Maths T Paper 1 2009 Johor - Marking Scheme

JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN 954/1 PERCUBAAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN STPM JOHOR 2009 JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN MATHEMATICS T(MATEMATIK T) JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PAPER 1(KERTAS 1)JABATAN PELAJARAN JOHOR JABATAN JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN PELAJARAN JOHOR JABATAN

No. 1 [4]

2 [5]

Marks x+y i + 3 − 4 i = 3 i 1 1

Using his in the modulus Equation correct.

x2 + 6x + 9 + y2 −8y +16 = 9

1

For squaring

x2 + y2 + 6x −8y +16 = 0

1

CA

( x + 3) 2 + ( y − 4) 2 = 3

f (x) = x 3 − 2

f ( x + δx ) = ( x + δx ) 3 − 2

f ( x + δx ) = ( x + δx ) 3 − 2 = x + 3 x δx + 3 x (δx ) + (δx ) − 2 3

2

2

( x + δx ) 3 − x 3 δx →0 δx

f ' ( x ) = lim

3 x 2 δx + 3 x (δx ) + (δx ) δx →0 δx 2 = 3x 2

1

correct formula substitute with limit

1

Expand

1

CA CA

3

= lim

f ' (a ) = 3a 2 3 [6]

1

3

1

For log 3 ( x 2 − 3 x + 2) and 2 log 3 ( 2 x − 1) to be defined, x 2 − 3 x + 2 > 0 and 2 x − 1 > 0 ⇒

x < 1 or x > 2 , and

⇒

1 < x < 1 or 2

x>

1

both conditions (with “and”)

1 2

x>2

log 3 ( x 2 − 3 x + 2) < log 3 2( 2 x − 1) 2

1

1

Correct way of using laws

x 2 − 3 x + 2 < 2( 2 x − 1) 2

1 x<0

∴

4 [5]

x>

or

5 7

1

5   the solution set is  x : < x < 1 or x > 2 7  

. − 8 + 16 + 2a + b = 0 a3 + 4 a2 − a2 + b = a3 2a − 3a2 + 8 = 0

1 1 1

CA CA

1

Solving

(3a + 4 ) ( a − 2 ) = 0 a=− a=2,

4 3

, b=−

b = − 12

16 3

1 1

5 [6]

Sn-1 =

a 3− n (b n −1 − a n −1 ) b−a

1

a 2− n (b n − a n ) a 3− n (b n −1 − a n −1 ) − b−a b−a 2−n n 2 3− n n −1 2 a b −a −a b +a = b−a 2 − n n −1 a b (b − a ) = b−a = a2− n bn−1.

Tn = Sn− Sn-1 =

Tn a 2− n b n −1 = 3− n n − 2 Tn −1 a b b = (independent of n) a ⇒ the series is a geometric series. 6 [9]

1

Use his Sn-1 In the formula

1

factorisation

1

1

use his to find r (do division)

1

with correct reason

f ( x) = x 3 − 10 x 2 + ax + b ∵ 3 + 2 is a zero, ∴ 3 − 2 ia also a zero.

[x − (3 + 2) ][x − (3 − 2) ] = x (

2

)

Equating coefficient of x :

b ) 7

b − 6 = −10 ⇒ b = −28 7 6b 7− = a ⇒ a = 31 7

a = 31, b = −28

(a)

Conjugate must be correct

1

Use factorisation or long division

1

Method to find a or b

1

both a and b correct

1

factorisation

1

* followed by **

1

Conclusion

− 6x + 7

x 3 − 10 x 2 + ax + b = x 2 − 6 x + 7 ( x + Equating coefficient of x 2 :

1

f ( x) = 12 − 3 x x 3 − 10 x 2 + 31x − 28 = 12 − 3 x ( x − 4)( x − 6 x + 10 ) = 0 2

x = 4 or x 2 − 6 x + 10 = 0 For x 2 − 6 x + 10 = 0 * Discriminant = ( −6 ) − 4(1)(10) = −4 < 0 * * 2

Hence, the only real root is 4.

(b)

(

)

x 3 − 10 x 2 + 31x − 28 = x 2 − 4 Q( x ) + Ax + B When x = 2,

1

Method

1

CA

2 = 2 A + B........(1)

When x = −2, − 138 = −2 A + B.....( 2) (1) + (2), 2B = −136 ⇒ B = −68 (1) − (2), 4 A = 140 ⇒ A = 35 ∴ the remainder is 35 x − 68. 7 [9]

1

1

7(a)

1 − 3 x = (1 − 3 x ) 2

1 1 1 1 3 ( )( − ) ( )( − )( − ) 1 2 ( −3 x ) 2 + 2 2 2 ( −3 x ) 2 + ... = 1 + ( −3 x ) + 2 2 2! 3! 3 9 2 9 3 = 1− x + x − x + ... 2 8 16

1 1  Range :  x : − < x <  3 3  (b)

1   x − 2  3x  

1, 1

2 correct (1) 3 correct(1+1)

1

1

24

 24  1   U r +1 =   x 24 −r  −  2  3x   r 

r

1

r

 24  r  1 =  (− 1)   x 24 −3 r r 3  

24 − 3r = 0 r =8

1

The term in x must be correct

1

His = 0

8

8 [10]

 24  8 1 The term independen t of x =  (− 1)   x 0 3 8 81719 = 729 x−3 x−3 y= = 2 (2 − x)( x + 1) − x + x + 2

(a)

Asymptotes : x = 2, x = −1, y = 0.

(

)

dy − x 2 + x + 2 − ( x − 3)(−2 x + 1) = 2 dx − x2 + x + 2

(

x − 6x + 5 2

=

(− x

2

+x+2

)

2

)

1

2

2 correct = 1 3 correc = 2

1

formula

dy =0 dx x 2 − 6x + 5

(− x

2

+x+2

)

2

=0 1

His = 0

1

Correct answer in coordinate form

1

method

1

nature of both the turning points correct

1

all parts correct

1

points

1

asymptotes and perfect

1

quadratic eq.

1 1

b2-4ac=0 or other method Conclusion

1

Equation of OR

1

substitution

x − 6x + 5 = 0 ( x − 1)( x − 5) = 0 2

x = 1, 5. 1 Turning points are (1, − 1) and (5, − ). 9

(b)

d 2 y ( 2 x − 6)( − x 2 + x + 2)2 − ( x 2 − 6 x + 5)(2)( − x 2 + x − 2)( −2 x + 1) = dx 2 ( − x 2 + x + 2 )4 =

(2 x − 6)( − x 2 + x + 2) − ( x 2 − 6 x + 5)(2)( −2 x + 1) ( − x 2 + x + 2)3

d 2y

Point (1, -1),

<0 ⇒

dx 2

d 2y

1  Point  5, − , 9 

dx

2

local maximum point

>0 ⇒

local minimum point

(c) y

-1 0 -3/2

9 [10]

2

3

(1, - 1)

5

x

1 (5, − ) 9

x(8m − m2 x ) = 16 m2 x2 – 8mx + 16 = 0 b2 − 4ac = 64m2 − 4 (m2)(16) =0 2 ∴ m x + y − 8m = 0 is a tangent to xy = 16. Eqn. of OR is y =

1 m2

m x + y − 8m = 0 2

(1) → (2) m2 x +

………….(2) 1 m2 3

x=

x …………(1)

8m 1+ m4

x − 8m = 0

8m 1+ m4 8m 3 8m R( , ) 4 1+ m 1+ m4 m2 = x

1

x or y correct

1 1

Coordinates R m or m2 in terms of x and y.

1

elimination of m

y=

y

(m2 x + y )2 = 64m2 ( x × x + y )2 = 64 ( x ) y

y

2

2 2

(x + y ) = 64xy 10 [11]

A  = k

1

−3 1 2 1 2 −3 −1 −1 1 2 2 2 2 1

= k (−6−1)−(4−2)−(2+6) = −7k −2 −8 = −7k −10 −7k −10 = −17 7k = 7 k=1 1 2 3

− 17 0

  BC =  2 3 4   − 66 − 38 3   1 5 6   51 34 0  0



 21 9 6  =  6 − 12 9     − 24 − 3 15 

1

1

*1

(

1

CA (if 1 is not obtained here, multiplication process must be seen to obtain *1)

=

1

2   − 24 − 3 15 

 51 0 0   0 51 0     0 0 51

1

A( *)

ABC = 51 I  21  6   − 24  7 / 17 =  2 / 17   − 8 / 17

A−1 =

1 51

9 6 − 12 9  − 3 15  3 / 17 2 / 17  − 4 / 17 3 / 17  − 1 / 17 5 / 17 

 1 1 − 1  x   −1 2 − 3 1    =      y   − 3  2 1 2   z   25   − 1 x A  y  =  − 3     z  25 

)(

) is seen



 1 1 − 1  21 9 6 ABC =  2 − 3 1   6 − 12 9      2

CA

1

1

His BC

1

CA

1

CA

* his BC

x  21  y = 1  51    6 z    − 24 102 1  = 51  255    408

9 6   − 1   − 12 9   − 3  − 3 15   25   2 =  5  8  

1

His inverse

1

CA

1

CA

1

Method to find A, B &C

1

CA

x = 2. y = 5, z = 8 11 [11]

5x + 4 x + 12 A Bx + C ≡ + 2 2 x+2 ( x + 2)( x + 4) x +4 2 2 5x + 4x + 12 = A(x +4) + (Bx+C)(x+2) x= −2, 20 − 8+12 = 8A A=3 2 coef. of x , 5 = A + B B=2 x = 0, 12 = 4A + 2C C=0 2 5x + 4 x + 12 3 2x ∴ ≡ + 2 2 ( x + 2)( x + 4) x + 2 x +4 2

a)

∫

5 x 2 + 4 x +12 ( x + 2 )( x 2 + 4 )

3

1

∫

dx

3

3 2x + 2 dx x+2 x +4 = [3 ln(x+2) + ln (x2 + 4 ) ] 3 =

1

1

= 3ln (5) + ln(13) − 3ln(3) − ln(5) 325 = ln 27 2 −1 b) h = = 0.25 4 x 1 1.25 1.5 1.75 2 2

∫ ln(1+x ) dx = 1

2

y 0.6931 0.9410 1.1787 1.4018 1.6094

0.25 [ 0.6931+1.6094 + 2

2(0.9410+1.1787+1.4018)] = 1.168

1, 1

First 1, ln is seen 2nd 1, all correct

1

his substitution

1

CA

1

CA

1

All correct

1

Correct formula with his value

1

CA

12 [14]

(a)

lim f ( x) = lim− (5 − x − 1 ) = 3

1

CA

lim f ( x) = lim+ ( x 2 − 6 x + 8) = 3

1

CA

1

With reason **

1, 1

Parts of y = 5 − x +1

1

parabola

1

points,  is marked at end-points

x→1−

x →1

x →1+

x →1

f (1) = 1 − 6 + 8 = 3 lim f ( x ) = lim+ f ( x ) = f (1). * *

x →1−

x →1

⇒ f is continuous at x = 1.

(b)

y 5 4 (1, 3) −6

(c)

-1 0 1 2 3 4

x

y = 9 − 18 + 8 = −1

1

Range of {y : −1 ≤ y ≤ 5}

1

(d) 5 − ( − x − 1) = 2 − x x − 2 = x 2 − 6x + 8

1 1

x = −2, 2, 3,

{x : −2 < x < 2} ∪ {x : 3 < x < 4} If other method is used, mark accordingly.

1

1,1

method to find minimum point

CA

CA CA

All correct

Correct line is see( if no line is seen, no mark) Ignore shading If 1 set is correct, I mark only.