Question ONE Evidence Achievement (u) Merit (r) Excellence (t) Apply the algebra of complex numbers in solving problems. Apply the algebra of complex numbers, using relational thinking, in solving problems. Apply the algebra of complex numbers, using extended abstract thinking, in solving problems. 1a f(1) = 0 so 3c – 4c – 8c +18 = 0 18 = 9c 2 = c Correct expression. 1b u = 1 + i = 2 ½ cis(450) u8 = 24cis3600 Correct solution. Degrees or rads 1c 4x – 8 = x2 – 4x + 4 0 = x2 – 8x + 12 0 = (x – 2)(x – 6) x = 2 or 6 check if x = 2 check if x = 6 lhs = 0, rhs = 0 lhs = 4 rhs = 4 so x = 2 is valid so x = 4 is valid Award u if minor error ( or no check done) Award r only for Correct solution with check. 1d Let z rcisθ z5 = 32ni r5cis 5θ = 32ncis(90 + 360n) r5 = 32n 5θ =90 + 360n r = 2n0.2 θ = 18 + 72n z1 = 2n0.2 cis 180 z2 = 2n0.2 cis 900 z3 = 2n0.2 cis 1620 z4 = 2n0.2cis2340 z4 = 2n0.2cis3060 Award u if changed correctly into polar form: r5cis 5θ = 32ncis(90 + 360n) either using degrees or radians. Correct solution either using degrees or radians.
N2: 1 u A3: 2 u A4: 3 u M5: 1 r M6: 2 r E7: 1 t with minor errors ignored E8: 1 t with full excellence criteria Judgement Statement Not Achieved Achievement Achievement with Merit Achievement with Excellence Score range 0 - 6 7 - 13 14 - 18 19 -24 Question ONE Evidence Achievement (u) Merit (r) Apply differentiation methods in solving problems. Apply differentiation methods, using relational thinking, in solving problems. Apply differentiation methods, using extended abstract thinking, in solving problems. 1a y = 8sin(4x) yꞌ = 32cos(4x) Correct derivative. 1b y = (x3 + 2x)2 yꞌ = 2(x3 + 2x)(3x2+2) = 2×3×5 = 30 so gradient of normal = -1 30 Correct answer = u 1c x = cos(2t) and y = 2sin(t) dx = –2sin(2t) dy = 2cos(t) dt dt dy = 2cos(t) = cos(t) = – 1 . dx –2sin(2t) –cos(t)sin(t) sin(t) Correct up to dy = 2cos(t) dx -2sin(2t) = u Correctly simplified = r 1d y = e 8 – 4x + 4x yꞌ= -4e8 – 4x + 4 = 0 at stat pts so 4 = 4e8 – 4x 1 = e8 – 4x 0 = 8 – 4x So x = 2 Correct dy for u dx and state that dx = 0 dt Correct answer for for r
1e If the slant height of a cone is 8√3 cm prove that the maximum volume is when h = 8 cm and r = 8√2 cm r2+ h2 = 192 r2= 192 – h2 h 8√3 r V = π (192 – h2)h = π (192h – h3) 3 3 dV = π (192 – 3h2) = 0 for max vol dh 3 192 = 3h2 so h2 = 64 h = 8 (not -8) and r2= 192 – h2 = 192 – 64 = 128 = 64×2 So r = 8√2 Correct Equ for V by eliminating r or h but preferably r Correct derivative and solving for derivative = 0 Correct r and h found for t Question TWO Evidence Achievement (u) Merit (r) Excellence (t) Apply differentiation methods in solving problems. Apply differentiation methods, using relational thinking, in solving problems. Apply differentiation methods, using extended abstract thinking, in solving problems. 2a y = ln(x2 + 1) 3x + 4 yꞌ =(3x + 4)× 2x – ln(x2 + 1) ×3 (x2 + 1) . (3x + 4)2 Correct derivative. 2b y = 3e2x – 8 where x = 4 yꞌ = 6e2x – 8 sub x = 4 yꞌ = 6e0 = 6×1 = 6 Correct derivative and sub x = 4 to get 6 for u
1c v = 2t – Find the distance moved from t = 1 to t = 4 x = t2 + e 1 – t + c sub t = 4 x4 = 16 + e – 3 + c sub t = 1 x1 = 1 + e0 + c = 2 + c x4 – x1 = 14 + e – 3 ≈14.05 metres Antiderivative correct. Correct integration with correct solution. 1d dm = dt t = 0 m = 8g and at t = 2 hr, m = 1 g. = = kt + c = kt + c Sub t = 0, m = 8 6 = c Sub t = 2, m = 1 3 = 2k + 6 so k = – 9 2 4 = + 6 If m = 0 t = 24 = hours 9 Correct diff equ with k and c found = u Correct time = r
1e The graphs have the equations y = (p2+1) – x2 and y = p2 where p > 1 x2 Intersection when (p2+ 1) – x2 = p2 x2 (p2 + 1)x2 – x4 = p2 0 = x4 – (p2 + 1)x2 + p2 (x2 – 1)(x2 – p2) = 0 x = ±1 and x = ± p area = = – = 2p3 – 2p2 + 2p – 2 3 3 Correct intersection for u Correct idea of area = area under cubic – area under parabola (and vice versa) shown with equations for r. Correct solution showing correct integrations and correct mathematical logic and statements.
Question TWO Evidence Achievement (u) Merit (r) Excellence (t) A =2 3+ 1 + 4(9 + 15 +5) + 2(13+11) 3 = 112 Apply integration methods in solving problems. Apply integration methods, using relational thinking, in solving problems. Apply integration methods, using extended abstract thinking, in solving problems. 2a Correct calculation. 2b y ꞌ = 4x3 + 3x2 – 4x + c sub x = 1, y ꞌ = 4 4 = 4 + 3 – 4 + c so c = 1 y = x4 + x3 – 2x2 + x + d sub x = 1, y = 2 2 = 1 + 1 – 2 + 1 + d so d = 1 y = x4 + x3 – 2x2 + x + 1 Correct integrations and constants evaluated and solution found = u. 2c A + B = dx = 8ln(8) So A = 4ln(8) A = dx = 8ln(k) Solving 8ln(k) = 4 ln(8) ln(k) = ½ ln(8) = ln( 8 ½) k = 8 ½ Correct value for total area for u. Correct integration with correct value for k. 2d x = 1 when y = 1, find y when x = 2 = = ln(y) = – 3 + c x2 sub x = 1 and y = 1 0 = –3 + c so c = 3 ln(y) = – 3 + 3 x2 sub x = 2 ln(y) = – ¾ + 3 = 2 ¼ y = e 2.25 ≈ 9.49 Correct integration with correct constant. Correct integration with correct constant and y value found.
Question TWO Evidence Achievement (u) Merit (r) Excellence (t) 2e = = = let u = ex + 1 du = exdx = = ln(u) + c = ln( ex + 1) + c - Correct manipulations of needed to integrate = r Correct solution = t Question THREE Evidence Achievement (u) Merit (r) Excellence (t) Apply integration methods in solving problems. Apply integration methods, using relational thinking, in solving problems. Apply integration methods, using extended abstract thinking, in solving problems. 3a = dx = x3 + 4x + 8lnx + c 3 Correct integration. x not essential 3b = = = Correct integration.
3c x – 3 . x + 3 x2 + 0x + 4 x2 + 3x – 3x + 4 – 3x – 9 + 13 = = x2 – 3x + 13 ln(x + 3) + c 2 Correct division or similar = u Correct Integration evaluated = r. 3d I = Let u = x + 2 so x = u – 2 and x – 2 = u – 4 du = dx I = equ A = = – 8 = 4 Uses substitution to obtain equ A Evaluates integral correctly
3e dR = kR dt ln(R) = kt + c sub t = 100, R = 150 : ln(150) = 100k + c equ1 sub t = 200, R = 90 : ln(90) = 200k + c equ2 subtracting : ln(90) – ln(150) = 100k so k = -0.005108 sub in equ 1 c = 5.5214 sub in the equ: ln(R) = kt + c ln(R) = –0.005108t + 5.5214 if R = 20 : ln(20) = –0.005108t + 5.5214 so t ≈ 494 days Obtains dR = kR dt and states equ 1 and equ 2 for grade u Obtains dR = kR dt and calculates c and k for grade r Obtains t ≈ 490 days with all correct mathematical statements for grade t.