2014 HSC Mathematics Extension 2 Marking Guidelines

Transcription

1 04 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer D A 3 B 4 C 5 C 6 D 7 B 8 B 9 A 0 D

2 BOSTES 04 HSC Mathematics Extension Marking Guidelines Section II Question (a) (i) Provides a correct solution Finds modulus or argument, or equivalent merit z + w ( i ) i i z + w, arg (z + w) 4 so z + w cos + isin 4 4 Question (a) (ii) Provides a correct solution Identifies conjugate, or equivalent merit z i 3 i w 3 + i 3 i 8 4i 0 4 i 5 4 i 5 5

3 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (b) Provides a correct solution 3 Correctly uses limits in one of two terms, or equivalent merit Attempts integration by parts, or equivalent merit ) sin x cos x 0 ( 3x ) cos xdx ( 3x sin xdx 0 3 3

4 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (c) Draws a correct sketch 3 Correctly shows region satisfying z z, or equivalent merit Correctly shows region satisfying 4 arg ( z ) 4, or equivalent merit 4

5 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (d) Draws correct sketch Sketches a graph with correct x-intercepts, or equivalent merit 5

6 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (e) Provides a correct solution 3 Attempts to evaluate the integral for the volume, or equivalent merit Applies a formula for the volume by cylindrical shells, or equivalent merit V 6 0 xy dy 6 y(6 y) ydy (6y y ) dy 6 3 y y4 4 0 ( ) Volume 6 units 3 6

7 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (a) (i) Draws a correct sketch Sketches graph symmetric with respect to y-axis or equivalent merit Question (a) (ii) Draws a correct sketch Shows asymptote at x, or y or equivalent merit 7

8 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (b) (i) Provides a correct solution Substituting x cos 8cos 3 6cos 3 3 so 4 cos 3 3cos 3 cos 3 Question (b) (ii) Provides a correct solution Finds one real solution, or equivalent merit 3 cos 3 so 3,, ,, Solutions are 3 x cos, cos, cos

9 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (c) Provides a correct solution 3 Finds the slope of one curve, and attempts to find slope of other curve, or equivalent merit Attempts to find slope at (x 0, y 0 ) of one curve using implicit differentiation, or equivalent merit x y 5 Differentiating with respect to x: dy x y 0 dx dy dx x y at point (x 0, y 0 ) slope is m x 0 y 0 xy 6 (x, y 0) Differentiating with respect to x: dy y + x 0 dx dy y dx x at point (x 0, y 0 ) slope is m y 0 x 0 If curves meet then x 0, y 0 0 x 0 y 0 and m m y 0 x 0 Then tangents are perpendicular. 9

10 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (d) (i) Provides a correct solution I 0 x + dx 0 4 tan x 0 Question (d) (ii) x n x n I n + I n + dx x + x + (x + ) 0 x n dx Provides a correct solution Writes the sum as a single integral, or equivalent merit 0 0 n x (x + ) n x n 0 dx ( 0) n n 0

11 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question (d) (iii) Provides a correct answer Attempts to apply the recursion relation from part (ii), or equivalent merit x 4 dx I x + 0 Now, I + I 3 I + I 0 Subtracting: I I I I

12 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (a) Provides a correct solution 3 Finds a correct primitive, or equivalent merit Attempts to obtain an integral in terms of t, or equivalent merit x t t t tan, sin x + t, cos x + t dt x + x sec tan dx ( + t ) dt dx + t when x, t and when x, t 3 3

13 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (a) Sample answer continued: I t + t dt 6t 4 + 4t t 3 dt 9t + 6t + 3 dt (3t + ) 3 3 3t + 3 t t ( + t ) dt

14 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (b) Provides a correct solution 4 Finds an expression for the volume of the solid 3 Finds the area of each cross-section, or equivalent merit Finds the height of each trapezium, or equivalent merit h x sin 3 x 3 Area of cross-section 3 x ( x + x ) 3 4 x 3x Volume of the slice x 4 x 4 x 3 3 Volume of the solid x 4 dx x ( 5 ) unit 3 5 4

15 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (c) (i) Provides a correct solution a b M t + t, t t Substituting for x and y, x y t + t a b 4 t 4 t t + + t t + t + 4 t t t t t 4 t So M lies on the hyperbola. 5

16 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (c) (ii) Provides a correct proof 3 Finds the slope of the tangent, or equivalent merit Finds slope of PQ, or attempts to find the slope of the tangent, or equivalent merit Slope of PQ is bt + b t at a t b t + t a t t Differentiating at M x a y with respect to x: b x a y b dy dt 0 dy xb dt ya dy dt a t + t b b t t a b t + t a t t slope of PQ so PQ is a tangent to hyperbola at M 6

17 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (c) (iii) Provides a correct solution Finds OP OQ in terms of a and b, or equivalent merit OP ( a + b ) t OQ a + b t OP OQ a + b OS a e b a + a a + b OP OQ 7

18 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 3 (c) (iv) Provides a correct solution Finds one coordinate of M in terms of a, b and e, or equivalent merit b Equations of asymptotes are y ± x a If at ae then t e. The coords of M are then slope of MS is a e + e, b e e b e e a e + e ae b e e ae + a e ae b e e a e e b a which is the slope of one of the asymptotes and so MS is parallel to it. 8

19 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 4 (a) (i) Provides a correct solution Shows that P () 0, or equivalent merit P(x) 5x 4 0x + 5 P(x) 0x 3 0 P() P() P() x is a root of multiplicity 3 Question 4 (a) (ii) Provides a correct solution Finds product and sum of remaining two roots, or equivalent merit Let other roots be, , ()()() Hence, and are the solutions to x + 3x x 3 ± 9 4()(6) 3 ± 5 3 ± 5i The complex roots are 3 + 5i, 3 5i 9

20 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 4 (b) (i) Provides a correct solution 3 Uses the slopes of OP and the normal in a correct expression for tan, or equivalent merit Finds the slope of the tangent at P, or equivalent merit x a y + b Differentiating with respect to x: x a y dy + 0 dx b dy xb dx ya slope of tangent at P is slope of normal at P is slope of OP is bcos asin asin m bcos bsin m acos m m Hence, tan + m m a sin b sin b cos a cos sin + cos sin a b cos b a cos + sin cos a b sin cos ab cos,a > b cos + sin b a sin cos ab 0

21 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 4 (b) (ii) a b sin tan ab Provides a correct solution Uses double angle formula, or equivalent merit as sin x has maximum at x tan has maximum at is a maximum when 4 Question 4 (c) (i) Provides a correct solution Finds mx F kv, or equivalent merit From Newton s second law mx F kv The terminal velocity (v 300) occurs when x 0, so ) F F k ( 300, or k ) ( 300 Hence mx F ( 300 F v ) v F 300

22 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 4 (c) (ii) F dt m Provides a correct solution 4 Obtains an expression for t in terms of v, or equivalent merit 3 Attempts to use partial fractions to find t in terms of v, or equivalent merit Obtains an equation relating t and v involving integrals, or equivalent merit m dv dt F v 300 dv v 300 F m t v v 300 dv + dv v v v 0 when t 0 c 0 v ln ln v c Thus t v + m ln F v m v ln F 300 v 50m Put v 00, t ln 5 F

23 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (a) Provides a correct solution Expands ( a + b + c) and attempts to use the relative sizes of a, b or c, or equivalent merit a + b + c, a b c So (a + b + c) LHS a + b + c + (ab + bc + ac) a + b + c + ( a + b + a ) (since b a,c b,c a) LHS 5a + 3b + c 5a + 3b + c 3

24 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (b) (i) Provides a correct solution Writes + i or i in modulus argument form, or equivalent merit Write ( + i) and ( i) in modulus-argument form: i i + i + and i cos + isin cos isin cos + isin 4 4 Using de Moivre s theorem ( + i) n cos + isin 4 4 n n ( ) n cos + isin 4 4 and ( i) n cos + isin 4 4 ) n n n ( cos + isin 4 4 n n n cos isin 4 4 adding ( + i ) n + ( i ) n ( ) n cos n 4 n n 4

25 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (b) (ii) n n n n Provides a correct solution 3 Attempts to add two binomial expressions, or equivalent merit Applies the binomial theorem to ( + i) n or ( i) n, or equivalent merit Using the binomial theorem n n ( + i ) n k and k0 i k n n n n n n i i n n n ( k i ) n ( ) k k0 i k n n n n n n i + i n The last terms are positive since n is divisible by 4 Adding the identities ( + i) n + ( i) n Combining this with the expressions from (i) for ( + i) n + ( i) n n n n ( ) n cos n n 4 n n () 4 n since cos 4 n for odd multiples of 4 odd for even multiples of n even 4 5

26 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (c) (i) Provides a correct solution 3 Resolves forces in both directions and eliminates r, or equivalent merit Resolves forces in one direction, or equivalent merit kv r Tcosφ φ Tsin φ T l mg Vertical kv T sin + mg T sin kv mg Horizontal T cos mv r mv cos T cos mv T sin kv mg T cos mv sin k g ie cos m v 6

27 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (c) (ii) Provides a correct solution Obtains a quadratic inequality in sin and attempts to solve it, or equivalent merit cos > 0 so k sin < cos m m,,k all positive k ( sin ) m m sin + sin < 0 k m sin < sin k m m + k k + 4 sin < as < m + k k m + 4 k sin < m + m + 4 k k 7

28 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 5 (c) (iii) Provides a correct solution Differentiates the given expression with respect to, or equivalent merit sin F ( ) cos F ( ) cos 3 + sin cos cos 4 > 0 as, for < <, cos 3 > 0, cos > 0, sin > 0 and cos 4 > 0 F is increasing. Question 5 (c) (iv) Provides a correct explanation As v increases g k decreases, so v m g v increases. That is, By (iii) sin increases. cos sin is an increasing function and so increases as v increases. cos 8

29 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (a) Question 6 (a) (i) Provides a correct solution Uses angles on alternate segment, or equivalent merit APX ADP (The between a tangent and a chord equals the angle in the alternate segment). ADP DPQ (Alternate s, AD PQ ) APX DPQ 9

30 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (a) (ii) Provides a correct solution 3 Makes significant progress towards the solution Observes result similar to that in part (i) in the other circle, or uses angle in semicircle, or equivalent merit Using a similar argument to that in part (i), it can be shown that BPY BCP CPR APD 90 (Angle in a semi-circle) Similarly BPC 90 XPQ YPR (Vertically opposite s) APX + APD + DPQ BPY + BPC + CPR.APX CPR APX CPR APC is a straight line (equal vertically opposite s) A, P, C are collinear. 30

31 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (a) (iii) Provides a correct solution From (i) and (ii) APX CPR implies that ADP BCP A, B, C and D are concyclic points (equal s subtended at C and D by interval AB on the same side) ABCD is a cyclic quadrilateral. Question 6 (b) (i) Provides a correct solution 3 Correctly sums and simplifies the middle term, or equivalent merit Applies formula for geometric series, or equivalent merit n ( n) + x ( ( ) ) x x x + x ( x ) n + ( x ) n, using sum of geometric series. + x + x + x n ( x ) + x ) n Since + x for all x, x n ( x + x x n We have n 4 n ( ) x ( x + x + ( ) x n ) x + x n 3

32 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (b) (ii) x ± x n dx ± ± n + n + Provides correct solution Integrates some terms in the inequality in (i), or equivalent merit Inequalities are preserved by integration: 0 n+ 0 dx tan x x n x x + ( ) n x dx n x n x ( ) x x 3 5 x 7 n 0 n ( ) n n so ( ) n n n + 3

33 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (b) (iii) Provides a correct explanation Since lim lim 0, x n + x n + then as n n ( ) n

34 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question 6 (c) Provides a correct solution 3 Makes substantial progress Does substitution u + lnx or u lnx, or equivalent merit I ln x ( + ln x ) dx dt Put t ln x, then dx x e t te t I ( + t ) dt ( + t)e t e t dt ) dt ( + t) ( + t t e e t e t dt + dt + t + t + t, by parts, t e + c + t x + c + ln x 34

35 BOSTES 04 HSC Mathematics Extension Marking Guidelines Mathematics Extension 04 HSC Examination Mapping Grid Section I Question Content Syllabus outcomes 7.6 E4 7.4 E E4 4.,.4 E3 5.7 E E E8 8.3 E3 9 8 HE3, E9 0 8 E8, HE6 Section II Question Content Syllabus outcomes (a) (i). E3 (a) (ii). E3 (b) 3 4. E8 (c) 3.5 E3 (d).,. E6 (e) 3 5. E7 (a) (i).3 E6 (a) (ii).5 E6 (b) (i) 8.0 E (b) (ii) 8.0 E (c) 3.8 E4, E9 (d) (i) 4. E8 (d) (ii) 4. E8 (d) (iii) 4. E8 3 (a) 3 4. E8 3 (b) 4 5. E7 3 (c) (i) 3. E4 3 (c) (ii) 3 3. E4, E9 3 (c) (iii) 3. E4 3 (c) (iv) 3. E4 35

36 BOSTES 04 HSC Mathematics Extension Marking Guidelines Question Content Syllabus outcomes 4 (a) (i) 7. E3 4 (a) (ii).,7. E3 4 (b) (i) 3 3. E3, E9 4 (b) (ii) 8.0 E 4 (c) (i) 6.. E5 4 (c) (ii) E5 5 (a) 8.3 E 5 (b) (i).4, 8.0 E, E3 5 (b) (ii) 3.4, 8.0 E 5 (c) (i) E5 5 (c) (ii) 8.3 E 5 (c) (iii) 8.0 E 5 (c) (iv) 6.3.3, 8.0 E5 6 (a) (i) 8. PE3, E, E9 6 (a) (ii) 3 8. PE3, E, E9 6 (a) (iii) 8. PE3, E, E9 6 (b) (i) E 6 (b) (ii) 8.0 E 6 (b) (iii) 8.0 E 6 (c) 3 4. E8 36