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8/7/2019 Perc STPM 2010 MTS1
1/9
SMJK Yu HuaSTPM Trial Examination 2010
Mathematics T/SPaper 1
Form : 6AS1, 6AS2, 6AS3, 6AK1, 6AK2 Prepared by : Pn RosmayaDate : 18th
October 2010 Checked by : Pn. Chew Aun SimTime : 10.10 am 1.10 am Verified by : En. Pradeep
Instructions to candidates :
Answer all questions. All necessary working should be shown clearly. Non-exactnumerical answers may be given correct to three significant figures, or one decimal place
in the case of angles in degrees, unless a different level of accuracy is specified in thequestion.
1. Use the laws of algebra of sets to prove that[A (A B )] - (A B ) = f [4 marks]
2. Given that z1 = -3 + i and z2 = 2 + 4i , find the modulus and argument of z1z2.[4 marks]
3. The sum of the first n terms of a series is given as
Show that the series is a geometric progression.[5 marks]
4. Given that the line y = mx cuts the circle x2 + y2 + 4x + 2y 20 = 0 at the point Aand B. If P is the midpoint of the chord AB, find the coordinates of P, in terms ofm. Hence, find the equation of the locus P, as m varies.
[6 marks]
5. Given that y = ln (sin3 2x) , find [3 marks]Hence , show that 3 + + 36 = 0 [3 marks]
6. The function g is defined by(x 3)
2
, x 3b , x > 3x
Given that g(x) is continuous at x = 3, find the value of b . [3 marks]
Sketch the graph of y = g(x). [4 marks]
7. Express in partial fractions. [5 marks]
g(x) =1 - _
dy
dxd
2y dy
2
dx2
dx
6x2
+ 3x - 1(x + 1)(2x - 1)
10 -2
n + 1
3n - 1
____
8/7/2019 Perc STPM 2010 MTS1
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Hence, find the value of dx , give your answer correct
to three decimal places. [4 marks]
8. Expand 1 + in ascending powers of x up to and including the term in x3.State the range of x for which this expansion is valid. [6 marks]
Without using calculators or mathematical tables, evaluate (1.004)3
correct tofive decimal places. [4 marks]
9. a) Given that P = 2 1 3 , find the matrix Q such that Q = P2 5P 3I,- 1 2 13 1 1
where I is a 3 3 identity matrix. [3 marks]
b) Find PQ and hence P-1
[4 marks]c) Use your answer in b) to solve the system of equations :
2x y + 3z = 23
-x + 2y + z = -463x + y + z = 69 [4 marks]
10.a) Find the area of the region bounded by the curve y = + 1and the graph y = x + 1 [5 marks]
b) Find the volume of the solid generated when the region bounded by the curve
y = + 1 and the graph y = x + 1 for x > 0 is rotated 360o about
the y-axis. [6 marks]
11.Using the substitution y = x + , express f(x) = 6x4 + 5x3 38x2 + 5x + 6 = 0as a polynomial in y. [3 marks]
Hence, solve the equation f(x) = 0. [10 marks]
12.Sketch on the same coordinate axes, the graphs of y = ex and y = 10x.Show that the equation e
x
10x = 0 has a root in the interval [0,1][4 marks]
Using the Newton-Raphson method and 0.2 as the first approximation, find the
root correct to three decimal places.[5 marks]
The other root of the equation lies in the interval [n, n + 1], where n is an integer.State the value of n.
[5 marks]
6x2
+ 3x - 1(x + 1)(2x - 1)
2
1
2x
3
3
2
x2
23
2
x2
2
3
2
1
x
8/7/2019 Perc STPM 2010 MTS1
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Mark Scheme for STPM Trial examination
1. [A (A B) ] ( A B )= [(A A) (A B )] ( A B ) Distributive Law= [f (A B ) ] (A B) De Morgan s Law
= (A B) (A B) Identity Law= f proven
2. z1z2 = (-3 + i)(2 + 4i)= -10 - 10i
z1z2 = (-10 )2 + (-10)2 = 200 = 10 2
3. Sn = 10 2n+13
n 1
Using Tn = Sn Sn 1
= 10 2n+1
- 10 2n3
n 13
n 2
= 2n
2n+1
3n 2 3n 1
= 2n
1 23
n 23
= 2n
13
n 23
= 2n
3n 1
= 3 2n
3
Tn = 3 2n
Tn 1 33 2
n 1
3= 2
3Therefore, it is a geometric progression with r = 2
3
-10
-10
arg (z1z2) = tan-1 -10
-10
= _ 3p rad4
________
8/7/2019 Perc STPM 2010 MTS1
4/9
4.
P = x1 + x2 y1 + y22 2
= ( X , Y )
Substitute y = mx into circle equation :x
2+ (mx)
2+ 4x + 2mx 20 = 0
(1 + m2) x
2+ 2(2 + m)x 20 = 0
SOR = x1 + x2 = _ 2(2 + m)
(1 + m2)
SOR 2 = x1
+ x2
= _ (2 + m) = X
2 (1 + m2)Substitute point P (X,Y) into y = mx,
Y = mX = m _ (2 + m) = _ m(2 + m)(1 + m
2) (1 + m
2)
Therefore point P = _ (2 + m) , _ m(2 + m)(1 + m
2) (1 + m
2)
From point P, x = _ (2 + m)
(1 + m2)
Substitute m = y into above equation : x = _ 2 + y
x x1 + y
2
x2
x + y2
= -2 yx
2x
x2
+ y2
+2x + y = 0
5. y = ln (sin3 2x)Using chain rule :dy = 1 . d (sin
32x)
dx sin3
2x dx
= 6 sin2
2x. cos 2xsin
32x
= 6 cos 2x = 6 cot 2xsin 2x
y = mxP
A
B
(x1,y1)
(x2,y2)
8/7/2019 Perc STPM 2010 MTS1
5/9
d2y = -12 cosec2 2x
dx2
= -12 (1 + cot2 2x)
d2y + 12 cot
22x + 12 = 0
dx2
3 d
2y + 36 cot
22x + 36 = 0
dx2
3 d2y + ( 6 cot 2x )
2+ 36 = 0
dx2
3 d2y + dy
2+ 36 = 0
dx2
dx
6. (x 3)2 , x 3g(x) = 1 a , x > 3
x
lim g(x) = lim (x - 3)2
= 0
x3- x3-
lim g(x) = lim 1 - a = 1 a
x3+ x3+ x 3
Given that g(x) is continuous at x = 3
Therefore : lim g(x) = lim
x3- x3+0 = 1 a
3a = 13
a = 3
9
1
0 3 x
y
8/7/2019 Perc STPM 2010 MTS1
6/9
7. 6x2 + 3x 1 = 6x2 + 3x 1 = 3 + 2 .
(x + 1)(2x 1) 2x2
+ x 1 2x2
+ x 1= 3 + 2 .
(x + 1)(2x 1)
2 . = A . + B .(x + 1)(2x 1) (x + 1) (2x 1)2 = A(2x 1) + B(x + 1)
When x = -1 , A = -23
When x = , B = 43
6x2
+ 3x 1 = 3 - 2 . + 4 .(x + 1)(2x 1) 3(x + 1) 3(2x 1)
26x
2+ 3x 1 =
23 - 2 . + 4 . dx
1 (x + 1)(2x 1) 1 3(x + 1) 3(2x 1)= 3x 2 lnx + 1 + 2 ln2x 1 2
3 3 1= 3x + 2 ln 2x 1
2
3 x + 1 1= 6 + 2 ln 1 3 + 2 ln 1
3 3 2= 3 + 2 ln 2
3= 3.462 (3 decimal places)
8. 1 + 2x = 1 + C1 2x + C2 2x 2 + C3 2x 3 + .3 3 3 3
= 1 + x + 1 x2
1 x3
(up to x3)
6 54For expansion to be valid :
2x < 13
\ 3 < x < 32 2
(1.004)3
= (1 + 0.004)3
= (1 + 0.004)
= 1 + 0.006 + 1 (0.006)2
1 (0.006)3
2x = 0.004
6 54 3= 1 + 0.006 + 0.000006 0.000000004 x = 0.006
= 1.00601 (5 decimal places)
3
2
3
2
3
2
3
2
32
8/7/2019 Perc STPM 2010 MTS1
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9. a)Q = P2 5P 3I= 2 1 3 2 1 3 2 1 3 1 0 0
1 2 1 1 2 1 5 1 2 1 3 0 1 03 1 1 3 1 1 3 1 1 0 0 1
= 14 1 8 10 5 15 3 0 01 6 0 5 10 5 0 3 08 0 11 15 5 5 0 0 3
= 1 4 7
4 7 57 5 3
b) PQ = 2 1 3 1 4 7
1 2 1 4 7 53 1 1 7 5 3
= 23 0 0
0 23 00 0 23
PQ = 23I
P1
PQ = 23P1
IIQ = 23P
1
_ 1 Q = P1
23
P1
= _ 1 Q23
= _ 1 1 4 723 4 7 5
7 5 3
c) 2 1 3 x 231 2 1 y = 46
3 1 1 z 69
x 2 1 31
23y = 1 2 1 46
z 3 1 1 69
_ 1 1 4 7 23= 23 4 7 5 46
7 5 3 691 4 7 1
= - 4 7 5 27 5 3 3
8/7/2019 Perc STPM 2010 MTS1
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28 28
= - 3 = 3 x = 28, y = 3, z = 1212 12
10.a)
The area is symmetrical about the y-axis :
HenceA = 2 [ ( x + 1) ( + 1) ] dx
= 2 [ x ] dx
= 2 [ x2
]
= 4 unit2
b) Volume = p x2 dy Volume of cone
= p 2(y 1) dy 1 p (32) (11 1)3 2
= p [(y 1)2] 27 p2
= 27 p unit3= 6 p unit3
11.
6x
4+ 5x
338x
2+ 5x + 6 = 0
6x2
+ 5x 38 + 5 + 6 = 0x x
2
6 (x2
+ 1 ) + 5 ( x + 1 ) 38 = 0x
2x
6 ( y2
2 ) + 5y 38 = 06y
2+ 5y 50 = 0
(3y +10)(2y 5 ) = 0
y
x0
1
A
y = x2
+ 1
y =3/2x + 1
Coordinates of A?x
2+ 1 = 3x + 1
2 2
\ x = 3, y = 112
A 3 , 11
2
3
0
3 x2
2 2
30
3 x2
2 23 x
2
4 6
3
0
11
/2
1
11
/2
1 11/2
1
y = x + 1x
y2
= x2
+ 2 + 1
x2
y2
2 = x2
+ 1
x2
8/7/2019 Perc STPM 2010 MTS1
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y = _ 10 or y = 53 2
x + 1 = _ 10 x + 1 = 5x 3 x 2
3x2
+ 10x + 3 = 0 2x2
5x + 2 = 0
(3x + 1)(x + 3) = 0 (2x 1)(x 2) = 0x = _ 1 , x = -3 x = , x = 23
12.
ex
10x = 0let f(x) = e
x10x
when x =1, f(1) = e1
10(1) = 7.28 negative
when x = 0, f(0) = e0
10(0) = 1 positive
f(x) = ex
10xf (x) = e
x10
Using Newton-Raphson method to find the root in the interval [0,1] :
First approximation xo = 0.2,x1 = 0.2 e
0.210(0.2) = 0.111307
e0.2
10x2 = 0.111307 e
0.11130710(0.111307) = 0.1118
e0.111307
10x3 = 0.1118 e
0.111810(0.1118) = 0.112
e0.1118 10\ the root is 0.112 ( 3 decimal places)
To find n if the other root is in the interval [n, n+1] :
f(2) = e2
10(2) = 12.61f(3) = e
310(3) = 9.91 negative
f(4) = e4
10(4) = 14.60 positive
Therefore , n = 3
0
1
y
x
y = 10x
y = ex
Therefore , there is a root
in the interval [0,1]