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7/29/2019 maths 2011 memo
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MARKS: 150
MATHEMATICS P1
NOVEMBER 2011
MEMORANDUM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
7/29/2019 maths 2011 memo
2/28
Mathematics/P1 2 DBE/November 2011NSC Memorandum
NOTE:
If a candidate answers a question TWICE, only mark the FIRST attempt. If a candidate has crossed out an attempt of a question and not redone the question, mark the
crossed out version.
Consistent Accuracy applies in all aspects of the marking memorandum.QUESTION 1
1.1.1
( )( ) 02306
6
6)1(
2
2
=+
=+
=+
=+
xx
xx
xx
xx
2or3=x
OR
062 =+xx
( )( )( )12
61411
2
4
2
2
=
=
a
acbbx
2or3=x
9 standard form
9 factors
9 answers
(3)
9 standard form
9 substitution
into correct
formula
9 answers
(3)
1.1.2
0843
843
2
2
=
=
xx
xx
( ) ( )( )( )
1124
6
96164
328344)4(
2
4
2
2
+=
=
=
a
acbbx
9 standard form
9 substitution intocorrect formula
9 112
Note:Answers by inspection:
award 3/3 marks
Note:
Answer only of 2=x :
award 1/3 marks
Note:
If candidate converts
equation to linear:
award 0/3 marks
Note:If candidate uses
incorrect formula:
maximum 1/4 marks
(for standard form)
Note:
If an error in subs and
gets:804
and
Note: Penalise
1 mark for
inaccurate
di ff
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Mathematics/PI 3 DBE/November 2011NSC Memorandum
OR
0843
843
2
2
=
=
xx
xx
( ) ( )( )( )
10,1or43,2
32
8344)4(
2
4
2
2
=
=
=
a
acbbx
9 standard form
9 substitution intocorrect formula
9 answer
9 answer
(4)
1.1.3
( )( ) 01140154
514
2
2
+
+
xx
xx
xx
4
1x or 1x OR [ )
;1
4
1;
OR
OR
NOTES:
If a candidate gives an answer of4
11 x then max 3/4 marks.
If a candidate gives an answer of 11
x then max 2/4 marks.
9 factors
9 both critical
values of41 and 1
9 orOR 9 answer
(4)
41 1
x
1
4
1
x
Note: If candidate giveseither of these correct
graphical solutions but
writes down the incorrect
intervals or uses AND:
max 3/4 marks
41 1
+ 0 +0
141
Note: Penalise 1 mark
for inaccurate
rounding off to ANYnumber of decimal
places if candidate
gives decimal answers
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Mathematics/PI 4 DBE/November 2011NSC Memorandum
1.2.1
( )( ) 023065 22
=++
=++
yxyx
yxyx
3
303
=
==+
y
x
yxyx
OR
2
202
=
==+
y
x
yxyx
9 factors
99 answers
(3)
OR
Let k=
y
x
0)2)(3(
065
065
065
2
2
22
=++
=++
=+
+
=++
kk
kk
y
x
y
x
yxyx
k= 3 or k= 2
3=y
xor 2=
y
x
OR
065 22 =++ yxyx
2
5
2
5
)1(2
)6)(1(4)5(5
2
22
yyx
yyx
yyyx
=
=
=
3
3
=
=
y
x
yx
or2
2
=
=
y
x
yx
OR
9 factors
99 answers
(3)
9 substitutes
correctly into
correct formula
99 answers
(3)
Note:
If a candidate gives
3=y
xor 2=
y
x
award 2/3 marks
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Mathematics/PI 5 DBE/November 2011NSC Memorandum
3
3
=
=
y
x
yx
or2
2
=
=
y
x
yx
OR
Let k=y
x
kyx =
( ) ( )
( ) 065065
065
065
22
2222
22
22
=++
=++
=++
=++
kky
ykyyk
ykyyky
yxyx
0)2)(3(
0652
=++
=++
kk
kk
k= 3 or k= 2
3=y
xor 2=
y
x
Note: (x;y) = (0;0) is also a solution, but in this casey
xis undefined
OR
Let 1=y ,
( )( ) 0320652
=++
=++
xx
xx
2=x or 3=x
2=
y
x
or 3=
y
x
99 answers
(3)
9 factors
99 answers
(3)
9 factors
99 answers
(3)
1.2.2
4
82
83
8
=
=+
=+
y
y
yy
yx
OR
8
8
82
8
=
=+
=+
y
y
yy
yx
9 substitution
x = 3y
9 subs yx 2=
7/29/2019 maths 2011 memo
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Mathematics/PI 6 DBE/November 2011NSC Memorandum
OR
xy
yx
=
=+
8
8
38
=x
xOR 2
8=
x
x
( )
4
12
242
324
83
=
=
=
+=
=
y
x
x
xx
xx
( )
8
16
16
216
82
=
=
=
+=
=
y
x
x
xx
xx
OR
( )( )
( )( ) 08288
8
032
=++=
=+
=++
yyyx
yx
yxyx
8=y or 4=y
16=x 12=x
OR
( ) ( )
( )( ) 04803212
064242
065401664
06858
8
2
2
222
22
=++=++
=++
=+++
=++
=
yyyy
yy
yyyyy
yyyy
yx
8=y or 4=y
16=x 12=x
9 xy = 8
9 substitution
99x values
correct
9 bothy values
correct
(5)
9 yx = 8
9 substitution
99y values
correct
9 bothx values
correct
(5)
9 yx = 8
9 substitution
9factors
9 bothy values
correct
9 bothx valuescorrect
(5)
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Mathematics/PI 7 DBE/November 2011NSC Memorandum
OR
( ) ( )
03212
064242
065401664
06858
8
2
2
222
22
=++
=++
=+++
=++
=
yy
yy
yyyyy
yyyy
yx
( )( )( )
2
1612
1232141212
2
=
=y
8=y or 4=y
16=x 12=x
OR
( ) ( )
( )
( )( ) 01216019228
0384562
016646540
08685
8
2
2
222
22
=
=+
=+
=+++
=++
=
xx
xx
xx
xxxxx
xxxx
xy
4
12
=
=
y
xor
8
16
=
=
y
x
OR
xy = 8
( ) ( )
( )0384562
016646540
08685
2
222
22
=+
=+++
=++
xx
xxxxx
xxxx
9 yx = 89 substitution
9 substitutes into
correct formula
9 bothy values
correct
9 bothx values
correct
(5)
9 xy = 8
9 substitution
9factors
9 bothx values
correct
9 both y values
correct
(5)
9 xy = 8
9 substitution
Note:
If a candidate uses the
formula and replacesx
fory and then answers
are swapped:
maximum 4/5 marks
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Mathematics/PI 8 DBE/November 2011NSC Memorandum
QUESTION 2
2.1.1
18
362
324
=
=
=
x
x
xx
OR
a = 4
a + 2d= 32
2d= 28
d= 14x = 14 + 4
x = 18
OR
18
2
324=
+=x
9 2312 TTTT =
9 answer
(2)
9 a + 2d= 32 and a = 4
9 answer
(2)
9 substitutes correctly
into arithmetic mean
formula i.e. 2
324 +
9 answers
(2)
2.1.2
128
128
32
42
=
=
=
x
x
x
x
28=x OR 31,11=x OR 27
2=x
ORa = 4
4432
44
4
2
22
=
=
=
x
xar
xr
92
3
1
2
T
T
T
T=
9 1282 =x
9 both answers(surd or decimal or
exponential form)(3)
9
2
4432
=
x
9 1282 =x
Note:
If only 128=x thenpenalty 1 mark
Note:
If answer only:
award 2/2 marks
Note:If candidate writes
4x x32 only(i.e. omits equality) :
0/2 marks
Note: If candidate
writes4
x
x
32only
(i.e. omits equality) :0/2 marks
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Mathematics/PI 9 DBE/November 2011NSC Memorandum
2.2
( )
81
797161
or81
40
9841or499841
13
133
3333
3333
3
134
8234
513535251
13
1
5
,
...
...
P
-
k
k
=
=
++++=
++++=
=
=
OR
81
797161or
81
409841or499841
6561...9
1
27
1
81
1
3333
3333
3
8234
513535251
13
1
5
,
...
...
P
-
k
k
=
++++=
++++=
++++=
=
=
9 43=a or81
1
9 3=r 9 subs into correct
formula
9 answer
(4)
99 expand the sum
9 13 terms in expansion
9 answer
(4)
2.3 [ ] [ ] ( )[ ] ( )[ ]
( )[ ] ( )[ ] [ ]( )[ ] ( )[ ] ( )[ ] ( )[ ]
( )[ ]
( )[ ]dnan
S
dnan
dnadnadnadnaS
adadnadnaS
dnadnadadaaS
n
n
n
n
122
12
1212...12122
...21
12...2
+=
+=
++++++++=+++++++=
+++++++++=
OR [ ] [ ][ ]
( )[ ]dnaanTaTaTaTaS
adadTTS
TdTdadaaS
nnnnn
nnn
nnn
1
...2
...)(
)(...2
++=
++++++++=
+++++=
+++++++=
9writing out Sn9
reversing Sn
9expressing 2Sn
9 grouping to get
( )[ ]dnanSn 122 += (4)
9writing out Sn9 reversing Sn
9expressing 2Sn9 grouping to get
( )[ ]
Note: If the candidate rounds
off and gets 9841,46 (i.e.
correct to one decimal place):
DO NOT penalise for the
rounding off.
Note:Correct answer only:
1/4 marks only
Note:
7/29/2019 maths 2011 memo
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Mathematics/PI 10 DBE/November 2011NSC Memorandum
QUESTION 3
3.1 21; 24 9 219 24
(2)
3.2 12 2.3
= kkT
and so 1006632962.3126
52 ==
T
36)1(6312 =+= kkT k
and so ( ) 153326651 ==T
100663143
1531006632965152
=
=TT
OR
Consider sequence P: 3 ; 6 ; 12 12.3 = nnP
1006632962.3 12626 ==
P
Consider sequence Q: 3 ; 9 ; 15
36 = nQn
( ) 153326626 ==Q
100663143
153100663296
26265152
=
=
= QPTT
912.3k
9 52T
9 36 k 9 51T
9 answer
(5)
912.3 = nnP
9 26P
9 36 = nQn
9 26Q
9 answer
(5)
Note:
If candidate writesout all 52 terms and
gets correct answer:award 5/5 marks
Note:
If candidate usedk= 52: max 2/5
Note: if candidate
interchanges order
i.e. does 5251 TT :
max 4/5 marks
Note: writes out all52 terms and
subtracts 5251 TT :
max 4/5 marks
Note:
If candidate writes 218 =T
247 =T : award 1/2 marks
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Mathematics/PI 11 DBE/November 2011NSC Memorandum
3.3 For all Nn , kn 2= or 12 = kn for some Nk
If kn 2= :1
2 2.3== kkn TT
If 12 = kn :
( )123
36
12
=
=
=
k
k
TT kn
In either case,n
T has a factor of 3,
so is divisible by 3.
OR
12.3 = nnP
Which is a multiple of 3
)12(3
36
=
=
n
nQn
Which is also a multiple of 3
Since 12 = kn QT or kn PT 2= for all Nn ,
nT is always divisible by 3
OR
The odd terms are odd multiples of 3 and the even terms are 3 times a
power of 2. This means that all the terms are multiples of 3 and are
therefore divisible by 3.
9 factors 12.3 k
9 factors ( )123 k
(2)
9 factors 12.3 n
9 factors ( )123 n (2)
9 odd multiples of 3
9 3 times a power of2
(2)
[9]
Note:
If a candidate only
illustrates divisibility
by 3 with a specific
finite part of the
sequence, not thegeneral term:
0/2 marks
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Mathematics/PI 12 DBE/November 2011NSC Memorandum
QUESTION 4
4.1 The second, third, fourth and fifth terms are 1 ; 6 ; T4 and 14
First differences are: 7 ; T4 + 6 ; 14 T4So T4 + 6 + 7= 14 2T4 6
T4 = 11
d = 11 + 6 + 7 = 2 or 14 + 22 6 = 2
OR
( ) ( ) ( )
( ) ( )
2
36
32115
772715
23344525
=
=
+=
++++=
++=
d
d
d
dd
TTTTTTTT
OR
1
66
14210
8216
14525
75
639
124
=
=
=+
=+
=++
=+
=++
=++
a
a
ba
ba
cba
ba
cba
cba
T2 T3 T4 T5
1 -6 -14
-7 -7+d -7+2d
d d
9 7
9T4 + 69 14 T4
9 setting up
equation( ) ( ) ( )
23344525 TTTTTTTT ++=
9 answer
(5)
9 7
9 7 + d
9 7 + 2d
9 setting upequation
( ) ( ) ( )23344525 TTTTTTTT ++=
9 answer
(5)
9 124 =++ cba 9 639 =++ cba
9 14525 =++ cba
9 solved
simultaneously
9 answer
(5)
Note: Answer only (i.e.
d = 2) with no working:
3 marks
Note: Candidate gives
114 =T and 2=d only:
award 5/5 marks
Note: Candidate uses trial
and errorand shows this:
award 5/5 marks
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Mathematics/PI 13 DBE/November 2011NSC Memorandum
OR
11
333
22013
=
=
=+
y
y
yy
Second difference = 2131113 =+=+y
T1 T2 T3 T4 T5x 1 -6 y -14
1 x -7 y + 6 -14 -y
- 8 +x y +13 - 20 2y
9 79 6+y
9 y14
9 setting up
equation
9answer
(5)
4.2
T1 = 10
OR
10
21)1(12)1(
2112
21
1)12(2)1(4
1
12
7)1(575
1
21
2
=
+=
+=
=
=++
=++
=
=+
=+
=
T
nnT
c
c
cba
b
bba
a
n
OR
T1 1 6
-9 -7
2
9 method
9T1 = 10
(2)
9 method
9T1 = 10
(2)
Note: Answer only:
award 2/2 marks
Note:
If incorrect din 4.1,
2/2 CA marks for
T1 = d+ 8 (since
dT = 71 1 )
7/29/2019 maths 2011 memo
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Mathematics/PI 14 DBE/November 2011NSC Memorandum
QUESTION 5
5.1.1
1
130
6
)0(
=
=
= fy
( )1;0 OR x = 0 andy = 1
9 1=y
9 0=x (2)
5.1.2
3
63
361
13
60
=
==
=
x
x
x
x
( )0;3
9y = 0
9 63 =x
9 answer
(3)
5.1.3
x
y
x = 3
y = 1
(0 ; 1)(3; 0)
0 3
1
9 shape
9 both intercepts
correct
9horizontal asymptote9 vertical asymptote
(4)
5 1 4 33
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Mathematics/PI 15 DBE/November 2011NSC Memorandum
5.1.5
5
2
)2(0
1
5
1
132
6
51
=
=
=
=
m
y
OR
5
2
20
1
)2(0
)2()0(
51
=
+
=
=
ffm
951
9 formula
9 substitution
9 answer
(4)
9 formula
9 = )2(f51
9 substitution
9 answer
(4)
5.20
2
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Mathematics/PI 16 DBE/November 2011NSC Memorandum
QUESTION 6
x
y
B
C(0 ; 4,5)
A
f
g
O
6.1
3
22
28
820
3
=
=
=
=
x
x
x
x
7
81
82)0( 0
=
=
=f
A(3 ; 0) B(0 ; 7)
9y = 0
9 answer for A
9x = 0
9 answer for B
(4)6.2 8=y OR 08 =+y 9 answer
(1)
6.3
( ) 8828)2()(
2 +=
+=
x
xfxh
x4= or x22
9 ( 822 x 9 answer of
( )=
xh
x
4 or
x2
2 (2)
6.4 yx 4= OR yx 22= xy 4log= xy 2log2 =
xy 2log2
1= OR xy 2log=
OR4log
logxy =
9 switch
x andy
9 answer in the
formy =
(2)
6 5 l)( O l)( 9
Note: no CA marks
Note: answer only:
award 2/2 marks
Note: answer only
award 2/2 marks
Note: candidate
works outf-1and gets
( )8log2 += xy award 1/2 marks
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Mathematics/PI 17 DBE/November 2011NSC Memorandum
6.6==
5
4
3
0
)()(kk
kgkg
)5()4()3()2()1()0( gggggg +++=
3=x is the axis of symmetry ofgby symmetry
)4()2( gg = and )5()1( gg =
Answer = )3()0( gg +
= 4,5 + 0
= 4,5
OR
==
5
4
3
0
)()(kk
kgkg
)3()2()1()0()(3
0
ggggkg
k
+++==
)5()4()(5
4
ggkgk
+==
3=x is the axis of symmetry ofgby symmetry
)1()5(
)2()4(
gg
gg
=
=
5,4
05,4
)3()0(
)()(5
4
3
0
=
+=
+=
==
gg
kgkgkk
OR
( )2
2
0)30(5,4
03)(
+=
+=
a
xaxg
9 )5()4()3()2()1()0( gggggg +++=
9 )4()2( gg = and )5()1( gg =
9 )3()0( gg +
9 answer
(4)
9 expansion
9 )4()2( gg = and )5()1( gg =
9 )3()0( gg +
9 answer
(4)
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Mathematics/PI 18 DBE/November 2011NSC Memorandum
5,225,0
)5()4()(5
4
=
+=
+==
ggkgk
5,4
5,27
)()(5
4
3
0
=
=
== kk
kgkg
OR
cbag
cbag
cbag
cg
cbkakkg
cbxaxxg
++=
++=
++=
=
++=
++=
39)3(
24)2(
)1(
)0(
)(
)(
2
2
cbakgk
4614)(3
0
++==
cbag
cbag
++=
++=
925)5(
416)4(
++==
5
4
2941)(k
cbakg
cbakgkgkk
2327)()(5
4
3
0
+===
( )
2
1
95,4
0)30(5,4
03)(
2
2
=
=
+=
+=
a
a
a
xaxg
9 5,27
9 answer
(4)
99 cba 2327 +
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Mathematics/PI 19 DBE/November 2011NSC Memorandum
QUESTION 7
7.1 ( )
( )
years55,9
93,0log
2
1log
93,0log21log
93,02
1
07,012
1
=
=
=
=
=
=
n
n
PP
iPA
n
n
n
OR
( )
( )
years55,9
2
1log
93,02
1
07,012
1
93,0
=
=
=
=
=
n
n
PP
iPA
n
n
n
92
PA =
9 subs into correct
formula
9log
9 answer
(4)
Note:If candidate interchangesA and P
i.e. uses2
AP = : max 2/4 marks
Note:
If candidate uses incorrect
formula: max 1/4 marks
for2
PA =
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Mathematics/PI 20 DBE/November 2011NSC Memorandum
7.2 Radesh:
( )( )5508
5085,0100061
=
+=
+=inPA
OR
5508
25506000
55106000
56000of%5,80006
=
+=+=
+=A
300
000605,0Bonus
=
=
8508R
3005508Received
=
+=
Thandi:
( )
68,9158R
4
08,010006
1
20
=
+=
+=n
iPA
Thandi's investment is bigger.
9 8 550
9 8508R
9n = 20
9i=4
08,0
9 answer
9 choice made
(6)
7.3 =vF initial deposit with interest + annuity
91,28215R
33,0324158,2501
12
15,0
112
15,0
1700
12
15,010001
18
18
=
+=
++
+=
OR
9 0125,0or80
1or
12
15,0=i
9n = 189n = 18
9
18
12
15,010001
+
9
+
1215,0
112
15,01
700
18
9 answer
(6)
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Mathematics/PI 21 DBE/November 2011NSC Memorandum
OR
91,28215R
74,9074117,375
12
15,0
1
12
15,01
7001215,01300
19
18
=
+=
+
+
+=vF
9 0125,0or80
1or
12
15,0=i
9n = 19 (correspondingto 700)
9n = 18 (corresponding
to 300)
9
18
12
15,01300
+
9
+
12
15,0
11215,01
700
19
9 answer
(6)
[16]
QUESTION 8
8.1( )
( ) ( )
( ) ( )
( )
( )
( )hxh
hxhh
hxh
h
xhxhx
h
xhxhx
h
xhx
h
xfhxfxf
h
h
h
h
h
h
h
48lim
48lim
48lim
4484lim
424lim
44lim
lim
0
0
2
0
222
0
222
0
22
0
0
=
=
=
+=
+++=
+=
+=
9 formula
9 substitution
9 expansion
9 hx 48
9 answer
(5)
Note:
Incorrect notation:
no lim written:
penalty 2 marks
lim written before
equals sign:
penalty 1 mark
Note:
A candidate who
gives 8x only:
0/5 marks
Note:
A candidate who omits
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Mathematics/PI 22 DBE/November 2011NSC Memorandum
x
hx
h
hxh
h
hxhxf
hxhxfhxf
hxhx
hxhxf
xxf
h
h
h
8
)48(lim
)48(lim
48lim)(
48)()(
484
)(4)(
4)(
0
0
2
0
2
22
2
2
=
=
=
=
=+
=
+=+
=
9 substitution
9 expansion
9 formula
9 hx 48
9 answer
(5)
8.2.1
21
2
2
1
2
3
22
3
xx
x
xy
=
=
xx
xxdx
dy
=
=
2
2
2
3
2
3
9 1
23 x
9 2
2
3 x
9 x
(3)8.2.2
112
14)1(98)1(
1498)(
11449
)17()(
2
2
=
+=
+=
++=
+=
f
xxf
xx
xxf
OR
2)17()( += xxf
( )( )7172)(f B h h i l
9multiplication
9 x98
914
9 answer
(4)
Note:
Incorrect notation in
8.2.1 and/or 8.2.2:Penalise 1 mark
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Mathematics/PI 23 DBE/November 2011NSC Memorandum
QUESTION 9
9.1 ( ) cbxaxxxf +++= 232
( )( )( )
( )60426
1076
256
26
2
2
2
+=
+=
=++=
xx
xx
xx
baxxxf
60
21
422
==
=
b
a
a
( ) ( ) ( )
43
2518
56052152)5(23
=
+=
++=
c
c
cf
OR
( ) ( ) ( )
43
529
26022122)2(23
=
+=
++=
c
c
cf
43;60;21 === cba
OR
9 ( ) baxxxf ++= 262
99 ( )( )256 xx
9b= 609 422 =a
9 subs (5 ; 18) or (2 ; -9)
9c = 43
(7)
( )
( )
)5(2)5(6)5(
424
4240
)2(2)2(62
26
2
2
2
++=
=
++=
++=
++=
baf
ab
ba
baf
baxxxf9
( ) baxxxf ++= 262
9 0)2( =f
9 0)5( =f
9
Note:
If derivative equal to
zero is not written:
penalize once only
Note:A candidate who substitutes
the values ofa, b and c and
then checks (by substitution)
that T ( )9;2 and ( )18;5S lieon the curve:
award max 2/7 marks
Note:A candidate who substitutes the values ofa, b and c into the
function i.e. gets 4360212)( 23 += xxxxf and then shows by
substitution that T ( )9;2 and ( )18;5S are on the curve and works
out the derivative i.e. gets ( ) 60426 2 = xxxf and shows (bysubstitution into the derivative) that the turning points are atx = 2
andx = 5 (assuming what s/he sets out to prove and proving what isgiven): award max 4/7 marks as follows:
9x = 2 from ( ) 0= xf OR subsx = 2 into the derivative and gets 09x = 5 from ( ) 0= xf OR subsx = 5 into the derivative and gets 09substitution ofx = 2 in fand gets 9
9substitution ofx = 5 in f and gets 18
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Mathematics/PI 24 DBE/November 2011NSC Memorandum
OR
9)2( =f i.e. 92416 =+++ cba
724 =++ cba 18)5( =f i.e. 18525250 =+++ cba
268525 =++ cba 261321 =+ ba
( ) baxxxf ++= 26 2 and ( ) 02 =f OR ( ) 05 =f 244 =+ ba 15010 =+ ba
21
9
189
1899
72312
=
=
=
=+
a
a
a
ba
OR
21
9
189
1899
450330
=
=
=
=+
a
a
a
ba
( )
60
1803
7232112
=
=
=+
b
b
b
( ) ( )43
7602214
724
=
=++
=++
c
c
cba
OR ( ) ( )43
2686052125
268525
=
=++
=++
c
c
cba
9 92416 =+++ cba
and 18525250 =+++ cba
9 ( ) baxxxf ++= 26 2 9 0)2( =f or ( ) 05 =f
9 1899 =a
9b = 60
9 subs (5 ; 18) or (2 ; -9)
9c = 43
(7)
9.2 60426)( 2 += xxxf
( ) ( )
24
60142162
tan
=
+=m
( ) ( ) ( ) ( )2
4316012112123
=
++=f
Point of contact is (1 ; 2)
( )1242 = xy )1(24224
+=
+=
c
cxy
9 60426)(2 += xxxf
9 subs ( )1f 9 24tan =m
9f(1) = 2
9 ( )1242 = xy O 2624
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Mathematics/PI 25 DBE/November 2011NSC Memorandum
2
7
2
52
=
+=
x
x
OR
( )
2
7
23
21
=
=x
92
7=x
(2)
9( )23
21
=x
92
7=x
(2)
[14]
QUESTION 10
x
y
4 0 1
y = f/(x)
10.1 x-value of turning point:
2
3
2
14
=
+=x
2
3>x OR
;
2
3x
92
3>x OR
;
2
3
(1)
10.2 f has a local minimum atx = 4 because:9 x = 4
99 graph
(3)
f' f
4 1
4
(1;y)
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Mathematics/PI 26 DBE/November 2011NSC Memorandum
OR
Gradient of f changes from negative to positive at 4=x
OR
0)4( =f
0)4( >f so graph is concave up atx = 4, sof has a local
minimum atx = 4.
9 x = 4
9gradient negative for
4
7/29/2019 maths 2011 memo
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Mathematics/PI 27 DBE/November 2011NSC Memorandum
QUESTION 12
Note: If the wrong inequality 50x + 25y 500 is used, candidate wrongly says that there are more
learners than available seats. Maximum of 10 marks.
12.1 Nyx,
8
5002550
15
+
+
y
yx
yx
OR
8
202
15
+
+
y
xy
xy
99 15+yx
99 8y
99 5002550 + yx
(6)
12.2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1
2
3
4
5
6
7
8
9
0
1
2
3
4
56
7
8
9
0
1
2
x
y
9 15+yx
9 5002550 + yx
9 8y
9 feasible region(4)
12.3 yxC 300600 += 9 answer
(1)12.4.1 (6 ; 8) ; (7 ; 6) ; (8 ; 4) ; (9 ; 2) and (10 ; 0)
NOTE: The gradient of the search line is1
2=m
3 marks for all correct
solutions
2 marks if only 3 or 4
correct solutions
1 mark if only 1 or 2
Red buses
Bluebuses
Note: If candidate
gives 5002550 =+ yx :
max 5/6 marks
Note: for the inequalitys marks to be awarded,
the LHS and the RHS must be correct
M th ti /P1 28 DBE/N b 2011
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Mathematics/P1 28 DBE/November 2011NSC Memorandum
Copyright reserved
QUESTION 12.2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
1
2
3
4
5
6
7
8
9
1011
12
13
14
15
16
17
18
19
20
2122
x
y
Red Buses
Blue Buses