GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPERS AND MEMOS MAY/JUNE2016

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GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPERS AND MEMOS MAY/JUNE2016

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Mathematics Paper 1
Grade 12
Senior Certificate Examinations
Past Paper (memorandum) 2016
NOTE:

If a candidate answers a question TWICE, only mark the FIRST attempt.
Consistent accuracy applies in ALL aspects of the marking memorandum.

MEMORANDUM

QUESTION 1

1.1.1

4x² - 25 = 0
(2x - 5)(2x + 5) = 0
x = ⁵/₂ OR x= -⁵/₂

4x2 = 25
x² = ²⁵/₄
x = ± √²⁵/₄
x = ⁵/₂ OR x= -⁵/₂

✓✓ factors

✓ answers (3)

✓ x² = ²⁵/₄

✓ x = ± √²⁵/₄

✓ answer (3)

1.1.2

x² - 5x - 2 = 0
x = -b ± √b² - 4ac
2a
= -(-5) ± √ (-5)² - 4(1)(-2)
2(1)
= (5) ± √33
2
x= 5,37 or x = -0,37

x² - 5x + ²⁵/₄ = 2 + ²⁵/₄
[x - ⁵/₂] ² =³³/₄x - ⁵/₂ = ± √³³/₂
x = 5 ± √33
2
x= 5,37 or x = -0,37

✔ correct substitution into correct formula

✔ answer

✔ answer (3)

✔ for adding ²⁵/₄ on both sides

✔answer

✔answer (3)

1.1.3

✔method

✔critical values in context of inequality

✔inequality or interval (3)

✔change of inequality

✔critical values in context of inequality

✔inequality or interval (3)

1.1.4

x - 3x^½ - 4 = 0
[x^½ - 4] [x^½ + 1] =0
x^½ = 4 or x^½ = -1
x = 16 N/A

x - 3x^½ - 4 = 0
let x^½ = k
k² -3k - 4 = 0
[k - 4] [k + 1] =0
x^½ = 4 or x^½ = -1
x = 16 N/A

x - 3x^½ = 4
x - 4 = 3√x Restrictions
9x = x² - 8x +16 x ≥ 4 and x ≥ 0
x² - 17x +16 = 0
[x -1] [x - 16] = 0
x = 1 or x = 16 N/A

✔ standard form
✔ factors
✔x^½ = 4
✔ x^½ = -1
✔ answer (5)

✔ standard form

✔ factors
✔ x^½ = 4
✔ x^½ = -1
✔ answer (5)

✔ isolating 3√x or 3x^½✔ standard form

✔ factors
✔ answers
✔ selection (5)

1.2

y = 2x +1
x² -3x - 4 - (2x + 1) = (2x + 1)²
x² -3x - 4 - 2x + 1 = 4x² + 4x + 1
3x² + 9x + 6 = 0
x² + 3x +2 = 0
[x + 2] [x + 1] = 0
x = -1 or x = -2
if x = -2 then y = -3
if x = -1 then y = -1

x = y - 1
2
[^{(y - 1)}/₂]² -3 [^{(y - 1)}/₂] - 4 -y =y²
y² - 2y +1 -3 (y - 1) - 4 - y = y²
4 2
y² - 2y +1 -6y + 6 - 16 - 4y = 4y²
3y² + 12y +9 =0
y2 + 4y +3 = 0
[y + 3] [y + 1] = 0
y = -3 or y = -1
if y = -3 then x = -2
if y = -1 then x = -1

✔ y subject of formula
✔ substitution

✔ standard form
✔ factors
✔ values of x
✔ values of y (6)

✔ x subject of formula
✔ substitution
✔ standard form
✔ factors
✔ values of y
✔ values of x (6)

1.3.1

2x + 1 ≥ 0
x ≥ -½

✔ answer (1)

1.3.2

f(x) = 2x -1
√2x +1 = 2x -1 Restrictions
2x + 1 = 4x² - 4x +1
x ≥ -½ and x ≥ ½
4x² - 6x = 0
x (4x -6) = 0
x = ³/₂ or x = 0
∴x = ³/₂

✔ √2x +1 = 2x −1

✔ standard form
✔ factors
✔ answers
✔ correct selection (5)
[26]

QUESTION 2

2.1.1

27 - b = b - 13
b = 27 +13
2
b = 20
27 - 20 = 13 - a
a = 6

27 - 13 = 2d
d = 7
b = 13 + 7 = 20
a = 13 - 7 = 6

✔ 27 − b = b −13

✔ 27 − 20 =13− a (2)

✔ d = 7 or 27 −13 = 2d

✔b = 13 + 7
a = 13 + 7 (2)

2.1.2

a = 6 d = 7
S_n = ⁿ/₂ [2a +(n - 1)d ]
S₂₀ = ²⁰/₂ [2(6) +(20 - 1)(7) ]
=1450

T₂₀ = a + 19(d)
T₂₀= 6 + 19 (7)
T₂₀=139
S_n = ⁿ/₂ [a + T_n]
S₂₀ = ²⁰/₂ [6 + 139]
=1450

✔ d = 7
✔correct substitution into correct formula

✔answer (3)

✔ d = 7
✔ T₂₀=139

✔answer (3)

2.1.3

T_n= 6 + (n -1)(7)
= 7n - 1
₂₀
∑ (6 + 7 (n -1))
^n-1
₂₀= ∑ (7n -1)
^n-1

✔T_n= 6 + (n -1)(7) or 7n - 1
₂₀
✔∑
^{n - 1}

(2)

2.2.1

r = (x -2) (x + 2)
x -2
or
r = (x² - 4) (x +2)
x² - 4
For convergence
-1 < r < 1
-1 < x + 2 < 1
-3 < x < -1

✔ (x -2) (x + 2)
x -2

or = (x² - 4) (x +2)
x² - 4

✔r = x + 2
✔ −1< r <1
✔ answer (4)

2.2.2

= - ⁷/₂
1- ½
=-7

= (x - 2)
1 - (x + 2)
= x - 2
-x - 1
= -³/₂ - 2
³/₂ - 1
= - ⁷/₂½
= -7

✔a = − ⁷/₂✔ substitution into correct formula
✔ answer (3)

✔ substitution into correct formula

✔ substitution of x = − ³/₂

✔answer (3)

[14]

QUESTION 3

3.1	2a = 4 a =2 3a + b =3 b = -3 a + b + c = -1 c = 0 T_n = 2n² - 3n OR T_n = T₁ + (n -1) d₁ + (n - 1)(n - 2) d₂ 2 =(-1) + (n - 1)(3) + (n - 1)(n - 2) (4) 2 = -1 +3n - 3 + 2n² - 6n + 4 =2n² -3n	✔2^nd difference = 4 ✔ a = 2 ✔b = −3 ✔T_n = 2n² - 3n (4) ✔ formula ✔ 2nd difference = 4 ✔simplifying ✔T_n = 2n² - 3n (4)
3.2	T_n = 2n² - 3n T₄₈ = 2(48)² - 3(48) =4464	✔ substitution ✔ answer (2)
3.3	3 + 7 +11.......... Sn = ⁿ/₂ [2a + (n -1) d] = ⁿ/₂ [2(3) + (n -1) 4] =ⁿ/₂ [6 + 4n - 4] =2n² + n	✔ a = 3 ✔ d = 4 ✔ substitution into correct formula (3)
3.4	S₆₉ = 9591 and T₁ = -1 (of the original sequence) 9591 + (-1) = 9590 S₆₉+ T₁= 9590 The 70th term of the original sequence will have a value of 9590 OR 2n² - 3n = 9590 2n² - 3n - 9590 = 0 (n - 70)(2n + 137) = 0 n = 70 T₇₀ = 9590	✔(9591)+ (−1) ✔70 (2) ✔ 2n² - 3n - 9590 = 0 ✔70 (2) [11]

QUESTION 4

4.1	(0 ; 3)	✔ (0 ; 3) (1)
4.2	x = - ^b /_2aor -2x -2 = 0 = - (-2) 2(-1) ∴x = -1 =-1 y = -(-1)² -2(-1) + 3 or y = 4ac - b² 4a = 4(-1)(3) - (-2)² 4(-1) c(-1 ; 4)	✔x = -2x -2 = 0 or - (-2) 2(-1) ✔ simplification ✔ in the context of a turning point (3)
4.3	B (1 ; 0) By symmetry A (-3 ; 0) OR x2 + 2x -3 = 0 (x + 3) (x -1) = 0 x = -3 or x = 1 A = ( -3; 0)	✔ A(− 3 ; 0) (1) ✔ A(− 3 ; 0) (1)
4.4	Equation of g: m = 4 - 0 -1 + 3 = 2 y = 2x + q or y - 0 = 2(x + 3) 0 = 2(-3) + q or 4 = 2(-1) + q y = 2x + 6 or y - 4 = 2(x + 1) q = 6 y = 2x + 6 E(0 ; 6) C( 1 ;4)	✔ m = 2 ✔subs of A(–3;0) or C (–1;4) ✔ y = 2x + 6 ✔ E(0 ;6) ✔substitution into distance formula ✔answer (6)
4.5	f'(x) = -2x -2 But m_tan = 2 -2x -2 = 2 x = -2 f(-2) = 3 y = 2x + k 3 = 2(-2) + k k = 7 OR -x² - 2x + 3 = 2x + k -x² - 4x + 3 - k = 0 x² + 4x - 3 + k = 0 For equal roots: Δ = b² - 4ac = 0 (-4)² - 4(-1)(3 - k) = 0 (4)² - 4(1)(k - 3) = 0 16 + 12 - 4k = 0 or 16 - 4k +12 = 0 k = 7 k = 7	✔ − 2x − 2 ✔ − 2x − 2 = 2 ✔ x = −2 ✔ y = 3 ✔answer (5) ✔ -x² - 2x + 3 = 2x + k ✔standard form ✔ b² - 4ac = 0 ✔substitution ✔answer (5)
4.6	g : y = 2x + 6 g^-1 : x = 2y + 6 2y = x - 6 y = x - 6 or y = x - 3 2 2	✔ x = 2y + 6 ✔ y = x - 6 or y = x - 3 2 2
4.7	g(x) ≥ g^-1 (x) 2x + 6 ≥ x - 6 2 4x + 12 ≥ x - 6 3x ≥ -18 x ≥ -6	✔2x + 6 ≥ x - 6 2 ✔ 4x +12 ≥ x − 6 ✔ x ≥ −6 (3) [21]

QUESTION 5

5.1	r = 2	✔r = 2 (1)
5.2	g(x) = 2^x + 2 g(0) = 2º + 2 = 3 B(0 ; 3) 3 = 3 + 2 0 - p p = −3	✔ g(0) = 2º + 2 = 3 ✔ y = 3 ✔substitute B(0 ; 3) and q = 2 ✔ p = −3 (4)
5.3	at A: x = −3 y = 2^-3 + 2 = 2¹/₈	✔ at A : x = –3 (p-value) ✔ substitute x = –3 into exponential eqution ✔ y-value (3)
5.4	− 3 < x ≤ 0 OR (− 3; 0]	✔ − 3 < x ✔ x ≤ 0 (2)
5.5	f (x) = 3 + 2 x + 3 f (x - 2) = 3 + 2 x - 2+ 3 h(x) = 3 + 2 x + 1	✔substitution of x – 2 ✔ h(x) = 3 + 2 x + 1 (2) [12]

QUESTION 6

6.1	A= P (1 -i )ⁿ ²/₃P = P( 1- 0,047)ⁿ ²/₃ = (1- 0,047)ⁿ log ²/₃ = nlog (1- 0,047) n = log²/₃ log (1- 0,047) n = 8, 42 years	✔A = ²/₃P ✔substitution into correct formula ✔logs ✔answer (4)
6.2.1	The book value of the tractor after 5 years book value = x(1 - 0,2)⁵ or x(0,8)⁵ =0, 32768x	✔x(1 - 0,2)⁵ or x(0,8)⁵ ✔ 0,32768x (2)
6.2.2	Price of new tractor after 5 years book value = x(1 + 0,18)⁵ or x(1,18)⁵ =2,28776x	✔ x(1 + 0,18)⁵ or x(1,18)⁵ ✔ 2,28776x (2)
6.2.3		✔ i = ^0,10/₁₂ ✔ n = 60 ✔ subst. into future value formula ✔ answer (4)
6.2.4	Sinking fund = New tractor price – Scrap value 619496,58 = x(1 +0,18)⁵ - (1 - 0,2)⁵ 619496,58 = x[(1,18)⁵ - (0.8)⁵] x = 619496,58 [(1,18)⁵ - (0.8)⁵] x = R316 057,15 x = R316 000 OR 619496,58 =x(2,28776) -x(0,32768) 619496,58 = x[1,96008] x = 619496,58 1,96008 x = R316 056,78 x = R316 000	✔ 619496,58 ✔ x(1 +0,18)⁵ - (1 - 0,2)⁵ ✔ common factor x ✔ R316 000 (4) ✔ 619496,58 ✔ x(2,28776) − x(0,32768) ✔ simplification ✔ R316 000 (4) [16]

QUESTION 7

7.1

f(x + h) = 3(x +h)² - 5 = 3(x² + 2xh +h²) - 5
=3x² + 6xh + 3h² -5
f(x + h) - f(x) = 3x² + 6xh + 3h² - 5 - 3x² +5
= 6xh + 3h²

✔3x² + 6xh + 3h² - 5
✔ 6xh + 3h²

✔ f(x + h) - f(x)
h

✔ common factor/ (6x + 3h)

✔ answer (5)

✔ f(x + h) - f(x)
h

✔ 3x² + 6xh + 3h² - 5

✔ 6xh + 3h²

✔ common factor/(6x + 3h)

✔ answer (5)

7.2.1

y = 2x⁵ + ⁴/x³
y = 2x⁵ + 4x^-3
^dy /_dx = 10x⁴ - 12x^-4

✔ 2x⁵ + 4x^-3

✔ 10x⁴

✔ - 12x^-4(3)

7.2.2

y = ( √x -x²)²
y = x^½ - x²)²
=x - 2x ^5/₂ + x⁴
^dy /_dx= 1 - 5x ^3/₂+ 4x³

✔ x - 2x ^5/₂ + x⁴✔1
✔ - 5x ^3/₂✔ 4x³(4) [12]

QUESTION 8

8.1	y = 12	✔answer (1)
8.2	12 = (0 - 2)²(0 -k) k = -3 (x - 2)²(x + 3) = 0 x = -3 OR y = 0 (x - 2)²(x + k) = 0 (x² - 4x + 4)(x + k) = 0 x³ - kx² - 4x² + 4kx +4x - 4k = 0 But -4k is the -intercept -4k = 12 k = -3 x = -3	✔ substituting (0;12) ✔ k = −3 ✔ x = −3 (3) ✔–4k ✔ − 4k = 12 or k = –3 ✔ x = −3 (3)
8.3	f(x) = x³ + 3x² - 4x² - 12x + 4x +12 f(x) = x³ - x² - 8x + 12 f'(x) = 3x² - 2x - 8 3x² - 2x - 8 = 0 (3x + 4)(x -2) = 0 x = -⁴/₃ or x = 2 y = ⁵⁰⁰/₂₇ or 18,52 or 18¹⁴/₂₇ C[ -⁴/₃ ; 18,52]	✔ f(x) = x³ - x² - 8x + 12 ✔derivative ✔derivative equal to 0 ✔factors or formula ✔x = − ⁴/₃✔y = ⁵⁰⁰/₂₇or 18,52 or 18¹⁴/₂₇ (6)
8.4	f''(x) = 6x - 2 6x - 2 < 0 x < ¹/₃ f is concave down when x < ¹/₃ OR f''(x) = 6x - 2 6x - 2 = 0 x = ¹/₃f is concave down when x < _¹/₃ OR x = x_e + x_d 2 = -⁴/₃ + 2 2 =¹/₃ or x = - ^b/_3a = - -1 3(1) =¹/₃f is concave down when x < _¹/₃	✔ 6x − 2 ✔✔x < ¹/₃ (3) ✔ 6x − 2 ✔✔x < ¹/₃ (3) ✔= -⁴/₃ + 2 2 or = - -1 3(1) ✔✔x < ¹/₃ (3) [13]

QUESTION 9

9.1

V = π r²h
π r²h = 340
h = 340
π r²

✔formula

✔equating to 340

✔ h = 340
π r² (3)

9.2

A = 2π r²+2π rh
= 2π r²+2π r [ 340 ]
[ π r² ]
= 2π r²+ ⁶⁸⁰/r
A'(r) = 4π r - ⁶⁸⁰/r²

A'(r) = 0 for minimum surface area
4π r - ⁶⁸⁰/r²= 0
r³ = ⁶⁸⁰/_4π = ¹⁷⁰/_π
= 54, 11268
r = 3,78 cm

✔ 2π r²+2π rh
✔substituting h
✔4π r - ⁶⁸⁰/r²✔A′(r) = 0
✔r³ = ⁶⁸⁰/_4π✔answer (6)
[9]

QUESTION 10

10.1.1 (a)	P(Female) = 70 = 7 = 0,47 150 15	✔ 70 ✔ answer (2)
10.1.1 (b)	P(Female playing tennis) = 20 = 2 = 0,13 150 15	✔ answer (1)
10.1.2	P(Female) = 70 150 P(Playing) = 70 150 P(Female playing tennis) = 20 = 2 = 0,13 150 15 P(Female) × P(Playing) = 70 × 70 = 4900 = 0,22 150 150 22500 P(Female playing tennis) ≠ P(Female) × P(Playing) Therefore the event of playing tennisis not independent of gender. OR P(Male) = 80 150 P(Playing) = 70 150 P(Male playing tennis) = 50 = 0,33333 150 P(Male) × P(Playing) = 80 × 70 = 5600 = 0,25 150 150 22500 P(Male playing tennis) ≠P(Male/Manlik) × P(Playing) Therefore the event of playing tennisis not independent of gender. OR P(Male) = 80 150 P(Not playing tennis) = 80 150 P(Male not playing tennis) = 80 = 0,53333 150 P(Male) × P(Not playing tennis) = 80 × 80 = 6400 = 0,28 150 150 22500 P(Male not playing tennis) ≠ P(Male) × P(Not playing tennis) Therefore the event of playing tennis in not independent of gender. OR P(Female) = 70 150 P(Not playing tennis) = 80 150 P(Female not playing tennis) = 50 = 0,33333 150 P(Female) × P(Not playing tennis) = 70 × 80 = 5600 = 0,25 150 150 22500 P(Female not playing tennis) ≠ P(Female) × P(Not playing tennis) Therefore the events of playing tennis and gender are not independent.	✔P(Play ten) = 70 150 ✔ = 70 × 70 = 4900 = 0,22 150 150 22500 ✔P(Female playing tennis) ≠ P(Female) × P(Playing) Not independent (3) ✔P(Playing) = 70 150 ✔ = 80 × 70 = 5600 = 0,25 150 150 22500 ✔P(Male playing tennis) ≠P(Male/Manlik) × P(Playing) Not independent (3) ✔ P(Not playing tennis) = 80 150 ✔ = 80 × 80 = 6400 = 0,28 150 150 22500 ✔P(Male not playing tennis) ≠ P(Male) × P(Not playing tennis) Not independent (3) ✔ P(not play ten) = 80 150 ✔ = 70 × 80 = 5600 = 0,25 150 150 22500 ✔P(Female not playing tennis) ≠ P(Female) × P(Not playing tennis) Not independent (3)
10.2	P(B) = 1 - P(B') = − 1 - 0,28 = 0,72 P(A or B) = P(A) + P(B) - P(A and B) 0,96 = 0,24 + 0,72 − P(A and B) 0,96 = 0,96 − P(A and B) P(A and B) 0 Events A and B are mutually exclusive	✔ P(B) = 0,72 ✔ P(A) = 0,24 ✔ substitution into correct formula ✔ P(A and B) = 0 (4) [10]

QUESTION 11

11.1

2 × 2! × 7! = 20 160

✔ 2 × 2!
✔ 7!
✔ 20 160 (3)

11.2

All seated in 9! = 362 880 ways

Girls seated together in 4! ways.
With the girls as one unit they can all be seated in
4! 6! ways = 17280

P(all girls seated together)
= 4!6!
9!
= 17280
362880
= 1
21
= 0,047619....
= 4,76%

✔ 9! or 362 880

✔ 4! 6! or 17280

✔ 17280 or
362880
1 or
21

0,047619 21

or
4,76% (3)
[6]

TOTAL

150

Last modified on Tuesday, 15 June 2021 08:20

Published in Grade 12 2016 June Exams, Past Papers

GRADE 12 MATHEMATICS PAPER 2 MEMORANDUM - SENIOR CERTIFICATE EXAMINATIONS PAST PAPERS AND MEMOS MAY/JUNE2016

MEMORANDUM

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