GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS FEBRUARY/MARCH 2017

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GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS FEBRUARY/MARCH 2017

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GRADE 12 MATHEMATICS
PAPER 1
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2017

MEMORANDUM

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 - 3)(x + 1) = 0 x = 3 or x = -1	✔ answer ✔ answer (2)
1.1.2	√x³ = 512 x^3/₂ = 512 [x^3/₂]^2/₃ = (8³)^2/₃ x = 64 OR √x³ = 512 x³= 262144 x³= 2¹⁸ x = 2⁶ x = 64	✔x^3/₂✔(8³)^2/₃✔ answer (3) ✔ squaring both sides ✔x³= 2¹⁸✔ answer (3)
1.1.3	x(x − 4) < 0 0 < x < 4 OR x ∈(0; 4)	✔critical values ✔inequality or interval (2)
1.2.1	x² - 5x + 2 = 0 x = 5 ± √(-5)² - 4(1)(2) 2(1) x = 5 ± √17 2 x = 0,44 or x = 4,56 OR x² - 5x + 2 = 0 x² - 5x = -2 x² - 5x + [-⁵/₂]² = -2 + [-⁵/₄]² [x - ⁵/₂]²= ¹⁷/₄ x = 5 + √17 or x = 5 - √17 2 2 x = 0,44 or x = 4,56	✔ subst correct formula ✔ answer ✔ answer (3) ✔[x - ⁵/₂]²= ¹⁷/₄✔ answer ✔ answer (3)
1.2.2	f(x) = x² - 5x + 2 x² - 5x + 2 = c x² - 5x + 2 - c = 0 b² - 4ac < 0 (-5)² - 4(1)(2 - c) < 0 25 - 8 + 4c < 0 4c < -17 c < -17 4	✔ standard form ✔b² - 4ac < 0 ✔ substitution ✔ answer (4)
1.3	x = 2y + 2 x² - 2xy + 3y² = 4 (2y + 2)² - 2y(2y + 2) + 3y² = 4 4y² + 8y + 4 - 4y² - 4y +3y² = 4 3y² + 4y = 0 y(3y + 4) = 0 y = 0 or y = - 4 3 x = 2 or x = - 2 3 OR x = 2y + 2 y = ½x - 1 x² - 2xy + 3y² = 4 x² - 2x[½x - 1] + 3[½x - 1]² = 4 x² - x² + 2x + 3[¼x² - x + 1] = 4 2x + ¾x² - 3x + 3 = 4 3x² - 4x - 4 = 0 (3x + 2)(x - 2) = 0 x = 2 or x = - 2 3 y = 0 or y = - 4 3	✔ substitution ✔ simplification ✔ standard form ✔ factors ✔ y = 0 or y = - 4 3 ✔ x-values (ca on both x-values) (6) ✔ substitution ✔ simplification ✔ standard form ✔ factors ✔x = 2 or x = - 2 3 ✔ y-values (ca on both y-values) (6)
1.4	S = 6 x² + 2 For S to be a maximum the denominator needs to be at a minimum. Minimum of x² + 2 is 2 Maximum of S = 6 x² + 2 = 6 2 = 3	✔ Minimum of x² + 2 is 2 ✔3 (2)
		[22]
QUESTION 2
2.1	For geometric: -¼; b ; -1 ; .... b = - 1 -¼ b b² = ¼ b = ±½ OR b = ±√[-¼ ][-1] b = ±½	✔ b = - 1 -¼ b ✔ b = ½ ✔b = -½ (3) ✔b = ±√[-¼ ][-1] ✔ b = ½ ✔b = -½ (3)
2.2	-¼ ; ½ ; -1 ; ........ r = -2 T₁₉ = ar¹⁸ = [-¼ ][-2]¹⁸ = [- 2¹⁸] 2² = -2¹⁶ = - 65536 OR T¹⁹ = ar¹⁸ == [-¼ ][-2]¹⁸ = [- 2¹⁸][2¹⁸] = -2¹⁶ = - 65536	✔r = – 2 ✔subst. into correct formula ✔–65536 / –2¹⁶ (3) ✔r = – 2 ✔subst. into correct formula ✔–65536 / –2¹⁶ (3)
2.3	The series is: -¼; ½ ; -1 ; 2 ; -4 ; 8 ; .............. The new positive term series: ½; 2 ; 8 ; 32 ; 128 ; .................. a = ½ r = 4 OR	✔ a = ½ ✔ r = 4 ✔ ✔ correct formula (4)
2.4	No, the series is not convergent r = 4 and for convergence –1 < r < 1	✔no ✔reason (2)
		[12]
QUESTION 3
3.1.1	24	✔24 (1)
3.1.2	2a = 3 3a + b = 0 a + b + c = 6 a = ³/₂ b = - ⁹/₂ c = 9 T_n = ³/₂ n² - ⁹/₂ n + 9 OR T_n = T₁ + (n - 1)d₁ + (n - 1)(n - 2)d₂ 2 = 6 + (n - 1)(0) + (n - 1)(n - 2)(3) 2 = 6 + n² - 3n + 2 [3] 1 [2] = 6 + [3]n² - 9n + 3 [2] 2 = 3n² - 9n + 9 2 2	✔a = ³/₂✔b = - ⁹/₂ ✔c = 9 ✔ T_n = ³/₂ n² - ⁹/₂ n + 9 (4) ✔formula ✔substitution ✔simplifying ✔T_n = ³/₂ n² - ⁹/₂ n + 9 (4)
3.1.3	3n² - 9n + 9 = 3249 2 2 3n² - 9n + 18 = 6498 3n² - 9n + 6480 = 0 n² - 3n + 2160 = 0 (n + 45)(n - 48) = 0 n ≠ −45 or n = 48 (4)	✔ equating general term to 3249 ✔ standard form ✔ factors ✔n ≠ −45 or n = 48 (4)
3.2	-1 ; 2sin3x ; 5 ; ............ 2sin3x + 1 = 5 - 2sin3x 4sin3x = 4 sin3x = 1 3x = 90º x = 30º	✔ 2sin3x + 1 = 5 - 2sin3x ✔ sin3x = 1 ✔ 3x = 90º ✔ x = 30º (4)
		[13]
QUESTION 4
4.1	U(1 ; 0)	✔(1; 0) (1)
4.2	x =1 y =1	✔x =1 ✔y =1 (2)
4.3	2 + 1 = 0 x - 1 2 = -x + 1 x = -1 T(-1; 0)	✔ y = 0 ✔x = −1 (2)
4.4	f(x) = log₅X h : x = log₅y y = 5^x	✔ change x and y ✔y = 5^x (2)
4.5	y = 0	✔ answer (1)
4.6	V(√2 + 1; √2 + 1) V(2,41 ; 2,41) OR x = 2 + 1 x - 1 x² - 2x = 2 + x - 1 x² - 2x - 1 = 0 x = 2 ± √4 - 4(1)(-1) 2 =2 ± √8 2 = 2 ± 2√2 2 = 1 ± √2 V(1 + √2 ; 1 + √2) OR x - 1 = 2 x - 1 (x - 1)² = 2 x = 1 ± √2 V(1 + √2 ; 1 + √2)	✔✔2 +1 ✔✔2 +1 (4) ✔ x = 2 + 1 x - 1 ✔ subs into correct formula ✔ x = √2 + 1 ✔ y = √2 + 1 ✔x - 1 = 2 x - 1 ✔ (x - 1)² = 2 ✔ x = 1 + √2 ✔ y = 1 + √2 (4)
4.7	T' (3; 2)	✔x = 3 ✔y = 2 (2)
		[14]
QUESTION 5
5.1.1	C(0 ; –3)	✔ C(0 ; –3) (1)
5.1.2	f(x) = x² - 2x -3 (x - 3)(x + 1) = 0 x = -1 or x = 3 AB = 3 - (-1) AB = 4 units	✔ factors ✔ x-value ✔ other x-value ✔ answer (4)
5.1.3	x = 2 or 2x - 2 = 0 or x = -1 + 3 2(1) 2 = 1 y = (1)² - 2(1) - 3 = -4 D(1 ; -4)	✔x = 1 ✔ y value (2)
5.1.4	C(0 ; -3) D(1 ; -4) Average gradient = -4 + 3 or -3 + 4 1 - 0 0 - 1 = -1	✔ -4 + 3 or -3 + 4 1 - 0 0 - 1 ✔ -1
5.1.5	OC = OB = 3 OCB= 45º isosceles right angled triangle OR tan β = m_g tan β = 1 β = 45º OBC = 45º OCB = 45º	✔equal lengths ✔45º (2) ✔tanβ =1 ✔45º (2)
5.1.6	− 4 < k < −3 OR (–4 ; –3)	✔–4 ✔ –3 ✔ notation (3)
5.1.7	f'(x) . f''(x) > 0 (2x - 2) . 2 > 0 2x − 2 > 0 x > 1	✔2x − 2 ✔ 2 ✔x > 1 (3)
5.2	f(x) = a(x - 1)(x - 5) 4 = a(3 - 1)(3 - 5) 4 = -4a a = -1 f(x) = -x² + 6x - 5	TP ✔ x = 3 ✔ y = 4 ✔ x – intercepts ✔ y-intercept ✔ shape (5)
		[22]
QUESTION 6
6.1.1	A = 150 000(1 - 0,2)² = R96 000	✔ n = 2 ✔ 150 000 in correct formula ✔ 96 000 (3)
6.1.2	150 000(1 - 0,2)ⁿ = 49152 (0,8)ⁿ = 1024 3125 n log (0,8) = log 1024 3125 n = 5 The machine will need to be replaced at the beginning of 2020 OR 150 000(1 - 0,2)ⁿ = 49152 (0,8)ⁿ = 1024 3125 n = log _0.8 1024 3125 n = 5 The machine will need to be replaced at the beginning of 2020	✔150 000(1 - 0,2)ⁿ = 49152 ✔n log (0,8) = log 1024 3125 ✔n = 5 ✔ 2020 (4) ✔150 000(1 - 0,2)ⁿ = 49152 ✔n = log _0.8 1024 3125 ✔n = 5 ✔ 2020 (4)
6.1.3	R280 000 - R49 152 = R230 848 x = R9 383,26	✔ R230 848 ✔ i = 0,085 = 0,02125 and n = 20 4 ✔ subs into correct formula ✔ R 9 383,26 (4)
6.2	= R791 837,43 Lerato qualifies for a loan of R791 000 under the given conditions	✔ i = 0,11 12 ✔ n = 180 ✔ substitution correct formula ✔R791 837,43 ✔ R791 000 (5)
		[16]
QUESTION 7 PENALISE ONLY ONCE for incorrect notation in this question.
7.1	f(x + h) = (x + h)² - 5 = (x² + 2xh + h²) - 5 = x² + 2xh + h² - 5 f(x + h) - f(x) = x² + 2xh + h² - 5 - (x² - 5) = 2xh + h² OR	✔simplifying ✔formula ✔subst. into formula ✔factorisation ✔answer (5) ✔formula ✔subst. into formula ✔simplifying ✔factorisation ✔answer (5)
7.2	g(x) = 5x² - 2x x³ = 5x² - 2x^-2 g'(x) = 10x + 4x^-3 =10x + 4 x³	✔ 5x² - 2x^-2 ✔ 10x ✔ 4x^-3 or 4 (3) x³
7.3	h(x) = ax² , x > 0 h^-1 : x = ay² y > 0 y = √^x/_a h^-1 (8) = √⁸/_ah'(x) = 2ax h' (4) = 2a(4) = 8a √⁸/_a= 8a 64a² = ⁸/_aa³ = ¹/₈ a = ¹/₂	✔ y = √^x/_a ✔ √⁸/_a ✔ h' (4) = 8a ✔ √⁸/_a= 8a ✔ a³ = ¹/₈ ✔ a = ¹/₂(6)
		[14]
QUESTION 8
8.1	f'(x) = 0 6x² - 10x + 4 = 0 3x² - 5x + 2 = 0 (3x - 2) (x - 1) = 0 x = ²/₃ or x = 1 y = 2[²/₃]³ - 5[²/₃]² + 4[²/₃] y = 2(1)³ - 5(1)² + 4(1) y = ²⁸/₂₇ or y = 1 Turning points are : [ 2 _; 28 ] and [1 : 1] [ 3 27]	✔ derivative ✔ derivative = 0 ✔ factors ✔ x-values ✔ y-values (5)
8.2	2x³ - 5x² + 4x = 0 x(2x² - 5x + 4) = 0 x = 0 or x = 5 ± √25 - 4(2)(4) 4 = 5 ± √-7 4 No real roots OR 2x³ - 5x² + 4x = 0 x(2x² - 5x + 4) = 0 x = 0 or b² - 4ac = 25 - 4(2)(4) = -7 < 0 No real roots	✔ x(2x² - 5x + 4) = 0 ✔ x = 0 ✔ 5 ± √-7 (3) 4 ✔ x(2x² - 5x + 4) = 0 ✔ x = 0 ✔ b² - 4ac < 0 (3)
8.3	f(x) = 2x³ - 5x² + 4x x(2x² - 5x + 4) = 0	✔(0 ; 0) ✔turning points ✔shape (3)
8.4	f(x) = 2x³ - 5x² + 4x f'(x) = 6x² - 10x + 4 f''(x) = 12x - 10 f''(x) > 0 12x - 10 > 0 x > ⁵/₆ OR x = b 3a x = - (-5) 3(2) x > ⁵/₆The function is concave up for x > ⁵/₆since a > 0 OR Point of inflection: x = ²/₃ + 1 2 x > ⁵/₆The function is concave up for x > ⁵/₆since a > 0	✔ 12x - 10 ✔ f''(x) > 0 ✔ answer (3) ✔ x = - (-5) 3(2) ✔ x > ⁵/₆ ✔ f''(x) > 0 (3) ✔ x = ²/₃ + 1 2 ✔ x > ⁵/₆ ✔ f''(x) > 0 (3)
		[14]
QUESTION 9
9	Length of one side of the square =^x/₄Length of the rectangle 2L + x + ^x/4 = 6 L = 6 - ^5x/₄ 2 = 24 - 5x 8 A = [x]² + x[24 - 5x] [4] 4 8 = x² + 24x - 5x² 16 32 = 24x - 3x² 32 A = 24x - 3x² 32 For minimum area dA = 0 dx dA = 24 - 6x dx 32 6x = 24 x = 4	✔ ^x/₄ ✔ 6 - ^5x/₄_{or [24 - 5x]} 2 8 ✔ [^x/₄ ]² ✔ x[24 - 5x] 4[ 8 ] ✔ dA = 0 dx ✔ 24 - 6x 32 ✔ x = 4 (7)
		[7]
QUESTION 10
10.1.1	P(S and T) = P(S) + P(T) ¹/₆ = [¹/₄] × P(T) P(T) = ²/₃	✔P(S and T) = P(S)×P(T) ✔ P(T) = ²/₃(2)
10.1.2	P(S or T) = P(S) + P(S and T) = [¹/₄] + [²/₃] - [¹/₆] =¾	✔ [¹/₄] + [²/₃] - [¹/₆] ✔ ¾
10.2.1	5! = 120	✔ 5 ✔ 5! or 120 (2)
10.2.2	5⁵ = 3125	✔5⁵ or 3 125 (1)
10.3	n(E) = 5! ×2! × 2! n(S) = 7! P(E) = 5! ×2! × 2! 7! = 2 21	✔5! ✔2! × 2! ✔5! ×2! × 2! 7! ✔ 2 (4) 21
		[11]
QUESTION 11
11	P(Win) = P(F and W) + P(not F and W) = 0,7 × 0,85 + 0,3 × 0,45 = 0,595 + 0,165 = 0,76 = 76% = ¹⁹/₂₅	✔ 0,3 ✔ P(F and W) = 0,7 × 0,85 = 0,595 ✔ P(not F and W) = 0,3 × 0,45 = 0,165 ✔ 0,595 + 0,165 ✔ 0,76 / 76% / ¹⁹/₂₅ (5)
		[5]
		[150]