TECHNICAL MATHEMATICS PAPER 1
GRADE 12
NOVEMBER 2021
NSC EXAMINATIONS

QUESTION 1
1.1.1
2x(x + 3) = 0
x=0 or x= -3
(2)

1.1.2
x(x+9) = 12
x² + 9x - 12 = 0
x = -b ± √b² - 4ac
               2a
= -(9) ± √(9)2 - 4(1)(-12)
                   2(1)
= -9±√129
        2
x ≈ 1,1,8 or x ≈ -10,18
(4)

1.1.3
x(6 - x) ≥ 0
Critical values: 0 and 6
0 ≤ x ≤ 6 OR x ∈ [0;6] OR
x ≥ 0 and x ≤ 6

• both critical values
• notation/notasie A
• number line representation CA (3)

Note: Award full marks if ONLY the correct number line representation is shown.

1.2
x = 1 - 2y and 3x² = 3 + x + y
3(1 - 2y) = 3+(1 - 2y) + y 
3 - 12y + 12y² = 4 - y 
12y² -11 - 1 = 0
(12y + 1)(y - 1) = 0 OR
y = -(11)±√(11)² - 4(12)(-1)
                  2(12)
y = -¹/₁₂ -0.08 or y = 1
x = 1 -2(-¹/₁₂) or x = 1 - 2(1)
x = ⁷/₆ ≈ 1,17 or x = -1

• substitution A
• S CA
• correct standard form CA
• factors/formula CA
• both y-values CA
• both x-values CA

OR
y = 1 - x and 3x² = 3 + x + y 
        2
3x² = 3 + x + 1 - x
                        2
6x² = 6 + 2x + 1 - x
6x² - x - 7 = 0 
(6x - 7)(x + 1) = 0 OR
x =  -(-1)±√(-1)² - 4(6)(-7)
                   2(6)
x = ⁷/₆ ≈ 1,17 or x = -1
y = 1 - 7/6 or y = 1-(-1)
         2                   2 
y = -¹/₁₂ ≈ -0.08 or y = 1

• substitution A
• S CA
• correct standard form CA
• factors/form CA
• both x-values CA
• both y-valuesCA

OR
x = 1 - 2y(1) and 3x² - 3 - x = y(2)
Sub(2) into (1)
x = 1 - 2(3x² - 3 - x)
x = 1 - 6x² + 6 - 2x
6x² - x - 7 = 0
(6x - 7)(x + 1) = 0
x = ⁷/₆ ≈ 1,17 or 
x =  -(-1)±√(-1)² - 4(6)(-7)
                   2(6)
y = -¹/₁₂ ≈ -0.08 or y = 1

• substitution A
• S CA
• correct standard form CA
• factors/form CA
• both x-valuesCA
• both y-valuesCA
  NPR
  (6)

1.3.1
T = 2π√^L/_g
^T/_2π = √^L/_g
(^T/_2π)² = (^L/_g)²
L = g.(^T/_2π)²
M squaring both sides A
L subject A
OR
T = 2π√^L/_g
T² = (2π√^L/_g)²
T² = 4π^2L/_g
L =  gT²
      4π²
M squaring both sides A
L subject A
(2)

1.3.2
L = g.(^T/_2π)²
L = 9.8.(^1.74/_2π)²
L = 0.75m

• SF CA
• value of L CA

OR
L = gT²
     4π²
L = (9.8)(1.74)²
           4π²
L = 0,75m

• SF CA
• value of L CA

OR
T = 2π√^L/_g
1.74 = 2π√^L/_9.8
L = 9.8.(^1.74/_2π)²

• SF A
• value of L CA
  NPR NPU
  (2)

1.4.1
1101100₂ - 11100₂ = 1010000₂
(1)
Note: No penalty if base 2 is omitted.

1.4.2

= 108 - 28 = 80

• M CA
• decimal value CA
  OR
• M A
• decimal value CA (2)

AO: full marks
[22]

QUESTION 2
2.1.1 Non-real (1)

2.1.2 Real, rational, equal(1)

2.2 
-x² + 2qx - 4 = 0 OR x² - 2qx + 4 = 0
b2 - 4ac < 0 
(2q)² - 4(-1)(-4) < 0 OR (-2q)² - 4(1)(4) < 0
4q² - 16 < 0 
q² - 4 < 0 
(q - 2)(q + 2) < 0 
-2 < q < 2 OR q (-2; 2) OR q > -2 and q < 2 

• SF A
• end points and correct notation CA (3)

[5]

QUESTION 3
3.1.1
(81a^-8) ^-¾
= (3⁴a^-8)^-¾ 
= 3^-3a⁶ OR ¹/₂₇a⁶ OR a⁶/27 
Prime base or exponential property(3)

3.1.2 
log₂16 + log₃4º
= log₂2⁴ + log₃1
= 4log₂2 + log₃1
= 4(1) + 0
= 4
OR
log₂16 + log₃4º
= log₂2⁴ + 0log₃4
= 4log₂2 + 0log₃4
= 4(1) + 0
= 4
OR
log₂16 + log₃4º
= log16 + log4º
    log2     log 3
= log2⁴ + log1
    log2    log3
= 4log2 +   0   
    log2    log 3
= 4 + 0 
= 4
(4)

3.1.3
√50x¹⁰ x √18x^-4
= √900x⁶
= 30x³

• Product of the surds
• 30 CA
• 3 x CA

OR
√50x¹⁰ x √18x^-4
= √25 x 2x¹⁰ x √9 x 2x^-4
= √52x⁵ x 3√2x^-2
= 30x³

• Product of perfect squares and prime number
• 30 CA
• 3 x CA
  (3)

3.2
log₃ (x + 2) = 2 +log₃x
log₃ (x + 2) - log₃x = 2
log₃x + 2 = 2
         x
x + 2 = 3² OR log₃x + 2 = 2log₃3
  x                            x 
x + 2 = 9x
x = ¼ OR x = 0.25

• log property A
• exponential CA
• S CA
• x – value CA

OR
log₃ (x + 2) = 2 + log₃x
log₃ (x + 2) = 2log₃3 + log₃x
log₃ (x + 2) = log₃9 + log₃x
log₃ (x + 2) = log₃9x
x + 2 = 9x
x = ¼ OR x = 0.25

• log property A
• log property A
• S CA
• x – value CA
  (4)

3.3.1
IzI = r = √x² + y²
2√5 = √(p)² + (4)²
20 = p² + 16
p² = 4 OR (p + 2)(p - 2) = 0 OR p ±√4
p = ±2
p = -2 , θ ∈(90º;180º)
(4)

3.3.2
tanθ = ^y/_x  OR cosθ = ^x/_r
tanθ = ⁴/_-2 OR cosθ =^-2/_2√5
ref .angle = 63,43º
θ =180º - 63,43º =116,57 OR 2.03rad
z = 2√5 cis 116,57º OR
z = 2√5 cis 2,03 rad

• ref. angle CA
• value of CA
• z in polar form CA

OR
sinθ = y/r
sinθ = ⁴/_2√5
ref .angle = 63,43º
θ = 180º - 63,43 = 116,57 OR 2,03 rad
z = 2√5 cis 116,57º OR
z = 2√5 cis 2,03 rad

• ref. angle A
• value of CA
• z in polar form CA (3)

3.4
2m -  ni - 6i = -3i (4i + 5)
2m -  ni = 12i² - 15i + 6i
2m -  ni = 12(-1) - 9i
2m -  ni =12 - 9i
2m = 12
m = 6
-ni = -9i
n = 9

• Product 
• substituting i² with – 1
• value of m
• value of n

OR
2m -  ni - 6i = -3i (4i + 5)
2m - (n + 6)i = 12i² - 15i
2m -(n + 6)i = 12(-1) - 15i
2m - (n + 6)i =12 - 15i
2m = 12
m = 6
-(n + 6)i = 15i
n = 9

• Product 
• substituting i² with – 1
• value of m
• value of n

OR
2m -  ni - 6i = -3i (4i + 5)
2m - (n + 6)i = 12i² - 15i
2m - ni - 6i = 12(-1) - 15i
2m - ni - 6i =12 - 15i
2m - 12 = ni + 6i - 15i
2m - 12 = 0 and ni-9i = 0
2m = 12
m = 6
ni = 9i
n = 9

• Product 
• substituting i² with – 1
• value of m
• value of n (4)

[25]

QUESTION 4
4.1.1

1. y = 6 OR (0;6) (1)
2. y = 5 (1)
3. k(x) = -2x² + 4x + 6 
   y-int: y = 6
   x-ints..:
   2x² + 4x + 6 = 0
   x² - 2x - 3 = 0
   (x + 1)(x - 3) = 0 OR x = -(-2)±√(-2)² -4(1)(-3)
                                                       2(1)
   
   • y-int
   • factors/form
   • both values of x (3)
4. TP
   x = ^-b/_2a = ^-(4)/_2(-2)
   =1
   y = -2(1)² + 4(1) + 6 = 8
   =(1;8)
   SF
   x-value
   y-value
   OR
   TP
   x = -1 + 3 = 1
   2
   y =-2(1)2 + 4(1) + 6 = 8
   = (1;8)
   M 
   x-value
   y-value
   OR
   k(x) = -2x² + 4x + 6
   k/(x) = -4x + 4 = 0
   x = 1
   y = -2(1)² + 4(1) + 6 = 8
   = (1;8)
   M
   x-value
   y-value
   OR
   (  -b   ; 4ac - b²)
      2a        4a
   [ -(4)   ; 4(-2)(6) - (4)²]
    2(-2)           4(-2)
   = (1;8)
   SF A
   x-value
   y-value (3)

4.1.2

h :

• shape
• y- int.
• asymptote

Related Items

k:

• shape
• x & y int..
• urning point(6)

4.1.3
p(x) = ^a/_x + q and (1; 8)
q = 5
8 = ^a/_-1 + 5
a = -3
value of q
SF 
value of a  (3)

4.2.1
C(5 ;0)
x-value of C 
y-value at C  (2)

4.2.2
D(0; -10) and B(0; -5)
BD = -5 -(-10) = 5 units
coordinates of B & D
length of BD (2)

4.2.3
f (x) = -√25 - x² OR f (x) = -√5² - x²
equation(1)

4.2.4
3< x <5 OR x ∈( 3;5) OR x > 3 and x < 5
critical values
correct notation (2)
[24]

QUESTION 5
5.1.1
⁴/₂₃ ≈ 17,39 % (1)

5.1.2
A = P(1 + in)
A = R63 150(1 + ⁴/₂₃ x 7) OR R63 150 (1 + 17,39 % x 7)
= R140028, 26
≈R140022,50
OR
SI = P × i × n
= R63 150 x ⁴/₂₃ x 7
OR
=R63150 x 17,39% x7
= R76 878,26 
≈R76 872,50
A ≈ R63 150 + R76 878,26
OR
R63150 + R76 872,50
»R140 028,26   ≈R140 022,50
(2)

5.2
A = P(1 + i)ⁿ
R274 000 = R726 900 (1 - 15,8%)ⁿ
2 740 = (1 - 15%)ⁿ
7 269
n = log(²⁷⁴⁰/₇₂₆₉)
         log(0.842)
n ≈ 5.67
n > 5.67 years
log form
value of (4)

5.3.1
A = P(1 + i)ⁿ
= R25 000(1 + 2,8%)⁴
= R27 919,81
(2)

5.3.2
Value of investment after 27 months
A = P(1 + i)ⁿ
= R15 000 (1 + 5.98%)²⁷
                            12
= R17154,59482
Value of investment after 21 months
P = R17154,59482 + R6823,54 ≈ R23978, 13482
Value of investment after 7 quarters
= R23 978, 13482( 1 + 7,78%)⁷
                                        4
= R27439, 55
R27439, 55 <  R27 919,81
He will not have enough money
OR
27
5,98%
A =R15 000(1 + 5.98%)²⁷ (1 + 7.78%)⁷
                             12                    4 
+R6 823,54( 1 + 7,78%)⁷
                               4
= R27439, 55
R27439, 55 < R27 919,81
He will not have enough money
OR
A = [R15 000 (1 + 5.98%)²⁷ + R6 823,54 (1 + 7.78%)⁷]
                               12                                         4 
≈ R27439, 55 < R27 919,81
He will not have enough money
Conclusion without calculation: 0 marks (6)
[15]

QUESTION 6
6.1
f(x) = -3x
f^/(x) = limf(x + h) - f(x)
                     h
= lim -3 (x + h) - (-3x)
                   h
= lim -3x - 3h + 3x
                 h
lim -3h 
       h 
= lim(-3)
f^/(x) = -3
definition
Penalty of one mark for incorrect notation (5)

6.2.1
D_x[p³x² - 7x + 10]
= 2p³x - 7 (2)

6.2.2
y = x - 3x²
         x⁷
y = x^-6 - 3x^-5
dy/_dx = -6x^-7 + 15x^-6
(3)

6.2.3
f(x) = ³√x² + 5x⁴
f(x) = x^2/₃ + 5x⁴
f/(x) =²/₃x^-1/₃ + 20x³ (3)

6.3.1
m = - 9
value of m (1)

6.3.2
y = x² + 3x - 2
^dy/_dx = 2x + 3
2x + 3 = -9
2x = -12
x = -6
y = (-6)² + 3(-6) - 2
y = 16
(-6 ; 16)

• derivative of  y
• equat.deriv
• – 9 
• value of x 
• value of y (4)

6.3.3
g(x) = x² + 3x - 2
g(x) = (-2)² + 3(-2) -2 = -4
g(x) = (3)² + 3(3) - 2 = 16
Ave. grad. = y₂ - y₁
                    x₂ - x₁
= 16 - (-4) 
3-(-2)
= ²⁰/₅
= 4
both values of y
M subst. into Ave. gradient Formula/verv in gem gra
m_ave value
OR
Ave. grad. = g(x₂) - g(x₁)
                        x₂ - x₁
= [(3)² + 3(3) - 2] - [(-2)² + 3(-2) -2]
                         3-(-2)
= 16-(-4)
    3-(-2)
= ²⁰/₅
= 4
OR/OF
both values of y
M subst. into Ave. gradient Formula/verv in gem gra
m_ave value (3)
[21]

QUESTION 7
7.1
f(x) = x³ - 2x² - 7x - 4
y = -4 OR (0;-4) (1)

7.2
f(4) = (4)³ - 2(4) - 7(4)
= 0
(x - 4) is a factor of f (x) (2)

7.3
x-intercepts; y = 0
(x - 4)(x² + 2x + 1) = 0
( x - 4) ( x + 1) ( x + 1) = 0
x = -1 or x = 4

• quadratic factor
• factors
• x-intercepts

OR
(x + 1)(x² - 3x - 4) = 0
( x + 1) ( x - 4) ( x + 1) = 0
x = -1 or x = 4

• quadratic factor
• factors
• x-intercepts

7.4
f^/(x) = 3x² - 4x - 7 = 0
(3x - 7)(x + 1) = 0 OR x = -(-4)±√(-4)² -4(3)(-7)
                                                     2(3)
x = ⁷/₃ or x = -1
f(⁷/₃) = (⁷/₃)³ - 2(⁷/₃)² - 7(⁷/₃) - 4 = -⁵⁰⁰/₂₇  ≈ -18.52
(⁷/₃ ; -⁵⁰⁰/₂₇) and(-1 ; 0)

• derivative
• quating derivative to 0
• factors/formula
• both values of x
• both values of y (5)

7.5

• shape 
• y-intercept
• both x-intercepts
• both turning points(4)
  AO: Full marks/Volpunte

7.6
-1 < x < ⁷/₃OR -1 < x < 2,33
crit. values
correct notation
OR
x ∈ (-1 ; ⁷/₃) OR x ∈ (-1 ; 2,33)
crit. values
correct notation
OR
x > -1 and x < ⁷/₃ OR x > -1 and x < 2,33
crit. values
correct notation(2 )
[17]

QUESTION  8

8.1
V = l x b x h
4000 = x2h 2 
∴ h =   4000 
            x²
✔ SF A (1)

8.2
Tot. Surface Area = length × breadth + 2 × length × height +2 × breadth × height
 Tot .Surface Area = x2 + 2xh + 2xh 
x2 + 4x (⁴⁰⁰⁰/_x²)
= x2 + 16 000 
              x
OR/OF

∴  Tot. Surface Area = Area of base + (perimeter of base × height)
Tot. Surface Area = x² + 4xh
x2  + 4x (4000/x²)
x² + 16 000  =
           x

8.3
∴ Tot = x² +  16 000
                         x
= x² + 16 000x^-1

d(Surface Area) = 2x - 16000x^-2
        dx
= 2x - 16000
              x²
2x³ - 16000 = 0
x³ = 8000
x = 20cm
h = 4000 = 10
      (20)²

QUESTION 9

9.1.1
∫x (x² x 6x)dx
= ∫(x3 + 6x2 )dx 
= x⁴/₄ + 2x³ + C
✔ S A
✔
(4)

9.1.2
∫ (3x + ¹/_x)dx
= 3^x /ln3 + ln x + C

9.2
A = ∫⁴_k g(x)dx
= ∫⁴_k 3x² dx
= x³]⁴_k 
= (4)3 - (k)3

∴(4)³ - (k)³ = 56

∴k³ = 8

∴k= 2

OR

A= ∫⁴₁₀ g(x)dx
=  ∫⁴₁₀ 3x² dx
= x³]40
= (4)³ - (0)³ = 64
A = ∫²₀ g(x)dx
= x³]²₀
= (2)³ - (0)³ =8

∴ 64 - 8 = 56

∴ k= 2

OR

Trial & Error Method
A= ∫⁴_k g(x)dx
= ∫⁴_k 3x² dx
= x³]⁴_k

Let k = 1
= x³]⁴₁
= (4)³ - (1)³
= 63

Let k = 2
= x³]⁴₂
=(4)³ - (2)³
= 56
∴ k = 2

[13]
TOTAL/TOTAAL: 150