f x( ) = 2x ( ( ) = 0

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