Sabah Trial Stpm 2012-Mathst Paper 1(Q&A)

CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL* TrialTrialTrialTrial STPMSTPMSTPMSTPM SABAHSABAHSABAHSABAH 2012201220122012 MathematicsMathematicsMathematicsMathematics TTTT PaperPaperPaperPaper 1111

954/1*This question paper is CONFIDENTIAL until the examination is over CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*

- 1 -

1111 Determine the set of values of x which satisfies the inequality

1302+

>+x

x [4 marks]

2222 If P and Q are arbitrary sets, by using the laws of algebra of sets, show that)()()()( QPQPPQQP ∩−∪=−∪− [4 marks]

3333 Use the trapezium rule with six ordinates to find an approximation for

θθπ

π d∫ 26

sin [5 marks]

4444 If the area of a circle of radius, r is A cm2, finddrdA . Given that the area

increases with time t (seconds), at a rate of 3)1(2+t

cm2s-1. Find the rate of

change in the radius of the circle, in terms of r and t.

If 2)1(11+

−=t

A , shows that when t = 1, the rate of change of radius is 0.081

(correct to 2 significant figures). [6 marks]

5555 Expandxx

−+

11 as a series in ascending powers of x. If x is too small that x3

and higher powers of x may be neglected, show that

2

211

11 xx

xx

++≈−+ [4 marks]

Hence, by substitute91

=x , find an approximation for 5 . [3 marks]

6666 A geometric progression has common ratio, r where | r | < 1. The sum of thefirst n terms is denoted by Sn, and the sum of infinity is S. Express r in termsof Sn, S and n. [4 marks]

Hence, show that the sum of first 2n terms isS

SSSS nn

n)2(

2−

= . [4 marks]

7777 Express)2)(1(

32 ++

−xx

in the form of)1()2( 2 +

+++

xC

xBAx , where A, B and C are constants.

[3 marks]

Hence, evaluate dxxx∫ ++

−1

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

8888 Show that 02222 =+−−+ cfygxyx is the equation of the circle with centre (g, f) and

radius cfg −+ 22 . [3 marks]Given that point A and point B have coordinates (2a, 0) and (-a, 0) respectively. The point P

moves such that AP = 2PB. Prove that locus of P is a circle, and state its centre and radius. [5 marks]



- 2 -

9999 The functions f and g are defined as

1: 2 −xxf ↦21:

−+

xxxg ↦

(a) State the domain f and g. [2 marks](b) Sketch the graph of function f and state its range. [2 marks](c) Determine the inverse function 1−g and state its domain and range.

[4 marks](d) Find the composite function gf � and state its domain. [4 marks]

10101010 Given polynomial 36)( 234 −+++= xbxaxxxf , where a and b are not dependent on x. If(3x – 1) is a factor of f(x) and (x + 1) is a factor of f ’’’’(x), find the values of a and b. [4 marks]

Using these values of a and b, factorise f(x) completely to obtain the roots of equation0)( =xf . [5 marks]

Substitutex

y 1= , find the roots of 0317196 432 =−+++ yyyy [3 marks]

11111111 Matrices A and B are given as⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −=

145232111

A ,⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−−=

117468555

B . Show that IAB 10= ,

where I is the 33× identity matrix. Hence, deduce 1−A . [6 marks]

A shop selling 3 special drinks, that is mango, strawberry and apricot. The price of eachvariety of drink is fixed. The price of an apricot drinks same as the sum of a mango and a strawberrydrinks. Ali paid for RM39 for 2 mango, 3 strawberry and 2 apricot drinks. Jenny paid RM46 for 5mango, 4 strawberry and 1 apricot drinks. Suppose RMx, RMy and RMz are the prices of mango,strawberry and apricot drinks per glass respectively. Find a system of linear equations base on theinformation given above. Rewrite the equations in matrix form and hence, solve by using matrixmethod. [6 marks]

12121212 The function f is defined by112)( 2

2

++

=xxxf .

(a) Find the value of )0(f . [1 mark](b) Show that the function f is an even function. [2 mark](c) State the equation of the asymptote of )(xf and the axis of symmetry. [2 marks](d) Find the coordinate of turning point on this curve and determine if this is a maximum

or minimum point. [6 marks]

(e) Show that there are points of inflexion at31

±=x . [2 marks]

(f) Sketch the graph of )(xfy = . [2 marks]



- 3 -

1.1

302+

>+x

x

01

302 >+

−+x

x

01

30)1)(2(>

+−++

xxx ------------------------------------------------------------------------------------------------------------------------------------ 1111

012832

>+−+

xxx

01

)4)(7(>

+−+

xxx ---------------------------------- 1

- 7 - 1 4

----------------------------- 1

{ }17,4,: −<<−>ℜ∈ xxxx-------------------------- 1

{ } { }17,:4,: −<<−ℜ∈∪>ℜ∈ xxxxxx

2. )()()()( QPQPPQQP ∩−∪=−∪−Left Hand Side = )'()'( PQQP ∩∪∩

= ]')'[(])'[( PQPQQP ∪∩∩∪∩= )]'('[)]'([ QPPQPQ ∩∪∩∩∪= )''()'[)]'()[( QPPPQQPQ ∪∩∪∩∪∩∪= )]''([])[( QPUUQP ∪∩∩∩∪= )''()( QPQP ∪∩∪= )'()( QPQP ∩∩∪= )()( QPQP ∩−∪= Right Hand Side

Right Hand Side = )'()( QPQP ∩∩∪= )''()( QPQP ∪∩∪= ]')[(]')[( QQPPQP ∩∪∪∩∪= )]('[)]('[ QPQQPP ∪∩∪∪∩= )]'()'[)]'()'[( QQPQQPPP ∩∪∩∪∩∪∩= ])'[()]'([ φφ ∪∩∪∩∪ QPQP= )'()'( QPQP ∩∪∩= )'()'( QPQP ∩∪∩= )()( PQQP −∪−= Left Hand Side

x + 7 - ve + ve + ve + vex - 4 - ve - ve - ve + vex + 1 - ve - ve + ve + ve

---- veveveve ++++ veveveve ---- veveveve ++++ veveveve



- 4 -

3. θθ

π

π

d∫2

6

sin

2094.0155

62 ==−

=π

ππ

h -------------------------------- 1

iiii xxxxiiii yyyyiiii

0 5236.06=

π300 7071.05.0 =

1 7330.0307

=π

420 8180.06691.0 =

2 9425.0103

=π

540 8994.08090.0 =

3 1519.13011

=π

660 9558.09135.0 =

4 3614.13013

=π

780 9890.09781.0 =

5 5708.12=

π900 0000.11 =

----------------------------------------- 1

θθ

π

π

d∫2

6

sin [ ])9890.09558.08994.08180.0(2)17071.0()15(

21

+++++≈π ------------ 1

[ ]662.327071.130

×+≈π

946.0≈ ----------------------------------------- 1Alternative :

θθ

π

π

d∫2

6

sin ⎥⎦⎤

⎢⎣⎡ +++++≈ )9890.09558.08994.08180.0()17071.0(21

15π

[ ]662.38536.015

+≈π

946.0≈

4. 2rA π=

rdrdA

π2= ------------------------------- 1

Given 123)1(

2 −

+= scmtdt

dA

dtdr

drdA

dtdA

×=



- 5 -

rtdAdr

dtdA

dtdr

π21

)1(2

3 ×+=×= ------------------------------- 1

13)1(

1 −

+= cms

trπ------------------------------- 1

Given 2)1(11+

−=t

A

When43

211,1 2 =−== At

432 =rπ

489.0@321@

43

ππ=r -------------------------------- 1

32431

⋅⋅=

ππ

dtdr -------------------------------- 1

= 081.028.121

= -------------------------------- 1

5. 21

21

)1()1(11 −

−+=−+ xxxx

21

)1( x+ = ...!2

)121(

21

211 2 +

−++ xx

= ...81

211 2 +−+ xx --------------------- 1

21

)1(−

− x = ...)(!2

)121(

21

211 2 +−

−−−++ xx

= ...83

211 2 +++ xx --------------------- 1

xx

−+

11 ...)

81

211( 2 +−+≈ xx ...)

83

211( 2 +++ xx

...81

41

21

83

211 222 +−++++≈ xxxxx --------------------- 1

...211 2 +++≈ xx -------------------- 1

When91

=x

xx

−+

11 5

21

45=≈ -------------------- 1



- 6 -

2)91(

21

9115

21

++≈ --------------------- 1

81181

≈

235.2@811815 ≈ --------------------- 1

6.rraSn

n −−

=1

)1( -------- (1)

raS−

=1

-------- (2)

)2()1( nn r

SS

−= 1 ------------------------------_ 1+1

SS

r nn −= 1 -------------------------------- 1

nn

SS

1

1 ⎟⎠⎞

⎜⎝⎛ −= @

nn

SSS

1

⎟⎠⎞

⎜⎝⎛ −

@ n n

SSS −

------------------------------- 1

rraS

n

n −−

=1

)1( 2

2

From (2), )1( rSa −= ------------------------------- 1

rrrSS

n

n −−−

=1

)1)(1( 2

2

)1( 2nrS −=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡⎟⎠⎞

⎜⎝⎛ −−=

× nn

SSnS

21

11 ------------------------------- 1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−=

2

11SSnS ------------------------------- 1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−−= 2

2211

SS

SS

S nn

SS

S nn

2

2 −=

⎟⎠⎞

⎜⎝⎛ −=

SS

S nn 2

SSSS nn −

=2(

------------------------------- 1

7.)1()2()2)(1(

322 +

+++

≡++

−xC

xBAx

xx)2()1)((3 2 ++++=− xCxBAx

)2()()(3 2 CBxBAxCA +++++=−



- 7 -

0=+CA --------- (1)0=+ BA --------- (2) --------------------------- 132 −=+ CB --------- (3)

(2) – (1) 0=−CB --------- (4)Solve (3) & (4): 33 −=C

1−=CFrom (1): 1=A --------------------------- 1From (2): 1−=B

)1(1

)2(1

)2)(1(3

22 +−

+−

≡++

−xx

xxx

--------------------------- 1

∫ ++−1

02 )2)(1(3 dxxx

∫ +−

+−

=1

02 )1(

1)2(

1 dxxx

x This is not integrable form

∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+

−+

−+

=1

022 )1(

12

12

dxxxx

x --------------------------- 1

= ∫ ∫ ∫ +−

+−

+

1

0

1

0

1

022 1

12

12

dxx

dxx

dxxx

= [ ] [ ]101

0

110

2 1ln(2

tan212ln(

21

+−⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+ − xxx --------------------------- 1

[ ] [ ]1ln2ln20tan

21tan

212ln3ln

21 11 −−⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⎟

⎠

⎞⎜⎝

⎛−−= −− -------------------------- 1

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−−−= −

21tan

212ln2ln

213ln

21 1

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−−= −

21tan

212ln

233ln

21 1

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−= −

21tan

21

83ln

21 1 ---------------------------- 1

8. 02222 =+−−+ cfygxyx0)()( 2222 =+−−+−− cffxggx ---------------------------- 1 + 1

cfgfxgx −+=−+− 2222 )()( ---------------------------- 1Hence, the equation of the circle with centre ),( fg

and radius cfg ++ 22 ---------------------------- 1@@@@

cfgfygx −+=−+− 2222 )()(



- 8 -

cfgffyyggxx −+=+−++− 222222 22 ---------------------------- 1 + 102222 =+−−+ cfygxyx ---------------------------- 1

@@@@222 )()( rfygx =−+−

022 22222 =−+−++− rffyyggxx022 22222 =−++−−+ rfgfygxyx ---------------------------- 1 + 1

Compare with 02222 =+−−+ cfygxyx222 rfgc −+=cfgr −+= 222

cfgr −+= 22 ---------------------------- 1------------------------------------------------------------------------------------------------------------

)0,2( aA , )0,( aB − and let ),( yxPGiven that AP = 2 PB,

[ ]2222 )(4)2( yaxyax ++=+− --------------------------- 1222222 448444 yaaxxyaaxx +++=++−

03123 22 =++ yaxx04 22 =++ yaxx --------------------------- 1

0)0(4)2( 222 =−+−+ yaax222 4)0()2( ayax =−++ -------------------------- 1 + 1

Hence, the locus of P is circle with)0,2( a− and radius, aar 24 2 == -------------------------- 1

@@@@04 22 =++ yaxx0422 =++ axyx then compare 02222 =+−−+ cfygxyx

to get ag 2−= , 0=f and 0=c ------------------------- 1 + 1

Conclusion : Centre )0,2( a− and radius, ar 2= ------------------------- 1

9. 1)( 2 −= xxf ;21)(

−+

=xxxg

(a) Domain f = { }ℜ∈xx : ---------------------------- 1Domain g = { }2,: ≠ℜ∈ xxx ---------------------------- 1

(b)

---------------------------- 1

-1 1-1 Range f = { }1,: −≥ℜ∈ yyy ---------------------------- 1



- 9 -

(c) Let yxg =− )(1

then xyg =)(

xyy

=−+21 ---------------------------- 1

112@

112

−+

−−−

=xx

xxy ---------------------------- 1

Domain = { }1,: ≠ℜ∈ xxx ---------------------------- 1Range f = { }2,: ≠ℜ∈ yyy ---------------------------- 1

(d) ⎥⎦⎤

⎢⎣⎡

−+

=21

xxfgf �

121 2

−⎟⎠⎞

⎜⎝⎛

−+

=xx ---------------------------- 1

2

22

)2()2()1(

−+−+

=x

xx

22 )2()12(3@

)2(36

−−

−−

=xx

xx ---------------------------- 1

Domain { })()(),(:)( fDxggDxxgf ∈∈=�{ })()(,2,: fDxgxxx ∈≠ℜ∈={ }ℜ∈≠ℜ∈= )(,2,: xgxxx ---------------------------- 1{ }2,: ≠ℜ∈= xxx ---------------------------- 1

10. 36)( 234 −+++= xbxaxxxf

0)31( =f , 03

31

31

31

316

234

=−+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛∴ ba

2770

927=+

ba

703 =+ ba ------------- (1)

12324)(' 23 +++= bxaxxxf0)1(' =−f -------------------- 1 + 1

01)1(2)1(3)1(24 23 =+−+−+− ba2323 =− ba ------------- (2)

Solve (1) & (2): 18711 =b17=b

-------------------- 1 + 1From (1): 7051 =+a

19=a

36)( 234 −+++= xbxaxxxf03)1()1(17)1(19)1(6)1( 234 =−−+−+−+−=−f



- 10 -

)1( +∴ x is a factor -------------------- 1 + 1

)352)(1)(13()( 2 +++−= xxxxxf ------------------- 1)1)(32)(1)(13( +++−= xxxx ------------------- 1

2)1)(32)(13( ++−= xxx

0)( =xf , 1,23,

31

−−=x ------------------- 1

0317196 432 =−+++ yyyy

,1x

y = 01311171196432

=⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+

xxxx0317196 234 =−+++ xxxx ------------------- 1

1,23,

31

−−=x

1,32,3 −−=y ------------------- 1 + 1

11.⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −=

117468555

145232111

AB

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

100001000010

--------------------- 1

I10100010001

10 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡= --------------------- 1

IAB 10=IAABA 1011 −− = --------------------- 1

IAIB 110 −=110 −= AB --------------------- 1

BA1011 =∴ − --------------------- 1

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−

101

101

107

52

53

54

21

21

21

117468555

101 --------------------- 1

IAB 10=11 10 −− = IAABA11 10 −− = ABAA ---------------------- 0



- 11 -

IAB 10=IAA =−1 --------------- 1 + 1 - 1

BA1011 =∴ −

Price : Mango – RM x; Strawberry – RM y; Apricot – RM z0@ =−+=+ zyxzyx -------------------- 1

39232 =++ zyx -------------------- 14645 =++ zyx -------------------- 1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

46390

145232111

zyx

-------------------- 1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−

−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

46390

101

101

107

52

53

54

21

21

21

zyx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

5.80.55.3

---------------------- 1

The price of Mango is RM 3.50;Strawberry is RM 5.00;Apricot is RM 8.50. ---------------------- 1

12.112)( 2

2

++

=xxxf

(a) 1)0( =f --------------------------- 1(b) Even function, )()( xfxf =−

1)(1)(2)( 2

2

+−+−

=−xxxf *Substitute*Substitute*Substitute*Substitute valuevaluevaluevalue isisisis notnotnotnot acceptedacceptedacceptedaccepted

112

2

2

++

=xx

)(xf= ---------------------------- 1(c) Asymptote : 2=y ---------------------------- 1

Axis of symmetry: axisyx −= @0 ---------------------------- 1

(d)112)( 2

2

++

=xxxf

22

22

)1(2)12(4)1()('

+⋅+−⋅+

=x

xxxxxf ---------------------------- 1



- 12 -

22

22

)1()1222(2

+−−+

=x

xxx

22 )1(2+

=x

x ---------------------------- 1

0)(' =xf

0)1(

222 =

+xx , 10 == yandx

Turning point )1,0(= ---------------------------- 1

42

222

)1(]2)12[(22)1()(''

+⋅+⋅−⋅+

=x

xxxxxf ---------------------------- 1

32

2

)1()31(2

+−

=x

x ---------------------------- 1

02)0(",0 >== fx ---------------------------- 1)1,0(∴ is minimum point. ---------------------------- 1

(e) 0)(" =xf

0)1()31(232

2

=+−

xx ---------------------------- 1

5774.0@31

±±=x ---------------------------- 1

(f) y

2

1

31

−31 x

1 mark for Asymptote with label (y = 2) and symmetry at y-axis(with graph)

1 mark for Graph passes through (0, 1),31

± showed and graph correct.

Sabah Trial Stpm 2012-Mathst Paper 1(Q&A)

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Transcript of Sabah Trial Stpm 2012-Mathst Paper 1(Q&A)