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MATHEMATICS PAPER 1 GRADE 12 ERRATA - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS

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QUESTION 1

 1.1.3 

 −x2 − x + 2 ≤ 0
x2 + x − 2 ≥ 0
(x+2)(x−1) ≥ 0
∴ x ≤ − 2 or/of x ≥ 1
Maths ans 1b 2020

( If learner treats inequality as and equal sign, i.e
x  ≥ −2 or x ≥ 1)      Max. (2/4)
Solution as a whole is incorrect


             OR
−x2 − x + 2 ≤ 0
(1−x)(x+2) ≥ 0
∴ x ≤ − 2 or/of x ≥ 1 
Maths ans 1c 2020

 ✓x2 − x + 2 ≤ 0
✓factorisation
✓ x ≤ − 2  ✓  x ≥ 1
(ACCURACY AS COMBO - NOT MARKED IN PARTS)
(4)

 

 

 

 

       OR

 

 

✓1−x
✓factorisation
✓x ≤ − 2 ✓ x ≥ 1
 (4)

 

 

1.3

 

 

 

CORRECTION 

 (x+m)(x+n) = 3p2
x2 + nx + mx + mn − 3p2 = 0
x2 + (m + n)x + (mn −3p2 = 0
For real roots
b2 − 4ac ≥ 0
∴ (m + n)2 − 4(1)(mn − 3p2) 
m2 + 2mn + n2 − 4mn + 12p2 
m2 − 2mn + n2 + 12p2 
(m − n)2 + 12p2 
Now: (m − n)2 ≥ 0 and 12p2 ≥ 0
∴ Δ ≥ 0 ⇒ roots are real		  

 
✓ x2 + (m + n)x + (mn − 3p2 = 0
✓ (m + n)2 − 4(1)(mn − 3p2) ≥ 0 

 

 

✓ (m − n)2 + 12p2 ≥ 0
✓ explanation

 (4)

 

QUESTION 2

 2.1.4   Pn =  Tn + k = −n2 + 36n + (51 + k)   ✓✓ answer
(ANSWER ONLY- FULL MARK)    (2)

 

QUESTION 3 

 3.1    1 + 4 + 42 + 43 + ... + 4n−1   
∴ Tn =  4n−1 
  ∴ Sum  Maths ans 3.1b 2020 OR Maths errata 3.1c 2020
 For original sequennce:
Sum : Maths ans 3.1a 2020
Sn : Maths ans 3.1a 2020− Maths ans 3.1b 2020OR Maths ans 3.1a 2020−  Maths errata 3.1c 2020		    ✓ Tn = 4n−1 
  ✓ Maths ans 3.1b 2020  OR Maths errata 3.1c 2020
 ✓ Maths ans 3.1a 2020
 ✓ answer

(4)

 

QUESTION 5

 ALTERNATIVE SOLUTION 5.3  g(x)=bx − 8

−5 = b1 − 8
 ∴ b = 3
ƒ(x) = ax2 = 3x + c 
ƒ' (x) = 2ax + 3 
            2ax + 3 = 0 (at turning point) 

2a (−¾) + 3 = 0
−3/2a = −3
a = 2
∴−5 = 2(1)2 + 3(1) + c
c = −10

  ✓substituting  (1; −5)

✓ b = 3

 

 

✓2ax + 3 = 0

✓2a (−¾) + 3 = 0

✓simplifying

✓ ƒ(1) = −5    (6)

 

 5.6   y = 3x or y = bx
 y = log3x or y = logbx 

 ✓finding h(x) 

✓answer
 Answer ONLY - Full Marks
(2) 

 

QUESTION 6

Related Items

 6.1    A = P(1−i)n
 1/3 x = x(1 − i)4 
 1/3 = (1−i)4
 3−¼ − 1 = − i
−0,24 01643143 = −i
∴  i = 0,2401643143
∴ r = 24,02% p,a

 ✓ 1/3 x = x(1 − i)4 

 

 

✓i = 0,2401643143 
✓answer   (3)

 

ACCEPT

 6.2.1    Monthly Instalment = R596458,10 
                                            72
                               =R8284,14

✓dividing by 72

✓answer    (2)
 6.2.2  Maths errata 6.2.2 2020  

✓n = − 72
✓ sustitution into correct formula

 

 

✓A = R338070,29
✓n = 5
✓sustitution into correct formula
✓answer   (6)

 6.2.1 allocate 2 marks
6.2.2 stick to original mark allocation of 6 marks
      Total of paper reduced to 147 (convert to 150)
NB - Learners that answered according to the original memo - mark out of 150

QUESTION 7

 IN THIS QUESTION WE PENALISE ONCE FOR INCORRECT NOTATION 

 

QUESTION 8

 8.3    For concave down
ƒ''(x)<0
ƒ'(x) = 6x2 + 2x − 12
ƒ''(x) = 12x + 2
∴12x + 2 < 0
x < −1/6   
IF LEARNER WORKS WITH EQUALITY SIGN AND CONCLUDES CORRECTLY  (4/4)
IF LEARNER WORKS WITH EQUALITY SIGN AND DOES NOT CONCLUDE CORRECTLY  (3/4)

✓ƒ''(x)<0
✓ƒ'(x) = 6x2 + 2x − 12
✓ƒ''(x) = 12x + 2

✓x < −1/6   

(4)
ALTERNATIVE SOLUTION  Point of inflection (x-coordinate)
ƒ(x) = 2x3 + x2 − 12x + 9
x = − b/3a 
   = − 1/3(2) 
   = − 1/6 
∴ x < −1/6    

 ✓ formula

✓substitution 

✓answer 

✓ conclusion
 8.4  6x2 + 2x − 12 ≤ 0 
Maths ans 8.4 2020
 =−2± √292
         12
=−1,59 or/of 1,26
∴ −1,59 ≤ x ≤ 1,26		  

✓substitution into correct formula
(BOTH MARKS CAN BE AWARDED FOR THE CORRECT X-VALUES IF THE LEARNER USED CALCULATOR)

✓ x - values
✓✓ answer (accuracy)

(4)

 

QUESTION 10

 10.1   Maths ans10 ERRATA 2020   

Correct entries of:

✓5; 12; 24
✓2y=3; y; y
✓x

 ✓12 / (91 − 4y − x )/ (60 −3y)

(4)
 10.3   P(M or (P and A)) = 2y + 3 + 5 + 12 + x + 24
                                                      135
                                 = 35 + 5 + 12 + 15 + 24 
                                                 135
                                = 91/135 or  0,67		  

✓2y + 3 + 5 + 12  ✓ x + 24

 

✓answer      (3)
  ALTERNATIVE SOLUTION  
P(M∩(P∩A)) = P(M) + P(P∩A) − P(M∩P∩A)
= 67/135 + 39/135 − 15/135 
= 91/135 		 ✓formula
✓substitution
✓answer    (3)

 

QUESTION 11

ALTERNATIVE SOLUTION

 11.2   Probability = 3 × 2 × 3! 
                            5!
                   =3/10   		  ✓ 3 × 2
  ✓ 3!
  ✓ 5!
  ✓ answer  (4)
Last modified on Friday, 18 February 2022 07:24
Published in Grade 12 September 2020 Preparatory Examinations

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