2017 HSC Mathematics Extension 1 Marking Guidelines

Transcription

1 07 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer A B 3 B 4 C 5 D 6 D 7 A 8 C 9 C 0 B

2 NESA 07 HSC Mathematics Extension Marking Guidelines Section II Question (a) Provides correct answer x value of P = + 3 = 0 5 = Question (b) Provides correct solution Attempts to use the chain rule, or equivalent merit Let y = tan ( x 3 ) y = + x 3 ( ) 3x 3x = + x 6

3 NESA 07 HSC Mathematics Extension Marking Guidelines Question (c) Provides correct solution 3 Correctly identifies and as important, or equivalent merit Attempts to deal with the denominator, or equivalent merit x > x + Multiply both sides by (x + ) x( x + ) > (x + ) x + x > x + x + x > x > or x < Question (d) Provides correct sketch Indicates correct range, or equivalent merit 3

4 NESA 07 HSC Mathematics Extension Marking Guidelines Question (e) Provides correct solution 3 Provides a correct primitive, or equivalent merit Attempts to use given substitution, or equivalent merit 3 x x = u dx 0 x + dx = u du dx = udu when x = 0 u = 0 u = when x = 3 u = 3 u = 3 x u dx =.udu x u 0 + = (u ) du u 3 = u 3 8 = = 3 Question (f) Provides correct primitive sin x cos xdx sin 3 x = + c 3 4

5 NESA 07 HSC Mathematics Extension Marking Guidelines Question (g) (i) Provides correct answer Question (g) (ii) Provides correct answer Question (g) (iii) Provides correct answer

6 NESA 07 HSC Mathematics Extension Marking Guidelines Question (a) Provides correct solution Attempts to relate two angles in the diagram, or equivalent merit Reflex AOC = (angle of revolution) = 60 ABC = reflex AOC (angle at centre equals twice the angle at circumference standing in the same arc) = (60 ) =30 6

7 NESA 07 HSC Mathematics Extension Marking Guidelines Question (b) (i) Provides correct sketch 3 Provides correct sketch of y = 3 x, or equivalent merit Provides correct sketch of y = x +, or equivalent merit Question (b) (ii) Provides correct answer x when x + = 3 x the graphs have points in common. ie x + + x = 3 7

8 NESA 07 HSC Mathematics Extension Marking Guidelines Question (c) (i) Provides correct solution 3 Equates the volumes in the correct ratio and attempts to evaluate an integral, or equivalent merit Provides correct integral for the volume of one solid, or equivalent merit V = π y dx h = π x dx h x 3 = π x 3 h h 3 = π π h 3 3 h 3 = π h h V = π x dx = π x x 3 h 3 h 3 = π h π 3 3 h 3 = π h Ratio V : V = : V = V h 3 h 3 h + = h h 3 = h

9 NESA 07 HSC Mathematics Extension Marking Guidelines 3h h 3 + = 4 6h + h 3 3h 3 9h + = 0 Question (c) (ii) Provides correct answer Let f( h ) = 3h 3 9h + h = 0 f( h ) f ( 0) h = h = 0 f h ( ) ( ) f 0 f( h )= f ( 0 ) = f ( h ) = 9h 9 f ( 0)= 90 ( ) 9 = 9 h = 0 9 = 9 9

10 NESA 07 HSC Mathematics Extension Marking Guidelines Question (d) Provides correct solution 3 d v Attempts to use x =, or equivalent merit dx dt Correctly finds, or equivalent merit dx t = 4 e x dt =+ e x dx dx = e x dt v = e x v = e 4x 4 d x = v dx d = e 4x dx 8 4 = e 4x 8 x = e 4 x 0

11 NESA 07 HSC Mathematics Extension Marking Guidelines Question (e) Provides correct solution Correctly uses double angle result, or equivalent merit cosx lim x 0 x sin x = lim since cosx= sin x x 0 x sin x = lim x 0 x sin x = lim x 0 x = =

12 NESA 07 HSC Mathematics Extension Marking Guidelines Question 3 (a) Provides correct solution 3 Correctly finds the value of n, or equivalent merit Correctly uses v = n ( a x ), or equivalent merit v = n ( a x ) 4 = n a n and 3 = n a 5 n 7 = 4n + 5n n = 7 n = 3 π Period, T = n n =, n > 0 3 = π 3 = π 3

13 NESA 07 HSC Mathematics Extension Marking Guidelines Question 3 (b) (i) Provides correct solution Correctly applies binomial theorem to one binomial expression, or equivalent merit n is even n n n n n r n ( + x ) = + x + x + + x + + x 0 r n n n n n n r n ( x ) = x + x + + ( ) x r n x n n n n ( x ) + ( x) n = + x n xn (odd terms have different sign and so cancel) n n n n n n ( + x) + ( x ) = + x + x 4 x n n n n Question 3 (b) (ii) Provides correct solution Differentiating with respect to x. n n n n n n( + x) n 3 n( x ) = + + n x + 4 x x 4 n ) n x) n n n n n n(( + x ( ) = x + 4 x3 + + n x 4 n 3

14 NESA 07 HSC Mathematics Extension Marking Guidelines Question 3 (b) (iii) Provides correct solution Uses x =, or equivalent merit Let x = n n n n n ( 0 ) = n 4 n n n n n n = n 4 n n n n 3 n n n n = n Question 3 (c) (i) Provides correct solution Attempts to solve y = 0, or equivalent merit Set 0 = y = Vtsinθ gt t = 0 or V sinθ = gt V sinθ t = g At this value V sinθ x = V cos θ g V = sin θ g 4

15 NESA 07 HSC Mathematics Extension Marking Guidelines Question 3 (c) (ii) Provides correct solution Range = V g sin θ 00gsin θ If v < 00g, then range g ie range < 00sinθ since sin θ then range < 00 Question 3 (c) (iii) Provides correct solution Provides an inequality in sin θ, or equivalent merit 00gsinθ If V = 00g, then range = 00 g ie 00sinθ 00 sinθ π Now, 0 θ or 0 θ π π 5π θ 6 6 π 5π θ 5

16 NESA 07 HSC Mathematics Extension Marking Guidelines Question 3 (c) (iv) Provides correct solution Obtains correct expression for the maximum height, or equivalent merit Maximum height occurs when dy 0 = = V sinθ gt dt V t = sinθ g At this time V V y = sin θ g sin θ g g V sin θ = g V = 00sin θ = 00 g π For 0 θ sin θ is an increasing function, 5π so maximum occurs when θ = 5π greatest height is 00sin 6

17 NESA 07 HSC Mathematics Extension Marking Guidelines Question 4 (a) Provides correct solution 3 Correctly shows Pn ( ) P( n+ ), or equivalent merit Establishes result for n =, or equivalent merit Let P(n) be the given proposition. P() is true since = 58 = 7 74 which is divisible by 7. Let k be an integer for which P(k) is true. That is 8 k+ + 6 k = 7m, for some integer m. Consider 8 (k+) (k+) = 8 8 k k = 8 8 k k 8 6 k k = 8 (8 k+ + 6 k ) + (6 8 )6 k = 64 7m 8 6 k = 7( 64m 4 6 k ) which is divisible by 7, since m and k are integers and 64m 4 6 k is an integer. P(k + ) is also true P(n) is true for all n by induction. 7

18 NESA 07 HSC Mathematics Extension Marking Guidelines Question 4 (b) (i) Provides correct solution Obtains equation of tangent at P, or equivalent merit The equation of the tangent at P is y = px p substituting into x = 4ay x = px p 4a x + 4apx 4ap = 0 Question 4 (b) (ii) Provides correct coordinates Correctly verifies one coordinate The sum of the roots of the equation is 4ap. x-coordinate of M is average of the roots x-coordinate of M is ap. Substituting into the tangent, y = p ap p = p (a + ) M is the point ( ap, p (a + )) 8

19 NESA 07 HSC Mathematics Extension Marking Guidelines Question 4 (b) (iii) Provides correct solution Eliminates the parameter, or equivalent merit Let x = ap, y = p (a + ) x y = a + a (a + ) = x 4a so x 4a = y a + a Hence = a + ie a a = 0 a = ± a = +, since a > 0 Question 4 (c) (i) Provides correct solution Correctly applies the product rule, or equivalent merit d (Ft () e 0.4t = t ) F ( t)e F()e t 0.4t dt = e 0.4t (50e 0.5t 0.4F()) t + 0.4F()e t 0.4t = 50 e 0.t 9

20 NESA 07 HSC Mathematics Extension Marking Guidelines Question 4 (c) (ii) Provides correct solution Correctly integrates the expression given in part (i), or equivalent merit Integrating, Ft ()e 0.4t = 50e 0.t dt = 500e 0.t + c Now F()= t 0 when t = 0 c =+500 Ft ()e ( ) 0.4t 0.t = 500 e Ft ()= 500(e 0.4t e 0.5t ) 0

21 NESA 07 HSC Mathematics Extension Marking Guidelines Question 4 (c) (iii) Provides correct solution Makes substantial progress F ( t)= 500( 0.4e 0.4t + 0.5e 0.5t ) = 0 for a maximum 0.4 e 0.4t = 0.5 e 0.5t 5 = e 0.t 4 5 t = 0ln 4 (.3 hours) Alternative F ( t)= 50e 0.5t 0.4F() t = 50e 0.5t ( e 0.4t e 0.5t ) 0.4t = 50e 0.5t 00e For maximum F ( t ) = 0 50e 0.5t = 00e 0.4t 0.4t = 0.5t 5 e 4 e 0.t = e 5 0.t = ln 4 5 t = 0ln 4

22 NESA 07 HSC Mathematics Extension Marking Guidelines 07 HSC Mathematics Extension Mapping Grid Section I Question Content Syllabus outcomes 6. PE3. HE7 3.0 PE PE HE PE HE HE HE HE3 Section II Question Content Syllabus outcomes (a) 6.7 PE6 (b) 8.8, 5.5 HE4 (c) 3.4 PE3 (d) 5.3 HE4 (e) 3.5 HE6 (f) 3.6E HE6 (g) (i) 8. HE3 (g) (ii) 8. HE3 (g) (iii) 8. HE3 (a).8 PE3 (b) (i) 3.4,. PE6 (b) (ii).4,. PE (c) (i) 3.4 H8, HE7 (c) (ii) 6.4 HE7 (d) HE5 (e) 5.7, 8., 3.4E HE7 3 (a) HE3 3 (b) (i) 7., 7.3 HE7 3 (b) (ii) 7., 7.3 HE7 3 (b) (iii) 7., 7.3 HE7 3 (c) (i) 4.3 HE3 3 (c) (ii) 4.3 HE3 3 (c) (iii) 4.3 HE3

23 NESA 07 HSC Mathematics Extension Marking Guidelines Question Content Syllabus outcomes 3 (c) (iv) 4.3 HE3 4 (a) HE 4 (b) (i) 9.6 PE3, PE4 4 (b) (ii) 9.6 PE3, PE4 4 (b) (iii) 9.6 PE3, PE4 4 (c) (i) 8.8,.,.5 HE3 4 (c) (ii).5 HE3 4 (c) (iii) 0.6,.5 HE3 3