stpm-trial-2012-mathst-qa-kedah.pdf

7/27/2019 stpm-trial-2012-mathsT-qa-kedah.pdf

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3

1 0 .

M a t r i x P i s gi v en b y

−

=

3

2

2

1

2

1

1

0

1

P

.

I f

−

−

−

−

−

=

+

+

2

2

2

2

5

1

2

2

4

2

b I

P

P

a

,wh er eI i s t h e 3

× 3 i d en t i t y m a t r i x ,f i n d

t h ev al u e s of a an d b .

[ 4 m ar k s ]

F i n d

)

(

2

b I

P

P P

+

+ a

.

[ 1

m ar k ]

H en c e , s ol v e t h e s i m ul t an e o u s e q u a t i on s

3

=

− z

x

2

2

+

+

z

y

x

5

3

2

2

=

+

+

z

y

x

[ 5 m ar k s ]

1 1 .

F i n d t h ev al u e of a an d of b i f

b

x

x

x

x

x f

+

−

+

+

=

1 2

1 3

)

(

2

3

4

a

i s ex a c t l y d i v i s i b l e

b y

2

3

)

(

2

+

−

=

x

x

x g

.

[ 2 m ar k s ]

H en c e ,

( a )

f i n d t h e s ol u t i o

n of

0

)

(

= x f

.

[ 3 m ar k s ]

( b )

f i n d t h e s e t of

v al u e s of ݔ wh i c h s a t i s f y

)

(

3

)

(

x

x f

g−≤

.

[ 5 m ar k s ]

( c )

u s i n g t h e s u b s

t i t u t i on

x

y

1

=

, s ol v e t h e e q u a t i on

0

1

6

1 3

1 2

4

2

3

4

=

+

−

+

−

y

y

y

y

[ 3 m ar k s ]

1 2 .

Gi v en t h a t f ( x ) =2 x 2

+l n ( 4 x + 5 ) ,h a s d om ai n { x : x ∈R ,– 5 4

<x ≤ 1 } .

( a )

S t a t e t h e a s y m

p t o t e of f .

[ 1

m ar k ]

( b )

F i n d al l ( l o c al )

m ax i m um an d mi ni m um p oi n t s of f .

[ 7 m ar k s ]

( c )

F i n d t h e c o or d

i n a t e of t h e p oi n t of i nf l ex i on.

[ 3 m ar k s ]

( d )

S k e t c h t h e gr a

ph of f .

[ 4 m ar k s ]

( e )

S t a t e t h em ax i m umv al u e of f f or t h e gi v en d om ai n.

[ 1

m ar k ]

http://www.chngtuition.blogspot.com



STPM KEDAH 2012-MT PAPER 1-MARKING SCHEME

1. 2x = 3y

2x = 31 – x

x ln 2 = (1 – x )ln 3

x (ln 2 + ln 3) = ln 3

x =ln 3ln 6 .

2.

3. θ θ

2sec=

d

d x,

tan 0 = 0, tanπ

4 = 1

(1 + x 2)2 = (1 + tan2θ)2 = sec4 θ

( ) ∫∫ ⋅=

+

4

0

2

4

1

022

secsec

1

1

1π

θ θ

dx x

θ d

= ∫ 4

0

2cos

π

θ θ d

( )( )∫ ∫ +=

+

1

0

4

022

2cos12

1

1

1π

θ θ d dx

x

=12

4

0

1sin2

2

π

θ θ +

=18 [π + 2].

B1

M1A1

A1

B1

M1

M1A1

M1

A1

A1

A1

M1

B1

2

4.

5. cos3−= x y ax

cos3= yx ax

dx

dy x x y

32))(3( + = axasin−

dx

dy x y x

323 + = axa sin−

( )( ) ( )( ) axa xdx

dy

dx

yd x x y

dx

dy x cos323

22

2

232

−=++

+

axadx

dy x

dx

yd x xy

dx

dy x cos363 22

2

232

−=+++

axadx

yd x xy

dx

dy x cos66 2

2

232

−=++

Sub into

32

2

232 66 yxadx

yd x xy

dx

dy x −=++

066 322

2

32 =+++ yxadx yd x xy

dxdy x

( ) 066 22

2

22

=

+++ y xa y

dx

dy x

dx

yd x x

( ) 066 22

2

22

=+++ y xadx

dy x

dx

yd x

6.

Let( )

( )4:

4:

2

1

+=

−=

xm yl

xm yl

D1: x y =

D1 : shape of 2−= x y

D1 : points (2, 0) , (0, 2)

B1 : x = 1 , 4

B1 : 1 < x < 4

The solution set is }{ ℜ∈<< x x x ,41: B1

1

2

1 2

M1A1

M1A1

M1

A1

B1 (l 1 or l 2)

3

intercept 2554 =+ y x at point F (x 1, y 1) and G (x 2, y 2).

( )

( )

m

m x

mm x

mmx x

xm x

54

2025

202554

252054

25454

1

1

11

11

+

+=

+=+

=−+

=−+

−

+

+= 4

54

20251

m

mm y

m

m y

54

91

+=

∴

++

+

m

m

m

mF

54

9,

54

2025

( )

( )

m

m x

mm x

mmx x

xm x

54

2025

202554

252054

25454

2

2

22

22

+

−=

+=+

=−+

=++

+

+

−= 4

54

20252

m

mm y

m

m y

54

412

+=

∴

++

−

m

m

m

mG

54

41,

54

2025

854

41

54

9

54

2025

54

202522

=

+−

++

+

−−

+

+

m

m

m

m

m

m

m

m

2

22

854

32

54

40=

+

−+

+ m

m

m

m

64254016

26242

2

=++ mm

m

01640162

=++ mm

02522

=++ mm

( )( ) 0212 =++ mm

2,2

1−−=m

(a)

(b)

M1

A1

A1

M1

A1

M1

A1

4

7. (a) |w| = 41 ,

arg w = tan –1

4

–5= 2.47 rad. (3 s.f.)

(b) z w =

3 + 2i–5 + 4i

x –5 – 4i –5 – 4i

= – 741

– 2241

i

(c) Z(3, 2), W(–5, 4), Z*(3, –2), P(x , y )

mid-pt of ZZ* = mid-pt of WP

(3, 0) = (x – 5

2,y + 4

2)

x = 11, y = –4

p = 11 – 4i.

8.

( )∫1

0

2225ln5 dx x

x

= ( )∫1

0

225ln525ln dx x

x

=

1

0

2225ln525ln5

−∫ dx x

x x

= [ ]1

022

525ln5x x

x −

= 50 ln5 – 24

Hence, m = 50 , n = 24

9. (a) Domain : }{ 5,: −≥ℜ∈= x x xg

Range : }{ 0,: ≥ℜ∈= y y yg

M1A1

M1A1

A1

A1

M1

B1

5ln2ln x y =

5ln21

=dx

dy

y

5ln2 ydx

dy=

25ln52 x

dx

dy=

M1

A1

B1B1

M1

M1A1

A1A1

B1

B1




5

(b) ( )3)( +−= x x f g o

( )( )3

2

532

+−=+−+

xb xa

( )( )[ ]2

2

32

53+−=

+−+ x

b xa

12=

a

a = 2

02

5=

+− b

b = 5

(c)

Since the line parallel to the x-axis intersects the curve y = f(x) once, therefore thefunction f is one-to-one function.

f(x) = 2(x + 3)2 – 5

2(y + 3)2 – 5 = x

2(y + 3)2 = x + 5

( )2

53

2 +=+x

y

2

53

+±=+

x y

2

53

+−=+

x y

2

53

+−−=

x y

5,2

53)(1

−≥+

−−=∴−

x x

x f

B1

M1

A1

A1

Shape D1All correct D1

since 3−≤ y

M1

A1

f(x )

x -3

-5•

6

10.

−−

−−

−

=

+

−

+

−

−

222

251

224

100

010

001

322

121

101

322

121

101

322

121

101

ba

−−

−−

−

=

+

−

+

−−−

222

251

224

00

00

00

322

2

0

91010

465

421

b

b

b

aaa

aaa

aa

−−

−−

−

=

++++

++++

−−−++−

222

251

224

39210210

4265

421

baaa

abaa

aba

6

15

−=

−=+

a

a

11

41

=

=++−

b

ba

( )

−−

−−

−

−

=++

222

251

224

322

121

1012

bI aPPP

=

600

060

006

( ) I bI aPPP 62

=++

−−

−−

−

=−

222

251

224

6

11P

=

−

5

2

3

322

121

101

z

y

x

−−

−−

−

=

5

2

3

222

251

224

6

1

z

y

x

M1

A1

A1

A1

B1

B1

M1

B1

7

−=

02

13

z

y x

0,2

1,3 =−== z y x

11.

A1

A1

ሺݔ െ 1ሻଶሺݔ െ 2ሻଶ 3ሺݔ െ 1ሻሺݔ െ 2ሻ 0

ሺ1ሻ ൌ 2 ൌ 0

ሺ2ሻ ൌ 8 44ൌ 0

ൌ െ6, ൌ 4

(a) ସݔ െ ଷݔ6 ଶݔ13 െ ݔ12 4 ൌ 0

ሺݔଶ െ ݔ3 2ሻሺݔଶ െ ݔ3 2ሻ ൌ 0

ሺݔ െ 1ሻଶሺݔ െ 2ሻଶ ൌ 0

ݔ ൌ 1, ݔ ൌ 2

(b) ሺݔሻ െ3 ሺݔሻ

ሺݔ െ 1ሻሺݔ െ 2ሻሺݔଶ െ ݔ3 5ሻ 0

Consider ଶݔ െ ݔ3 5,

Since ൌ ܦ, 1 െൌ 11 ൏ 0

Hence ଶݔ െ ݔ3 5 0, ݔ א

Hence, ሺݔ െ 1ሻሺݔ െ 2 ሻ 0

1 ݔ 2

The solution set is ሼݔ : 1 ݔ 2, ݔ א ሽ

M1

A1

M1A1

A1

M1A1

M1

A1

A1

ݕ ൌ ݕ, 1 ൌ1

2

c) ସݕ4 െ ݕ12 ଷ ଶݕ13 െ ݕ6 1 ൌ 0

011611311214

234

=+

−

+

−

x x x x

ݔ ସ െ ݔ6 ଷ ݔ13 ଶ െ ݔ12 4 ൌ 0

ሺݔ െ 1ሻଶሺݔ െ 2ሻଶ ൌ 0

ݔ ൌ 1, ݔ ൌ 2

B1

M1 refer to (a)

A1

8

12. (a) the asymptote is x = – 54

.

(b) f ′ (x ) = 4x +4

4x + 5.

=4(4x + 1)(x + 1)

4x + 5

f ′′ (x ) = 4 + 4(–1)(4x + 5) –2(4)

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

(4x + 5)2

When f ′ (x ) = 0, (x , y) = (–1, 2) or (– 14 ,

18 + ln 4)

local max imum at (–1, 2)

local minimum at (– 14 ,

18 + ln 4)

(c) When f ′′ (x ) = 0 , x = – 34

{ – 74

is not in the given domain}

inflex ion at (– 34

,98

+ ln 2)

(e) max imum f occurs at x = 1

max . f is 2 + ln 9

sign of f ′ ( x):

sign of f ′′ ( x):

–1–1

4

x

–3

4

+ +–

–5

4

–

+ +––

B1

M1 for f’ and f’’

M1 A1 for stationarypoints

M1 determine natureof stationary points

A1

A1

M1 A1

A1

B1

A1 for f’ or f’’

D1 for shape

D1 for asymptote

D1 for points in (b) & (c)

D1 for end point (1, 2 + ln 9)

xO

y

(1, 2 + ln 9)

(–1

4,

1

8+ ln 4)

(–1, 2)

(–3

4,9

8+ ln 2)

4

5−= x

(d)




STPM KEDAH 2012-MT PAPER 2-MARKING SCHEME CHU/SMKK

1. LHS = ( ) ( ) A A A +°−° 60sin60sinsin

= ( )

−°− A A 2cos120cos

2

1sin

=

−−

− A A 2cos

2

1

2

1sin

=

−+ A A

2sin21

2

1sin

2

1

= [ ] A A 2sin43sin4

1−

= [ ] A A 3sin4sin3

4

1−

= A3sin4

1

2. (a)−1

2k − 1=

k

−6or vector method M1

k = − 32

or k = 2 A1

(b) (−1

2k − 1) x (

k

−6) = − 1 or vector method M1

k =613 A1

3.

(a) Given ∠RQS = ∠QPS = α

∠QRS = ∠PQS = β ( Angle in alternate segments) B1

2α + 2β = 180o (Sum angles of ∆PQR)

α + β = 90o.

Therefore ∠QSR = 180o − α − β = 90o (Sum angles of ∆QRS)

M1

A1

M1

A1

PQ

R

S

2

⇒ Chord QR is the diameter of the circle (Angle in semi-circle) B1

The length of chord QR is twice the radius of the circle. B1

(b) Let O be the centre of circle

∠ROS = 60o (Angle at the centre is twice angle at the circumference) B1

OR = OS (radius of circle)

∆ROS is isosceles

∠OSR = ∠ORS = 60o = ∠ROS

∆ORS is an equilateral triangle B1

Therefore OR = OS = RS

Length of RS equal to the radius B1

4. 25 0.

dy xy y

dx+ − =

௬

௬మିହ ݀ݕ = ଵ௫ ݀ݔ B1

ଵଶ ln (y2 – 5) = ln Ax M1

y2 – 5 = Cx2 A1

x = 2, y = 1 C = –1 M1

y2 – 5 = – x2

ଶݔ ଶݕ ൌ 5 A1

x = 1, y = 0 C = –5

ଶݔ5 ଶݕ ൌ 5 M1

A1

D1

x

y

1-1

√5

െ√5

O

െ√5

x

y

√5

√5

െ√5 O

3

5.

(a) = AC BC AB + = a + b M1

AC AP λ = = λ( a + b ) A1

(b) DP AD AP +=

= b + µ DX M1

= b + µ

− ba

2

1

= a µ 2

1 + ( )b µ −1 A1

Comparing; λ( a + b ) = a µ 2

1+ ( )b µ −1 , we get M1

λ= µ 2

1and λ= 1− µ M1

Solving: λ =3

1and µ =

3

2

AC AP3

1= and DX DP

3

2= or equivalent ly XD XP

3

1=

Thus, the point P trisects AC and XD. A1

6. (cos x + 1)2 + (sin x + 3 )2 = 5 + 2 [ cos x + 3 sin x ]

= 5 + 2 [ b cos (x − α) ] M1

b = 2, α =π3 M1M1

= 5 + 4 cos (x − π3 ) A1

(a) (cos x + 1)2 + (sin x + 3 )2 = k2

M1

A1 A1

P

X

D C

BA

P

4

5 + 4 cos (x − π3 ) = k2

cos (x − π3

) =k2 − 5

4

−1 ≤ k2 − 5

4 ≤ 1 M1

1 ≤ k2 ≤ 9

{ k : k ∈ R : −3 ≤ k ≤ −1 or 1 ≤ k ≤ 3 } A1A1

(b) (cos x + 1)2 + (sin x + 3 )2 = 5 + 2 2

cos (x − π3

) =1

2M1

x − π3

= −π4

,π4

M1

x =

π

12 ,

7π

12 A1

(c) 1 ≤ 5 + 4 cos (x − π3 ) ≤ 9 M1

29 ≤

2

5 + 4 cos (x − π3 )

≤ 2 M1

p =29

and q = 2 A1

p =29

, corresponding value of x =π3

B1

q = 2, corresponding value of x =4π3

B1

7. n(S) = 3125

C

(a) Required probability =

3125

345

C

C

= 0.0447 A1

(b) Required probability =

3125

135

120

125

C

C C C

M1

B1M1

B1




5

= 0.0551 A1

8. (a) P(X = 3) + P(X = 4) + P(X = 5) Three correct terms M1

= 10(0.7)3(0.3)2 + 5(0.7)4(0.3) + (0.7)5. Binomial distribution M1

= 0.8369 A1

(b) P( he miss the target ) = P( M ) = 0.05

np = 2.5 M1

P( M ≤ 4) = P(M = 0) + P(M = 1) + P(M = 2) + P(M = 3) + P(M = 4)

sum of 5 terms M1

= e−2.5 { 1 + 2.5 +2.52

2+

2.53

6+

2.54

24}

Use Poisson distribution M1

= 0.8912 A1

9. (a) P(X = 1) = P(WR)

=5

3

6

3×

=10

3B1

P(X = 0) = P(R) =

2

1

6

3=

P(X = 2) = P(WWR) =20

3

4

3

5

2

6

3=××

P(X = 3) = P(WWWR) =20

1

3

3

4

1

5

2

6

3=××× B1B1

(b) E(X) =

+

++

20

13

20

32

10

30 M1

=4

3A1

6

E(X2) =20

27

20

19

20

34

10

30 =

+

++

Var(X) =2

4

3

20

27

−

M1

=80

63A1

10.

x f

4.7 24.9 75.1 165.3 215.5 125.7 2

Mean =∑∑

f

fx

=60

6.311M1

= 5 hr 12 minutes A1

S. deviaton = ට ∑ ௫మ∑ െ ቀ∑ ௫

∑ ቁଶ

=2

60

6.311

60

36.1620

− M1

= 11 minutes A1

New mean =100

406.56.311 ×+M1

= 5 hr 21 minutes A1

2

2

)6.5(40

−∑ fx

= 0.3

∑ ݂ݔ ଶ = 1258 M1

New s. deviation =

2

100

406.56.311

100

125836.1620

×+−

+M1

7

= 19 minutes A1

11. ሺ 4ሻ = 0

a (16) – 40a – 24 = 0 M1

a = –1 A1

(a) +−

=,0

,102)(

x x f

(b) E(X) = ݔሺെ2ݔ 10ሻହସ dx

= െ2ݔ ଶ ହݔ10ସ dx

=

5

4

2

3

53

2

+− x

xM1

= ଵଷଷ A1

(c) P(X<4.5) = െ (4.5)2 + 10(4.5) െ 24 =4

3B1

Prob =ଷ!ଶ! [P(X<4.5)]2[P(X>4.5)]

=4

1

4

3

4

3

!2

!3××× M1

=64

27A1

12.

(a) Probability = ( )14.2< X P

=

−<

03.0

2.214.2 Z P M1

= ( )2−< Z P A1

= 0.0228

Percentage = 2.28 % A1

(b) Probability = ( )2.21.2 <<Y P

=

−<<

−

02.0

15.22.2

02.0

15.21.2 Z P M1

otherwise

54 ≤≤ x

B1

M1A1

8

= ( )5.25.2 <<− Z P A1

= 0.9876

Percentage = 98.8% A1

(c) X – Y ∼ N( 2.2 –2.15, 0.032

+ 0.022)

X – Y ∼ N( 0.05, 0.0013) B1

P ( rod will not pass through tube)

= P ( X < Y ) B1

= P ( X – Y <0)

=

−<

0013.0

05.00 Z P

= ( )387.1−< Z P A1

= 0.08272 ≈ 0.0827 A1

(d) P ( rod will pass through tube) = 1 – 0.0827 = 0.9173 B1

P ( two packets out of three selected packets of rod –tube where rod will pass

through tube)

= 3 ( 0..9173)2(0.0827) M1

= 0.2088 ≈ 0.209 A1

M1


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