GRADE 12 TECHNICAL MATHEMATICS PAPER 2 MEMORANDUM - 2018 JUNE EXAM PAST PAPERS AND MEMOS

ElimuZA Access to Education

Thursday, 12 August 2021 11:28

GRADE 12 TECHNICAL MATHEMATICS PAPER 2 MEMORANDUM - 2018 JUNE EXAM PAST PAPERS AND MEMOS

More in this category: « GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - 2018 JUNE EXAM PAST PAPERS AND MEMOS GRADE 12 TECHNICAL MATHEMATICS PAPER 2 QUESTIONS - 2018 JUNE EXAM PAST PAPERS AND MEMOS »

Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

TECHNICAL MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE 2018

NOTE:

Continuous accuracy (CA) applies in ALL aspects of the marking guideline.
After two mistakes, do not apply CA marking.
Assuming values/answers in order to solve a problem is unacceptable.

Symbol	Explanation
M	Method
MA	Method with accuracy
A	Accuracy
CA	Consistent accuracy
S	Simplification or Statement
R	Reason
SR	Statement and correct reason

QUESTION 1


1.1	BC =√( x₂ - x₁)² + (y₂ - y₁)² = √(8-0) ² + (0+6) ² =√100 = 10	MA formula ✓ substitution ✓ CA Answer ✓ (3)
1.2		A x-coordinate✓ A y-coordinate✓ (2)
1.3	m_BC = y₂ - y₁ x₂ - x₁= 0 + 6 8 - 0 = 6 8 = 3 4 = 0,75	MA formula and substitution ✓ CA Simplification ✓ (2)
1.4	y - y₁ = m ( x - x₁ ) y - 2 = 0, 75( x - 2) y = 0, 75x + 0, 5	M Correct formula✓ A Correct substitution pt D or E ✓ CA Simplification ✓ CA Standard form ✓ (4)
1.5	inclination of DE = inclination of BC (lines; same gradient) m_DE = m_BCtan incl DE = 0, 75 incl DE = 36,87° m_AE= 4 + 6 = - 5 -4 - 0 2 tan incl AE =- 5 2 incl AE = 180° - 68,198...° = 111,8° AED = 111,8° - 36,87° = 74, 93°	M m_DE = m_BC ✓ A tan incl DE = 0, 75 ✓ CA incl DE = 36,87° ✓ MA m_AE =- 5✓ 2 CA incl AE = 111,8°✓ CA AED = 74,93°✓ (6)
		[17]

QUESTION 2

2.1.1	r² = x² + y² = (-5)² + (-12)² = 169 ∴ Equation of circle: x² + y² = 169	M Correct formula✓ A Correct substitution✓ CA Value of r²✓ (3)
2.1.2	m_radius = 0 +12 = 12 0 + 5 5 m_tangent =- 5 ≈ -0, 42 12 Equation of tangent: y - y₁ = m ( x - x₁) y +12 = -0, 42( x + 5) y = -0, 42x -14,1 OR m_radius = 0 +12 = 12 0 + 5 5 m_tangent =- 5 12 y - y₁ = m ( x - x₁ ) y +12 = - 5 ( x + 5) 12 y +12 = - 5 x - 25 12 12 y = - 5 x - 169 12 12 OR x₁ x + y₁ y = r ² -5x -12 y = 169 y = - 5 x - 169 12 12	MA Gradient of radius✓ CA Gradient of tangent✓ A Subst. P into equation✓ CA Equation of line✓ CA Gradient of tangent✓ MA Gradient of radius✓ CA Gradient of tangent✓ A Subst. P into equation✓ M Correct formula✓ A Correct substitution of P✓ A Correct substitution of grad.✓ CA Equation of line✓ (4)
2.2	4x² + 9 y² = 36 x ²+ y ² = 1 9 4 x ² + y ² =1 3² 2²	M Rewrite the equation RHS = 1✓ M Rewrite into a² & b²✓ CA x-intercepts✓ CA y-intercepts✓ CA shape – clearly showing the horizontal major axis✓ (5)

[12]

QUESTION 3
3.1
3.1.1	13sinθ = 12 sinθ = 12 13 x = -√13² -12² (Pyth) x = -5 y = 12 P(-5;12)	M sin θ the subject✓ M Pythagoras✓ A x-value✓ A y-value✓ (4)
3.1.2	tanθ + secθ = 12 + 13 -5 -5 = 25 -5 = -5	CA correct value: tan✓ CA correct value: sec✓ CA answer✓ (3)
3.1.3	sinθ = 12 13 Ref ∠ = sin^-¹(¹²/₁₃ ) = 67, 38...° θ = 180° - 67, 38...° θ = 112, 6°	M 180° -✓ CA 112,6°✓ (2)
3.2		A - tanq✓ A cos θ✓ A (-cosq )²✓ A (-sinq )²✓ A sinθ✓ cosθ A cos² θ +sin² θ =1✓ CA -sinθ✓ (7)
3.3	RHS = 2 sin a LHS = 1+ cos a + sin a sin a 1+ cos a = (1+ cos a )² + sin² a sin a(1+ cos a) = 1+ 2 cos a + cos² a + sin² a sin a (1+ cos a ) = 1+ 2 cosa +1 sin a (1+ cosa ) = 2 + 2 cos a sin a (1+ cos a ) = 2(1+ cos a) sin a (1+ cosa ) = 2 sin a = RHS	A RHS✓ A denominator✓ A numerator✓ A cos² a +sin² a =1✓ CA expansion✓ CA factorising✓ (6)
		[22]

QUESTION 4

4.1	sinθ = 9, 48 = 0, 606... 15, 64 θ = 37, 3°	A trig ratio✓ CA value for θ ✓ (2)
4.2	Area Δ PSR = ½ PS´SQ sin PSˆQ = 0, 5´ 4, 95´15, 64´sin 52, 69° = 30, 79 cm²	A area-rule✓ CA value for PSˆQ✓ A correct substitution✓ CA Area of ∆DEF✓ (4)
4.3	PSˆQ = 90°- 37,31° = 52, 69° PQ² = PS ² + SQ² - 2PS ´ SQ cos PSˆQ = 4, 95² +15, 64² - 2´ 4, 95´15, 64 cos 52, 69° = 175, 261... PQ = √175, 261... » 13, 24 m²	S ✓ A cos-rule✓ A correct substitution✓ CA simplification✓ CA value of✓ (5)
		[11]

QUESTION 5
5.1.		A x-int’s 120° and 300°✓ A TP (30°;1)✓ A TP (210°;-1)✓ A shape✓ (4)
5.2.1	Amplitude of f = 2	A ✓ amplitude = 2 (1)
5.2.2	Range g: –1 ≤ y ≤ 1	A y- values✓ CA interval✓ (2)
5.2.3	x = 90°	CA ✓ (1)
5.2.4	f (x) ≤ g(x) ⇔ x Î[210°;360°] OR 210° ≤ x £≤ 360°	CA x = 210°✓ CA x = 360°✓ A notation✓ (3)
		[11]

QUESTION 6
6.1	Proportionally(1)
6.2
	Let QS = x SC = 4 – x CS = CR (line \|\| to one side of D) SA RB 4 - x = 1 x + 2 3 12 - 3x = x + 2 10 = 4x x = 2,5 ∴ QS = 2,5 units	S SC = 4 – x ✓ S R ✓ S SA = x + 2 ✓ A Setup equation ✓ CA Simplify equation✓ CA value✓ (7)
		[8]

QUESTION 7
7.1
7.1.1	AB = 8,36 cm (sides opp = ∠ s) BD = 5,91 cm	S R ✓✓ S ✓ (3)
7.1.2	AB = 8, 36 ≈ 1, 41 BD 5, 91 BC = 8, 36 ≈ 1, 41 DC 5, 91 AC = 11,82 ≈ 1, 41 BC 8, 36	S ✓ S ✓ S ✓ (3)
7.1.3	triangles are similar	S similar triangles ✓ (1)
7.2
7.2.1(a)	Rˆ₃ = Tˆ₂ = 56° (∠s in same segment)	S R(2) ✓✓
7.2.1(b)	Rˆ₄ = T₁ = 56° (tan-chord)	S R ✓✓ (2)
7.2.2	Aˆ = Mˆ (∠s in same segment) Aˆ = Rˆ₁ (tan-chord)	S ✓ S ✓ (2)
7.2.3	AP = PR (Δ s /// ) MP PT 8, 29 = 11,11 MP 12, 01- 8, 29 MP = 8, 29 x 3, 72 11,11 = 2, 78 cm	S ✓ S ✓ S PT = 12,01 – 8,29 = 3,72 ✓ S ✓ S value of ✓ (5)
		[18]

QUESTION 8
8.1
	PTˆR = 90° (line from centre to midpt of chord) TR² = PR² - PT ² (Pythagoras) = 9, 47² - 8, 43² = 18, 616 TR = 4, 31... WR ≈ 8, 63 units	S ✓ R ✓ S Pythagoras ✓ A Simplification✓ CA Value of TR✓ CA Value of WR✓ (6)
8.2
	ABO = 40° (radii; equal ∠ s opp equal sides) AOB = 100° (int ∠s of D) D = 50° (∠ at centre = 2 ∠ at circumf.) x = 130° (opp ∠s of cyclic quad)	S ✓ S ✓ S ✓ R ✓ S ✓ R ✓ (6)
		[12]

QUESTION 9
9.1.1	108, 5° = 108, 5° x π 180° = 217 π 360 = 1,89 radians	A Multiply with factor✓ CA Answer✓ (2)
9.1.2	Radius = 10,84 = 5, 42 units 2	A Answer✓ (1)
9.1.3	s = rθ d = 5, 42 ´1,89 = 10, 24 units	A correct✓ CA ✓ Substitution CA answer ✓ (3)
9.1.4	Area =r ² θ 2 = 5, 42 x 1,89 2 = 27,76 units²	A Correct formula✓ CA Substitution✓ CA answer✓ (3)
9.2	4h² - 4dh + x² = 0 d = x ²+ h 4h = 8, 76 + 3,15 4 x 3,15 = 9, 24 units	A Correct formula✓ A Making d the subject✓ A Substitution✓ CA answer✓ (4)
9.3
		A Correct formula✓ A Convert to cm ✓ A Substitution✓ CA 109,8✓ CA 658,8✓ (5)
		18]

QUESTION 10
10.1.1	n = 15 rev/s ω = 2πn = 2p (15) = 30π = 94, 25 rad/s	A Correct formula ✓ A Substitution✓ CA value of✓ (3)
10.1.2	d = 570mm = 0, 57m υ = πDn = π(0, 57)(15) = 26,86 m/s	A Convert diameter to metre✓ A Correct formula✓ A Substitution✓ CA value of v✓ (4)
10.1.3	s=vt = 26,86...x(2 x 60) = 3223 m	A Correct formula✓ A Multiply t = 2, with 60✓ CA value of s✓ CA rounded answer✓ (4)
10.1.4	θ = ωt = (30π )(0, 3) =9π rad	A Correct formula✓ Substitution✓ CA value of θ✓ (3)
10.2	Vol of cylinder = πr²h	A Formula of cylinder ✓ A Divide diameter by 2✓ A Substitution✓ CA Vol of cylinder✓ A Formula of hemisphere✓ A Substitution ✓ CA Vol of hemisphere✓ CA Adding the volumes✓ (7)
		[21]

TOTAL;150

Last modified on Thursday, 12 August 2021 13:46

Published in 2018 June Common Examination Papers and Memos Grade 12

Related items

More in this category: « GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - 2018 JUNE EXAM PAST PAPERS AND MEMOS GRADE 12 TECHNICAL MATHEMATICS PAPER 2 QUESTIONS - 2018 JUNE EXAM PAST PAPERS AND MEMOS »