sabah trial stpm 2012-mathst paper 1(q&a)
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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]
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954/1*This question paper is CONFIDENTIAL until the examination is over CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*
- 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]
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*
- 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
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*
- 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
×=
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- 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
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- 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 +++++=−
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954/1*This question paper is CONFIDENTIAL until the examination is over CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*CONFIDENTIAL*
- 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 )()(
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- 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
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- 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
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- 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
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- 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
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- 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.
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