NCEA Level 3 Calculus (91579) 2013 — page 1 of 7
Assessment Schedule – 2013
Calculus: Apply integration methods in solving problems (91579)
Evidence Statement
ONE
(a)
Expected Coverage
-2x -1 +
x -2
+c
2
Merit
Excellence
Correct answer.
(b)
(c)
Achievement
Correct answer.
v = 6d=
5
t +1
æ
Correct integration.
Correct answer.
Correct DE and
integration.
Correct DE,
integration and
general solution to
DE with logical
working.
5 ö
ò v dt = ò çè 6 - t + 1÷ø dt
= 6t - 5ln(t + 1)
t = 4 d = 24 - 5ln5 = 15.95
t = 3 d = 18 - 5ln 4 = 11.07
Distance travelled = 4.88 m
(d)(i)
(ii)
dV
= kV
dt
dv
ò V = ò k dt
ln aV = kt
1 kt
e
a
= Ae kt
V=
t = 0 months in May 2007
So t = 60 months in May 2012
And t = 78 months in Nov 2013
365000 = Ae k (60)
A=
365000
e60k
382000 =
365000
´ e78k
e60k
æ 382000 ö
ln ç
= ln(e18k )
è 365000 ÷ø
18k = 0.045523
k = 0.002529
A=
365000
e0.002529t
= $313 611 when t = 0
Correct answer
including correct DE,
integration and general
solution to DE with
logical working. Units
not required in answer.
NCEA Level 3 Calculus (91579) 2013 — page 2 of 7
(e)
One correct
integration.
-
Correct energy for
conical tank.
Both integrals correct
and correct solution.
NCEA Level 3 Calculus (91579) 2013 — page 3 of 7
NØ = No response / no relevant evidence
N1 = ONE answer demonstrating limited knowledge of integration techniques
N2 = ONE correct integration
A3 = TWO of Achievement (u)
A4 = THREE of Achievement (u)
M5 = ONE of Merit (r)
M6 = TWO of Merit (r)
E7 = Excellence with minor errors ignored (t)
E8 = Excellence correct (t)
NCEA Level 3 Calculus (91579) 2013 — page 4 of 7
TWO
(a)
Expected Coverage
ò
2
Achievement
1
´ 0.2(1.41+ 2.24
2
+2(1.56 +1.72 +1.89 + 2.06))
f (x) dx =
1
Merit
Correct answer.
= 1.81
(b)
=
ò
k
1
Correct answer.
3(x) 2 dx
1
k
= é 2x 2 ù
êë
úû1
3
3
= 2k 2 - 2
(c)
1
y = x 2 = 2x
3
2
x - 6x = 0
Correct integral (or
difference of two
correct integrals).
Correct solution.
Correct integration.
Correct solution.
x = 0 or 6
Area =
ò
6æ
0
1 2ö
çè 2x - 3 x ÷ø dx
6
é
1 ù
= ê x 2 - x3 ú
9 û0
ë
= 12
(d)
dV
400
=
= 400(t + 2)-2
dt (t + 2)2
ò
V = 400(t + 2)-2dt
=
-400
+c
(t + 2)
t = 0, V = 20 Þ 20 =
-400
+c
2
c = 220
At t = 6, V =
-400
+ 220 = 170 L
8
Excellence
NCEA Level 3 Calculus (91579) 2013 — page 5 of 7
(e)
1
f '(x) = 1- x 2
3
x3
f (x) = x - + c
9
x = 0, y = 0,\c = 0
Correct equation of
tangent.
2
-8
and f (x) =
3
9
Tangent has equation
At x = -1, f '(x) =
2
2
x3
9
Intersection at
y=
x3 2
2
= x9 3
9
x = -1,2
x-
Area between curve and tangent
=
æ
x3 æ 2
2ö ö
x - - ç x - ÷ ÷ dx
ç
-1 è
9 è3
9øø
ò
2
ù
1 é 3x 2 x 4
= ê
+ 2x ú
9 êë 2
4
úû
=
2
-1
3
4
NØ = No response / no relevant evidence
N1 = ONE answer demonstrating limited knowledge of integration techniques
N2 = ONE correct integration
A3 = TWO of Achievement (u)
A4 = THREE of Achievement (u)
M5 = ONE of Merit (r)
M6 = TWO of Merit (r)
E7 = Excellence with minor errors ignored (t)
E8 = Excellence correct (t)
Correct expression
for area required
(could be difference
of two integrals).
Correct solution.
NCEA Level 3 Calculus (91579) 2013 — page 6 of 7
THREE
(a)
Expected Coverage
Achievement
1
- cosec 2x + c
2
Correct integration.
(b)
(c)
ò
2m
(
2x + 5
)
ln
4m 2 + 10m
m 2 + 5m
m 2 - 5m = 0
) (
Correct integration.
Correct answer.
Correct integration.
Correct answer.
Correct integration.
Correct solution.
Correct DE and
integration.
Correct DE,
integration and
general solution to
DE, with logical
working and correct
value of k.
Excellence
2m
= éê ln x 2 + 5x ùú = ln3
ë
ûm
m x 2 + 5x
é ln 4m 2 + 10m - ln m 2 + 5m ù = ln3
êë
ûú
(
Merit
)
= ln3
m = 0,5
(d)
ò v = ò -k dt
dv
2
1
- = -kt + c
v
When t = 0, v = u
c=
-1
u
-1
1
= -kt v
u
When t = 1, v =
u
1+ ku
(e)(i)
dT
= k(T - 18)
dt
(ii)
ò T - 18 = ò k dt
dT
ln T - 18 = kt + c
At t = 0, ln 65 - 18 = c = ln 47
At t = 2, ln 62 - 18 = 2k + ln 47
k = -0.03297
ln 54 - 18 = -0.03297 ´ t + ln 47
t = 8.08 minutes
Correct answer
including correct DE,
integration and general
solution to DE with
logical working. Units
not required in answer.
NCEA Level 3 Calculus (91579) 2013 — page 7 of 7
NØ = No response / no relevant evidence
N1 = ONE answer demonstrating limited knowledge of integration techniques
N2 = ONE correct integration
A3 = TWO of Achievement (u)
A4 = THREE of Achievement (u)
M5 = ONE of Merit (r)
M6 = TWO of Merit (r)
E7 = Excellence with minor errors ignored (t)
E8 = Excellence correct (t)
Marking codes
Codes that may have been used in marking this examination paper have meaning as follows.
#
Hash - where a candidate obtains a correct answer but continues with further, unnecessary, material that is incorrect
but does not show that the candidate shows a lack of understanding or a contradiction.
C
Consistency - where a candidate has obtained an incorrect value within a question and subsequently used that value.
NC
Non-consistency - where a candidate has obtained an incorrect value or expression within a question and has not
used that value or expression where it was subsequently required.
RAWW
Right answer, wrong working - where a candidate presents a correct answer but the working or reasoning leading to
it is incorrect, incomplete or contains one or more errors.
R
Rounding error - where a candidate produces a correct sequence of calculations, but the answer does not agree to 2
significant figures with the answer given in the assessment schedule as a result of rounding a number in the sequence
of calculations.
MEI
Minor error ignored - where a candidate has made a minor error and this has been ignored.
Two ans
Two answers given - where a candidate writes two answers, one correct and the other incorrect, and neither has been
deleted (the correct answer is not accepted).
Judgement Statement
Score range
Not Achieved
Achievement
Achievement
with Merit
Achievement
with Excellence
0–6
7 – 12
13 – 20
21 – 24