excel_math answer_ 2011 stpm 2011 trial sabah

8/4/2019 Excel_math Answer_ 2011 Stpm 2011 Trial Sabah

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SULIT*

JABATAN PELAJARAN NEGERI SABAH

EXCEL TINGKATAN 6 ATAS 2011 (STPM 2011)

SIJIL TINGGI PERSEKOLAHAN MALAYSIA

[ SKEMA JAWAPAN]

___________________________________________________________________________

This question paper consists of 6 printed pages.

(Kertas soalan ini terdiri daripada 6 halaman bercetak.)

Jabatan Pelajaran Negeri Sabah 2011

954/2 [Turn over (Lihat sebelah)

* This question paper is CONFIDENTIAL until the examination is over.

* Kertas soalan ini SULIT sehingga peperiksaan kertas ini tamat.

CONFIDENTIAL *

SULIT*

JABATANPELAJARANSABAHJABATANPELAJARANSABAH JABATANPELAJARANSABAH









954/2 STPM

MATHEMATICS T

PAPER 2

Three Hours ( Tiga jam)


2/10

Answer scheme

1

Known that BCD is an equilateral triangle, BCD = 60o.

= =

Therefore,

=

()

Because and

= =

( )

=

D1

M1

M1

A1

B1

M1

4a

4a4a

4a

A

D

B

C

N

PO


3/10

=

=

=

= (showed)

The angle that the line AP makes with the plane BCD is APO =

=

= (showed)

M1

A1

M1

A1

2

* ||+

* ||+

|| , k=

B1

A1 + M1

M1

M1


4/10

||

|

|

A1

3

Diagram 1

(a) BMQ = PQN + QPN

BMQ = MNQ ( alternate angle of circumference )

MNQ = QPN + PQN (exterior angle equal to sum of two opposite interior

angles of a triangle)

BMQ = QPN + PQN (proven)

(b) RMQ = PQN

RMQ = BMQ - BMR (complementary angle)

Known that BMA is the tangent to circle PMR

BMR = MPR ( alternate angle of circumference )

Known that MPR = NPQ

RMQ = PQN + NPQ - NPQ

RMQ = PQN (proven)

(c) RMQ = QMP.

PR is the tangent to the circle MQN

PQN = QMN ( alternate angle of circumference )

Known that QMN = QMP

From (b) RMQ = QMP (proven)

D1

B1

M1

A1

B1

M1

A1

B1

M1

A1

R

Q

P

N

BA M


5/10

4 (a)

OP = x + y

OQ = 3x3yOR = 4x + ky

PQ = mQR

OQOP = m(OROQ)

3x3y - ( x + y ) = m [ 4x + ky(3x3y)]2x4y = mx + m(k+3)yCompare:

2 = m

-4 = m(k+ 3)

-4 = 2(k+ 3)

-2 = k+ 3

k= -5

(b) If PQ is perpendicular to QR, then

PQQR = 0

[2x4y][x + (k + 3)y] = 0

Known that x y = 0, x x = 1, y y = 1

2x x4(k + 3) y y = 024(k + 3) = 024k3 = 0- 4k = 1

k =

B1

M1

M1

A1

B1

A1 + M1

M1

A1

5 3 sin x + 4 cos x = r sin x cos + r cos x sin

r cos = 3, r sin = 4

r = 5 or42

43

tan =3

4

= 53.1o

Hence,

3 sin x + 4 cos x = 5 sin (x + 53.1o

)

6 sin x + 8 cos x + 5 = 10 sin (x + 53.1o

) + 5

-1 10 sin (x + 53.1o

) + 5 11

Therefore, Max value = 11

Min value = -1

B1

M1

M1

A1

M1

A1

6 (a)

B1

M1


6/10

= 0.00642

=

=

(b)

=

=

A1

B1

M1

A1

7 Given

, A = constantIf When

B1

M1

M1

M1


7/10

When t= 20

Therefore, the percentage of the material left after 20 years is 16.0%.

A1

M1

A1

B1

8 Let X be the number of workers taking a day off,.

(a) =1 -

= 0.2424

(b)

Let the number of consecutive days off be n and the number of workers taking

days off be as Y, .

n = 4

B1

M1

A1

B1

B1

M1

M1

A1

9

= - 0.1505

1a = 0.7071

B1 + M1

M1

M1


8/10

a = 0.2929

= 0.293 A1

10 (a)

| |

| | | | | |

(b)

X= k 0 1 2

P(X= k)

(c)

= =

= = 1

=

=

(d)

M1

A1

A1

M1

M1

A1

D1

A1

0

1

P(X=k)

X


9/10

11 Let , and (a)

Let T = A + B + C

= 8 + 7 + 10

= 25

= 1 + 2 + 5

= 8

= = 0.0385

(b)

New

= 710=3

= 2 + 5

= 7

= = 0.1284

= 0.128

B1

B1

M1

A1

B1

M1

M1

A1

12

(a)

Mean =

=

= 56.29

= 56.3

Median =

=

= 53

Mode = 52

M1

A1

M1

A1

A1


10/10

(b)

= = 17.475

= 17.5

(c)

= 44

= 63

Semi-interquartile range

=

= = 9.5

(d)

(e) the data is skew to the right because .

M1 + M1

A1

M1

M1

A1

M1

A1

D1

A1

A1

32 99

excel_math answer_ 2011 stpm 2011 trial sabah

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