ADDITIONAL MATHEMATICS

SPM 2012

PAPER 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)

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 ADiagram 1 1 15

6

731 1 3 4

Given that f(x) = 3x + 4 and 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 26x + 33= 2x + 1 f(x) = 3x + 4fg(x) = 6x + 7

Given that f : x x + 5, find (a) f(3)(b) the value of h such that 2 f -1(h) = f(3), Answer:

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

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

2(h 5) = 8Using the principle: ax + b ccx b a= 1h = 9 2 M1

It is given that 3 and m + 4 are the roots of the quadratic equation x2 + (n 1)x + 6 = 0, where m and n are constants. Find the value of m and of n. Answer: (a) (x 3) [x (m + 4)] = 0 x2 3x (m + 4)x + 3(m + 4) = 0 x2 (m + 7)x + 3(m + 4) = 0 Compare to x2 + (n 1)x + 6 = 0. 3m + 12 = 6 ; (m + 7) = n -1 m = 2 5 = n 1 n = 4 If both correct 2 M1 M1 if only one correct

A quadratic equation x(x 4) = p 2q, where p and q are constants, has two equal roots.Express p in terms of q. Answer: x2 4x p + 2q = 0 a = 1, b = 4, c = p + 2q (or) 2q p b2 4ac = 0 ( 4)2 4(1)(2q p) = 0 16 8q + 4p = 0 p = 2q 4 3 M1 M2In general formroots condition

6. Given that f(x) = 3x2 + 2x + 13, find the range of values of x for f(x) 5 Answer:

3x2 + 2x + 13 5 3x2 + 2x + 8 0 3x2 2x 8 0 Let 3x2 2x 8 = 0 (3x + 4)(x 2) = 0 x = 4/3 ; x = 2 ?

x 4 3 ; x 2

- 4/3 2 xa = 3 > 0 minimum graph M1 M2 3 M2

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

33 (3 2x + 4) = 1

32x + 7 = 1 2x + 7 = 0 x = 7/2

30 = 1 M1 M2 3Use the concept of the rules of equation of indicessame baseuse law of indices No. 1

8. Solve the equation: 1 + log2 (x 2) = log2 x Answer:

log2 2 + log2 (x 2) = log2 x OR log2 2(x - 2) = log2 x 2x 4 = x x = 4

Use the concept of the rules of equation of logarithms. M1 M1M2 3Converting log to indices (x 2)log2 1 = x (x 2)x 21 = M2 M1

The first three positive terms of a geometric progression are 2, p and 18. Find the value of p and the common ratio of the progression. Answer: = 36

d = 3

M1 3 M2p18 2 p = p2 p = 6 use the concept of common ratio

It is given that 11, y + 4 and 3y x are three consecutive terms of a arithmetic progression. (a) Express y in terms of x. (b) Find the common difference if x = 8

Answer:

(a) (y + 4) 11 = (3y x) (y + 4) y = x 3

(b) x = 8 then y = 5 The three terms now are: 11, 9, 7

d = 2

M1 2 2 M1use the concept of common difference

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 T3 T2 = 12a ar2 ar = 12a a(r2 r) = 12a r 2 r 12 = 0 (r + 3)( r 4) = 0 r = 3, r = 4 3 M2 M1use formula Tn = ar n - 1

12. The variables x and y are 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 x21/y1/x2(5, 6)0 2 M1 M1pyy11x22Diagram 121px2yp q1p1 = 2 p =

pq= gradient = q = 2/5 5 0 6 2 2 =

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 = 4 Eqn: y 5 = 4(x 7) y = 4x + 33 M2 3 M1xy 15 -1, 3 +70B(15, 7)A( 1, 3) 2215 + 17 3 M1Use the concept of m1m2 = 1

14. Diagram 14 shows a straight line PQ with the equation

(a) x - intercept: Answer: y - intercept: 2k = 8

k = 4 1 1x= 1. Determine the value of2k 10 y + h

(b) k5h = 10 h = 2

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 a and b. Answer:

(a) OP = 32 + 32 = 32 OR 4.243 (b) OP = 2b a 1 1OabAPBDiagram 15bause triangle law

16. The following information refers to the vectors a and b. a = ( ) , b = ( )

It is given that a = kb, where a is parallel to b and k is a constant. Find the value of (a) k (b) m Answer:

( ) = k ( )

(a) 6 = 2k (b) m 4 = 5k k = 3 m = 19 M1 16m 4 2 5 6m 4 25 1

Solve the equation tan2 3tan + 2 = 0 for 0 360o. Answer: (tan 1) (tan 2) = 0 tan = 1; tan = 2 = 45o , 135o; = 63.43o , 243.43 o = 45o , 63.43o, 135o , 243.43 o M1 3 M2

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 = 1.125CBAO=4 M1 2 2DDiagram 18OD32 M1

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

(a) h(x)(b) the value of k if h(1) = 4 ,

Answer:

(a)

(b)

1 3h(x) = 6kx 8 h(1) = 6k(1) - 8h(x) = 3kx2 8x + 56k 8 = 4 k = 2 M2 M1basic differentiationapply 2nd derivative

The gradient of the tangent to the curve y = x2(2 + px) at x = 2 is 7. Find the value of p

Answer:

dy/dx = 4x + 3px2 M2 M1 3y = 2x2 + px3 At x = 2; 4( 2) + 3p( 2)2 = 7 12p = 15

p = 5/4

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 = 2510 1 M1 27 22 77 27 27 210 k [ x ] = 257 2 k [ 7 2 ] =15 k = 3

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 the students. Answer:

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

32 = (32 + 402) (6)6 402 = 6954 x2 1 M1 2x2 = manipulating the formula of meanmanipulating the formula of variance

There are 10 different coloured marbles in the box.Find(a) the number of ways 3 marbles can be chosen from the box.(b) the number of ways at least 8 marbles can be chosen from the box. Answer:

(a) 10= 120C(b) C + C + C M1 2 M13 1010108109= 56 2Applying the concept of combination

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(black 1st round) = 5/20 = , 5/19 2nd round P(white 1st round) = 15/20 = , 14/19 2nd round

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

(b) ( x 15/19) + ( x 5/19) = 15/38 M1 2 2 M1

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 the test.

Answer: p = 0.6, q = 0.4, n = 8, r > 6P( r > 6) = 1C(0.6)878(0.4)C+(0.6) (0.4)7880 = 0.1064 M1 M2 3 M1Applying the principle of binomial distribution probability

THE END

Those who have worked hard with the correct method of learning Additional Mathematics surely will receive their rewards