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

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NOTE:

If a candidate answers a question TWICE, mark the FIRST attempt ONLY.
Consistent accuracy applies in ALL aspects of the marking guideline.
If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.
The mark for substitution is awarded for substitution into the correct formula.

QUESTION 1

1.1.1	2x² + x − 3 = 0 (2x+3)(x−1)=0 ∴ x = −³/₂ or x=1	✓ factorisation ✓ x = −³/₂ ✓ x = 1 (3)
1.1.2	x(7x+2)]=1 7x² + 2x − 1 = 0 =− 2±√32 14 =0,26 or −0,55	✓standard form ✓ Substitution ✓0,26 ✓−0,55 (4)
1.1.3	−x² − x + 2 ≤ 0 x² + x − 2 ≥ 0 (x+2)(x−1) ≥ 0 ∴ x ≤ − 2 or/of x ≥ 1 OR −x² − x + 2 ≤ 0 (1−x)(x+2) ≥ 0 ∴ x ≤ − 2 or/of x ≥ 1	✓x² − x + 2 ≤ 0 ✓x² + x − 2 ≥ 0 ✓ x ≤ − 2 ✓ x ≥ 1 OR ✓1−x ✓factorisation ✓x ≤ − 2 ✓ x ≥ 1 (4)
1.1.4	2^x + 2²^−x = ¹⁷/₂ 2.2^x + 2.2²^−x = 17 2.2^x + 2³ = 17 2^x 2.2²^x − 17.2^x + 8 = 0 let k = 2^x ∴ 2k² − 17k + 8 = 0 (2k −1)(k −8) = 0 k = ½ or/of k = 8 ∴ 2^x = 2⁻¹ or/of 2^x= 2³ x = −1 or/of x = 3 OR 2^x + 2²^−x = ¹⁷/₂ 2.2^x + 2.2²^−x = 17 2.2^x + 2³ = 17 2^x 2.2²^x − 17.2^x + 8 = 0 (2.2^2x − 1)(2^x − 8)=0 ∴ 2^x = 2⁻¹ or/of 2^x= 2³ x = −1 or/of x = 3	✓ standard form ✓ substitution ✓factorisation ✓2^x = 2⁻¹ and 2^x ≈ 3 ✓both answers OR ✓standard form ✓factorisation ✓2^x = 2⁻¹ and 2^x = 3 ✓ both answers (5)
1.2	(x − 2)² + y² = 25 (1) x + 3 − 3y = 0 (2) x = 3y − 3 (3) (3y − 3 − 2)² + y² = 25 (3y − 5)² + y² = 25 9y² − 30y + 25 + y² = 25 10y² − 30y = 0 10y(y−3) = 0 ∴ y = 0 or/of y = 3 x = −3 or/of x= 6 ∴ A(−3;0) and B(6;3) OR (x − 2)² + y² = 25 (1) x + 3 − 3y = 0 (2) y = ¹/₃ (x + 3) (3) (x−2)² + (^x/₃ + 1)² = 25 x² − 4x + 4 + x² + 2x + 1 − 25 = 0 9 3 9x² − 36x + 36 + x² + 6x − 216 = 0 10x² − 30x − 180 = 0 x² − 3x − 18 = 0 (x+3)(x−6) = 0 ∴ x = −3 or/of x= 6 ∴ y = 0 or/of y = 3 ∴ A(−3;0) and B(6;3)	✓ x = 3y − 3 ✓ substitution ✓standard form ✓factorisation ✓x and y values ✓both sets of coordinates OR ✓y = ¹/₃ (x + 3) ✓substitution ✓standard form ✓factorisation ✓x and y values ✓both sets of coordinates (6)
1.3	(x+m)(x+n) = p²x² + nx + mx + mn − p² = 0 x² + (m + n)x + (mn − p² = 0 For real roots b² − 4ac ≥ 0 ∴ (m + n)² − 4(1)(mn − p²) ≥ 0 m² + 2mn + n² − 4mn + 4p² ≥ 0 m² − 2mn + n² + 4p² ≥ 0 (m − n)² + (2p)² ≥ 0 Now: (m − n)² ≥ 0 and (2p)² ≥ 0 ∴Δ ≥ 0	✓ x² + (m + n)x + (mn − p² = 0 ✓ (m + n)² − 4(1)(mn − p²) ≥ 0 ✓ (m − n)² + (2p)² ≥ 0 ✓ explanation (4)
		[26]

QUESTION 2

2.1

2.1.1	206; 231	✓206 ✓ 231 (2)
2.1.2	2a = −2 ∴a = −1 3a + b = 33 −3 + b = 33 ∴ b = 36 a + b + c = 86 −1 + 36 + c = 86 ∴ c = 51 T_n = −n² + 36n + 51	✓a = −1 ✓b = 36 ✓ c = 51 ✓ T_n = −n² + 36n + 51 (4)
2.1.3	326 = −n² + 36n + 51 n² − 36n + 275 = 0 (n − 11)(n − 25) = 0 ∴ n = 11 or/of n = 25	✓substitution ✓method ✓answers (3)
2.1.4
	2a = − 2 ∴ a = −1 3a + b = 33 −3 +b = 33 ∴ b = 36 a + b + c = 86 + k −1 + 36 + c = 86 + k ∴ c = 51 + k ∴T_n + k = −n² + 36n + (51 + k)	✓ a = −1 and b = 36 ✓ c = 51 + k (2)
2.2.1	2y − 1 ; 4y − 1 ; 6y − 1 d = 4y − 1 − (2y − 1) = 2y T_n = 2yn − 1 T₃₀ = 2y(30) − 1 = 60y − 1 OR T_n = a + (n−1)d T₃₀ = (2y − 1) + (30 − 1) (2y) = 2y − 1 + 58y = 60y − 1	✓ d = 2y ✓2yn ✓ 60y − 1 OR ✓ d = 2y ✓ substitution ✓ answer (3)
2.2.2	S_n = ⁿ/₂ [2a + (n − 1)d] S₃₀ = ³⁰/₂[2(2y − 1) + 29(2y)] =15(4y − 2 + 58y) = 15 (62y − 2) ∴ −2820 = 15(62y − 2) −188 = 62y − 2 −186 = 62y ∴ y = −3 OR S_n = ⁿ/₂ (a+l) S₃₀ = ³⁰/₂ (2y − 1 + 60y − 1) −2820 = 15(62y − 2) −188 = 62y − 2 −186 = 62y ∴ y = −3	✓ d = 2y ✓ substitution into correct formula ✓ equating to − 2820 ✓ answer OR ✓ d = 2y ✓ substitution a and l ✓ equating to − 2820 ✓ answer (4)
		[18]

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)
3.2	S_∞ = 1 a − r ∴ 1 + x + x² + x³ + ... = 1 and 1 − x 1 − x + x² − x³ + ... = 1 1 + x Sum : 1 + 1 1 − x 1 + x = 1 + x + 1 − x (1 − x)(1 + x) = 2 1 − x² ∴ 2 = 8 1 − x² 8 − 8x² = 2 −8x² = − 6 x² = ¾ x = ± ^√3/₂	✓ 1 1 − x ✓ 1 1 + x ✓ 2 1 − x² ✓ equating sum to 8 ✓ x² = ¾ ✓ answer (6)
		[10]

QUESTION 4

4.1	ƒ (x) = a + 3 x − 1 x = 1; y = 3	✓ x = 1 ✓ y =3 (2)
4.2	y = a + 3 0 − 11 = 3 − a 0 = a + 3 x − 1 − 3 = a x − 1 −3x + 3 = a x = 1 − ^a/₃ or /of 3 − a 3	✓ 3 − a ✓ y = 0 ✓ x = 1 − ^a/₃ or /of 3 − a 3 (3)
4.3		✓ asymptotes ✓ y - intercepts ✓ x - intercepts ✓ shape (4)
4.4	ƒ (x) = a + 1 or/of ƒ(x) = −1 +1 x + 2 x + 2	✓ x + 2 ✓ +1 (2)
		[11]

QUESTION 5

5.1	0; −7	✓y=−7 (1)
5.2	q = −8	✓ q = −8 (1)
5.3	g(x)=b^x − 8 −5 = b¹ − 8 ∴ b = 3 At turning point of ƒ: x = −^b/_2a = − ¾ ⇒−³/_2a = −¾ ∴−6a = −12 a = 2 ∴−5 = 2(1)² + 3(1) + c c = −10	✓substituting (1; −5) ✓ x = −^b/_2a ✓−^b/_2a = −¾ ✓ b = 3 ✓ Simplifying ✓ ƒ(1) = −5 (6)
5.4	y > −8	✓✓ y > −8 (accuracy) (2)
5.5	y + 9x = − 28 y = −9x − 28 y = 2x² + 3x − 10 ∴ 2x² + 3x − 10 = − 9x −28 2x² + 12x + 18 = 0 x² + 6x + 9 = 0 (x + 3) (x + 3) = 0 ∴ x = −3 y = 2(−3)² + 3(−3) − 10 =−1 ∴ T = (−3; −1) OR y = −9x − 28 ∴m=−9 ƒ(x) = 2x² + 3x − 10 ƒ' (x) = 4x + 3 = −9 ∴ 4x = −12 x = −3 y = −9(−3)−28 = −1 ∴T(−3;−1)	✓ 2x² + 3x − 10 = − 9x −28 ✓ standard form ✓factorisation ✓x =− 3 ✓y =− 1 OR ✓ m =− 9 ✓ƒ'(x) = 4x+3 ✓equating gradients ✓x = −3 ✓y =− 1 (5)
5.6	y = log₃x	y = log₃x (2)
5.7	p(x) = ƒ(x) + 1 = 2x² + 3x − 10 + 1 =2x² + 3x −9 =(2x−3)(x+3) ∴ when: y = 0 x = ³/₂ or/of −3 ∴ x < −3 or/of 0 < x < ³/₂	✓p(x) =2x² + 3x −9 ✓x-intercepts ✓x < −3 ✓0 < x < ³/₂ (accuracy) (4)
		[21]

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 = −i ∴ i = 0,24 ∴ r = 24%	✓ ¹/₃ x = x(1 − i)⁴ ✓i = 0,24 ✓answer (3)
6.2.1		✓ i = 0,095 12 ✓n = 72 ✓F =R596458,10 ✓Substitution into correct formula ✓answer (5)
6.2.2	=R325 000	✓n = − 72 ✓ sustitution into correct formula ✓A = R338070,29 ✓n = 5 ✓sustitution into correct formula ✓answer (6)
		[14]

QUESTION 7

7.1	OR ƒ(x) = −2x² ƒ(x+h)−ƒ(x) = −2(x+h)²−(−2x²) =−2x² − 2xh − 2h² + 2x² = −4xh − 2h² =−4x	✓ substitution ✓ expansion ✓ −2x² − 4xh − 2h² + 2x² ✓factorisation − 4x OR ✓−2x² − 2xh − 2h² + 2x² ✓−4xh − 2h² ✓substitution ✓factorisation ✓−4x (5)
7.2	y = 7x⁴ − 2 √x³ = 7x⁴ − ∴ ^dy/_dx =	✓ ✓✓ (3)
7.3	D_t (½gt² − ⁵/_t + 3g) = D_t (½gt² − 5t⁻¹ + 3g) =gt + 5t⁻²	✓5t⁻¹ ✓gt ✓+5t⁻² ✓D_t (3g) = 0 940
		[12]

QUESTION 8

8.1	(0;9)	(0;9)
8.2	ƒ(x) =2x³ + x² − 12x + 9 ƒ(1) = 2(1)³ + (1)² − 12(1) + 9 =0 ∴ ƒ(x) = (x−1)(2x² + 3x − 9) ∴ 0 = (x−1)(x+3)(2x−3) ∴ x=1 or/of x=−3 or/of x=³/₂ D(−3;0), E(1;0), F(³/₂;0)	✓method ✓2x² + 3x − 9 ✓factorisation ✓D(−3;0), ✓E(1;0), ✓F(³/₂;0) (6)
8.3	For concave down ƒ''(x)<0 ƒ'(x) = 6x² + 2x − 12 ƒ''(x) = 12x + 2 ∴12x + 2 < 0 x < −¹/₆	✓ ƒ''(x)<0 ✓ƒ'(x) = 6x² + 2x − 12 ✓ƒ''(x) = 12x + 2 ✓x < −¹/₆ (4)
8.4	6x² + 2x − 12 ≤ 0 =−2± √292 12 =−1,59 or/of 1,26 ∴ −1,59 ≤ x ≤ 1,26	✓substitution into correct formula ✓ x - values ✓✓ answer (accuracy) (4)
		[15]

QUESTION 9

Area of Δ PBA = ½bh = ½ × t × 12 = 6tcm²
Area of Δ ACS = ½t(7 − t) = 7 t− t² cm²
2 2
Area of ΔBQC = ½(12 −t)7 = 42 − ^7t/₂ cm²
Area of rectangle : PBQS = 12 × 7 = 84cm²
∴ Area of ΔABC:
A(t) = 84 − 6t − 7t + t² − 42 + 7t
2 2 2
=½t² − 6t + 42
a>0, so A (t) is a min graph
A'(t) = t −6
At min : A'(t) = 0
∴ t − 6 = 0
t = 6
Smallest area
A(6) = ½(6)² − 6(6) + 42
=24cm²

✓Area of ΔPBA
✓Area of Δ ACS

✓Area of ΔBQC

✓subtracting from 84

✓A'(t) = 0

✓smallest area

(6)

[6]

QUESTION 10

10.1		Correct entries of: ✓12 ✓5; 12; 24 ✓2y=3; y; y ✓x (4)
10.2	x + y + 5 + 24 = 60 x + y = 31 2y + 3 + x + 5 + 12 = x + y + 12 + 24 2y + x + 20 = x + y + 36 ∴ y = 16 and x = 15	✓x + y + 5 + 24 = 60 ✓2y + 3 + x + 5 + 12 = x + y + 12 + 24 ✓y = 16 and ✓ x = 15 (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)
		[11]