2013 HSC Mathematics Extension 1 Marking Guidelines

Transcription

1 03 HSC Mathematics Extension Marking Guidelines Section I Multiple-choice Answer Key Question Answer C D 3 C 4 D 5 A 6 B 7 A 8 D 9 B 0 C

2 03 HSC Mathematics Extension Marking Guidelines Section II Question (a) Correct answer d For any cubic αβγ =, so a 7 αβγ =. Question (b) Correct primitive Attempts to use standard integral, or equivalent merit 49 4x dx = 49 4 x 4 dx = dx 7 x x = sin + c 7 = sin x + c 7

3 03 HSC Mathematics Extension Marking Guidelines Question (c) Correct answer Identifies correct binomial coefficient, or equivalent merit The probability is ! 3 3 C 7 = 4 4 7!3! = = For each question, the probability of choosing the correct option is and the 4 3 probability of choosing an incorrect option is. 4 The probability of choosing the correct option for exactly 7 out of 0 questions is the term in + containing Question (d) (i) Correct solution Finds Domain: all real x ± Using the quotient rule ƒ ( x) = ( 4 x ) x ( x ) ( 4 x ) = 4 x + x ) ( 4 x = 4 + x ( 4 x ) ƒ ( x ) > 0 since 4 + x > 0 for all x, and ( 4 x ) > 0 for all x ± Hence ƒ ( x) > 0 for all x ±. 3

4 03 HSC Mathematics Extension Marking Guidelines Question (d) (ii) Correct graph Finds both vertical asymptotes, or equivalent merit y 0 x y = ƒ ( x) Question (e) Correct answer x x sin sin lim = lim x 0 3x x 0 6 x x sin = lim 6 x 0 x sin t = since lim =. 6 t 0 t 4

5 03 HSC Mathematics Extension Marking Guidelines Question (f) Correct solution 3 Finds a primitive of the form of a tan u Attempts to use given substitution du 3x 3x 6x u = e, = 3e, u = e dx If x = 0, then u = e 0 = If x = 3 Hence 3 3 e 3x dx 0 e 6x +, then u = e 3 = e e = du 3 u + e = tan u 3 = 3 ( tan e tan ) π = tan e 3 4. Question (g) Correct solution Uses correct derivative of, or equivalent merit Using the product rule ƒ x ( ) = x 5 + x sin 5x 5x 5x = + x sin 5x. 5x 5

6 03 HSC Mathematics Extension Marking Guidelines Question (a) (i) Correct answer We want 3 cos x sin x = cos ( x + α ) 3 cos x sin x = cos ( x + α ) = cos x cosα sin x sinα, 3 So cosα = and sinα =. Hence tanα = 3. π π π α = and 0 < < 6 6 Question (a) (ii) Correct solution Obtains an equation of the form, or equivalent merit Rearrange the equation and use part (i): π 3 cos x sin x =, ie cos x + = 6 π cos x + = 6 π π π 5π x + = or x + = Hence the solutions satisfying 0 < x < π are π π π x = = π π 3π and x = =

7 03 HSC Mathematics Extension Marking Guidelines Question (b) Correct solution 3 Finds correct primitive, or equivalent merit Finds correct integrand, or equivalent merit y O 3π x Volume of solid is given by V = π y dx 3π x V = π 3 sin dx 0 3π = 9π sin x dx 0 3π V = 9π 0 ( cos x ) dx 3π 9π = ( x sin x ) 0 9π 3π 3π = sin ( 0 sin 0 ) 9π 3π = + 9π 3π Note: cos x = cos x sin x = sin x sin x = ( cos x ) The volume is + units 3. 7

8 03 HSC Mathematics Extension Marking Guidelines Question (c) Correct solution 3 Finds values for A, B and k, or equivalent merit Finds value for A, or equivalent merit As t, T approaches the temperature of the room, and so A =. When t = 0, T = + B Hence = 80 B = 58. kt T = + 58e To find k : After 0 minutes the temperature is 60 (t = 0, T = 60) k 0 60 = + 58e 0 k 38 = 58e 38 log = 0 k, e so k = log e 0 58 To determine time to drop to 40 C: 40 = + 58e 38 8 log t 0 e log t 0 e 58 = e log = log t e 58 0 e 58 8 log e 58 t = 0, 38 log e 58 t = temperature drops to 40 after 8 minutes (to the nearest minute). 8

9 03 HSC Mathematics Extension Marking Guidelines Question (d) (i) Correct solution Uses an appropriate method to attempt to find Using the perpendicular distance formula from P ( t,t + 3) to the line y = x : D(t) = = = = t ( t + 3) 4 + t + t 4 5 ( t t + 4) 5 t t Now t t + 4 = ( t ) + 3 > 0 for all t, so t t + 4 D(t) = is the perpendicular distance. 5 Question (d) (ii) Correct answer To find minimum distance, D (t) = 0. D (t) = ( t ) 5 t = = ( t ) 5 = 0 Minimum because D(t) as t ±. ) Alternative: D(t) is a quadratic with minimum at t = since D(t) = (( t + 3 ). 5 9

10 03 HSC Mathematics Extension Marking Guidelines Question (d) (iii) Correct solution t = corresponds to x =, on the curve given by y = x + 3. y = x, so the slope of the tangent at x = is. Also, the slope of the line y = x is. the line and the tangent are parallel, since their gradients are equal. Question (e) Correct solution Correctly differentiates v with respect to x, or equivalent merit v + 9x = k v = k 9x 9 v = k x d x = v = 9x dx For SHM, x = n ( x b ). Now, x = 9x, is of this form with n = 3 and b = 0 π As n = 3, the period is. 3 0

11 03 HSC Mathematics Extension Marking Guidelines Question 3 (a) (i) Correct solution Using V = 4 πr 3 and A = 4πr, 3 dv 0 4 A = dt dv dr = dr dt dr = 4πr dt dr = A dt Hence dr = 0 4, which is constant. dt Question 3 (a) (ii) Correct answer Finds initial radius, or equivalent merit From part (i) dr = 0 4 dt To determine the initial radius of the raindrop: 4 πr 3 = 0 6, 3 3 so r 3 = 4π 0 6 r = 3 3 4π = 0 4π Time to evaporate; r t = = 0 4π time = 6 seconds (to the nearest second).

12 03 HSC Mathematics Extension Marking Guidelines Question 3 (b) (i) Correct solution Writes an unsimplified expression for the x- or y- coordinate of G, or equivalent merit The point G divides NT in the ratio : externally means that T is the midpoint of NG. Hence if (x 0, y 0 ) are the coordinates of G, then 0 + x 0 = ap (x-coordinate of T) a + ap + y 0 = 0 (y-coordinate of T ) x 0 = ap and y 0 = a ap ie G is the point (ap, a ap ), as required

13 03 HSC Mathematics Extension Marking Guidelines Question 3 (b) (ii) Correct solution for G Shows that G lies on a parabola, or equivalent merit The original parabola has focal length a and directrix y = a. To show that G (ap, a ap) lies on a parabola, eliminate p from x = ap, y = a ap x Now, p =, so a x y = a a a y = a ax 4a = a x 4a y + a = x 4a x = 4a( y + a) This is a parabola with focal length a and vertex (0, a). So the directrix is y = a + a = a Hence focal length and directrix are the same for both parabolas. Alternative: (ap, ap ) is the parabola obtained if the original parabola is reflected about the x-axis. It has focal length a and directrix y = +a. Now G(ap, a ap ) is obtained by translating the reflected parabola down by a units. Translation preserves the focal length a, and the directrix becomes y = a a = a. 3

14 03 HSC Mathematics Extension Marking Guidelines Question 3 (c) (i) Correct solution Obtains an expression for, or equivalent merit We need to find the time when the quadratic y = ut sinα g t is a maximum. dy = usinα gt dt = 0 for max/min usinα t = g This will be at the maximum height since y = ut sinα g t is a concave down parabola. Question 3 (c) (ii) Correct solution Similarly from part (i), the projectile fired from B reaches its maximum height when wsin β t =. g The projectiles collide when they both reach their maximum height. usinα wsin β ie = t =, g g so usinα = wsin β. 4

15 03 HSC Mathematics Extension Marking Guidelines Question 3 (c) (iii) Correct solution Finds the distance one projectile has travelled When the projectiles collide, the sum of the two distances travelled must be d. Both projectiles have travelled for time t. usinα wsin β t = =, so g g usinα wsin β d = u cosα + w cosβ g g wsin β usinα = u cosα + w cosβ g g uw = (sin β cosα + sinα cosβ) g uw = sin (α + β). g 5

16 03 HSC Mathematics Extension Marking Guidelines Question 3 (d) Correct solution 3 Proves that QBT and are equal AND uses the properties of the common tangent, or equivalent merit Proves that QBT and are equal or uses the properties of the common tangent, or equivalent merit The circles C and C have a common tangent at T. Choose points X and Y on that tangent as marked on the diagram above. YTP = PAT = α (angle in alternate segment, chord PT and tangent XY ) XTQ = QBT = β (angle in alternate segment, chord QT and tangent XY ) But PAT = QBT (alternate angles equal, BQ PA) ie α = β YTP = XTQ Hence QP is a straight line making equal vertically opposite angles with line XY at T. P,T,Q are collinear. 6

17 03 HSC Mathematics Extension Marking Guidelines Question 4 (a) (i) Correct solution k + k + = ( ( ) + k + ) k k + ( k + ) k k + k ( k + ) = k + k + k < 0 (since k + > k and k > 0). Question 4 (a) (ii) Correct solution 3 Uses correct assumption in an attempt to prove that Establishes initial case, or equivalent merit 5 3 Case n = : LHS + = + = RHS: = 4 4 LHS < RHS Hence true for n = Assume true for some k k ie < p k p= Need to show true for n = k + k + ie < p k + p= Now k + k LHS = = + ie p= p p= p ( k + ) LHS < + (by induction assumption) k ( k + ) < + using part (i), ie < k k k + ( k + ) k k + k + < p k + p= By the principle of Mathematical Induction, it is true for all n. 7

18 03 HSC Mathematics Extension Marking Guidelines Question 4 (b) (i) Correct answer From the binomial theorem 4n ( + x ) 4n = 4n x 4n k k =0 k The coefficient of x n 4n is, choosing k = n. n Question 4 (b) (ii) Correct solution Writes as or uses the binomial expansion, or equivalent merit From the binomial theorem ( ) n + x + x = ( + x( x + )) n n n k = (x( x + )) n k k =0 n n n k = ( x + ) n k. x k =0 k 8

19 03 HSC Mathematics Extension Marking Guidelines Question 4 (b) (iii) Correct solution 3 Substitutes the given expression into part (ii) and makes progress towards the solution Connects and, or equivalent merit ( + x + x) = ( + x ), so ( ) n + x + x = ( + x ) 4n By part (i), 4n is the coefficient of xn in ( + x + x) n. n From part (ii) these are n n k ) n k n the terms in ( x + for k = 0,, n. If k > n there is no term involving x. x k Using the given identity the coefficients of x n are: n n 0 k = 0 : 0 0 n n k = : n n k = : n n n k = n : n n Hence n 0 n n n n n 4n n n n n n n = n + n + + n 0 0 n n n n n k = n k. k k k =0 0 9

20 03 HSC Mathematics Extension Marking Guidelines Question 4 (c) (i) Correct solution Correctly applies Newton s method Take ƒ ( ) Then ƒ ( ) t = e t t t = e t + t Apply Newton s method ( ) f t 0 t = t 0 ƒ (t 0 ) 0.5 = 0.5 e 0.5 e + 4 = 0.56 Hence t = 0.56 is another approximate solution. 0

21 03 HSC Mathematics Extension Marking Guidelines Question 4 (c) (ii) Correct solution 3 Uses part (c) (i) to find an approximate solution to merit, or equivalent Establishes, or equivalent merit We need to determine r and x so that rx rx e = log x and re = e x (common point and equal slope of tangents). From the equal slopes rx e = rx Setting rx = t and using part (i), we have an approximate solution rx = 0.56 Since the curves intersect at x, log x = e e 0.56 e x = e rx 0.56 = e now rx = r = = r = = 5.758

22 Mathematics Extension 03 HSC Examination Mapping Grid Section I Question Content Syllabus outcomes 6. PE3 5. HE4 3.8 PE PE3 5.5 HE HE H5, PE H5, PE HE4 0.4E PE3 Section II Question Content Syllabus outcomes (a) 6.3 PE3 (b) 5.5 HE4 (c) 8. HE3 (d) (i) 8.8 P7, HE4 (d) (ii) 0.5 H6, HE4 (e) 3.4 H5, HE7 (f) 3.5, 5.5 HE4, HE6 (g) 8.8, 5.5 P7, HE4 (a) (i) 5.7, 5.9 PE (a) (ii) 5.7, 5.9 PE (b) 3.4, 3.6E H5, H8, HE7 (c) 3 4.E HE3 (d) (i) 6.5 P4 (d) (ii) 8.5, 0.6 P4, H5 (d) (iii) 8.5 P6 (e) 4.4 HE3 3 (a) (i) 4. HE5 3 (a) (ii) 4. HE5

23 03 HSC Mathematics Extension Mapping Grid Question Content Syllabus outcomes 3 (b) (i) 6.7E, 9.6 PE3 3 (b) (ii) 9.6 PE3, PE4 3 (c) (i) 4.3 HE3 3 (c) (ii) 4.3 HE3 3 (c) (iii) 5.7, 4.3 PE, HE3 3 (d) 3.3,.9 P4, PE, PE3, PE6 4 (a) (i).4e PE3 4 (a) (ii) HE 4 (b) (i) 7.3 HE3 4 (b) (ii) 7.3 HE3 4 (b) (iii) HE3 4 (c) (i) 6.4 HE7 4 (c) (ii) 3 8.6,.,.4,.5 P6, H3, HE7 3