PREPARATORY EXAMINATION 2018 - Crystal Math

PREPARATORY EXAMINATION 2018

MARKING GUIDELINES

10611

MATHEMATICS (PAPER 1)

14 pages

GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION

MATHEMATICS (Paper 1)

MARKING GUIDELINES

10611/18

1.1 1.1.1

1.1.2

QUESTION 1

x2 x 30 0

(x 6) x 5 0

x 6 or x 5 3x2 8x 4 0

b b2 4ac x

2a (8) (8)2 4(3)(4) x

2(3) x 3,10 or x 0,43

1.1.3

5 x x 1

5 x 1 x ( 5 x)2 (1 x)2

x2 3x 40 (x 4) (x 1) 0 x 4 or x 1 NA

1.1.4

6x2 3x 3x2 3

2x2 x 3x2 2x2 3x2 x 0 x2 x 0 x (x 1) 0 x 0 or x 1

OR

correct factors both answers correct

(2) standard form

correct subst. in the correct formula

x 3,10 x 0,43 Penalise for incorrect rounding off here only

(4)

isolate square root square both sides standard form

both answers correct choose correct answer

(5)

simplify

standard form correct factors critical values correct inequalities

2

6x2 3x 3x2 3

6x2 3x 9x2 6x2 9x2 3x 0 3x2 3x 0 3x(x 1) 0 x 0 or x 1

1.1.5

2x2 7. 2x 2

2 2.2 x

x

7.2 2

2

0

4.2 x

x

7.2 2

2

0

x

x

(4.22 1) (22 2) 0

2 x 2

1

4

x

or 22 2

x

22

22

NA

x 2 2 x 4

1.2 6x2 2 px 3x p 0 6x2 x(2 p 3) p 0 b2 4ac (2 p 3)2 4(6)( p) 4 p2 12 p 9 (2 p 3)2 = perfect square rational roots

10611/18

simplify

standard form correct factors critical values correct inequalities

(5)

standard form factors

x

22

22

NA

x = -4 (5)

coefficient of x

correct substitution in correct formula

factors = perfect square

(conclusion) (4)

[25]

3

QUESTION 2

2.1

Tn a (n 1) d

239 6 (n 1) (5)

239 6 5n 5

250 5n

n 50

2.2

T3 = ar2 = 18 ...

T5 = ar4 = 162 ...

? : r2 = 9 (r < 0)

r = 3 ar2 = 18

a(9) = 18

a = 2

Sn

a(r r

n 1) 1

2((3)7 1) S7 3 1

2(2187 1) 4

1 094

2.3 3; x; 11; 21; 35 x ? 3; 11 x; 10; 14 2x = 4 ? 14 2x = 10 x = 5

10611/18

Tn =?239 d = ?5

answer (3)

eq eq

r = 3

a = 2

correct substitution in correct formula

answer (6)

first difference d1 equating second

difference answer

(3)

4

2.4

S8 : S4 97:81

S8 97

S4 81

a1 r8 a1 r 4 97

1 r 1 r 81

a 1 r8 1 r 97

1 r a 1 r 4 81

1 r8 97 1 r 4 81

1 r 4 1 r 4 97

1 r 4

81

1 r 4 97 81

r 4 16 81

r 4 2 4 3

r 2 3

r 2 3

First three terms: 9; 6; 4

2.5.1

2( p 5) 2( p 5)2 2( p 5)3 ... r p5 Convergent: 1 r 1 1 p 5 1 4 p 6

2.5.2

S

a 1 r

1

1

(

1 2

)

2 3

5

10611/18

a1 r8 a1 r 4 97

1 r 1 r 81

simplify 1 r8 97 1 r 4 81

factors

r 4 2 4 3

r 2 3

first three terms (6)

r = p ? 5

1 p 5 1

4 p 6 (3)

substitute a = 1

r

1 2

answer (3) [24]

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