Mathematics P1 Grade 12 Memorandum - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations

More in this category: « Mathematics P2 Grade 12 Questions - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations Mathematics P2 Grade 12 Memorandum - NSC Exams Past Papers and Memos September 2019 Preparatory Examinations »

Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

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.

MARKING SCHEMES

QUESTION 1
1.1.1	x2 - 3x - 4 = 0 (x + 1)(x - 4) = 0 x = -1 or 4 Answerd only: (2/3) OR can use quadratic formula ∴ x = 4 or x = -1	√ Factors √ x = -1 √ x = 4 √ correct substitution √ √ answers (3)
11.2.2	2x² - x - 7 = 0 penalise 1 mark for incorrect rounding off x = 2,14 or -1,64	√ substitution √ x = 2,14 √ x = -1,64
1.1.3	5^x+1 - 5^x = 2500 5^x.5¹ - 5^x = 2500 5^x(5 - 1) = 2500 5^x.4 = 2500 5^x = 625 5^x = 5⁴ ∴ x = 4	√ factorisation √ 5^x = 625 √ x =4
1.1.4	(x - 3)(x + 1) < 12 x² - 2x - 3 - 12 < 0 x² - 2x - 15 < 0 (x - 5)(x + 3) < 0 OR -3 < x < 5 OR x∈(-3;5)	√ standard form √ factorisation √√ -3 < x < 5 (accuracy)
1.2	y = 2x - 1 .......... (1) 3x² - xy - y² = 1 ......(2) (1) into (2) 3x² - x(2x - 1) - (2x - 1)² = 1 3x² - 2x² + x - (4x² - 4x + 1) = 1 3x² - 2x² + x - 4x² + 4x + 1 - 1 = 0 -3x² + 5x - 2 = 0 (3x - 2) (x - 1) = 0 ∴ x = ²/₃ or x=1 y = 2(²/₃) - 1 or y = 2(1) - 1 y = ¹/₃ or y = 1 OR x = y + 1 .......(1) 2 3x² - xy - y² = 1 .....(2) (1) into (2) 3y² + 6y + 3 - 2y² - 2y - 4y² - 4 = 0 -3y³ + 4y - 1 = 0 3y² - 4y + 1 = 0 (3y - 1)(y - 1) = 0 ∴ y = ¹/₃ or y = 1	√ y = 2x - 1 √ substitution √ standard form √ factorisation √ x-values √ y-values OR √ x = y + 1 2 √ substitution √ standard form √ factorisation √ y-values √ x-values
1.3	f(x) = x² - 2px + 8 + 2p For equal roots: b² - 4ac = 0 (-2p)² - 4(1)(2p + 8) = 0 4p² - 8p - 32 = 0 p² - 2p - 8 = 0 ∴ p = -2 or p = 4 but : p < 0 ⇒ p = -2 so, f(x) = x² + 4x + 4 ∴ h(x) = x² + 4x + 1 = x² + 4x + 4 - 4 + 1 = (x + 2)² - 3 ∴ TP:(-2;-3) OR b² - 4ac = 0 (-2p)² - 4(1)(8 + 2p) = 0 4p2 - 8p - 32 = 0 p² - 2p - 8 = 0 (p - 4)(p + 2) = 0 ∴ p # 4 or p = -2 ∴ turning point of (-2;0) ∴ turning point of (-2;-3)	√ b² - 4ac = 0 √ substitution √ p-values √ h(x) = x² + 4x + 1 √ answer in coordinate form √ b² - 4ac = 0 √ substitution √ p values √ (-2;0) √ (-2;-3) √ b² - 4ac = 0 √ substitution √ p values √ (-2;0) √ (-2;-3)
2.1.1	-17;-2	√ both terms
2.1.2	2a = -2 3a + b = -2 a + b + c = 3 ∴ a = -1 3(-1) + b = -2 -1 + 1 + c = 3 ∴ b = 1 ∴ c = 3 Tn = -n² + n + 3	√ a = -1 √ b = 1 √ c = 3 √ Tn = -n² + n + 3
2.1.3	-n² + n + 3 = -809 n² - n - 812 = 0 (n - 29)(n + 28) = 0 ∴ n = 29	√ equating Tn to -809 √ factors √ choosing n = 29
2.2.1	T_n = 2n - 3 T₅₃ = 2(53) - 3 = 103 OR T₅₃ = a + 52d = -1 + 52(2) = 103	√ substitution into T₅₃ √ answer √ substitution into T₅₃ √ 103
2.2.2		√ substitution into correct formula √ 783
2.2.3		√ √ 2n - 3
2.3	T₄ = a + 3d and T₁₀ = a + 9d ∴ T₁₀ - T₄ = 6d 6d = (8x - 2y) - (2x + y) = 6x - 3y ∴ d = x - ½y T₄ = a + 3d 2x + y = a + 3(x - ½y) 2x + y = a + 3x - ³/₂y ∴ a = ⁵/₂y - x	√ T₁₀ - T₄ = 6d √ 6d = (8x - 2y) - (2x + y) √ d = x - ½y √ substitution √ value of a
QUESTION 3
3.1	T₁ = (x - 1) T₂ = (x - 1)2 ∴ r = x - 1 for convergence: -1 < r < 1 ∴ -1 < x -1 < 1 0 < x < 2	√ -1 < r < 1 √ answer
3.2	When:	√ substituting for x √ values for a and r √ substituting into S∞ formula √ answer
QUESTION 4
4.1	x = -3 y = 1	√ x = -3 √ y = 1
4.2	1 + 2 = 0 x + 3 2 = -1 x + 3 2 = -x - 3 x = -5 y = 1 + 2 0 + 3 = ⁵/₃	√ substitution √ x -intercept √ y-intercept
4.3		√ asymptotes √ x -intercept √ y-intercept √ shape
4.4	h(x) = -2 - 1 x + 3 point of intersection of asymptotes (-3;-1) or y = -(-x - p) + q y = (x - (-3)) - 1 or y = -(-x - 3) - 1 y = x + 2 OR h(x) = -2 - 1 x + 3 point of intersection of asymptotes (-3;-1) y = x + k -1 = -3 + k k = 2 ∴ y = x + 2	√ h(x) = -2 - 1 x + 3 √ substitute point of intersection of asymptotes √√ answer √ h(x) = -2 - 1 x + 3 √ substitute point of intersection of asymptotes √√ answer
QUESTION 5
5.1	(0;-8)	√ answer
5.2	y = mx + c y = mx - 8 10 = 9m - 8 m = 2 ∴ y = 2x - 8 OR mTQ = 10 - (-8) 9 - 0 m = 2 ∴ y = 2x - 8	√ c = -8 √ substituting T(9;10) into equation of line √ equation
5.3	y = x² - 7x - 8 = x² - 7x + (-⁷/₂)² - 8 - (-⁷/₂)² = (x - ⁷/₂)² - ⁸¹/₄	√ completing the square √ equation
5.4	(⁷/₂ ; -⁸¹/₄)	√ x-coordinate √ y-coordinate
5.5	Ave gradient y - 10 = 1 x - 9 y - 10 = x - 9 y = x + 1 f(x) = x² - 7x - 8 x + 1 = x² - 7x - 8 0 = x² - 8x - 9 0 = (x - 9)(x + 1) ∴ x = 9 or -1 y = 10 or 0 ∴ w(-1 ; 0) OR x² - 7x - 8 - (10) = 1 x - (9) x² - 7x - 18 = x - 9 x² - 8x - 9 = 0 (x - 9)(x + 1) = 0 x = 9 or x = -1 y = 10 or y = 0 ∴ w(-1 ; 0) OR f'(x) = 2x - 7 f'(9) = 2(9) - 7 = 11 f'(9) + f'(x) = 1 2 11 + 2x - 7 = 1 2 2x + 4 = 1 2 x + 1 = 1 x = -1 y = 0 ∴ w(-1 ; 0)	√ method √ making y the subject √ equating 2 equations √ factors √ specifying coordinates for W √ x² - 7x - 8 - (10) = 1 x - (9) √ equating to 1 √ standard form √ factors √ specifying coordinates for W. √ f'(x) = 2x - 7 √ f'(9) = 2(9) - 7 √ average gradient = 1 √ substitution √ coordinates of W
5.6	x² - 7x - 8 = 0 (x - 8)(x + 1) = 0 ∴ P(-1;0) and R(8;0) y = 2x - 8 0 = 2x - 8 ∴ V(4;0) ∴ x < -1 or 4 < x < 8 OR x∈(-∞;-1)∪(4;8)	√ x intercepts of f √ x intercept of g √ x < -1 accuracy √ 4 < x < 8 accuracy
QUESTION 6
6.1	f(x) = log_mx 3 = log_m64 m³ = 64 m³ = 4³ ∴ m = 4	√ substitution √ answer
6.2	f(x) = log₄x ∴ f^-1 : x = log₄x y = 4^x	√ interchanging x and y √ answer
6.3		√ y-intercept √ shape and asymptote
6.4	y > -2 OR y∈ (-2; ∞)	√ answer OR √ answer
QUESTION 7
7.1		√ substitution into correct formula √ simplification √ value for r
7.2.1	= R14 018,28	√ i = ^0.115/₁₂ and n =180 √ substituting into correct formula √ answer
7.2.2 (a)	OR = 2 453 828, 34 - 1528 392,76 = R 925 435, 58 Outstanding balance after 80 months; = 925 435,58 [1 + 11,5%]5 12 = R 970 637,48	√ ? = 105 for P and √ ? = 5 for A √ substituting into correct P formula √ substituting into correct A formula √ ?(1 + 0,115)⁵ 12 √ answer √ ? = 75 for both formulae √ substituting into correct F formula √ substituting into correct A formula √ P(1 + 0,115)⁵ 12 √ answer
7.2.2 (b)	∴ n = 115 months	√ P = R970 637,89 √ substituting into correct formula √ correct use of logs √ final answer
QUESTION 8
8.1		√ 3 - 2x² - 4hx - 2h² √ substitution √ simplification √ factorisation √ answer
8.2.1		√ x³ - 4x² + 4x √ 3x² √ - 8x √ +4
8.2.2		√ -2x^½ √ 3/7ax^4/₇ √ -x^-½ (derivative of constant must be zero to get 3rd mark)
QUESTION 9
9.1	x = -¹/₃ and x = 1	√ x = -¹/₃ √ x = 1
9.2	x = (1 + (-¹/₃)) ÷ 2 = ¹/₃	√√ answer
9.3	g(x) is increasing when g'(x) > 0 -¹/₃ < x <1 OR x∈ (-¹/₃ ;1)	√√ answer
9.4	OR	√ substituting all intercepts √ a = -3 √ y = -3(x + ¹/₃)(x -1) √ g'(x) = -3x² + 2x +1 √ substituting all intercepts √ a = -1 √ y = -(3x + 1)(x - 1) √ g'(x) = -3x² + 2x + 1
9.5	g(x) = ax³ + bx² + cx + d g'(x) = 3ax² + 2bx + c = -3x² + 2x + 1 ∴ 3a = -3 2b = 2 c = 1 ∴ a = -1 b = 1 ∴ y = -x³ + x² + x + d + 1 0 = -0³ + 0² + 0 + d + 1 ∴ d = -1	√ g'(x) = 3ax² + 2bx + c √ 3a = -3 √ 2b = 2 √ a = -1; b = 1; c = 1 √ substitute (0;0) into g(x) + 1
QUESTION 10
10.1	Let the two numbers be x and y : x + y = 18 ∴ y = 18 - x Product :P(x) = yx² = (18 - x)x² = 18x² - x³ Product is maximum when: P'(x) 0 P'(x) = 36x - 3x² 36x - 3x² = 0 3x(12 - x) = 0 ∴ x = 0 or x = 12 ∴ y = 18 - 0 = 18 or y = 18 - 12 = 6 P is maximum when x = 12 ∴ the two numbers are: 12 and 6 OR let the two numbers be x and y x + y = 18 ∴ y = 18 - x product: P(x) = xy² =x(18 - x)² =x(324 - 36x + x²) = 324x - 36x² + x³ Product is maximum when: P'(x) = 0 P'(x) = 324 - 72x + 3x² 3x² - 72x + 324 = 0 x² - 24x + 108 = 0 (x - 18)(x - 6) = 0 ∴ x = 18 - 18 = 0 or y = 18 - 6 = 12 ∴ The two numbers are 12 and 6	√ x + y =18 √ yx² √ substitution and simplification √ P'(x)and equating to0 √ x-values √ y-values √ selection of the 2 numbers (if x = 0, Product = 0) √ x + y =18 √ xy² √ substitution and simplification √ P'(x)and equating to 0 √ x-values √ y-values √ selection of the two numbers (P = 0 when x = 18)
QUESTION 11
11.1.1	a = 111 b = 106	√ answer √ answer
11.1.2 (a)	P(a boy who plays crickcet) = 108 or 54 530 265	√ numerator √ denominator
11.1.2 (b)	P(A or B) = P(A) + P(B) - P(A and B) P(girl or not a tennis player) = 288 + 445 - 231 530 530 530 = 502 530 or 251 265 or 94,72% OR P(girl or not tennis) = 1 - P (boy and tennis) = 1 - 28 530 = 502 530 or 251 265 or 94,72%	√ formula √ substitution into correct formula √ answer √ method √ substitution √ answer
11.2.1	9⁹ or 387 420 489	√ 9⁹ (1)
11.2.2	If vowels are together: 6! × 4! ∴ if vowels are not together: 9! - (6! × 4!) = 345 600	√ 6! × 4! √ subtracting from 9! √ answer (3)
11.2.3	Vowels in odd spaces: = 4 × 5 × 3 × 4 × 2 × 3 × 1 × 2 = (4 × 3 × 2 × 1) × (5 × 4 × 3 × 2) = 4! × 120 = 2880 ∴ probability = 2880 (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2) = 2880 362880 = 1 126	√ 4! × 120 √ vowels in odd spaces (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2) √ answer (4) [15]