NCEA Level 2 Mathematics (90286) 2005 — page 1 o f 2 Assessment Schedule – 2005 Mathematics: Find and use straightforward derivatives and integrals (90286) Evidence Statement Achievement Criteria Find and use straightforward derivatives and integrals. Q 1 Evidence Code dy = 6x + 5 dx Gradient = 17 A1 1 2 1 "0 ( # x4 x2 & 2x ! x + 2x dx = % ! + 2x ( $2 2 '0 3 ) 2 Judgement Notation not required. Candidate must give the derived or integrated function and answer the question. Sufficiency Achievement: 3 × code A including at least 1 × code A1 and 1 × code A2. Achievement = ( 0.5 ! 0.5 + 2 ) ! ( 0 ) A2 =2 3 3 2 f ( x ) = 2x ! 2x + 5x + c x = 1, y = 3 gives c = !2 3 2 y = 2x ! 2x + 5x ! 2 4 dy dx A2 2 = x ! x +1= 3 2 x !x!2=0 ( x ! 2 ) ( x + 1) = 0 x = 2, !1 A1 Both needed. © New Zealand Qualifications Authority, 2005 All rights reserved. No part of this publication may be reproduced by any means without prior permission of the New Zealand Qualifications Authority. NCEA Level 2 Mathematics (90286) 2005 — page 2 o f 2 Achievement Criteria Apply calculus techniques to solve straightforward problems Q 5 Evidence dy Code Judgement Notation not required. Candidate must give the derivative or integral and answer the question. 2 = 1.2x + 0.8x ! 4.8 dx m = 10.8 + 2.4 ! 4.8 = 8.4 At (3,0) 0 = 8.4 × 3 + c Or equivalent. c = –25.2 A1 or M Achievement wijth Merit y = 8.4x – 25.2 6 Achievement with Merit: EITHER As for Achievement plus 2 × code M OR 3 × code M. ds = 36 ! 0.2t dt 36 ! 0.2t = 0 t = 180 s = 3240 m 7 Sufficiency 2 "0 ( x + 2 4 "2 ( x A1 M No alternative. Units not required. ) + x ! 6 dx 2 ) + x ! 6 dx 2 # x3 x2 & = % + ! 6x ( $3 2 '0 4 # x3 x2 & +% + ! 6x ( $3 2 '2 = 20 Achievement with Excellence Apply calculus techniques to solve problems. 8 1 area correct A2 Or M 2 v = 3t ! 16t + a A2 Candidates need to: A2 integrate and solve the resulting equations, If t = 0, v = 5, s = 3 2 v = 3t ! 16t + 5 3 Or equivalent. Ignore rounding errors and units. 2 s = t ! 8t + 5t + 3 For maximum distance and correctly answer the question. v=0 2 3t ! 16t + 5 = 0 ( 3t ! 1) (t ! 5 ) = 0 t = 5 or t = 1 3 s ( 5 ) = !50 cm E or M Accept 50 cm. Achievement with Excellence: As for Merit plus code E. NCEA Level 2 Mathematics (90286) 2005 — page 3 o f 2 Judgement Statement Achievement Achievement with Merit Achievement with Excellence Find and use straightforward derivatives and integrals. Apply calculus techniques to solve straightforward problems. Apply calculus techniques to solve problems. 3×A Achievement plus Merit plus including at least 1 each of A1 and A2. 2×M Code E or 3×M
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