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

Tuesday, 04 January 2022 08:41

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

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

Marking Codes
A	Accuracy
CA	Consistent accuracy
M	Method
R	Rounding
NPR	No penalty for rounding
NPU	No penalty for units omitted
S	Simplification
SF	Substitution in the correct formula
AO	Answer only

NOTE:

If a candidate answers a question TWICE, only mark the FIRST attempt.
If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed-out version.
Consistent accuracy (CA) applies to ALL aspects of the marking guideline.
Assuming answers/values to solve a problem is NOT acceptable.

QUESTION 1
1.1	1.1.1	x(x - 5) = 6 x² - 5x = 6 x² - 5x - 6 = 0 (x - 6)(x + 1) = 0 x = 6 or x = -1	√ Expansion √standard form √factors √x = 6 or x = -1	(4)
	OR
		x(x - 5) = 6 x² - 5x = 6 x² - 5x - 6 = 0 x = 6 or x = -1	√expansion √standard form √substitution √x = 6 or x = -1
	1.1.2	-2x² - 4x - 1 = 0	√ formula √substitution √both values of x	(4)
	1.1.3		√ critical values √ correct notation or √ number line	(4)
1.2	1.2.1	20 seconds	√ 20	(1)
	1.2.2	20 seconds = 20 hours 60 × 60	√ 20 hours 60 × 60	(1)
	1.2.3	20 hours = 5,55 × 10^-3 hours 60 × 60	√Notation	(1)
1.3		^x/_y = 2 ...............................(1) x² + xy + y = 2 ..................(2) x = 2y ................................(3) Then (2y)2 + (2y)y + y = 2 4y² + 2y² + y - 2 = 0 6y² + y - 2 = 0 (2y - 1)(3y + 2) = 0	√ x- the subject √ substitution by √ simplification/std form √factors √ y-values √ x-values	(6)
	OR
		^x/_y = 2 ...............................(1) x² + xy + y = 2 ..................(2) ^x/₂ = y ................................(3) Then, x² + x(^x/₂) + ^x/₂ = 2 2x² + x² + x = 4 3x² + x - 4 = 0 (3x + 4)(x - 1) = 0 x = -⁴/₃ or x = 1 therefore: y = -²/₃ or y = ¹/₂	√ y-the subject √ substitute ; ^x/₂ = y √ simplification / std form √factors √ x - values √ y- values
1.4	1.4.1	b = 2a = 2 × 2 = 4 and c = 2b = 2 × 4 = 8	√ b = 4 √ c = 8	(2)
	1.4.2	b² - 4ac = (4)² - 4(2)(8) = 16 - 64 b² - 4ac = -48 therefore; roots are non-real imaginary	√ Δ = -48 √ imaginary / non-real	(2)
1.5		41 = 1 × 2⁵ + 0 × 2⁴ + 1 × 2³ + 0 × 2² + 0 ×2¹ + 1 × 2⁰ = 101001₂ OR 41 = 32 + 8 + 1 = 101001₂ -1 mark if base 2 is omitted AO = 2/2	√ method √ 101001₂	(2)
				[26]
2.1	2.1.1		√ Prime factors √ CF numerator √ CF denominator √ ³/₂	(4)
	OR
			√ same base √ CF numerator √ CF denominator √ ³/₂
	2.1.2		√ log8^-1 √ same base rule √ -³/₅	(3)
	OR
			√ log 1 - log 8 √ same base rule √ -³/₅
2.2		2(x - y)(x - y) = 2(√2 - √3)(√2 + √3) = 2 (2 - 3) = -2 OR 2(x - y)(x + y) = 2(x² - y²) = 2 [ (√2)² - (√3)²] = 2(2 - 3) = -2	√ substitution √ expansion √ simplification
2.3			√ 3^x+1 = 1 √ 3^x/₂ = 2 √ 0 exponent rule √ log form 3^{x + 1} = 1 √ same base √ x = -1 3^x/₂ = 2 √ x = 2log₃2 √ x = 1,26
		OR
			√ 3^x+1 = 1 √ 3^x/₂ = 2 √ log₃1 √ log₃2 √ x + 1 = 0 √ x = 2log₃2 √ x = -1 √ x = 1,26
2.4		3x = -3 and -5y = 0 x = -1 and y = 0	√ equating parameters √ x = -1 √ y = 0
2.5			√ finding r √ reference angle √ θ = 310,88º √ correct form	(6)
				[27]
QUESTION 3
3.1		y = -4	√ y = -4
3.2		3^x - 4 = 0 3^x = 4 x = log₃4 x = 1,26	√ f(x) = 0 √ log form √ x = 1,26	(3)
3.3		f(x) = 3^x - 4 f(0) = 3º - 4 y = -3	√ subst. x = 0 √ y = -3	(2)
3.4			√ shape √ both intercepts √ asymptote	(3)
3.5		y > -4 or y∈(-4;∞) or -4<y<∞	√ -4	(1)
				[10]
QUESTION 4
4.1		B(0;2)	√B(2;0)	(1)
4.2			√ substitution √ g(x) = √4 - x²	(2)
4.3		x∈[-2;2] OR -2 ≤ x ≤ 2 OR x ≥ -2 and x ≤ 2	√ critical values from 4.1 √ notation	(2)
4.4		x∈(-2;0) OR -2 < x < 0	√ critical values from √ notation	(2)
				[7]
QUESTION 5
5.1		h(0) = -(0)² + 4(0) - 3 = -3 c(0;-3)	√ 0 √ -3	(2)
5.2		m(x) = ^p/_x - 3 -1 = ^p/₁ - 3 p = 2	√ substitution √ p = 2	(2)
5.3		m(x) = ²/_x - 3	√ substitute 2 and -3	(1)
5.4		x = ^-b/_2a = -4 2(-1) = 2 h(2) = -(2)² + 4(2) - 3 = 1 D(2;1)	√ substitution √ x = 2 √ h(2) √ D(2;1)	(4)
		OR
		h'(x) = -2x + 4 = 0 therefore x = 2 h(2) = -(2)² + 4(2) - 3 = 1 D(2;1)	√ h'(x) = -2x + 4 √ x = 2 √ h(2) √ D(2;1)
		OR
		(-x + 1) (x - 3) = 0 x = 1 or x = 3 therefore axis of symmetry x = 1 + 3 = 2 2 h(2) = - (2)² + 4(2) - 3 = 1 D(2;1)	√ x-intercepts √ x = 2 √ h(2) √ D(2;1)
5.5		DE = 1 + 3 = 4 units	√ OE + OD √ 4
QUESTION 6
6.1			√ formula √ substitution √ i_eff≈ 6.4%
6.2		A = P (1 - in) 40 000 = P(1 - 0,1 × 3) P = 40000 = R57 142,86 (1 - 0,1 × 3)	√ substitute i and n √ substitution √ P subject √ simplification
6.3		P₆ = R502 839, 7713 - R150 000 P₆ = R352 839,7713 A₁₀ = R 352 839, 7713 (1 + 0,07)⁴ A₁₀ = R 462 500,96	√ values of i and n √ substitute P, i and n √ A₆ √ P₆ √substitution P₆, i and n √ value of A₁₀	(6)
		OR
		A₆ = R502 839, 7713 A₆(1) = R 502 839, 7713 (1 + 0,07)⁴ A₆(1) = R 659 120,3659 A₇(1) = R 150 000(1 +0,07)⁴ A₇(1) = R 196 619, 4015 A₁₀ = R 659 120, 3659 - R 196 619, 4015 A₁₀ = R 462 500, 96	√values of i and n √ substitute P, i and n √ A₆ √ P₆ √ A₆ √ P₆ √ A ₂(1) √ A₆ √ P₆ √ value of A₁₀
				[13]
QUESTIONS 7
7.1			√ definition √ substitution √ simplification √ simplification √ f'(x) = -2	(5)
7.2			√ (x - 2)(x² + 2x + 4) √ simplification √ 12	(3)
7.3	7.3.1	y = - 3 - 2x² + 2x 5x² y = 3x^-2 - 2x² + 2x 5 dy = 6x^-3 - 4x + 2 dx 5	√ - 3 5x² √ 6x^-3 5 √ - 4x √ 2	(4)
	7.3.2		√ x ^2/₃ √ ²/₃x^-1/₃ √ 6x^-4	(3)
				[15]
QUESTION 8
8.1	8.1.1	y = -4	√ y = -4	(1)
	8.1.2	f(x) = (1 - 2x)(x² - 4) 0 = (1 - 2x) (x² - 4) x = ½ or x = 2 or x = -2	√ f(x) = 0 √ x = ½ √ x = ± 2	(3)
	8.1.3	f(x) = -2x³ + x² + 8x - 4	√ f(x) = -2x³ + x² + 8x - 4	(1)
	8.1.4	f(x) = -2x³ + x² + 8x - 4 f'(x) = -6x² + 2x + 8 0 = -6x² + 2x + 8 0 = 3x² - x - 4 0 = (x + 1) (3x - 4) x = ⁴/₃	√ f(x) CA from 8.3 √ f'(x) = 0 √ factors √ x = ⁴/₃ √ [100] [ 27 ]	(5)
	8.1.5		√ x-intercept √ y-intercept √ turning points √ shape	(4)
8.2		Average gradient = g(a) - g(4) = 9 a - 4 (2a² - a - 3) - 25 = 9 a - 4 2a² - a - 28 = 9a - 36 2a² - 10a + 8 = 0 a² - 5a + 4 = 0 (a - 4) (a - 1) = 0 a = 1	√ g(a) - g(4) a - 4 √ = 9 √ STD form √ factors √ a = 1	(5)
				[19]
QUESTION 9
9	9.1	V(t) = t² - 9t + 35 V(0) = (0)² - 9(0) + 35 = 35L	√ substitution √ V(0) = 35L	(2)
	9.2	V(t) = t² - 9t + 35 V(1) = (1)² - 9(1) + 35 = 27L	√ substitution √ V(1) = 27L	(2)
	9.3	V(t) = t² - 9t + 35 V'(t) = 2t - 9 V'(t) = 2t - 9 = 0 t = ⁹/₂	√ V'(t) √ V'(t) = 0 √ t = ⁹/₂	(2)
	9.4	maximum amount of fuel leaked	√ substitution √ V ≈ 14,75L
QUESTION 10
10.1			√ x³ 3 √ 3inx √ x √ c
10.2	10.2.1		√ integral function √ correct lower and upper bounds √ simplification √ substitute 3,41 √ substitute 0,59 √ 5,64	(6)
	10.2.2	Striped A = A under f (x) – A under g(x) Striped area =5,64−3,5 Striped area =2,13 square units	√ difference √ substitution √ 2,13 square units	(3)
				[13]
			TOTAL	150