# Assessment Schedule – 2012 Mathematics and Statistics: Apply calculus methods in solving problems (91262) Evidence Statement

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NCEA Level 2 Mathematics (91262) 2012 — page of

Assessment Schedule – 2012

Mathematics and Statistics: Apply calculus methods in solving problems (91262)

Evidence Statement
 Q Evidence Achievement (u) Merit (r) Excellence (t) Apply calculus methods in solving problems. Apply calculus methods, using relational thinking, in solving problems. Apply calculus methods, using extended abstract thinking, in solving problems. ONE (a) f ' (x) = 6x2 – 10 f ' (2) = 14 Correct. (b) or p(x) = 4x – 2x3 + 13 Correct. (c) v = 0.2 t + 5 When v = 8 8 = 0.2t + 5 t = 15 secs Finding expression for v and equating to 8. Finding the value of t. (d) So v = 25 turning point Shape of parabola or f " > 0 minimum Finding expression for f . Substitute 25 for v and show the value of the expression is 0 or find v = 25 from f  = 0 Justification. (e) t = 0, v = 0 c = 0 t = 0, s = 0 c = 0 When t = 30 s = 1125 metres Finding expression for v. Finding expression for s. Finding the distance when t = 30. NØ N1 N2 A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u 2 of u 3 of u 1 of r 2 of r 1 of t 2 of t

 Q Evidence Achievement (u) Merit (r) Excellence (t) TWO (a) x intercept aligns with turning point Correct (b)(i) h' (x) = 2x – 12 = 4 x = 8 Correct x co-ordinate found. (ii) h' (1) = 2 – 12 = –10 y + 11 = –10(x – 1) y = –10x – 1 Gradient found. Equation correct. (c) g' (x) = 3x2 – 18x + 24 x2 – 6x + 8 = 0 2 < x < 4 Justification: Shape of cubic OR check gradient at a point. Derivative found and set equal to 0. Solution without justification. Justified solution. (d) Derivative found. x found. NØ N1 N2 A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u 2 of u 3 of u 1 of r 2 of r 1 of t 2 of t

 Q Evidence Achievement (u) Merit (r) Excellence (t) THREE (a) f(x) = – x2 f(6) = 72 – 36 Hence (6,36) Anti-differentiation found. Coordinates found. (b) Rate of change of volume found. (c) Positive cubic drawn with minimum on x-axis. Minimum at (a , 0) and maximum aligned to x-intercept on h'(x) Positive cubic drawn. Cubic drawn with min at (a,0) and max aligned with x-intercept. (d) f ' (x) = 6x2 + A f ' (–2) = 6  4 + A = 10 A = –14 33 = 2  –8 – 14  –2 + B B = 21 f (x) = 2x3 – 14x + 21 f (4) = 93 Hence (4, 93) f ' (x) found and set equal to 10. B found. Coordinates found. (e) A =(1+2t )2 = 2 + 4t = 60 t = 4.27 Width = 9.55 m Derivative found. A ' = 60 Width found. NØ N1 N2 A3 A4 M5 M6 E7 E8 No response; no relevant evidence Attempt at one question 1 of u 2 of u 3 of u 1 of r 2 of r 1 of t 2 of t

## Judgement Statement

### Score range

0 – 7

8 – 14

15 – 20

21 – 24

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