education
MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA
NATIONAL
SENIOR CERTIFICATE
MARKS: 150
TIME: 3 hours
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Mathematics Grade 12 Paper 1 MP Mock 2023 Paper & Memo
Grade 12 Mathematics Paper 1 question paper and memo from the MP Mock Exam 2023, 22 pages. Read online or get the free PDF download.
Mathematics · Grade 12 · MP Mock Exam · 2023. Question paper and memorandum, 22 pages. Read online or download the PDF.
- Subject
- Mathematics
- Grade
- Grade 12
- Document type
- Question paper and memo
- Year
- 2023
- Exam period
- MP Mock Exam
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 22
- File size
- 3.9 MB
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MP Mock Exam 2023
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Mathematics P| 3
Downloaded from Stanmorephysics.com
QUESTION 1
1.1 Solve for x
LL. @-x)2-x)=0
Tele ax? 2 (Correct to 2 decimal places)
113 44 6x°
1.1.4 9° +9=10,3*
1.2 Solve for x and y:
y-2x=-1
y? —3xy=-2
12 ZS uf , calculate two values of cm
y =x 4 y
QUESTION 2
2.1. Consider the following quadratic sequence:
319 ; 280 ; 243 ; 208; 175; 144;....
2.1.1. Show that the n' term of the sequence is T, =n” —42n +360.
2.1.2 Determine which terms has a value of 0)?
2.1.3. Which term in the sequence will have the lowest value?
2.2 3r: 41-1; 23 are the first three terms of an arithmetic sequence.
2.2.1 Prove that 1=5.
2.2.2 Calculate the sum of the first 50 terms of the series.
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(4)
(4)
(4)
(6)
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(4)
(3)
(2)
(2)
(4)
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QUESTION 3
3.1 The following geometric series is given:
44+12+36+...to 15 terms
3.1.1 Write the series in sigma notation. (2)
349 ate the series. (2)
3.1.3 'S Series convergent? Provide a reason for your answer. (2)
3.2. The sum of an infinite, geometric series is 13,5(r #1).
The sum of the same series, calculated from the third term is 1,5.
3.2.1 Calculate r if r>0. (4)
3.2.2 Hence, determine the first THREE terms of the series. Q)
[12]
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QUESTION 4
In the diagram, A(—2; 0) is the x-intercept of h(x) =< 2.
=
di -....------.
h
ee ee ee fone nna Ee
! _4
i oO
!
1
H
4.1 Calculate a. (2)
4.2 Calculate the av een X=—4 and x=-2. (3)
4.3. For which values (2)
44 Determine the e (2)
your answer in
4.5 Write down the equations, S i i i (2)
the x-axis.
[11]
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QUESTION 5
Sketched below is f(x) = “58 shrek and h(x) = q*. The x-intercepts of f are BandC. h and f
intersect the y-axis . Aisa point on h vertically above C witha y-value of % PR is perpendicular
to the x-axis.
| “y
P x
R
5.1 Calculate the coordinates of D. (lb
52 Write down the value of k. ()
53 Calculate the length of BC. (4)
5.4 Calculate the length of PR if OP is 2 units. (2) ;
5.5 Consider h(x) = q*
§.5.1 Calculate the value of g. (3)
5.5.2 Hence, determine the equation of A‘, in the form y=. (2)
5.6 Write down the range of h. ()
5.7 Determine the equation of |g, the tangent to f at B. (5)
5.8 For which values of x is x.f/(x) <0? (2)
(21]
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QUESTION 6
6.1 Jane deposits Rx into an investment account. How long will it take for the value of the
investme' double if the interest rate is 5,4% p.a. compounded annually? (3)
6.2 Thabo a printing company and needs to borrow money for start-up costs. He can
make equal monthly payments of R3 300. What amount can Thabo borrow if the
interest rate on the loan is 12% p.a. compounded monthly and the loan is granted over 5
years? (4)
6.3 A group of investors consider investing in a fund that promises growth at a rate of
5% p.a. compounded quarterly. Calculate the effective annual percentage rate of the
growth promised. (3)
6.4 Sarah is 18 years old and wishes to accumulate R10 000 000 by the month before her
50" birthday. She will deposit equal monthly payments into an account that pays
15% p.a. compounded monthly. The first payment starts on her 18" birthday and the
last payment one month before her 50" birthday.
Calculate the monthly instalment that Sarah will make. (3)
{13}
QUESTION 7
7.1 Given: f(x)=-2x7 +1
Determine f'(x) from first principles. (5)
7.2 Determine:
721 Fist f@)=5* : (3)
wer gle 2x? + 4x
a ek (3)
{11}
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QUESTION 8
In the diagram, the graph of f(x) =— x? +5x" +8x-12 is drawn. A,B and C are the
x-intercepts of f. a turning point of f. T is a point on fand G is a point on the x-axis such
that TG is perpe lar to the x-axis. D is the y-intercept of f-
ny E
T,
A lo) B(I ; 0) G
D
8.1 Calculate the coordinates of C if B( ; 0). (4)
8.2 Determine the coordinates of E. (5)
8.3 For which values of x will f be concave up? (2)
8.4 Calculate the length of OG if the tangent to the curve at T is parallel to the tangent to (5)
the curve at D.
8.5 Determine the value of m if y=mx+c intersects f perpendicularly at x=5. GB)
[19]
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QUESTION 9
A farmer has a piece of land in the shape of a right-angled triangle OMN, as shown in the
figure below. He allocates a rectangular piece of land PTOR to his daughter, giving her the
freedom to choos nywhere along the boundary MN. Let OM=a, ON=b and P(x; y)
be any point on A
y
(0 :b)
1 P(x;y)
< >
fe) R M(a;0) *
Y
9.1 Determine an equation for MN in terms of a and b. (2)
9.2 Prove that the daughter’s land will have a maximum area if she chooses P at the (6)
midpoint of MN.
[3]
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NSC
QUESTION 10
10.1 A language survey was done with 45 grade 12 learners in Mbombela Secondary School.
The ai e study was to find out what language learners speak in their homes.
The re f the survey are listed below:
e 22 speak Afrikaans
e 24 speak English
e 11 speak SiSwati and Afrikaans
© 5 speak all three languages
e 3 speak only Afrikaans
e 38 learners speak at least one of the languages at home
18 speak SiSwati
10.1.1 Determine the number of learners that do not speak any of the named languages
at home. ()
10.1.2 If x learners speak English and SiSwati but not Afrikaans, draw a Venn diagram
to represent the information, (4)
10.1.3. Solve for x. (2)
10.1.4 What is the probability that a learner speaks only two of the three languages at
home? (2)
10.2 Three brothers Owen, Richard and Robert are to run a race which has 8 runners in total.
The eight competitors line up one to a lane in the lanes numbered 1 to 8.
10.2.1 Write down the total number of possible arrangements of the runners at the (D
starting line.
10.2.2 Calculate the total number of arrangements in which the three brothers are all (3)
next to each other.
10.2.3 Calculate the probability that Owen is in lane 1, Robe! lane 2 and Richard (2)
is in lane 3.
(15)
TOTAL: 150
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MPUMALANGA PROVINCE
NATIONAL
SENIOR CERTIFICATE
MARKS: 150
This marking guideline consists of 13 pages.
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QUESTION 1 / VRAAG 1
Lil (3-x)(2-x)=0
v Yanswers
x=3 or daar?
nnn Q)
1.1.2 2x? + "i
2x’ +7x=2% 0 Y standard form
—(N+ V7? -4@2)\(-2
x= ~M#NT ~4QV2) Y substitution into correct formula
2(2)
_ -74V65
4
Penalise 1 mark for || v V answers (4)
x=0,27 or x =—3,77 incorrect rounding
1.1.3 445x > 6x’
6x? —5x-4<0
(3x 4)(2x11)<0 Y factors
On ©)
<+—_}+——_+—> _ OR
it KOR
-= 4 2 3 Y critical values
2 3
1
-a<x<n
2 3 vv answer
OR
1.4
xel—5 55 (4)
1.1.4 9*+9 =10.3*
3" -10.3" +9 =0 vs
Y fact
(3° -9\3"-1)=0 “ses
; , v exponential law
3* -3° or 3* -3 Y answ (4)
x=2 or x=0 Q
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12 y-2x=-]
y= 2x-1....() v equation |
y —3xy=-2.... (2)
(2x1) —3x(@x-1) =-2 v substitution
4x° —4x 06x? +3x=-2
fanny
— 2x" — xk O Y standard form
2x +x
v
(2x+3)(x-1)=0 factors
3 v x-values
x=-> or x=1
2 v y-values (6)
y=f-3 1 or y=2(1)-1
y=-4 or y=1
OR
y-2x=-1
\ v equation |
yt
ve )
y’ —3xy=-2.... (2) V cubstiut
substitution
yo (21\y =-2
2
2yv -3y? -3y+4=0
ma ~sy+4=0 Y standard form
y +3y-4=0 ¥ factors
(y+4)(y-1=0
y=-4 or y= v y-values
Vx
_ -; or x=l x-values (6)
Multiply every term by LCD 4xy
13 4x? +4y’ =17xy v¥ multiply by LCD
4x’ -1Txy+4y’ =0 Y standard form
4x — y)(x-4y) =0
‘ p\Mx-4y) Y 5B
x-y=0
xl Y apver
y 4 y 5
x—4y=0 answer (5)
*_4
y
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OR
xy a
y * Y fraction
x tL
~yis
y x
y
x 17x Y standard form
= -==.4_.1=0
x 4y
2
7 + (2) -4(1)(1) Y substitution into correct formula
Em.
20)
17 . 225 Y simplification
_A4 ~\V 16
2
v
x x 1 answers (5)
~=4 or X=
y y 4
[25]
QUESTION 2 / VRAAG 2
2.1 | 319; 280; 243 ; 208; 175; 144;....
2.1.1 319 280 243 208
-39 -37 —35
2 2
2a=2 3a+b=-39 a+b+c=319
a=1 3(1)+b=-39 1-42 +e =319 ¥ 3(1)+b=-39
b=-42 c= 360
T, =n’ —42n+360
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v 1-42+c=319
(4)
2.1.2 n? —42n+360 =0 Y T,=0
(n-12\n-30)= 0 v factors
n=12 or 0 v answer (3)
On — Y derivative =
2.1.3 on 42 rim oR n= (42) derivative = 0
2()) v answer (2)
n=21 n=21 OR
Tar is lowest Y substitution into x = oa
a
Y answer (2)
2.2 3t 3 4r—-1; 23
2.2.1 3t 3 4t—-1; 23
4t -1—3t = 23-(4t-1) ¥ subtraction
4t-1-3t=23-4¢+1
5t=25 Y simplification (2)
t=5
2.2.2 15:19; 23 v a=15
a=15 d=4 v value of d
500
Sy = “7 PCS) +(49)4] ¥Y substitution into correct formula
=5650 v answer (4)
[15]
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QUESTION 3
3.1 x=4+12+36+....to 15 terms
T, =ar" | —~
3.1.1 a ir iat
T, =43"" T v T, = 43"
_——
(43 at v answer (2)
n=l
3.1.2 | S,= ale" =)
rl Y — substitution into the correct
15
S = 4G -1) formula
" 3-1
=28697812 V answer (2)
3.1.3 r=3 and not between —] and 1 ¥ r=3notin-I<r<l
.. not converging Y answer (2)
301 S.= 133 Y — substitution into correct formula
-r
a =13,5(I-r)
S,= _ =1,5 Y a=135(1-r)
ar? =1,5(1-r)
1,5(1-r)
=
15(1-r
13,5(1-r) = oe v equating
pea AS”
135 3,5
-l
9
r= +l
3
Butr>0 1 lear
Yo pe
= r= tal 4)
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3.2.2
v value of a
v answer Q)
[12]
QUESTION 4
4.1 0= a@_ 45 ¥ substitution
—2+1
v answer (2)
a=2
—2)- Y h(-4) and A(-2
4.2 Average gradient = M2) h-4) (A) and A-2)
-2-=(-4)
2 4+2- 2 +2) Y substitution
oe i —4+1
2
ew v answer (3)
3
43 -l<x<0 ¥ critical values
v answers (2)
44 y=xt+l+2 Y gradient
; ypax+3 v answer (2)
oR OR
yexte
2=-1he Y value of c
c=3 ¥ answer (2)
yaxt+3
-2
4.5 | g@=—- 2 ;
xt ¥ horizontal asymptote
x=-1 _
2
y=-2 (2)
ay
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QUESTION 5
hay=ge f(x)=—-48 45x +k g(x) =mx+e
S| AG) ="
=1 => D(0 3 1) v answer (1)
2 fran? v 1
5. k=1 fact answer (1)
53° | fa)=-deh tes
v =0
x -x-2=0 y Je)
(x-2)(x+1)= 0 factors
x=2 orx=-l v¥ both x values
BC = 3units v answer (4)
5.4 f2= 5027 +hC2+1 ¥ substitution of —2
=-2
PR =2 units v answer (2)
5.5.1 | hx) = va(t]
Y substitution of (5]
q=t 4
2 ¥ value of g (3)
5.5.2 1\ v equation of h
we A(x) -(5} . 4
2 Y equation of h (2)
h7'(x) =log, x
5.6 y>o v answer (1)
M(x) --x 41 v derivative
7 P@)=-e+ 2 v derivative x =-1
feyn=-Cy+h=3 Y f(=3
vr = m(x—x,)
B(-1 ; 0) ¥ substitution of B
3
yo0= 3 -CD) Y tangent__
parryi Oa]
J 7 x+ im (5)
5.8 x<-lor x>2 vv ansiaa (2)
OR
xe(-0 5; IUQ;%)
[21]
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QUESTION 6
6.1 | 4=P(l+i)’
dy = x(1 +0 ¥ substitution into correct formula
2=(1, Y correct use of logs
log(2)
a2
~ Tog(,054)
v answer (3)
n=13,18
62 |p-% [b+ i)
i
v .
012 60 value of i
3300 | 1 1+ 1D v value of n
p=—4—__*_t
0,12 Y substitution into correct formula
12
= R148351,63 Y answer (4)
; 0,05)" Y value of n
6.3 ig =| 1+ 4 -l ¥ substitution into correct
formula
— 0,05094533691
Rate = 5,09% Y rate (3)
0.15)"
x Ito al v value of n
6.4 10000000 = Js v substitution into correct formula
12
125000 = x(116,9484518)
v answer (3)
x =R1068,85
[13]
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=)
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MDE/MDOSeptember 2023
QUESTION 7 Penalty for notation error in 7.1 only.
71 | f(x)=— 2x7 41
¥ —2x? —4xh- 2h? +1
S(x+A)= +hy +1
S (x + h) =/20#? — 4xh-2h? +1
He) — fin SERS =F (2)
(0)= fig EL)
' . —2x? — 4xh— 2h? 4+1-(-2x? +1
f (x) = him I ( ) Y correct substitution into
formula
_ 2 _ 2 2
(x)= lim 2x° —4xh —2h? +14+2x° -1
0 h Y simplification
, . —4xh = 2h?
1()= fig 2
, . h(-4x—2h
S')= fim, Ce) h ) ¥ common factor
2 py, v answer (5)
721 | Se)= 42" 5x7 5x7
vx
Sf'(x) =x 45x? v 5x? (3)
244
7.2.2 | D 2x? +4x
x x
2x? x8 ¥ eax!
=D) = Paps a
xX ae -2
¥-2+x7
=D,|-2+x* |
7 -u =
= -—x" v i
4 x cere (3)
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i m
oJ
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QUESTION 8
8.1 1 _|-l 5 8 —12
00 \-a [4 [2
( i im 2 4 0 |e ¥ quadratic factor
x1 - hANX + =
: Y factors
(x-1(x?—4x-12)=0 , “e he
(x-1x -6\x+2)=0 “ewaiues
x=1or x=6 or x=-2 Vv" answer (4)
- (650)
8.2 —3x? +10x+8=0 ¥ derivative
3x? -10x-8 =0 S'(x)=0
(3x+2\x-4)=0 ¥ factors
x=-2 orx=4
3 ¥ x-values
Ff (4) =-(4)° +5(4)* +8(4) -12
= 36 v answer (5)
83 | -—6x+10>0 Y inequality
5
*s 3 Y answer (2)
8.4 | f(x) =-3x? +10x4+8
f'()=8 Y f'(0)=8
—3x? +10x+8=8 v .
—3x? +10x = 0 equating
x(—3x +10)=0 Y factors
x=0 or x= » v x-values
v¥ value of OG (5)
5 _10
“OG= 3
8.5 | f(x) =-3x? +10x+8 Y¥ substitution
S/(5) =-3(5)° +1065) +8 ;
=-17 Vv f'6)s=17
1 raat
me TT v answer (3)
loon 19]
+
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QUESTION 9
9.1 Mim = a Y m.,
b
yo-x
a
v answer (2)
9.2 | Area of PTOR =TP x PR
=o. + bx
a
x=
2
--2(2)+0=$
a\2 2
O+a 0+b) (ab
Midpoint of MN: |~9 °° J (92
«. Max at midpoint
v substitution into area formula
v simplification
v derivative =0
v value of x
Y value of y
Y midpoint (6)
[8]
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QUESTION 10 ;
0ar
10.1.1 | 45-38 = 7 Hla]
Janay v answer dd)
10.1.2 S=
v8
v6
Vil-x
¥7-x
(4)
10.1.3 | 3+84+64+5+11—x+x+7—-x+7=45 Y calculation
x=2
v answer Q)
6+8+2
10.1.4 | P(Qlanguages) = 7 Y 16
_16 v 16 (2)
a5 45
10.2.1 | 8! = 40320 v8 ()
10.2.2 | 3! x 5! x 6 = 4320 v3!
OR v 5!
3! x 6! = 4320 v 3x Sx 6 (3)
v5!
10.2.3 = = = a ;
" v= OR — (2)
8! 336
[15]
TOTAL: 150
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