Skema Pemarkahan Mathematics T 1/S l Pepcriksaan Percubaan STP:Vl 2011
l.
3x
Expre ss
.
-2x+1
.
x+
----.:1
[5 marks]
m parllal fracttons.
Ml
-2x+1 . x -2x+1 A +1 Let (x -1Xx -1)"' (x
81
x+l=A(x-l)+B
Ml
2;
Ml
X=
1,
0, "'
2. By using the substitution x
u,
2,
ll
AI
(x -1)
-2x+1
1, du
= ];
dx
5,
[5] 5
u, find the exact value
f.
- d x .
[5 marks]
Bl
Ml
4 4[(-
21n u
,,
21n4 4)- (-
15+161n7.
AI 21111+
:)l
I\11
AI
[5]
3. If jz[= 15.
find find the modulus u t ' z + ~ , u where t ' z + ~ , is the complex conjugate o f z . [6marks]
Let
bi
.Ja':b
/zl
1
15 .
b1
.
a +bi
=a+bi+--
Ml Ml
a2
bi + - (a
bi)
AI
bi)
El_(a
Ml Al
[6
kx + 15x + 50, where is constant and that 4. Given the polynomial f(x) Jx (x -· 5) is factor of f(x). Find the value and hence facto rise f(x) completely into marks] exact linear factors. f(5)=0
(5) . J , s /
:=:,
(-2) -3(-2)
AI
15(-2/
(x
2)
f(x) 5)(x 5)
•= (x
2)
5)(x - --/S)(x
Ml
15(-2) + 5 0 =
AI
io a factor
:.
f(x)
5
Ml
15(5) + 5 0 =
k=-15
:::;>
f(-2)
k(5)
Ml AI
-iS)
Use the binomid theorem to expand ~ L \ + x as a series to and including the term in
Find the set
valid.
values
ascending powers
= i , + i ( ~ } i ' l ; : _ 2 2 ( ~ r I - X ~ ~ 64
-i
..
x up
x such that the expansion is [7 marks]
(4+x)l =2(1+)(l)z \
r ~ y _ q r ~ y _ q
[6]
Ml
M1 A1
M1 55 "'2+-X+-X 64
Set of values of
6.
Given that function
AI
for the expansion to be valid is 11
(
xj if
,
I}
[7]
:<::
zf
x-
(a) Sketch the graph
x: -I
J(x). State the range of f(x).
(h) Hence, find the set of values
fx forwhichf(x)-1?:
[7 marks]
.,
x[
[1
D1
= [ I + x[
=_L_ __ -----].
D all correct
Range off= {y:
f(x)-1::>0 Solving y So!vmg
Tl'" set
2':
0)
Bl
=:>f(x)::>1 andy= -(x
I and
values
I)
xx.
is {x
=:>X=
M1
-2
Ml
=> X=
x:::;
-2 or x
::>
1 or x
0)
AI
[Tj
7.
Find the set
of
values
of
x which
[7 marks]
satisfles2 + I : :
Ml
2 ( 2 - x)
3x(2- x)
Ml
x ( 2 - x) 3x
7x
;::
Ml
x ( 2 - x) (3x-
) ( x - 1)
Ml
x ( 2 - x)
I The solution set
is
Ml
2
x S I or
2}
[7]
8.
Given that matrices A = [ !
and B
(a) Show that A s a non-singular matrix. (b) Find matrix AB and deduce A_,.
(c) Given matrix C =
ind matrix
(a) d e t A = J ( l - 6 ) - 2 ( 2 - 1 2 ) + Since IAI
0, :.
2
is
(4-4)=5
non-singular matrix.
~][~~ I
in terms
0
1
rs lo
0
[9 marks]
if AX = C.
Ml AI
s oJ
AI
0
AI
5
(c)X=k
--
--
l ~ l l ~ l AI
[9
A circle touches the straight line 4y at the point P(4, 1) an passes through another point Q(S, 3). Find the equation oC the circle an show that it touches th y-axis. [1 marks] the circle= C (a, b) Let the centre 9.
CP=CQ _+_(_b ___ :::; ) ~ ( a---4-)-;:
2a
=>
Equation o f C P
Solving ( 1)
a=%,
=>
(b
a
-;c'
4
y-1=
3)
rvl
17 .. .. ............. 1)
Al
- ~ ( x - 4 )
=>
3y
=>
4a+3b=l9
19
................. (2)
AI
(2),
or c(%· 3J
AI ,----:----
Radiusofthecircle,
Equation
th
eire! e,
At y-axis,
= 0, (y 3
(4-%r +(1-3) =%
(y 3 +
x-%)
(-%
2 equal roots, th circle touches y-axis. y-axis.
=>
AI
(y
J
0
Ml [1 OJ
0. Find the coordinates
x'
the stationary points on the curve
and determine
their nature. Sketch the curve. dy dx
xe
-X
[I
marks]
MI
-X
=xe-x(z-x)
Let dy
0,
dx
xe -x
x)
MI
or
MI
or
:. the stationary points are (0, 0) an (2,
A MI
e-x(2- 2x)- e-x(zx-
'y dx
4)
e·x(2- 4x
_c:i_'y
' dx
x=2,
--=--"
dx
e"
MI
<0
:. (0, 0) is a minimum point and (2, ~ - ) is a maximum point.
AI
AI
oo,
(2, 4/e
DI (shape) 01 (critical (critical points) p oints) D 1 (all (all correct) corre ct)
- - - 4 - L . __ _ _ _ _ _ _· ~ = - - - = - - o _ ; , , · ~ = - - - = - - o _ ; , ,
·)I
[II]
Sketch on the same coordinate axes, the curve (a) Calcula te the area
2x and the ellipse x
the region bounded by the curve y
y_·_
2x and the line
1.
11_
(b) find the volume the solid formed when the region bounded by the curve and 12 marks] the ellipse is rotated through 180° abou t the y-axis. y-axis. 4x .[3 +-=1=>X=±--=>y=9 2
D I (parabola) Dl(ellipse)
-1
(a)
y' X=12 Area
12
y2
jdy
12
4(3[3_
12
--
=1 (b) Volume=
fn(±y fY
H ~ - ~ - } y
' ' +l{y---1 [-ny] 4 9
16 19 16
J3
+l3-1- -+
·:.:· \.L;
+2
Jl
MJ
AI
AI
12. (a) Express (:lk-=2X:Jk "
in partial fractions.
1
X3k + 1
3k
k=l
and find lrm
for
series 0.75
(a) Let
+ ...
its su
to infinity.
1
A
(3k-2X3k+1)
3k-2
A(Jk
=>
1 (3k- 2XJk
1)
the sum
(0.75i
(0.75)
is greater than
[8 marks]
HW
the
(b
Hence, obtain an expressron for
[7 marks] M1
3k+1
B ( J k - 2)
J) 1
Al
3(3k + 11
3(3k- 21
" 3k 2X3k
"1 3k 12
1)
Ml
3k
_311(1--41)+ (-41- _71)
-71
11o)
lim
(b) S
im[·!J1--
n--+aJ
3l
~ _ 1 _ ~ _ 1 +_ ( ~ L ____1 _ ~ 1 _ ~ 3n
__
3n -·
3n
Ml
) ] = ~
1
3n
3n ..
AI
3 n ~ 1 J = 3nn+1 n--+w
...( - 1 - -
[8]
1(1-0.75") 0. 75
0.75" 0.25
AI
=-=-
Ml
= ~ 1 ~ = 4
Al
1-0.75
0.25
~ ( 4 )
1-0.75" =>
M1
0.25
=>
1- 0.75"
=>
0.75"
0.75 0.25
4.819 The least value
fn is 5.
AI
[7]
