Grade 12 Mathematics - Western Cape · Grade 12 Mathematics Paper 1 MEMORANDUM Other products for Mathematics available from Learning Channel: 2 2009 Mathematics Grade 12 Paper 1:

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National Senior CertificateGrade 12

MathematicsPaper 1

MEMORANDUMOther products for Mathematics available from Learning Channel:

2 2009 Mathematics Grade 12 Paper 1: Memorandum

MATHEMATICS EXEMPLAR EXAMINATION GRADE 12 PAPER 1

MEMORANDUM

QUESTION 1

1.1.1 x2 − 1 _

x + 1 = 2

\ x2 − 1 = 2(x + 1)

\ x2 − 1 = 2x + 2

\ x2 − 2x − 3 = 0

\ (x − 3)(x + 1) = 0

\ x = 3 or x = −1

But x ≠ −1

\ x = 3 is the solution

OR

x2 − 1 _

x + 1 = 2

\ (x + 1)(x − 1)

__ (x + 1)

= 2

\ x −1 = 2 provided x ≠ −1

\ x = 3

standard form = 0 ✓factorisation ✓both answers ✓excluding ✓ x = −1

factorisation ✓simplification correct answer

excluding ✓ x = −1

(4)

1.1.2 8 − ( x − 2)( x − 3) = 0

\ 8 − (x2 − 5x + 6) = 0

\ 8 − x2 + 5x − 6 = 0 – 1 for inaccurate rounding off for both answers.

\ − x2 + 5x + 2 = 0

\ x2 − 5x − 2 = 0

\ x = −(−5) ± √

______________ (−5)2 − 4(1)(−2) ____

2(1)

\ x = 5 ± √___

33 __ 2

\ x = 5, 37 or x = −0, 37

simplification ✓standard form ✓substitution into ✓formula

correct answers ✓

(4)

1.2.1 4x2 < 9

\ 4x2 − 9 < 0

\ (2x + 3)(2x − 3) < 0

\ – 3 _ 2 < x < 3 _

2

factorisation ✓endpoints ✓inequality notation ✓

(3)

– 3 _ 2 3 _

2

2009 Mathematics Grade 12 Paper 1: Memorandum A

1.2.2 x ∈{−1; 0 ;1} correct answer ✓ (1)

1.3 3x − y = 2

\ 3x − 2 = y

\ 3(3x − 2) + 9x2 = 4

\ 9x − 6 + 9x2 = 4

\ 9x2 + 9x −10 = 0

\ (3x + 5)(3x − 2) = 0

\ x = − 5 _ 3 or x = 2 _

3

\ y = −7 or y = 0

3 ✓ x − 2 = y

substitution ✓standard form ✓factorisation ✓both ✓ x-values

y ✓ = −7

y ✓ = 0

(7)

[19]


QUESTION 22.1 Area of triangle 1: 1 _

2 (2 cm)(2 cm) = (1)(2) cm2

Area of triangle 2: 1 _ 2 (4 cm)(3 cm) = (2)(3) cm2



The areas form the following pattern:

(1)(2) ; (2)(3) ; (3)(4) ; (4)(5) ; …

Area of triangle n: (n)(n + 1) cm2

Area of triangle 100: (100)(100 + 1) cm2 = 10 100 cm2

OR

Area of triangle 1: 1 _ 2 (2 cm)(2 cm)




The bases are in arithmetic sequence:

2 ; 4 ; 6 ; 8 ; 10 ; …

General term is 2 + (n −1)2 = 2n

The heights are in arithmetic sequence:

2 ; 3 ; 4 ; 5 ; …

General term is 2 + (n − 1)(1) = n + 1

Therefore, the general term for areas is:

Area of triangle n: 1 _ 2 (2n)(n + 1) cm2

= n(n + 1) cm2

Area of triangle 100: (100)(100 + 1) cm2 = 10 100 cm2

determining areas ✓establishing pattern ✓obtaining general term ✓area of 100 ✓ th triangle

determining areas ✓general terms of ✓arithmetic sequences obtaining area of nth triangle

area of 100 ✓ th triangle


OR

Area of triangle 1: 1 _ 2 (2 cm)(2 cm) = 2 cm2




The areas form a quadratic number pattern:

2 ; 6 ; 12 ; 20 ; …

a + b + c 2 6 12 20

3a + b 4 6 8

2a 2 2

2a = 2 3a + b = 4 a + b + c = 2

a = 1 \3(1) + b = 4 \1 + 1 + c = 2

\ b = 1 \ c = 0

Area of triangle n: (n2 + n) cm2

Area of triangle 100: [(100)2 + 100] cm2 = 10 100 cm2

determining areas ✓quadratic pattern ✓obtaining general term ✓area of 100 ✓ th triangle

(4)

2.2 n(n + 1) = 240

\ n2 + n − 240 = 0

\ (n + 16)(n −15) = 0

\ n = −16 or n = 15

But n ≠ −16

\ n = 15

The 15th triangle will have an area of 240 cm2

equating general term ✓to 240

factorising ✓obtaining 15 triangles ✓

(3)

[7]

QUESTION 3

3.1.1 1 _ 181

+ 2 _ 181

+ 3 _ 181

+ 4 _ 181

+ … + 180 _ 181

a = 1 _ 181

d = 1 _ 181

n = 180

\ S180 = 180 _ 2 [ 1 _

181 + 180 _

181 ] = 90[1] = 90

OR

S180 = 180 _ 2 [2 ( 1 _

181 ) + (179) 1 _

181 ] = 90[1] = 90

correct ✓ a and d

correct ✓ n

S ✓ n formula

correct answer ✓

(4)


3.1.2 ( 1 _ 2 ) + ( 1 _

3 + 2 _

3 ) + ( 1 _

4 + 2 _

4 + 3 _

4 ) + … + ( 1 _

181 + 2 _

181 + … + 180 _

181 )

= 1 _ 2 + 1 + 1 1 _

2 + 2 + … + 90

a = 1 _ 2 d = 1 _

2 Tn = 90

\ 1 _ 2 + (n – 1) 1 _

2 = 90

\ 1 + n – 1 = 180

\ n = 180

\ S180 = 180 _ 2 [ 1 _

2 + 90] = 90 [90 1 _

2 ] = 8 145

OR

S180 = 180 _ 2 [2 ( 1 _

2 ) + (179) ( 1 _

2 ) ] = 90 [90 1 _

2 ] = 8 145

simplifying fractions ✓to get series

✓ 1 _ 2 + (n −1) 1 _

2 = 90

n ✓ = 180

substitution into S ✓ n formula to get 8145

(4)

3.2 ar5 = √__

3

ar7 = √___

27

\ ar7 _

ar5 = √

___ 27 _

√__

3

\ r2 = √___

27 _ 3

\ r2 = √__

9

\ r2 = 3

\ r = √__

3 (terms are positive)

\ a( √__

3 )5 = √__

3

\ a = √__

3 _ ( √

__ 3 )5

\ a = 1 _ ( √

__ 3 )4

\ a = 1 _ ( 3

1 _ 2 ) 4

\ a = 1 _ 9

ar ✓ 5 = √ _ 3 ; ar7 = √

_ 27

dividing ✓r ✓ = √

_ 3

correct working with ✓surds

a ✓ = 1 _ 9

(5)

3.3.1 ∑ n = 1

∞ 2 ( 1 _

2 x) n

= 2 ( 1 _ 2 x) 1 + 2 ( 1 _

2 x) 2 + 2 ( 1 _

2 x) 3 + 2 ( 1 _

2 x) 4 + …

= x + 1 _ 2 x2 + 1 _

4 x3 + 1 _

8 x4 + …

The series converges for:

−1 < 1 _ 2 x < 1

\ −2 < x < 2

r ✓ = 1 _ 2 x

−1 < ✓ 1 _ 2 x < 1

−2 < ✓ x < 2

(3)

3.3.2 a = 1 _ 2 r = 1 _

2 ( 1 _ 2 ) = 1 _

4

\S∞ = 1 _ 2 _

1 – 1 _ 4 = 2 _

3

a ✓ and r

S ✓ ∞ formula

✓ 2 _ 3 (3)


3.4 4 + 6 + 9 + 13,5 + ..........

a = 4 r = 3 _ 2 Sn = 2 000 000

\ 2 000 000 = 4 [ ( 3 _

2 ) n – 1] __

3 _ 2 – 1

\ 2 000 000 = 8 [ ( 3 _ 2 ) n – 1]

\ 250 000 = ( 3 _ 2 ) n – 1

\ 250 001 = ( 3 _ 2 ) n

\ log 3 _ 2 (250 001) = n

\ n = 30, 65422881

Malibongwe will be able to pay off the R2 000 000 on the last day of March (31 days).

constant ratio ✓correct substitution ✓into the Sn formula

use of logs ✓n ✓ = 30, 65422881

31 days ✓

(5)

[24]

QUESTION 4

4.1.1 vertical: x = −1

horizontal: y = 0

vertical asymptote ✓horizontal asymptote ✓ (2)

4.1.2

–3 –2 –1 1 2 3

–3

–2

–1

1

2

3

x

y

x =

–1

(–3; –1)

(–2; –2)

(0; 2)

(1; 1)

0

x ✓ = −1

y ✓ = 0

left branch ✓coordinates on left ✓branch

right branch ✓coordinates on right ✓branch

(6)

4.1.3 y = 2 __ x + 1 – 3

+ 2

\ y = 2 _ x − 2

+ 2

denominator: ✓ x − 2

+2 ✓ (2)


4.1.4

–3 –2 –1 1 2 3

–3

–2

–1

1

2

3

x

y

x =

–1

(–3; –1)

(–2; –2)

(0; 2)

(1; 1)

0

y = 1

Therefore 2 _ x + 1

≥ 1 for −1 < x ≤ 1

−1 < ✓ x

x ✓ ≤ 1

(2)

4.2.1 ƒ(x) = 2x2 where x ≥ 0

OR

ƒ(x) = 2x2 where x ≤ 0

x ✓ ≥ 0 OR x ≤ 0

(1)

4.2.2

x

yƒ

y = x

ƒ–1

ƒ ✓ƒ ✓ −1


OR

x

y

ƒ y = x

ƒ–1

(2)

4.2.3 y = ( 1 _ 2 ) x

x = ( 1 _ 2 ) y

\ log 1 _ 2 x = y

\ g–1(x) = log 1 _ 2 x

x ✓ = ( 1 _ 2 ) y

g ✓ –1(x) = log 1 _ 2 x

(2)

4.2.4

x

y

(1; 0)

shape ✓(1; 0) ✓

(2)

4.2.5 log 1 _ 2 x < 0 for x > 1 x ✓ > 1 (1)

[20]


QUESTION 5

5.1 y = −2ƒ(x)

\ y = −2(2 cos x)

\ y = −4 cos x

90° 180° 270° 360°

–4

–2

2

4

x

y

–90°

amplitude ✓domain ✓

(2)

5.2 Amplitude is 4 amplitude ✓ (1)

5.3 y = ƒ ( x _ 2 )

\ y = 2 cos ( 1 _ 2 x)

period is 360° _ 1 _ 2 = 720°

720° ✓

(1)

5.4 g(x) = ƒ(x) − 2

g(x) = 2 cos x − 2

maximum is 0

max ✓

(1)

[5]

QUESTION 6

6.1 y = a(x + p)2 + q

\ y = a(x + 1)2 + 6

Substitute (0 ; 2) :

\ 2 = a(0 + 1)2 + 6

\ 2 = a + 6

\ 2 − 6 = a

\ a = −4

\ ƒ(x) = −4(x + 1)2 + 6

y ✓ = a(x + 1)2 + 6

Substitute (0 ; 2) ✓a ✓ = −4

ƒ( ✓ x) = −4(x + 1)2 + 6

(4)


6.2 Range: y ∈ (−∞ ; 6] y ✓ ∈ (−∞ ; 6] (1)

6.3 y = −4(x + 1)2 + 6 (ƒ)

\ − y = −4(x + 1)2 + 6 (g)

\ g(x) = 4(x + 1)2 − 6

g ✓ (x) = 4(x + 1)2 − 6

(1)

[6]

QUESTION 7

7.1.1 ieƒƒ = (1 + 0,08 _

2 ) 2 –1

\ ieƒƒ = 0,0816

formula ✓0,0816 ✓ (2)

7.1.2 P = 100 000(1,0816)−4

\ P = R73 069,02

OR

P = 100 000 (1 + 0,08 _

2 ) –8

\ P = R73 069,02

formula ✓answer ✓

(2)

7.2 90 000 = 200 000 (1 − 0,08)n

\ 9 _ 20

= 0,92n

\ log0,92 ( 9 _ 20

) = n

\ n = 9,576544593

9 years and 7 months

correct substitution ✓into formula

use of logs ✓answer ✓

(3)

[7]

QUESTION 8

8.1 2 500 000 = x [(1,0075)361 – 1]

___ 0,0075

\ 2 500 000 × 0,0075 ___

[(1,0075)361 – 1] = x

\ x = R1 354,67

correct formula ✓n ✓ = 361

✓ 0,09 _

12 = 0,0075

F = 2 500 000 ✓answer ✓

(5)

8.22 500 000 =

x [1 – (1 + 0,07 _

12 ) –240

] ___

( 0,07 _

12 )

\ 2 500 000 × ( 0,07

_ 12

) ___

[1 – (1 + 0,07 _

12 ) –240

] = x

\ x = R19 382,47

correct formula ✓n ✓ = 240

✓ 0,07 _

12

P = 2 500 000 ✓answer ✓ (5)

[10]


QUESTION 9

9.1 ƒ′(x) = lim h → 0

ƒ(x + h) − ƒ(x)

___ h

\ ƒ′(x) = lim h → 0

−2(x + h)2 + 1 − (−2x2 + 1)

____ h

\ ƒ′(x) = lim h → 0

−2(x2 + 2xh + h2) + 1 + 2x2 − 1

_____ h

\ ƒ′(x) = lim h → 0

−2x2 − 4xh − 2h2 + 1 + 2x2 − 1 _____ h

\ ƒ′(x) = lim h → 0

h(−4x − 2h)

__ h

\ ƒ′(x) = lim h → 0

(−4x − 2h)

\ ƒ′(x) = −4x − 2(0)

\ ƒ′(x) = −4x

−2( ✓ x + h)2 + 1

−(−2 ✓ x2 + 1)

−2 ✓ x2 − 4xh − 2h2

✓ h(−4x − 2h)

__ h

–4 ✓ x

(5)

9.2 y = (2 √__

x – 1 _ 3x

) 2 \ y = 4x – 4 √

__ x _

3x + 1 _

9x2

\ y = 4x – 4 x 1 _ 2 _

3x + 1 _

9 x–2

\ y = 4x – 4 _ 3 x

1 _ 2 + 1 _

9 x–2

\ dy

_ dx

= 4 – 4 _ 3 × – 1 _

2 x

– 3 _ 2 + 1 _

9 × –2x–3

\ dy

_ dx

= 4 + 2 _ 3 x

– 3 _ 2 – 2 _

9 x–3

\ dy

_ dx

= 4 + 2 _ 3x 3 _ 2 – 2 _ 9x

3

4 ✓ x − 4 _ 3 x

− 1 _ 2 + 1 _

9 x −2

✓ ✓ ✓ 4 + 2 _ 3x

3 _ 2 − 2 _ 9x

3

(4)

[9]

– 1 for inaccurate notation

– 1 for inaccurate notation


QUESTION 10

10.1 ƒ(x) = ax3 + bx

\ ƒ ′(x) = 3ax2 + b

\ ƒ ′(−1) = 3a(−1)2 + b

\ ƒ ′(−1) = 3a + b

Now y = x + 4

mt = 1

1 = 3a + b

Now at x = −1

y = −1 + 4 = 3

\Substitute ( − 1; 3) into the equation of ƒ :

ƒ(−1) = a(−1)3 + b(−1)

\ 3 = −a − b

\ a + b = −3

Solving simultaneously:

a = 2 and b = −5

ƒ ✓ ′(x) = 3ax2 + b

mt ✓ = 1

1 = 3 ✓ a + b

a ✓ + b = −3

a ✓ = 2

b ✓ = −5

(6)

10.2.1 y-intercept: (0 ; − 4)

x-intercepts:

0 = 2x3 − 6x − 4

\ 0 = x3 − 3x − 2

\ 0 = (x + 1)(x2 − x − 2)

\ 0 = (x + 1)(x − 2)(x + 1)

\ x = −1 or x = 2 (−1; 0) (2 ; 0)

y ✓ -intercept

0 = 2 ✓ x3 − 6x − 4

( ✓ x + 1)(x2 − x − 2) = 0

( ✓ x + 1)(x − 2)(x + 1)

x ✓ -intercepts

(5)

10.2.2 ƒ(x) = 2x3 − 6x − 4

\ ƒ ′(x) = 6x2 − 6

\ 0 = 6x2 − 6

\ 0 = x2 −1

\ x = ±1

ƒ(1) = −8

ƒ(−1) = 0

Turning points are (1; − 8) and (−1; 0)

ƒ ✓ ′(x) = 6x2 − 6

0 = 6 ✓ x2 − 6

x ✓ = ±1

(1; − 8) and (−1; 0) ✓

(4)


10.2.3

–3 –2 –1 1 2 3

–8

–6

–4

–2

2

4

x

y

0

intercepts with the ✓axes

turning points ✓shape ✓

(3)

10.2.4 ƒ′(x) = 6x2 − 6

\ ƒ″(x) = 12x

\ 0 = 12x

\ x = 0

ƒ(0) = −4

Point of inflection at (0 ; − 4)

ƒ ✓ ″(x) = 12x

x ✓ = 0

(0 ; − 4) ✓

(3)

10.2.5 p > 0 or p < −8 p > 0 ✓p < −8 ✓ (2)

QUESTION 11

11.1 A = (2x)(3x) + 2(y)(3x) + 2(y)(2x)

\ 200 = 6x2 + 6xy + 4xy

\ 200 = 6x2 + 10xy

\ 100 = 3x2 + 5xy

\ 100 − 3x2 = 5xy

\ 100 _ 5x

– 3x2 _

5x = y

\ y = 20 _ x – 3x _ 5

6 ✓ x2 + 6xy + 4xy

200 = ✓arriving at answer ✓

(3)

11.2 V = (2x)(3x)(y)

\ V = (2x)(3x) ( 20 _ x – 3x _ 5 )

\ V = (6x2) ( 20 _ x – 3x _ 5 )

\ V = 120x – 18x3 _

5

V = (2 ✓ x)(3x)(y)

V = 120 ✓ x − 18x3 _

5

(2)


11.3 V(x) = 120x – 18 _ 5 x3

\ V′(x) = 120 – 18 _ 5 × 3x2

0 = 120 – 18 _ 5 × 3x2

\ 0 = 120 – 54 _ 5 x2

\ 0 = 600 – 54x2

\ 54x2 = 600

\ x2 = 600 _ 54

\ x2 = 100 _ 9

\ x = 10 _ 3

V ✓ ′(x)

V ✓ ′(x) = 0

x ✓ = 10 _ 3

(3)

[8]

QUESTION 12

12.1 x + y ≥ 10

y ≥ 1 _ 2 x

y ≤ 8

x ✓ + y ≥ 10

y ✓ ≥ 1 _ 2 x

y ✓ ≤ 8 (3)

12.2

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

1

2

3

4

5

6

7

8

9

10

11

x

y

Blue

Red

A

B

(2; 8)

(4; 6)

(6; 4)

x ✓ + y ≥ 10

y ✓ ≥ 1 _ 2 x

y ✓ ≤ 8

feasible region ✓

(4)

12.3 C = 40x + 40y

\ 40x + 40y = C

\ 40y = −40x + C

\ y = −1x + C _ 40

C = 40 ✓ x + 40y

search line on diagram ✓(2 ; 8) ✓(4 ; 6) ✓(6 ; 4) ✓ (5)

[12]

Grade 12 Mathematics - Western Cape · Grade 12 Mathematics Paper 1 MEMORANDUM Other products for Mathematics available from Learning Channel: 2 2009 Mathematics Grade 12 Paper 1:

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