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

More in this category: « MATHEMATICS PAPER 1 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS MATHEMATICS PAPER 2 GRADE 12 QUESTIONS - NSC PAST PAPERS AND MEMOS SEPTEMBER 2020 PREPARATORY EXAMINATIONS »

Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

QUESTION 1

1.1.3

−x² − x + 2 ≤ 0
x² + x − 2 ≥ 0
(x+2)(x−1) ≥ 0
∴ x ≤ − 2 or/of x ≥ 1

( 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

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

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

1.3

CORRECTION

(x+m)(x+n) = 3p²x² + 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

P_n = T_n + k = −n² + 36n + (51 + k)

✓✓ answer
(ANSWER ONLY- FULL MARK) (2)

QUESTION 3

3.1

1 + 4 + 4² + 4³ + ... + 4ⁿ⁻¹
∴ T_n = 4ⁿ⁻¹
∴ Sum

For original sequennce:
Sum :

S_n :

−

✓ T_n = 4ⁿ⁻¹
✓

✓

✓ answer

(4)

QUESTION 5

ALTERNATIVE SOLUTION 5.3

g(x)=b^x − 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 = 3^x or y = b^x
y = log₃x or y = log_bx

✓finding h(x)

✓answer
Answer ONLY - Full Marks
(2)

QUESTION 6

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

✓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

=−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		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

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

Related items