2016 HSC Mathematics Marking Guidelines

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1 06 HSC Mathematics Marking Guidelines Section I Multiple-choice Answer Key Question Answer B C 3 B 4 A 5 B 6 A 7 A 8 D 9 C 0 D

2 Section II Question (a) Provides correct sketch Identifies radius, or equivalent merit Question (b) Provides correct derivative Attempts to use quotient rule, or equivalent merit d x + (3x 4 )() (x + )( 3) dx 3x 4 (3x 4) 3x 4 3x 6 (3x 4) 0 (3x 4)

3 Question (c) Establishes that x 5, or equivalent merit x 3 3 x 3 x 5 Question (d) Correct primitive, or equivalent merit ( x + ) 3 dx 0 4 ( x + )4 8 ( 3)4 8 ()

4 Question (e) Provides correct solution 3 Obtains x x 8 0 and solves for x, or equivalent merit Attempts to eliminate x or y, or equivalent merit 5 4x 3 x x x x 8 0 (x 4)(x + ) 0 x 4 or Subst x 4 into y 5 4(4) Subst x into y 5 4( ) 3 points of intersection are (4, ) and (,3). Question (f) π Obtains sec, or equivalent merit 8 y tan x dy sec x dx π dy π When x, sec 8 dx 8 π cos 8.7 Gradient of tangent is.7. 4

5 Question (g) Obtains one correct answer, or equivalent merit x sin for 0 x π Note x 0 π x π 5π, 6 6 π 5π x, 3 3 Question (a) (i) Obtains correct slope, or equivalent merit y 4 4 x 6 y 4 3 x 4 4( y 4) 3(x ) 4y 6 3x + 6 3x + 4y 0 5

6 Question (a) (ii) Attempts to calculate perpendicular distance of a point from a line, or equivalent merit BC :3x + 4y 0 3 ()+ 40 ( ) AD Question (a) (iii) Finds distance from B to C, or equivalent merit BC ( 6) + ( 4 ) Area ABC square units 6

7 Question (b) (i) Provides correct explanation BOA x CBO x ( s opposite equal sides in ABO) (exterior of ABO equal to sum of two opposite interior s) Question (b) (ii) Attempts to use the fact that triangle OBC is isosceles, or equivalent merit OC OB (radii) BCO CBO ( s opposite equal sides in CBO) x Then COD CAO + BCO (exterior of CAO equals sum of two opposite interior s) 87 x + x 87 3x x 9 7

8 Question (c) Provides correct solution 3 Substitutes correctly into cosine rule, and finds angle or equivalent merit Identifies the complementary angle to θ, or equivalent merit cosα α 0.77 θ 69 to nearest degree. Question (d) (i) Provides correct derivative 3x y xe dy dx 3x 3xe 3x + e 3x e ( + 3x) 8

9 Question (d) (ii) Attempts to use part (i) or equivalent merit Find exact value of 0 3x e (3 + 9x)dx e 3x 3x (3 + 9x )dx 3e ( + 3x)dx x 3 xe 3x e ( + 3x)dx 0 ( ) ( 0e 0 ) 3 e 6 6e 6 9

10 Question 3 (a) (i) Provides correct solution 4 Finds the x-values at which the stationary points occur and verifies the 3 maximum turning point, or equivalent merit Finds the x-values at which the stationary points occur, or equivalent merit Attempts to solve dy dx 0, or equivalent merit 4 y 4x 3 x Find stationary points and determine nature. dy x 4x 3 dx dy Need 0 x 4x 3 0 dx when x 0 y 0 when x 3 y 7 4x (3 x) 0 when x 0or x 3. Checking the gradients for x 0, x y +ve 0 +ve horizontal point of inflexion at x 0, ie at (0, 0) x 3, x y +ve 0 ve local maximum at x 3, ie at (3, 7) 0

11 Question 3 (a) (ii) Correct solution Locates the stationary points on the sketch of the curve, or equivalent merit Question 3 (b) (i) Attempts to complete the square, or equivalent merit x 4x y + 8 x 4x + 4 y + x 4x + 4 (y + ) (x ) 43 ( )(y + ) focal length 3

12 Question 3 (b) (ii) Provides correct solution focus (,) Question 3 (c) (i) Provides correct answer A 0 Question 3 (c) (ii) Obtains 5 0e 63k, or equivalent merit kt M() t 0e M (63) 5 63k 5 0e 63k e 63k log e log e k

13 Question 3 (d) Provides correct solution 3 Finds the area under the cosine curve, or equivalent merit Finds the area under y x, or equivalent merit π Area cos x dx π sin x π π sin sin 0 π π 4 square units π or 8 π π Question 4 (a) Provides correct solution 3 Attempts to find a difference in areas or equivalent merit Applies Simpson s Rule to the existing heights or equivalent merit The increase in area can be approximated using Simpson s Rule. 3

14 Question 4 (b) (i) Obtains a correct expression for A, or equivalent merit A A A 0.65 A ( ) Question 4 (b) (ii) Provides correct solution A A 0.65 n (0.65 n n + +) n A 0.65 n n ( 0.65 n ) S n 0.35 ( 0.65 n ) 0.35 Question 4 (b) (iii) Provides correct answer ( ) A

15 Question 4 (c) (i) Provides correct solution If w is the total width then Area 70 x w 70 w x Perimeter 5 x + w 70 5x + x Question 4 (c) (ii) Provides correct solution 3 Finds length at stationary point, or equivalent merit d Finds, or equivalent merit dx Stationary points occur when dl 0 dx 70 5 x 70 5 x x 44 x (x is length so ignore ) d 440 dx x 3 d 440 at x > 0 dx 3 so minimum at x

16 Question 4 (d) Sums the series, or equivalent merit + x + x + x 3 x + x 4 a, r x x x 5 for x x x 5 lim lim + x + x + x 3 + x 4 x x x 5 5 Question 4 (e) Expresses terms involving powers of, or equivalent merit log + log4 + log8 + + log5 log + log + log log 9 log + log + 3log + 9log 9 (log + 9log ) 9 0log 45log 6

17 Question 5 (a) Provides correct solution 4 Evaluates the correct integral for the volume when C is revolved, or equivalent merit Recognises the sum of two volumes is required and makes progress with both, or equivalent merit Attempts to find a volume using integration, or equivalent merit Volume of C rotated: 4 V π 3 3 6π 3 3 Volume of C rotated: 3 0 V π y dx 3 x π 4 dx 9 0 x 3 3 4π x 7 0 4π π 6π V + 8π 3 40π 3 7

18 Question 5 (b) (i) Provides correct tree diagram, or equivalent merit 7 7 P (game ends before 4th roll) Alternative solution: P (ends after roll) 8 7 P (ends after rolls) P (ends after 3 rolls) P (ends before 4th roll)

19 Question 5 (b) (ii) Provides correct solution 3 Finds an equation or inequality of probabilities that can be solved for n, or equivalent merit Finds, as a series, the probability that the game ends before the nth roll, or equivalent merit P (ends before nth roll) n n n n 8 7 n 3 Let n (n ) log log 8 4 log n + 4 log For probability of more than, we require n 4 9

20 Question 5 (c) (i) Identifies one pair of equal angles, giving reason(s) In s FCB, BAT FCB BAT (both 90 angles in square ABCD) Now AB DC (opposite sides of a square) CFB ABT (alternate s, AB DC ) FCB BAT ( pairs of equal s). Question 5 (c) (ii) Shows that TAS and BAE are complementary, or equivalent merit TSA BAD AEB 90 TAS + BAE 90 ( sum, straight line SAE) ABE + BAE 90 ( sum ABE) TAS ABE In s TSA, AEB TAS ABE (above) TSA AEB 90 (given) TSA AEB ( pairs of equal s). 0

21 Question 5 (c) (iii) Obtains AT, or equivalent merit x TS TA (matching sides in similar s) AE AB h TA y h y.ta TA BA Also (matching sides in similar s) BC FC TA x TA x Hence h y x y h x Question 6 (a) (i) Provides correct answer When t 0 4 v 0 +

22 Question 6 (a) (ii) Finds the value of t for which v 0, or equivalent merit The particle is stationary when v 0. 4 So v 0 0 t + 4 t + 4 (t + ) 4 t + t t So particle is stationary when t. dv acceleration dt dv 4(t + ) dt 4 (t + ) dv 4 4 when t dt ( + ) 4 acceleration is ms when particle is stationary.

23 Question 6 (a) (iii) Provides correct graph Describes the behaviour of v for large t, or equivalent merit 4 As t, 0 t + v 3

24 Question 6 (a) (iv) Provides correct solution 3 7 Correctly evaluates vdt, or equivalent merit Recognises the particle changes direction at t, or equivalent merit Particle changes direction when t travelling in negative direction for 0 t <, so 7 Distance travelled vdt + vdt dt + dt t + t [t 4log(t + )] + [t 4log(t + 0 )] ( 4log 0 + 4log) + (4 4log8 + 4log) + 4log + 4 4log8 + 4log 0 + 8log 4log 3 ( 0 + 8log log) ( 0 4log) 4

25 Question 6 (b) (i) Attempts to find dy dt, or equivalent merit y 00 ( + 9e 0.5t ) dy Rate of growth dt 0.5t 0.5t 9e 00 ( + 9e ) 900 e 0.5t ( ) + 9e 0.5t Question 6 (b) (ii) Provides correct solution with justification Provides range, or equivalent merit dy All terms in are positive so y is increasing. dt 00 when t 0 y so y 0 for t 0. As t e 0.5t 0 00 so y hence 0 y < 00. 5

26 Question 6 (b) (iii) Provides correct solution y ( y) t + 9e ) + 9e 0.5t ( e 0.5t + 9e + e 0.5t ( 0.5t ) 9 900e 0.5t ( + 9e 0.5t ) dy dt Question 6 (b) (iv) Recognises the importance of the vertex of the parabola in part (iii), or equivalent merit dy y (00 y) dt y y 400 which is a quadratic in y with roots at y 0 and y 00. Since the coefficient of y is negative, the quadratic has a max at y 00. Population growing fastest when population is y 00. 6

27 06 HSC Mathematics Mapping Grid Section I Question Content Syllabus outcomes 4. P5 3. H5 3 9., 9.5 P P4 5.5, 3.5 H H H5 8., 3. H5 9.,. H8 0. H3 Section II Question Content Syllabus outcomes (a) 4.3 P5 (b) 8.9 P7 (c). P3 (d). H5 (e) 3.4, 6.3, 9. P4 (f)., 8.4, 3.5 P6, H5 (g) 3., 3. H5 (a) (i) 6. P4 (a) (ii) 6.5 P4 (a) (iii).3, 6.5 P4 (b) (i).3 P4 (b) (ii).4 P4 (c) P4 (d) (i) 8.8,.4 H3, H5 (d) (ii). H3, H5 3 (a) (i) 4 0., 0.4 H6 3 (a) (ii) 0.5 H6 3 (b) (i).3, 9.5 P5 3 (b) (ii) 9.5 P5 3 (c) (i) 4. H4 3 (c) (ii) 4. H4, H5 7

28 Question Content Syllabus outcomes 3 (d) 3.4, 3.6 H8 4 (a) 3.3 H5, H8 4 (b) (i) 7.5 H4, H5 4 (b) (ii) 7.5 H4, H5 4 (b) (iii) 7.5 H4, H5 4 (c) (i) 0.6 H4, H5 4 (c) (ii) H4, H5 4 (d) 7., 8. P8, H5 4 (e) 7.,. H3, H5 5 (a) 4.4 H8 5 (b) (i) 3.3 H5 5 (b) (ii) 3 3.3, 7.,. H3, H5 5 (c) (i).3,.5 H5 5 (c) (ii).3,.5 H5 5 (c) (iii).3,.5 H5 6 (a) (i) 4.3 H4 6 (a) (ii) 4.3 H4, H5 6 (a) (iii) 4., 0.5 H4, H5, H6 6 (a) (iv) 3.5, 4.3 H4, H5 6 (b) (i).5, 4. H5 6 (b) (ii) 4.,.5 H5 6 (b) (iii).3, 4. H3 6 (b) (iv) 9.,.5, 4. H5, H7 8

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