NCEA Level 2 Mathematics (91262) 2012 — page 1 of 4
Assessment Schedule – 2012
Mathematics and Statistics: Apply calculus methods in solving problems (91262)
Evidence Statement
Q
Evidence
Achievement (u)
Apply calculus methods
in solving problems.
ONE
(a)
f ' (x) = 6x2 – 10
Merit (r)
Excellence (t)
Apply calculus methods,
using relational thinking,
in solving problems.
Apply calculus methods,
using extended abstract
thinking, in solving
problems.
Correct.
f ' (2) = 14
(b)
p(x) = 4x -
6x3
+13
3
Correct.
or
p(x) = 4x – 2x3 + 13
(c)
v = 0.2 t + 5
When v = 8
8 = 0.2t + 5
Finding expression for v
and equating to 8.
t = 15 secs
(d)
Finding the value of t.
2v
250
2 ´ 25
f ' = -0.2 +
=0
250
So v = 25
 turning point
f ' = -0.2 +
Finding expression for f .
Substitute 25 for v and
show the value of the
expression is 0
or find v = 25
from f  = 0
Shape of parabola or
f " > 0  minimum
(e)
Finding expression for v.
1
a = 5- t
4
t2
v = 5t - + c
8
t = 0, v = 0
 c =0
s=
Justification.
5t 2 t 3
- +c
2 24
t = 0, s = 0
Finding expression for s.
 c=0
When t = 30
Finding the distance when
t = 30.
s = 1125 metres
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
NCEA Level 2 Mathematics (91262) 2012 — page 2 of 4
Q
Evidence
Achievement (u)
TWO
(a)
Merit (r)
Excellence (t)
Correct
x intercept aligns with turning point
(b)(i)
h' (x) = 2x – 12 = 4
Correct x
co-ordinate found.
x=8
(ii)
h' (1) = 2 – 12 = –10
Gradient found.
y + 11 = –10(x – 1)
y = –10x – 1
(c)
Equation correct.
g' (x) = 3x2 – 18x + 24
Derivative found
and set equal to 0.
x – 6x + 8 = 0
2
Solution without
justification.
2<x<4
Justification:
Shape of cubic OR
check gradient at a point.
(d)
Justified solution.
1 2 1 3
ax - x
2
2
dA
3 2
= ax - x
dx
2
dA
=0
dx
2a
x=
3
2a3
max A =
27
A=
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
NCEA Level 2 Mathematics (91262) 2012 — page 3 of 4
Q
Evidence
THREE
(a)
f(x) =
Achievement (u)
Merit (r)
Anti-differentiation
found.
x3 2
–x
3
f(6) = 72 – 36
Hence (6,36)
(b)
Coordinates
found.
Rate of change of
volume found.
dV
= 3x2
dx
When x = 5,
(c)
Excellence (t)
dV
= 75
dx
Positive cubic drawn with minimum on x-axis.
Positive cubic
drawn.
Minimum at (a , 0) and maximum aligned to
x-intercept on h'(x)
Cubic drawn
with min at (a,0)
and max aligned
with
x-intercept.
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)
(d)
(e)
A=
f ' (x) found and set
equal to 10.
B found.
Coordinates
found.
1
 (1+2t )2
2
dA
= 2 + 4 t
dt
Derivative found.
dA
= 60
dt
A ' = 60
t = 4.27
Width found.
Width = 9.55 m
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
NCEA Level 2 Mathematics (91262) 2012 — page 4 of 4
Judgement Statement
Score range
Not Achieved
Achievement
Achievement
with Merit
Achievement
with Excellence
0–7
8 – 14
15 – 20
21 – 24