1.1.1

x ² - 5x - 6 = 0
(x - 6)(x +1) = 0
x = 6 or x = –1

OR
x ² - 5x - 6 = 0
x = 5 ± √(-5)² - 4(1)(-6)
                2(1)
x = 5 ± √49
           2
x = 6 or x = –1

1.1.2

(3x -1)(x - 4) = 16
3x ² -13x -12 = 0
x = 13 ± √(-13)² - 4(3)(-12)
                  2(3)
x = 13 ± √313
           6
x = 5,12 or x = -0,78 

OR
x ² = 3x ² -13x -12 = 0
 - 13 x = 4
    3
x² = 13x + (13)² = 4 + (13)²
         3        6                 6
(x - 13 )² = 313
       6          36 
x = 13 ± √313
            6
x = 5,12 or x = -0,78

🗸standard form
🗸correct subst into correct formula
🗸🗸answers
(4)

OR
🗸standard form
🗸adding (13)² both sides 
                6
🗸🗸answers
(4)

x(4 - x) ≥ 0
- x(x - 4) ≥ 0    or     - x(x - 4) ≥ 0
x(x - 4) ≤ 0

0 ≤ x ≤ 4      or   x ∈  [0 ; 4]

factorisation 
0 ≤ x ≤ 4
(3)

5^2x - 1 =4
5^x+ 1 
(5^x+ 1)(5^x- 1) = 4
       5^x + 1
5^x - 1 = 4
5^x= 5
x = 1 

OR
5^2x - 1 = 4
5^x+ 1 
5^2x - 1 = 4.5^x+ 4
5^2x - 4.5^x- 5 = 0
(5^x- 5)(5^x+ 1) = 0
5^x = 5 or   5^x ≠ -1
x = 1

🗸factors in numerator
🗸5^x - 1 = 4
🗸answer
OR

🗸standard form
🗸factors
🗸answer
(3)

y = 2  - x................................................. (1)
     3     3
x ² + 4xy - 5 = 0..................................... (2)
Substitute (1) in (2):
x ² + 4x ( 2  - x ) - 5 = 0
               3      3 
3x ² + 8x - 4x ² - 15 = 0
- x ² + 8x -15 = 0
x ² - 8x + 15 = 0
(x - 5)(x - 3) = 0
x = 3 or x = 5
y = - 1 or   y = -1
        3

🗸 y = 2 - x 
         3    3

🗸correct subst into correct formula
🗸either standard form 

🗸x – values
🗸y – values                         (5)

OR

🗸 y = 2 - x 
         3    3

🗸correct subst into correct formula
🗸either standard form 

🗸x – values
🗸y – values                         (5)

ab = 2√10
bc = 3√2
ac = 6√5
ab.bc.ac = 2√10.6√5.3 √2
(abc)² = 36√100
abc =360 = 6√10

OR
ac = 6√5   a = 6√5
                         c
bc = 3 2   b = 3√2 
                        c
ab = 2√10
(6√5)(3√2) = 2√10
   c      c
18√10 = 2√10.c²
c² = 9
c = 3
Volume = abc = 2√10.3 = √360 = 6√10

🗸 volume = abc 
🗸🗸 ab.bc.ac = 2√10.6√5.3√2
🗸 (abc)² = 36√100
🗸 answer
(5)

OR
🗸 a = 6√5
           c
🗸 b = 3√2
            c

🗸 value of c
🗸 Volume = abc
🗸 answer (5)
[22]

2.1.1

59

🗸 answer (1)

2.1.2

2a = -2
a =-1
3(- 1) + b = 14
b = 17
(- 1) + (17) + c = 15
c = -1
Tn = -n² + 17n - 1

🗸second difference of – 2
🗸a
🗸b
🗸 c
(4)

2.1.3

T₂₇ = -(27)²  + 17(27) - 1
= -271

🗸 substitution
🗸 answer
(2)

2.2.1

r = - 18 = - 1
       36      2

answer (1)

2.2.2

T_n = 36 (-½)^n-1
   9    = 36(-½)^n-1 
4096
    1     = (-½)^n-1
16384 
(-½)¹⁴ = (-½)^n-1  

n = 15 

OR
36 ; - 18 ; 9 ; - 9 ; 9 ; - 9 ; ... ;    9   
                      2    4     8         4096
If you look only at the denominator: 2 ; 4 ; 8 ; … ; 4096

2^k= 4096
2^k = 2¹² 
k = 12
n = 15 terms

🗸 T_n = 36 (-½)^n-1
🗸    1     = (-½)^n-1
   16384 
🗸 answer
(3)

OR

🗸 2^k= 4096

🗸 k = 12

🗸 answer

(3)

S =   a   
      1 - r
=  36   
1 - (-½)
= 24

🗸correct subst into correct formula with – 1 < r < 1 
🗸answer if – 1 < r < 1
(2)

T₁ + T₃ + T₅ + T₇ + ... + T₄₉₉
T2 + T4 + T6 + T8 + ... + T₅₀₀
= a + ar ² + ar ⁴ + ... + ar ⁴⁹⁸
    ar + ar ³ + ar ⁵ + ... + ar ⁴⁹⁹
=   a + ar ² + ar ⁴ + ... + ar ⁴⁹⁸
     r(a + ar ² + ar ⁴ + ... + ar ⁴⁹⁸ )
= 1 
   r
= -2

🗸r = ¼ and n = 250
🗸 S 250 even = -24
🗸S 250 odd = 48
answer
(4)

🗸a + ar ² + ar ⁴ + ... + ar ⁴⁹⁸
🗸ar + ar³ + ar⁵ + ... + ar ⁴⁹⁹
🗸r(a + ar ² + ar ⁴ + ... + ar ⁴⁹⁸)
🗸answer  (4)
[17]

3.1.1

p + 6 – (2p + 3) = p – 2 – (p + 6)
– p + 3 = – 8
p = 11

🗸equating i.t.o p
🗸simplifying
(2)

3.1.2

Tn = 25 + (n -1)(-8) = 33 - 8n
33 - 8n < -55
- 8n < -88
n > 11
Term 12 will be the first term smaller than  –55

🗸subst into T_nformula
🗸n > 11
🗸n = 12
(3)

3.2

S = n [a + l] = 6 [(x - 3) + (x -18)]⁶ 
      2              2 
= 6x - 63
S = n [a + l] = 9 [(x - 3) + (x - 27)]⁹ 
      2              2 
9x -135
6x - 63 = 9x -135
3x = 72
x = 24
S₁₅ = n [a + l] = 15 [(x - 3) + (x - 45)]
         2                2 
= 15 [2x - 48]
    2
= 15 [2(24) - 48]= 0 = RHS
    2

OR

= 21+18 +15 +.....+ -21.

S_n = n [a + l)]
        2
= 15 [21- 21]
    2
= 0 = RHS

OR
(x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18)
= (x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18)
+ (x - 21) + (x - 24) + (x - 27)
3x - 72 = 0
3x = 72
x = 24

= 21 + 18 + 15 +.....+ -21.
S_n = n [a + l]
        2
= 15 [21 - 21]
    2
= 0 = RHS

🗸6x - 63
🗸9x -135
🗸24
🗸15 [(x - 3) + (x - 45)]
    2
🗸substitution of x
(5)

OR
🗸expansion
🗸3x - 72 = 0
🗸24
🗸substitution of x
🗸sum of 15 terms
(5)

OR

🗸expansion
🗸3x - 72 = 0
🗸24
🗸substitution of x
🗸sum of 15 terms  (5)

[10]

4.1

y > 0
OR
y ∈ (0 ; ∞)

🗸answer (1)
OR
🗸answer (1)

4.2

y = log_½x   or   y = - log₂x  or y = log₂¹/_x

🗸x = (½)^y 
🗸equation   (2)

4.3

Yes. The vertical line test cuts g^–1 once
OR
Yes. For every x-value there is a unique y-value
OR
Yes. g is a one-to-one function
OR
Yes. The horizontal line cuts g only once

🗸yes
🗸valid reason  (2)
OR
🗸yes
🗸valid reason  (2)
OR
🗸yes
🗸valid reason  (2)
OR
🗸yes
🗸valid reason  (2)

y = -log₂x
2 = -log₂a
a = 2^-2 = ¼ or a = (½)² = ¼

🗸correct subst into correct formula (a ; 2)
🗸answer   (2)

M^/(2;¼) or M(2;a)

🗸answer (1)

M^//(-1; ⁹/₄) 

🗸-1
⁹/₄ 🗸🗸
(3)
[11]

5.1.1

x = -2
y = 3

🗸answer
🗸answer

(2)

5.1.2

🗸answer
(1)

5.1.3

  1    + 3 = 0
x + 2
1 + 3(x + 2) = 0
3x = -7
x = - 7
       3
x-intercept (-⁷/₃ ; 0)

5.1.4

🗸asymptotes at y = 3 and x = – 2
🗸intercepts at y = 3,5 and x = – 2,3
🗸shape (reasonable representation in correct quadrants)
(3)

- 2x + 4 = 0
2x = 4
x = 2
S(2 ; 0)

🗸y = 0

🗸x = 2
(2)

Equation of k:
y = a(x +1)²  +18
0 = a(2 +1)²  +18 or 0 = a(- 4 + 1)² + 18
9a = -18
a = -2
y = -2(x +1)²  +18

🗸y = a(x +1)2  +18
🗸substitute (2 ; 0) or (– 4 ; 0)
🗸a
(3)

🗸equating
🗸standard form
🗸factors
🗸choosing x = –3
🗸answer
(5)

OR
(-∞ ; - 1)

6.1.1

A = P(1 - i)ⁿ
79866,96 = 180 000(1 - 0,15)ⁿ
(1 - 0,15)ⁿ = 79866,96
                    180 000

n = 4,999… years
n » 5 years

🗸substitution
🗸use of logs
🗸answer
(3)

6.1.2

A = P(1 + i)ⁿ
= 49 000(1 + 0.1 )²⁰
                      4
= R80 292,21
The money will be enough to buy the car.

6.2.1

P = R793 749,25

OR
Balance Outstanding 

= 841 885,56 - 48 136,62
= R793 748,94

🗸n = 234 🗸 i = 0,1025
                        12
🗸substitution in present value formula
🗸answer
OR
🗸n = 6 in both
🗸 i = 0,1025
            12
🗸 A – F
R793 748,94

A = P(1 + i)ⁿ
= 793749,25(1 + 0.1025)³
                              12
New instalment

🗸= 793749,25(1 + 0.1025)³
                              12
🗸n = 231
🗸substitution of new P
🗸substitution of n and i into formula
🗸answer   (5)

[15]

7.1

f (x) = x² + 2
f (x + h) = (x + h)²  + 2
= x² + 2xh + h² + 2
f (x + h) - f (x) = x² + 2xh + h² + 2 - (x² + 2)
= 2xh + h²
f ^/( x) = lim f (x + h) - f (x)
h→0                h
=lim 2xh + h²
h→0      h

= lim h(2x + h)
 h→0      h

= lim (2x + h)
  h→0
= 2x

OR

f ^/( x) = lim f (x + h) - f (x)
h→0                h
= lim x + 2xh + h² + 2 - (x² + 2)
h→0            h
=lim 2xh + h²
h→0      h

= lim h(2x + h)
 h→0      h

= lim (2x + h)
  h→0
= 2x

🗸x ² + 2xh + h² + 2
2xh + h²

🗸=lim 2xh + h²
h→0      h

🗸= lim (2x + h)
  h→0
🗸answer
(4)

OR
🗸x² + 2xh + h² + 2
🗸=lim 2xh + h²
  h→0      h
🗸 = lim h(2x + h)
   h→0      h
🗸answer
(4)

8.1

h(x) = -2(x + 3 )(x -1)(x + 3)
                    2
h(x) = -(2x + 3)(x² + 2x - 3)
h(x) = -2x³ - 7x² + 9

OR
h(x) = -(2x + 3)(x -1)(x + 3)
h(x) = -(2x + 3)(x² + 2x - 3)
h(x) = -2x³ - 7x² + 9

🗸🗸 – 2 (x + 3 )(x -1)(x + 3)
                    2
🗸 correct simplification
(3)

OR

🗸🗸 - (2x + 3)(x -1)(x + 3)
🗸 correct simplification
(3)

8.2

h ^/(x) = -6x² -14x
- 6x ² -14x = 0
- 2x(3x + 7) = 0
x = 0 or x = - 7
                    3

🗸first derivative
🗸= 0
🗸both answers (3)

8.3

x < - 7 or  x > 0
       3
OR

x ∈ (- ∞ ; - 7) ∪ ( 0 ; ∞)
                3

🗸 🗸answer
(2)

OR
🗸🗸answer (2)

y = 4x + 7
- 6x ² -14x = 4
0 = 6x² +14x + 4
0 = 3x² + 7x + 2
0 = (3x +1)(x + 2)
x = - 1 or x = -2 
       3

🗸y = 4x + 7
🗸h ^/ (x) = 4
🗸 standard form
🗸 both answers

(4)

[12]

9.1

Volume of Sphere
= 4 π(8)³ or =2048π  or     = 2144,66
   3                    3 

🗸 answer (1)

9.2

r ² + x² = 8²      (Pythagoras)
r ² = 64 - x²

🗸 substitution or reason
Pythagoras (1)

9.3

Vcone = ¹/₃ πr²h
=¹/₃ π(64 - x² )(8 + x)
= π (512 + 64x - 8x² - x³ )
   3
dV = 64π - 16π x - 3πx²
dx      3        3          3
0 = 64 -16x - 3x²
0 = (8 - 3x)(x + 8)
x = 8           x ≠ 8
      3

🗸h = 8 + x
🗸 ¹/₃ π(64 - x² )(8 + x)
🗸expansion
🗸 dV = 64π - 16π x - 3πx²
    dx      3        3          3
🗸 x = 8 
          3
🗸 volume of the cone
🗸  8 or 0,3
    27
(7)

[9]

10.1

P(One Red and One Blue)
= P(Red, Blue) + P(Blue, Red)

= 

🗸 addition of products
🗸 answer
(4)

10.2.1

a = 0,48 x 250
a = 120

🗸 answer                            (1)

10.2.2

b = 150 P(S) × P(F)
= 200 x 150
   250    250
= 0,48
= P(S and F)
These events are independent

🗸 b
🗸 P(S) × P(F)
🗸 200 and 150
    250        250
🗸 conclusion
(with realistic probabilities)

(4)

[9]

11.1

10 × 9
= 90

🗸🗸10 × 9
(2)

11.2.1

10!
=3 628 800

🗸 10!
(1)

11.2.2

2! × 2! × 2! × 2! × 2! ×4!
= 768

🗸 2! × 2! × 2! × 2! × 2!
🗸 4!
🗸 2! × 2! × 2! × 2! × 2!×4!
or 768
(3)

[6]