Mathematics - HG - Nov 2002 National Paper 1 Memorandum [Grade 12 Mathematics - HG]

Question 1 Final Version 8/11/2002 1 1 1 1

(4) correct expansion of LHS std form (-1 if not = 0) both factors

both answers, none rejected

1 2

(5) factorisation, M/A

[if ] both answers

rejecting one & accept the other

OR

(5) squaring both sides

std form factors x = 4 rejecting x = 16

OR

(5)

Let factorisation 2 y- values rejecting x = 16 accepting x = 4

If 1 3

(2)

interpretation answer

If or 1 – (2 marks) If or 1 – (1 mark) If (0 marks) Answer only x = 1 (2 marks)

1 4

(4) removing | | transposing 4 changing inequality Sign correct values

If or / , -1

(4) squaring both sides

std form factors answer

(4) definition accuracy two inequalities answer

1 2

removing square & setting

inequality

start with

* = 0 1 mark both 3 and –7 1 mark

no T≥T 0, but use a number line, x > -7 and x

T≤T 3

– x > 0 & x + 7 > 0 [BD ]

(5)

critical values (for both) each inequality

1 3

(5) [25]

T∆ T < 0 b² – 4ac sustitution in T∆

T critical values & ienquality signs

Question 2 2 1 1 1

(3)

formula substitution in formula

equation

1 2

(3)

formula substitution equation

OR

(3) formula substitution

or equation

2 2

(4) with y = x – p and then substitutes

(4,2)

subsitution both equation both values of p selecting p = 2

OR

form point C & value of m

simplification

2 3

(6)

equating the equations squaring both sides

or std form factors correct values of x

answer (answer only )

2 4

(2) [18]

answer (1 mark) ( )

Question 3 3 1 y = 2(x - 1)²

Turning point (1,0) x-intercept 2(x - 1)² = 0 x = 1

y-intercept y = 2(0 - 1)² = 2 (0,2)

graph not drawn, i e only

calculations max

(4)

y-intercept turning point

(1 mark: x value, 1 mark equal roots) shape

OR

3 1

3 2 2 1

k T≥T 0

(2) 2 is C/A marked form 3 1

answer, k > 0

3 2 2 2 k < -2 / accept k T≤T -2 (2) answer 3 3 off at A and B

[Calculation, i e not using graph max ] Draw line y = 4½ only Calculate x values and show A and B on x-axis

(2) [10]

on graph

at points of intersection max

Question 4 4 1 (5)

& knowing rem thm substitution simplification

answer

4 2

(5) [10]

knowing factor theorem substitution

knowing remainder theorem substitution

answer

Question 5 5 1

(3) change of base (once only)

simplification write in terms of log 2 & log 3

for -1

OR

OR

If start with = 0 max 5 2 2 1

(6)

exponential law

multiplying by LCD & std factorizing or both equations or 2k = 1 only x = value

2 2

(4)

log law

single log

common base answer

OR

(4)

log law removing logs 50 = 30 answer

OR

(4) factorisation

single log factorisation

x-value

OR

(4) factorisation

single log factorisation x-value

OR

2 3

single log removing of logs

if

if

*A *B

(10) std form factorization

each answer

use of definition if T≥T is used –1

values of x

solution writes down final solution from *A to *B full marks

OR

6 marks

+ last 4 marks

5 3 3 1

OR

(2) writing in log form

equation

3 2

(4) curve of f / shape y-intercept curve of f-1 / shape x-intercept

no or incorrect label: if f incorrect max 2 marks

3 3 Q on graph (1) answer

no mark if no intersect 5 4

(3)

expressing in log from exponential form

answer

OR

(3) log form log law

answer

OR

(3)

[33] Question 6 6 1

1 1

(5) substitution Sn = 425

standard form factors or

accepting n = 10 rejecting the other solution

correct answer only 1 2

(4) interpretation if d = S2 – S1 = 18 (max 1)

substitution

answer

OR

(4) calculating term 1

calculating 2nd term common difference

answer

OR

6 2 with a = 400, r = 1,1 values for a and r 2 1

(4)

formula substitution answer

OR , 440, 484, 532,40, 585,64, 644,20,

708,62 marks

2 2

(3) formula

substitution

answer

OR + 440 + 484 + 532,40 + 585,04 + 644,20

= 3794,86 all the terms ,

answer

2 3

in the 15th month

(5) substitution

log law simplification value of n & answer

if n = 14,9 max

OR 7 = 708,62, 779, 857, 943, 1037, 1141,

1255, 1380, T15 = 1518

In the 15th month

Full marks

OR

6 3 a + 1, a – 1, 2a - 5 3 1

*A *B

(7) equation

std form factors both values of a sequence when a = 3 sequence when a = -2 value of a

if goes directly from *A to *B full marks

OR only 4, 2, 1 r = ½ and series convergent

marks

a = 3 OR

r only a = 3 max

3 2

(4) formula & value for r substitution answer

If working with a = -2 or with both

values of a max

[32] Question 7 7 1

(6) definition/formula substitution

simplification/expansion simplification factorization

answer

answer only no marks

7 2 2 1

(2) y subject with negative exponent derivative

2 2

(4)

simplification derivative of each term

C/A if simplifying incorrectly, but 3rd mark for similar difficulty of 3rd term Notation -1

7 3

(6)

derivative & = -3

value of x

value of y

substitution equation

7 4

factor or (x - 3) quadratic factor

y = 0 factorization both values

definition = 0 derivative factorization both values y-value/TP y-value/TP

(17)

each turning point y-intercept on graph or in

calculations x-intercept shape

[35] Question 8 1 1

(4) equation on A (2 or 0)

equation

substitution

1 2

(4) derivative = 0

correct calculation of derivative

value for R value for r

1 3R = r = 100 one will not get the

desired shape but a shape with Utwo equal circleU which Utouch externally U. Equal radius 1 mark If a diagram is drawn, showing 2 touching cicles (2 marks)

(2) valid explanantion

8 2 2 1

(3) P = 0

factorization both values of x

2 2

if units is left out in 8 2 2

interpretation / = 0 derivative

value of x

OR

(5) substitution

value for P

OR

formula

substitution

speed substitution & answer

8 2 2 3loss P < 0

(3) setting up the inequality

[21] accept any of these 2 possible answers ignore x T≥T 30 (given)

OR 40 < x < 120 or

If 40 T≤T x T≤T 120

Question 9

9 1 1 1 x T≤T 60 y T≤T 100

(2)

each inequality if x = M, y = B in inequalities – if = sign left out –1 once

1 2 x + y T≥T 80 (1) inequality 1 3 (2) LHS RHS + inequality 9 2 = 40x + 80y = P (1) equation 9 3 graph below (1) either dotted line 9 4 (accept any x between 15 and 20) (3) profit if x = 15 and y = 100 or

0

9 5

(2)

substitution answer

9 6

max P now if x = 0 and y = 100 i e 0 type M, 100 type B

(4)

values of x & y If m = -2 feasible region is lin segment Answer (60, 20) Pmax = R4000

answer

[16]: 200