) C = ( α ) ( ) α α n α n+ = α α α n α (n+) α (n+) + α n =, using α = ( α )( α ) α ( ( ) (α n + α n ) (α n+) + α n+ ) = ( α )( α ) 36

37 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (a) (iii) Provides correct solution Correctly matches appropriate terms and attempts to use the result from part (i), or equivalent merit Write C as C = + (α + α ) + (α + α ) + + (α n + α n ) Also, C = = + cosθ + cosθ + + cos(nθ ) by part (i) = + ( cosθ + cosθ + + cos(nθ )) = = = α n + α n (α n+ + α (n+) ) ( α )( α ) cosnθ cos (n + )θ (α + α ) cosnθ cos (n + )θ cosθ cosnθ cos (n + )θ cosθ 37

38 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (a) (iv) Provides correct solution π Put θ =, then n nπ cos cos (n + )π π π nπ n n + cos + cos + + cos = n n n π cos n RHS = = π cos π + n π cos n π + cos n π cos n = π π nπ LHS = + cos + cos + + cos n n n = π π nπ cos + cos + + cos = n n n Which is independent of n. 38

39 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (b) Provides correct solution Finds one correct value of a, or equivalent merit Since the eccentricity is the conic is a hyperbola The focus F is (ae, 0) = (a, 0) VF =, so a a = a = FV = a + a = a = 3 39

40 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (c) (i) Provides correct solution Obtains correct count for one case, or equivalent merit A B C D Case : Tiles B and C have the same colour. There are ( ) ways of choosing tile D. Case : Tiles B and C have different colours. There are ( ) choices of colour for tile C and there are ( ) choices of colour for tile D giving ( ) choices in total. There are ( ) + ( ) = choices in total. Question 6 (c) (ii) Provides correct proof Demonstrates true for n = Let P(n) be the given proposition. P() is true since this was given. Let k be an integer for which P(k) is true. That is, there are ) 3 + 3) k ( ( choices for the colours. Consider a (k + ) grid. A B C D k As in part (i) there are ( 3 + 3) choices for the colours of tiles C and D. There are ( )( 3 + 3) k choices for the colours. P(k + ) is true, so P(n) is true for all n by induction. 40

41 NESA 07 HSC Mathematics Etension Marking Guidelines Question 6 (c) (iii) Provides correct solution Finds the number of ways if not all colours need be used, or equivalent merit Let G ( )= ( )( ) 4 The number of ways to colour the grid with 3 colours, each used at least once, is G(3) 3 G() (there are 3 choices for which colour is left out) = = 480 4

42 NESA 07 HSC Mathematics Etension Marking Guidelines 07 HSC Mathematics Etension Mapping Grid Section I Question Content Syllabus outcomes.4 E3 3.4 E3 3. E3 4 E E E3 7.9 E6 8 E E5 0 8 E6 Section II Question Content Syllabus outcomes (a) (i). E3 (a) (ii). E3 (b) 3. E3 (c).5 E3 (d) 3 4. E8 (e) 5. E7 (f) 3 4. E8 (a) (i) 8 E6 (a) (ii) 8 E6 (a) (iii) 8 E6 (a) (iv) 8 E6 (a) (v).5 E6 (b) 3. E3 (c) 3 4. E8 (d) (i) 7. E4 (d) (ii) 7. E4 3 (a) 8.3 E 3 (b) (i) 7.5 E4 3 (b) (ii) 8.3 E4 4

43 NESA 07 HSC Mathematics Etension Marking Guidelines Question Content Syllabus outcomes 3 (c) E5 3 (d) 4 5. E7 3 (e).3 E3 4 (a) (i) 7.6 E8 4 (a) (ii) 4. E8 4 (a) (iii) 4. E9 4 (b) (i) 8.(.9) E 4 (b) (ii) 8.(.9) E 4 (b) (iii) 3 8.(.9) E 4 (c) (i) E5 4 (c) (ii) E5 4 (c) (iii) 6.3.3, E5 5 (a) (i) 4. E8 5 (a) (ii) 3 4. E8 5 (a) (iii) 4. E8 5 (b) (i).8 E6 5 (b) (ii) 3 8 E6 5 (c) (i) 3., 3. E4 5 (c) (ii) 3 3., 3. E4 6 (a) (i).4 E3 6 (a) (ii) 3., 8 E3 6 (a) (iii)., 8 E3 6 (a) (iv) 8 E3 6 (b) 3. E3 6 (c) (i) 8 E 6 (c) (ii) 8. E 6 (c) (iii) 8 E 43