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.



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.



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.





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

Not Achieved

Achievement

Achievement
with Merit

Achievement
with Excellence

Score range


0 – 7

8 – 14

15 – 20

21 – 24


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