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ADDITIONALMATHEMATICS

SPM 2012PAPER 1

MY ANSWER & MARKING(This is my own answer and marking scheme and it has

nothing to do with any scheme created by anyone else)

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PAPER 1

1. Diagram 1 shows the relation between set A and

set B. State

(a) the object of -1.the relation,

(b) the range of the relation.

Answer:

(a) 5

(b) { -3, -1, 1, 3}

Set BSet A

Diagram 1 1

1

5

6

7

3

11

3

4

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2. Given that f(x) = 3x + 4and fg(x) = 6x + 7, find

(a) fg(4),

(b) g(x).

Answer:

(a) fg(4) = 6(4) + 7 = 31

(b) f[g(x)] = 3g(x) + 4

3g(x) + 4 = 6x +7

g(x) =

1

M1

2

6x + 3

3

= 2x + 1

f(x) = 3x + 4

fg(x) = 6x + 7

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3. Given that f : x x + 5, find

(a) f(3)

(b) the value of hsuch that 2 f -1(h) = f(3),

Answer:

(a) f (3) = 3 + 5 = 8

(b) f -1(h) = h 5

2(h 5) = 8

Using the principle:

ax + b

ccxb

a=

1

h = 9 2

M1

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5. A quadratic equationx(x4) = p2q, wherepand q

are constants, has two equal roots.

Expresspin terms of q.

Answer:

x2

4xp + 2q = 0a = 1, b =4, c =p + 2q (or) 2qp

b24ac = 0

(4)24(1)(2qp) = 0

16 8q + 4p = 0

p = 2q4 3

M1

M2

In general form

roots condition

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6. Given that f(x) =3x2+ 2x + 13, find the range of

values ofxfor f(x) 5

Answer:

3x2 + 2x + 13 5

3x2 + 2x + 8 0

3x22x8 0Let 3x22x8 = 0

(3x + 4)(x2) = 0

x =4/3; x = 2 ?

x 43;x 2

- 4/3 2x

a = 3 > 0minimum graph

M1

M2

3

M2

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7. Solve the equation: 27(32x + 4) = 1

Answer:

33 (32x + 4) = 1

32x + 7 = 1

2x + 7 = 0

x =

7/2

30 = 1

M1

M2

3

Use the concept of the rules of

equation of indices

same base

use law of indices No. 1

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8. Solve the equation: 1 + log2 (x2) = log2x

Answer:

log2 2 + log2 (x2) = log2 x ORlog2 2(x - 2) = log2 x

2x4 = x

x = 4

Use the concept of the rulesof equation of logarithms.

M1

M1

M2

3Converting log to indices

(x2)log21 =

x

(x2)x

21 = M2

M1

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9. The first three positiveterms of a geometric

progression are 2, pand 18. Find the value of

pand the common ratio of the progression.

Answer:

= 36

d = 3

M1

3

M2

p 18

2 p

=

p2

p = 6

use the concept of common ratio

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10. It is given that 11, y + 4and 3yxare three

consecutive terms of a arithmetic progression.

(a) Express y in terms of x.

(b) Find the common difference ifx = 8

Answer:

(a) (y + 4)11 = (3yx)(y + 4)y = x3

(b) x = 8 then y = 5

The three terms now are: 11, 9, 7

d =2

M1

2

2

M1

use the concept of common difference

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11. In a geometric progression, the first term is a and the

common ratio is r. Given that the third term of the

progression exceeds the second term by 12a, find

the values of r.

Answer:

T3= ar2

T3T2= 12a

ar2 ar = 12a

a(r2 r) = 12a

r 2r 12 = 0(r + 3)(r4) = 0 r =3, r = 4 3 M2

M1

use formula Tn = arn - 1

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12. The variables xand yare related by the equation

= 1 . Diagram 12 shows the straight line graph

obtained by plotting against . Find the value of

(a) p,

(b) q.

Answer:

(a)

(b)

q

x

2

1/y

1/x2

(5, 6)

0

2

M1

M1

p

y

y

1 1

x2

2

Diagram 12

1

px2y p

q1

p

1= 2 p =

p

q= gradient = q = 2/5

5062

2

=

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13. Diagram 13 shows a straight line AB. Find

(a) the midpoint of AB,

(b) the equation of the perpendicular bisector of AB.

Answer:

(a) Midpoint AB = ( )

= (7, 5)

(b) MAB= =

Mbisector AB =4Eqn: y5 =4(x7)

y =4x + 33 M2

3

M1

x

y15 -1, 3 +7

0

B(15, 7)

A(1, 3)

22

15 + 1

73

M1

Use the concept of m1m2 =1

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14. Diagram 14 shows a straight line PQwith the equation

(a) x - intercept:

Answer:

(b) y - intercept: 2k =8

k =4

1

1

x= 1. Determine the value of

2k10

y+

(a) h

(b) k

5h = 10

h = 2

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15. Diagram 15 shows the vectors OA, OB and OP

drawn on a grid of equal squares with sides of 1 unit.

Determine

(a) OP,

(b) OP in terms of aand b.

Answer:

(a) OP= 32+ 32= 32 OR 4.243

(b) OP = 2ba 1

1

O

a

bA

PB

Diagram 15

b a

use triangle law

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16. The following information refers to the vectors

aand b.

a= ( ) , b = ( )

It is given that a = kb, where ais parallel to band k

is a constant. Find the value of

(a) k(b) m

Answer:

( ) = k( )

(a) 6 = 2k (b) m4 = 5k

k = 3 m = 19

M1

1

6m4

25

6

m4

2

5

1

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17. Solve the equation tan23tan+ 2 = 0for 0

360o.

Answer:

(tan 1) (tan 2) = 0

tan = 1; tan = 2= 45o , 135o; = 63.43o , 243.43o= 45o , 63.43o, 135o , 243.43o

M1

3

M2

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18. Diagram 18 shows sectors OAB and ODC with centre

O. It is given that OA = 4 cm, the ratio of OA : OD = 2 :

3 and the area of the shaded region is 11.25 cm2. Find(a) the length, in cm, of OD,

(b) , in radians.

Answer:

(a)

OD = 6

(b) (6)2 (4)2 = 11.25= 1.125

CB

A

O

=4

M1

2

2

D

Diagram 18

OD 3

2

M1

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19. Given the function h(x)= kx3 4x2+ 5x, find

(a) h(x)

(b) the value of k ifh(1)= 4 ,

Answer:

(a)

(b)

1

3

h(x)= 6kx8 h(1)= 6k(1) - 8

h(x) =3kx28x + 5

6k8 = 4k = 2

M2 M1

basic differentiation

apply 2nd

derivative

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20. The gradient of the tangent to the curve

y = x2(2 + px)atx =2is 7. Find the value ofp

Answer:

dy/dx= 4x + 3px2

M2

M1

3

y = 2x2+ px3

At x =2; 4(2) + 3p(2)2 = 7

12p = 15

p = 5/4

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21. Given that f(x)dx= 10,Find

(a) the value of f(x)dx,

(b) the value of k if [ f(x)k ] dx = 25.

Answer:

(a)

(b) f(x) dx k dx = 25

10 1

M1

2

7

22

7

7

2

7

2

7

2

10k [ x ] = 2572

k [ 72 ] =15

k =3

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22. The mass of a group of 6 students has a mean of 40

kg and a standard deviation of 3 kg. Find

(a) the sum of the mass of the students,

(b) the sum of the square of the mass of thestudents.

Answer:

N = 6 (a) x = x (N)= 3 = 40 (6) = 240x = 40

(b) 32=

(32+ 402) (6)6

402

= 6954

x2

1

M1

2

x2 =

manipulating the

formula of mean

manipulating

the formula

of variance

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23. There are 10 different coloured marbles in the box.

Find

(a) the number of ways 3 marbles can be chosen

from the box.(b)thenumber of ways at least 8 marbles can be

chosen from the box.

Answer:

(a)10 = 120C

(b) C + C + C

M1 2

M1

3

1010108 109

= 56 2

Applying the concept

of combination

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24. A box contains 20 chocolates. 5 of the chocolates

are black chocolates flavour and the other 15 are

white chocolates flavour. Two chocolates are taken

at random from the box. Find the probability that(a) both chocolates are black chocolates,

(b) the chocolates taken are of different flavour.

Answer:P(black1stround) = 5/20 = , 5/192ndround

P(white1stround) = 15/20 = , 14/192ndround

(a) (x 4/19) = 1/19

(b) (x 15/19) + (x 5/19) = 15/38

M1 2

2 M1

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25. In a test, 60% of the students has passed. A sample of

8 students is chosen at random. Find the probability

that more the 6 students from the sample passed thetest.

Answer:

p = 0.6, q = 0.4, n = 8, r > 6

P( r > 6) = 1C (0.6)8 7 8 (0.4)C+(0.6) (0.4)7 8

8 0

= 0.1064

M1 M2

3

M1

Applying the principle

of binomial distributionprobability

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THE END

Those who have worked hard

with the correct method of learning

Additional Mathematics

surely will receive their rewards