Pahang New Stpm 2012 Math t

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PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA BAHARU

NEGERI PAHANG DARUL MAKMUR

2012

Instruction to candidates:

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE TOLD TO DO SO.

Answer all questions in Section A and any one question in Section B. Answers may be written in either English or Bahasa Malaysia. All necessary working should be shown clearly. Scientific calculators may be used. Programmable and graphic display calculators are prohibited. A list of mathematical formulae is provided on page of this question paper.

This question paper consists of 5 printed pages.

954/1 STPM 2012

One and a half hours

MATHEMATICS T

PAPER 1

http://edu.joshuatly.com/ http://fb.me/edu.joshuatly

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Section A [45 marks] Answer all questions in this section.

1. The function f is defined by f : x ⟼ , x 1.

(a) Sketch , on the same axes of coordinates, the graphs of f and f 1 .

State the relation between the graphs of f and f 1 [4 marks] (b) Find f 1 , and state its domain and range . [4 marks]

2. Express in partial fractions. [6 marks]

3. The 33 matrix M is given as

013401312

M .

(a) Show that M is a non singular matrix (that is M has an inverse, M –1 ). [2 marks] (b) Determine the matrix M –1 using elementary row operation. [2 marks] (c) Solve the following system of linear equations by method of matrices that that involves

M and M –1.

20x + 10y + 30z = –100 – 10x + 40z = 300 30x + 10y = 200 [4 marks] 4. The complex number z is defined by sincos iz .

(a) By De Moivre’s theorem, state z1 , 2z and 2

1z

in terms of sine and cosine of or 2 .

Hence, find z

z 1 in terms of cos and show that 2cos21

22

zz . [5 marks]

(b) Use the results of z

z 1 and 2

2 1z

z , prove that 12coscos2 2 . [2 marks]


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5. A curve is represented by the equation 07682 yxy . (a) Show that the curve is a parabola with vertex )3,2( . [3 marks] (b) Find the focus and the directrix of the parabola. [3 marks] (c) Latus rectum of a parabola is the chord of the parabola passing through the focus and

perpendicular to the axis of the parabola. Find the length of the latus rectum of the parabola 07682 yxy . [2 marks]

6. A straight line l has equation r = (1, 2, -5) + t (2, -3, 1), the planes π1 and π2 have equations

2x + 5y – 3z = 6 and 4x + 3y = 8 respectively. (a) Find the position vector of the point of intersection of l and π1. [4 marks] (b) Find the angle between the planes π1 and π2. [4 marks]


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Section B [15 marks] Answer any one question in this section.

7. Write cos x – 2 sin x in the form R cos ( x α ), with R > 0 and 0° < α < 90 ° , and give the

value α correct to nearest 0.1°. Find the maximum and minimum values for y = cos x – 2 sin x and give, to the nearest 0.1°, the values of x in the range 0° < x < 360° where the maximum and minimum values exist. [5 marks]

(a) Sketch the curve y = cos x – 2 sin x for 0° ≤ x ≤ 360°. [2 marks] (b) Find the range of values x between 0° and 360° that satisfy the inequality,

cos x – 2 sin x ≥ 2 [4 marks]

(c) Find the largest value and least value of . [4 marks] 8. The position vectors a, b and c of three points A, B and C respectively are given by

a = 2i + j − k, b = 3i + j + 2k, c = 2i – 2j +2k.

(a) Find a unit vector parallel to 2a − b + c. [3 marks] (b) Calculate the acute angle between b and 2a − b + c. [3 marks] (c) Find the vector of the form λi + μj − k perpendicular to both a and b. [2 marks] (d) Determine the position vector of the point D which is such that ABCD is a parallelogram

having BD as a diagonal. [3 marks] (e) Calculate the area of the parallelogram ABCD. [4 marks]

===END OF QUESTION PAPER===


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MATHEMATICAL FORMULAE

(RUMUS MATEMATIK)

Binomial expansions

(Kembangan binomials)

Conics

(Keratan kon)

Parabola with vertex ( h, k ), focus ( a + h , k ) and directrix x = −a + h Parabola dengan bucu ( h, k ), fokus ( a + h , k ) dan direktriks x = −a + h

Ellipse with centre ( h, k ) and foci ( −c + h, k ), ( c + h, k ), where c2 = a2 − b2 Elips dengan pusat ( h, k ) dan fokus ( −c + h, k ), ( c + h, k ), dengan c2 = a2 − b2

Hyperbola with centre ( h, k ) and foci ( −c + h, k ), ( c + h, k ), where c2 = a2 + b2 Hiperbola dengan pusat ( h, k ) dan fokus ( −c + h, k ), ( c + h, k ), dengan c2 = a2 + b2


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1

Marking Scheme PEPERIKSAAN PERCUBAAN STPM BAHARU NEGERI PAHANG 2012

SEMESTER 1 2012 Mathematics T Paper 1 (954/1)

`No. SCHEME MARKS

1.(a)

The graphs of f and f 1 are reflections in the line y = x .

D1 graph f Shape and (1 , 3 )

D1 graph f 1

Shape and ( 3, 1 )

D1

f and f 1 must intersect

at point on line y = x

B1

1.(b) Let y = y =

y =

y = y =

x = 1 +

∴ f 1 : x ⟼ 1 +

Domain of f 1 is Range of f 1 is

M1 completing the squares

A1

B1 B1

8 marks


```

2

`No. SCHEME MARKS 2

= x +

= + +

= A + B + C

x = 2 , 20 = 16 A , A =

x = − 2 , −12 = − 4 C , C = 3

x = 0 , 0 = 4 A – 4 B − 2 C , B =

= − +

= x + − +

B1

B1

M1 substitution

A1

3 values correct

A1

A1

6 Marks


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3

`No. SCHEME MARKS 3.(a)

(a)

1

)]1)(1)(0()2)(4)(1()3)(0)(3[()1)(1)(3()3)(4)(1()0)(0)(2[(

M

Since 0M , M is not singular

(b) From

100010001

013401312

use the correct row operations to

change to

11111912434

100010001

and conclude

11111912434

M 1

(c)

203010

013401312

zyx

or

200300100

0103040010301020

zyx

203010

zyx

M

203010

11111912434

zyx

60610

210

x = 210 , y = –610 , z = 60

M1

A1

M1

A1

B1

M1

A1

A1 [8 marks]


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4

NO. SCHEME MARKS 4

(a) sincos1 iz

2sin2cos2 iz

2sin2cos12 i

z

cos2

)sin(cos)sin(cos1

iiz

z

2cos2

)2sin2(cos)2sin2(cos12

2

iiz

z

Square both sides of cos21

zz

12coscos2cos422cos2

cos421

)cos2(1

2

2

22

2

22

zz

zz

B1

(one correct) B1

(all correct)

M1

(find z

z 1

and 22 1

zz )

A1A1

M1

A1

[7 marks]


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5

NO. SCHEME MARKS 5.

(a) 07682 yxy 7862 xyy 78)3()3( 22 xy )2(8)3( 2 xy )2)(2(4)3( 2 xy Since the equation 07682 yxy can be transformed into the form )(4)( 2 hxaky which represent a parabola with vertex

),( kh , the curve 07682 yxy is a parabola with vertex )3,2( . (b) Focus = )3,0(),( kah Directrix : 4) ahx (c) Length of latus rectum = 84 a

M1

A1

A1

M1A1 (either one

correct) or

M1A1A1 (Both correct)

M1A1

[7 marks]


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6

NO. SCHEME MARKS 6.(a) Let T(x, y, z) be the point on the line r.

Then, OT = (1, 2, -5) + t (2, -3, 1), or

zyx

=

ttt

53221

.

That is, x = 1 + 2t, y = 2 – 3t, z = – 5 +t

The point T also lies on the plane. So (x, y, z) must satisfy the plane equation 2x + 5y –3z = 6. Thus, 2(1 + 2t) + 5(2 – 3t) – 3(– 5 +t) = 6. This gives t = 3/2 = 1.5. x = 1+2(1.5) = 4; y = 2 – 3(1.5) = – 5/2; z = – 5 + 1.5 = – 7/2

So, position vector of the point of intersection is (4, 25

, 27

).

M1

A1

M1

A1

(b) Vector equation of π1: r.(2i + 5j – 3k) = 6

Unit normal vector, 1n̂ = 232522

352

kji = 38

352 kji

Vector equation of π2: r.(4i + 3j) = 8

Unit normal vector, 2n̂ = 2324

34

ji = 5

34 ji

cos θ = 1n̂ • 2n̂ = 38

352 kji •

534 ji =

385)0(3)3(5)4(2

cos θ = 385

23 , θ = 41.74o.

The angle between the planes π1 and π2

is 41.74o or 41.7o.

M1 A1 (both

1n̂ and 2n̂ )

M1

A1


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7

NO. SCHEME MARKS

7.

Let cos x – 2 sin x ≡ R cos (x + α) ≡ R cos x cos α – R sin x sin α Hence, R sin α = 2 R cos α = 1 tan α = 2 , α = 63.43° R2 ( sin2 α + cos2 α) = 22+ 12

R2 = 5 R = Hence, cos x – 2 sin x = cos ( x + 63.4°) Maximum value of y is when x+ 63.4° = 360° i.e. x = 296.6° Minimum value of y is when x+ 63.4° =180° i.e. x =116.6 °

M1 2 equations

M1

and R values

A1

B1

B1

7.(a) When x = 0° , y =1 When x = 360°, y = 1

y = 2 1 116.6° 296.6° 360°

D1 shape

D1

Max & min point

Start and end points


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NO. SCHEME MARKS 7.(b) Solve cos x – 2 sin x ≥ 2

Let cos ( x + 63.43°) = 2

cos ( x + 63.43°) = x + 63.43° = 26.57° , 333.43° , ( 386.57° ) x = 270° , 323.14° Hence the range of values x such that is { x : 270° ≤ x ≤ 323.1°}

D1 draw

line y = 2 on

graph

M1 A1 A1

7.(c)

From )43.63cos(5sin2cos 0 xxx and 1)43.63cos(1 0 x 5sin2cos5 xx 533sin2cos53 xx

53

13sin2cos

153

1

xx

the greatest value of f(x) =

=

=

the least value of f(x) =

=

=

M1

M1

A1

A1

[15 marks]


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9

No. Mark Scheme Marks 8 (a) 2a – b + c = 2

112

–

213

+

22

2;

2a – b + c =

21

3

222 )2)132 (( cba ; cba 2 = 14

Unit vector parallel to 2a – b + c = 143 i –

141 j –

142 k

M1 M1 A1

(b) b • (2a − b + c) = b . cba 2 cos θ

cos θ = cbabcbab

2.)2.( ; cos θ =

222222 )2()1(3.213

)23).(23(

kjikji

cos θ =14.14

)2(2)1(1)3(3 = 144 =

72 .

the acute angle, θ = 73.40o or 73.4o or 1.281 rad or 1.28 rad or 0.4078π or 0.408 π.

M1 M1 A1

(c) The vector perpendicular to both a and b is the cross product a b.

a b = 23112

1

kji = (2+1)i – (4+3)j + (2–3)k

= 3i –7j – k.

M1 A1

(d) If D is the point (x, y, z), then the position vector OD = xi + yj + zk. For parallelogram ABCD, AB = CD (or AD = BC ).

AB = AO + OB =

211132

=

301

. CD = DO + OC =

222

zyx

AB = CD ⇒

301

=

222

zyx

. x = 1, y = –2, z = –1

The position vector of D is xi – 2j – k.

M1 (obtain AB and CD ) M1 A1


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No. Mark Scheme Marks 7(d) Alternative:

AD = BC ; AD = AO +OD =

zyx

112

. BC = BO + OC =

222123

=

031

AD = BC ⇒

zyx

112

=

031

. x = 1, y = –2, z = –1.

The position vector of D is xi – 2j – k.

M1 (obtain AD and CB ) M1 A1

(e) Area of //gram ABCD = ADAB .

AB AD =

301

031

.

AB AD = 01301

3

kji = (0 + 9)i – (0 + 3)j + (–3 – 0)k = 9i – 3j – 3k

Area of ABCD = ADAB

= 222 ))39 3(( = 99 = 3 11 unit2

Alternative: Area of //gram ABCD = ACAB .

AB AC =

301

1212

22 =

301

33

0

AB AC = 330301

kji = (0 + 9)i – (3 – 0)j + (–3 – 0)k = 9i – 3j – 3k

Area of ABCD = ACAB = 222 ))39 3(( = 99 = 3 11 unit2

M1 M1 M1 A1 M1 M1 M1A1


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