Thursday, 10 June 2021 08:57

GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2016

MATHEMATICS PAPER 1
GRADE 12
MEMORANDUM
NOVEMBER 2016
NATIONAL SENIOR CERTIFICATE

NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• Consistent accuracy applies in all aspects of the marking memorandum.

QUESTION 1

1.1.1 x(x-7)=0
x=0 or x=7

✓x = 0

✓x= 7
(2)

1.1.2

x-6x + 2 = 0

1.12

x = 0,35 or x= 5,65

OR

x2 - 6x + 2 = 0

x2 - 6x + 9 = 2+9

(x-3)2 = 7

x-3 = ± √7

x= 0,35 or x=5,65

✓correct substitution into
✓correct formula
✓x = 0,35
✓x = 5,65
(3)

✓(x-3)2 = 7

✓x= 0,35

✓x=5,65 (3)

1.1.3

√x-1 + 1 = x

√x-1 = x-1

x-1 = x2 - 2x + 1

x2 - 3x + 2 = 0

 (x+2)(x-1) = 0

 x=2 or x=1

Both  answers are valid

OR

or 2

OR

or

By inspection:

x-1=0 or x-1=1

x=2 or x=1

 

✓ isolate √x-1
✓ x2 - 2x+1
✓ standard form
✓ factors
✓ both answers

(5)

✓ isolate √x-1
✓ k2
✓ standard form
✓ factors
✓ both answers
(5)

✓ isolate √x-1
✓ x-1 = 0
✓ x-1 = 1
✓ x = 2 
✓ x=1
(5)

1.1.4

3x+3 - 3x+2 = 486

3x33-3x32=486

3x(33-32)=486

3x=27

3x=33

x=3

OR

3x+3 - 3x+2=486

3x+2(31-1) = 486

3x+2 = 35

x+2 = 5x+2 = 5

x= 3

✓expansion
✓common factor

✓3x=27

✓x=3 (4)

✓common factor

✓(31-1)

✓3x+2=243

✓x+3

(4)

1.2.1

ƒ(x) = x2 + 3x - 4

0=(x+4)(x+1)

x=-4 or x=1

✓factors
✓both answers
(2)
1.2.2

x2 + 3x - 4 < 0

(x+4)(x-1) < 0

1.22

-4 < x< 1 OR X∈(-4;1) 

✓✓-4 < x< 1 (2)
1.2.3

2x+3≥0

x≥-3/2

ƒ(x)≥0 when ƒ is increasing

The turning point occurs at x = (-4+1)/2

x≥-3/2

✓2x+3

✓x≥-3/2 (2)

✓x = (-4+1)/2

✓x≥-3/2 (2)

1.3

x=2y and x2 - 5xy = -24

(2y)2 - 5(2y)(y) = -24

4y2-10y2=-24

-6y2 = -24

y2=4

y=-2 or y=2

x=-4 or x=4

OR

1.3

OR

1.3b

✓substitution of 2y
✓-6y2=-24
✓ both y – values
✓ both x– values
(4)

✓ substitution of x/2
✓ (-3/2)x2=-24
✓ both x – values
✓ both y – values (4)

✓ equating x=x2+24
                   2      5x
✓ 3x2=48
✓ both x – values
✓ both y – values (4)

    [24]


QUESTION 2

2.1  T4=-7  ✓-7 (1)
2.2

Tn = a+(n-1)d

-87=5+(n-1)(-4)

-87=5-4n+4

4n=96

n=24

OR

-4n+9=-87

-4n=-96

n=24

✓a=5 and d=-4

✓-87 = 5+(n-1)(-4)

✓n=24 (3)

✓-4n+9

✓-4n+9=-87

✓n=24 (3)

 2.3

 -3;-7......;-87

2.3

OR

2.3 new

OR

All negative terms can be written down and added to get the answer of –990

OR

2.3b

 ✓n=22

✓a=-3

✓answer (3)

✓n=22

✓a=-3

✓answer(3)

✓a=-3

answer(3)

2.3b1

✓-6

✓answer(3)

 2.4

5;-15;-35......

d=-20

tn=-20n +25

Last term in the sequence divisible by 5 is:

-4187+4(3)

=-4175

Tn =-20n+25

-4175 =-20n+25

20n=4200

n=210

There will be 210 terms in the sequence that is divisible by 5.

OR

5 ; 1 ; –3 ; ... ; –83 ; –87; ……; – 4187

Tn=-4n+9

-4187 =-4n+9

4n=4196

n=1049

There are 1049 terms in the sequence

T1;T6;T11;T16 are divisible by 5

The largest integer value of k such that

5k-4≤1049

5k ≤ 1053

k≤210.6

k=210

OR

5 ; 1 –3 ; –7 ; … ; –4175; –4179 ;–4183 ; –4187

Tn=a+(n-1)d

-4175 = 5+(n-1)(-4)

-4180 = -4(n-1)

n=1046

Number of terms divisible by 5

=1046 - 1 +1
        5

=210

✓d=-20

✓Tn=-20n+25

✓-4175 =-20n+25

✓n=210 (4)

✓-4n+9=-4187

✓n=1049

✓5k-4 ≤1049

✓k=210 (4)

✓d=-4

✓-4175=-4n+9

✓1046

✓n=210 (4)

[11]

 

QUESTION3 

3.1.1

– 1 ;   x ;   3   ;   x + 8 ; ...

3.11

- 2x + 2 = 2x + 2

4x = 0

x = 0

 ✓x + 1; 3 - x and x

✓calculating second differences

✓- 2x + 2 = 2x + 2

 

✓x = 0(4)

3.1.2

    

First differences 1 ; 3 ; 5 ; ……

Sn = n [2(1) + (n - 1)(2)] 

        2

= n2

250 < n 2

n > √ 250

∴ n > 15,8

The sum of the 16 first differences will be greater than 250. Therefore the 17th term of the quadratic number pattern is the first satisfying this condition.

 

✓S = n 2

✓Sn > 250

✓n > 15,8

✓n = 17(4)

3.2.1

21 + 21(0,85) + 21(0,85)2  + .........

T = ar n-1 n

T   = (21)(0,85)9

= 4,86 cm

✓n = 10 ; r = 0,85 or 17

                                  20

✓substitution into correct formula

✓answer

(3)

 

3.2.2

Sn a(1 - r n )

         1 - r

S1521(1 - (0,85)15 )

              1 - 0,85

= 127,77

 Area of the page = 30 x 21 = 630 Percentage of paper covered in grey ink:

127,77 x 100%

      630

= 20,28%

✓n = 15

✓127,77 

✓630

✓20,28

(4)

[15]

 

QUESTION 4 

4.1

y = 0

✓y = 0 (1)

4.2

R(0 ; 1)

✓answer (1)

4.3

y = ax

9 = a2

∴ a = 3

✓substitution

✓a = 3

(2)

4.4

DP = 2 – b

y = 3x

1 = 3b

81

3-4 = 3b

b = - 4

DP = 2 - (-4)

= 6 units

1 = 3b

  81

✓3-4 or use of logs

✓b = - 4

✓DP = 6 units

(4)

4.5

h(x + 2) + k = 0

h(x + 2) = -k

0 < -k < 1

             81

- 1 < k < 0

 81

✓✓- k < 1    or k > - 1

        81              81

✓- 1 < k < 0

    81

(3)

[11]

 

QUESTION 5

5.1

f (x) = -x 2 + 4x - 3

f '(x) = 0      or      x = -    4 2(- 1)

- 2x + 4 = 0                     x = 2

x = 2

y = -(2)2  + 4(2) - 3

= 1

B(2 ; 1)

OR

 - x 2 + 4x - 3 = 0

x 2 - 4x + 3 = 0

(x - 3)(x - 1) = 0

x = 3 or x = 1

x 3 + 1

        2

x = 2

y = -(2)2  + 4(2) - 3

= 1

B(2 ; 1)

✓- 2x + 4 = 0 or

x = -    4  

         2(-1)

✓y

-(2)2  + 4(2)-3

(2)

✓x 3 +1

         2

 

✓y =

-(2)2  +4(2)-3

(2)

5.2

Range: y ≤ 1

OR

Range: y ∈ (-∞ ; 1]

✓y ≤ 1 (1)

✓(-∞ ; 1] (1)

5.3

x ≤ -1    or     x > 2

OR

( -∞; -1]∪(2 ;-∞)

✓critical values

✓x ≤-1  or x>2

(2)

✓critical values

✓x ≤-1  or x>2

(2)

5.4

(x - p) (y + t ) = 3

Vertical asymptote of h(x)  at x = 2

Translation 4 units to the left x = 2 – 4 = –2 is the equation of the vertical asymptote of h(x + 4)

OR

✓x = -2 (1)

 

 

h(x) =         3       + 1

             x - 2 + 4

    3    + 1

   x + 2

x = -2

is the equation of the vertical asymptote 

✓x = -2 (1)

5.5

(x - p)(y + t ) = 3

(y + t ) =    3   - t

               x - p

y =    3    - t 

      x - p

B(2 ;1)

Point of intersection of the asymptotes

p = 2

- t = 1

t = -1

  3    

    x - p

✓- t

✓p = 2

✓t = -1

(4)

5.6

x-intercepts of f :

x 2 - 4x + 3 = 0

(x - 3)(x -1) = 0

x = 1 or     x = 3

g / (x) < 0 for x ∈ R ; x ≠ 2

Hence f (x) < 0

x ≤ 1 or x ≥ 3       OR    (- ∞ ; 1]∪[3 ; ∞)

✓both critical values

✓x ≤ 1

✓or

✓x ≥ 3

(4)

[14]

 

QUESTION6

6.1

 6.1

g

✓shape: increasing curve

✓(1 ; 0):

only on log graph

f:

✓(3 ; 0)

✓(0 ; 3)(4)

6.2

y = log2 x

g -1 : x = log2 y

y = 2x

✓interchange x and y

✓y = 2 x

(2)

 

6.3

log2 (3 - x) = x

2 x = 3 - x

2 x = -x + 3

Reflect the graph of g about the line y = x to obtain g -1 and determine the point of intersection of f and g -1

✓✓2x = -x + 3

✓point of intersection of

f and g -1

(3)

6.4

x = 1

✓answer(1)[10]

 

QUESTION 7

7.1

A = P(1 + i )n

7.1

🗸substituting i and values in correct formula

 🗸 answer(2)

7.2

              
7.2

✓i = 0,15

        12

✓n = 46

✓substitution into correct formula

✓answer(4)

✓i 0,15

      12

✓n = 46

✓substitution into correct formula

✓answer

(4)

7.3

7.3

n = 35,41872568 months

∴ 36 payments are required

∴Thabiso will pay his loan off 10 months sooner

OR

✓x = 9 000

✓substitute into correct formula

✓use of logs

✓n =35,42

✓10 months

(5)

 

 

7.3b

n = 35,41872568

∴ 36 payments are required

∴Thabiso will pay his loan off 10 months sooner.

✓9 000

✓substitute into correct formula

✓use of logs

✓n =35,42

✓10 months

(5)

7.4

The balance of his loan after the 35th payment was made:

7.4

OR

7.4b

 OR

  7.4c

(4)

 

  7.4d 7.4d

 ✓✓✓✓ (4)

[15]

 

QUESTION 8

8.1

f (x + h) = 3(x + h) 2

= 3(x 2 + 2xh + h 2 )

= 3 x 2 + 6xh + 3h 2

f (x + h) - f (x) = 3 x 2 + 6xh + 3h 2 - 3x 2

= 6xh + 3h 2

OR

8.1b

8.1c

(5)

 

8.2

8.2

✓answer

✓answer

(2)

8.3

8.3 8.3b

(4)

8.4

f (x) = x3 + ax2 + bx +18

f / (x) = 3x2 + 2ax + b

At x = 1, mtan = -8

f / (1) = -8

3(1)2  + 2a(1) + b = -8

3 + 2a + b = -8

2a + b = -11........... (1)

y = f (1)

= g(1)

= -8(1) + 20

= 12

1 + a + b + 18 = 12

a + b = -7............ (2)

a = -4

b = -3

✓3x2 + 2ax + b

✓f / (1) = -8 or

3(1)2  + 2a(1)+ b = -8

✓1+ a + b +18 = 12

✓a = -4

✓b = -3

(5)

[16]

 

QUESTION 9

9.1

f'(x) = 3x2 + 8x - 3 = 0

(3x - 1)(x + 3) = 0

x = 1     or    x = -3

      3

✓equating derivative to zero

✓factors

✓x – values (3)

9.2

f // (x) = 6x + 8 6x + 8 < 0

x < - 4

        3

OR

9.2

✓6x + 8

✓✓x < - 4

        3

9.2b(3)

 (3)

9.3

x ≤ -3   or     x ≥ 1

                          3

OR

9.3

✓x  -3

✓x  1

        3 (2)

✓[-∞;-3]

9.3b(2)

9.4

f (0) = -18

d = -18

f (x) = ax3 + bx 2 + cx -18

f'(x) = 3ax 2 + 2bx + c

f'(x) = 3x 2 + 8x - 3

3a = 3              2b = 8

a = 1                  b = 4           c = -3

f (x) = x3 + 4x 2 - 3x -18

OR

f'(x) = 3x2 + 8x - 3

By integration

f (x) = x3 + 4x 2 - 3x + d

f (0) = d = -18

a = 1

b = 4

c = -3

✓d = -18

✓f '(x) = 3ax2 + 2bx + c

✓a = 1

✓b = 4

✓c = –3

(5)

 ✓f (x) = x3 + 4x2 - 3x + d

✓d = -18

✓a = 1

✓b = 4

✓c = –3

(5)

   

[13]

  

QUESTION 10

10.1

M(t ) = -t 3 + 3t 2 + 72t

M(3) = -(3) +  3(3)2  + 72(3)

= 216

216 molecules

M(3) =

✓-(3)3+3(3)2+72(3)

✓216

(2)

10.2

M (t ) = -t 3 + 3t 2 + 72t

M / (t ) = -3t 2 + 6t + 72

M / (2) = -3(2)2  + 6(2) + 72

= 72

72 molecules per hour

✓M/ (t ) = -3t 2 + 6t + 72

✓M/ (2)

✓72

(3)

10.3

M (t ) = -t3 + 3t 2 + 72t

M /(t) = -3t 2 + 6t + 72

M // (t) = 0

- 6t + 6 = 0

t = 1

Maximum rate of change of the number of molecules of the drug in the bloodstream is after 1 hour

✓M// (t )

✓M // (t ) = 0

✓answer

(3)

[8]

 

QUESTION 11

11.1

 

✓a = 20

✓b = 128(2)

 

Watches TV during exams

 

Do not watch TV during exams

Male

80

a

Female

48

12

Total

b

32

a +12 = 32

a = 20

b = 80 + 48

= 128

11.2

No

P(M and not watching TV) = 20  ≠0

                                             160

✓No

✓reason(2)

11.3.1

P(watching TV) = 128 = 4 = 0,8 = 80%

                             160    5

✓128

✓160(2)

11.3.2

P(female and not watching TV) = 12 = = 0,075 = 7,5%

                                                    160   40

✓12

✓160

(2) [8]

 

QUESTION 12

12.

              

We want to create codes that are even numbers greater than 5000. The digit 6 can be used in one of two places in these codes and therefore this presents two scenarios.

CASE 1: The first digit is a 6.

   6                                                                2

                                                                     4

   1         ×         5         ×         4         ×         2    

Number of codes starting with 6.

= 1 × 5 × 4 ×2 = 40

CASE 2: The first digit is a 5 or 7

   5                                                                2

   7                                                                4

                                                                     6

   2         ×         5         ×         4         ×         3                                                                                                                               

Number of codes not starting with 6.

begin = 2 × 5 × 4 ×3 = 120

Therefore total number of possible codes.= 40 + 120 = 160.

OR

(3 ´ 5 ´ 4 ´1) + (3 ´ 5 ´ 4 ´1) + (2 ´ 5 ´ 4 ´1)

=      60         +        60            +         40

= 160

OR

(3 ´ 5 ´ 4 ´ 3) - (1´ 5 ´ 4 ´1)

= 180 - 20

= 160

✓1 × 5 × 4 ×2

✓40

✓2 × 5 × 4 ×3

✓120

✓160

[5]

✓(3×5×4×1)

✓(3×5×4×1)

✓(2×5×4×1)

✓160

[5]

✓(3×5×4×3)

✓(1×5×4×1)

✓160

[5]

 

 

 

 

 

 

 

 

 

 
 

 

TOTAL

150

Last modified on Tuesday, 15 June 2021 07:13