Download - Overall Summary for a Level Maths 1_2
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1. AP,GP, series (summation), MOD, MI, partial fractions
-be aware of formulas for =
n
r
r1
, =
n
r
r1
2
=
n
r
r1
3and correction measures to be taken
when lower limit r is not equals to 1.
-be aware of formulas for nth term and sum to n terms of an AP as well as a GP.
-ability to appreciate the concept of sum to infinity for a GP with criteria that|r|
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d. Evaluate rkrn
nr
+++=
32
1
2
e. Use induction to prove that = +=++n
rnnrrr
2
22 .4!)1(!)1(
*f. At the end of a month, a customer owes a bank $1500. In the middle of themonth, the customer pays $x to the bank where x
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Therefore, (1500-x)(1.04)=x 71.764$= x (shown)
(iii) After the second payment of $x, amount owed at beginning of 3rd month is
[(1500-x)1.04 x] (1.04)= xx 04.104.1)04.1(1500 22
After the second payment of $x, amount owed at beginning of 4th month is
[ xx 04.104.1)04.1(150022 - ]x (1.04)
= xxx 04.104.104.1)04.1(1500233
After nth payment of $x, the amount still owed at the beginning of the (n+1)th month
= xxxxnnn 04.104.1...........04.104.1)04.1(1500 21
= )........(15002 nn
rrrxr +++ where r=1.04
At the (n+1)th payment,
=x )........(1500 2 nn rrrxr +++
nn rrrrx 1500).....1( 2 =++++
1)1(15001500)
11(
1
1
== ++
n
n
n
n
rrrxr
rrx (shown)
g. 61019020610])120(2[2
20=+=+ dada -------------(1)
Since the first, third and eleventh term of the AP forms a GP,
then222 4410
10
2
2dadaada
da
da
da
a++=+
++
=+
Simplifying gives da 23 = --------------------------------------(2)
Solving (1) and (2) gives 3,2 == da first term of GP=eleventh term of AP=2+10(3)=32 (shown)
common ration4
1
62
2
2=
+=
+=
da
ar (shown)
=
=n
n
nS4
11
3
128
4
11
4
1132
(shown)
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3
128
4
11
32
1=
=
=
r
aS
(shown)
2. Binomial series expansion
-be aware of basic structural expansion of a binomial series(Note that the term independent of x in the original compressed
structuren
axk )( + must be maintained at a value of 1, ie k=1)
-ability to compute the coefficient of the rth term within the series
-ability to consolidate coefficients of various individual terms throughmultiplication with other series based on question requirements
eg finding series up to and including term in 5x based on ( ) )43(1 423
xxx ++
-ability to use binomial series to make suitable approximations
-ability to obtain range of values of x for which the expansion is valid.
PREDICTED QUESTION STRUCTURES :
*a. Given the first three terms in the expansion ofbax )1( in ascending powers of x
are .....2461 2 +++ xx , where a and b are positive constants, find the values of a
and b. Show that the coefficient of nx is 12)2)(1( ++ nnn for n=0,1,2.
b. Find the coefficient of rx in the series3)1( x .
*c. Expand 1x
1x
n
in ascending powers of x up to and including the term in x2
.
State the set of values of x for which the expansion is valid. Hence find an
approximation to the fourth root of19
21in the form
p
q, where p and q are
positive integers.
SOLUTIONS FOR QUESTIONS MARKED WITH ASTERIX:
a. .........)(!2
)2)(()(1)1( 2 +
++= ax
bbxabax b
By comparison, 6=ab and 24)(2
22
=+
abb
Hence, 24)36
(2 2
2
=+
b
bb
22 24)(18 bbb =+
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03433 222 ==+ bbbbb 2,3 == ab (shown)
Coefficient of nx is =na
n
nbbbb)(
!
)1.......().........2)(1)((
+
=n
an
n)(
!
)2..().........5)(4)(3(
+ b=3
= nnn
an
n)1(
!
)2..().........5)(4)(3()1(
+
=12)2)(1(
!2
)!2( ++=+ nn nnan
n(shown) a=2
c.
( )( )
( )( )
+
+
+
+=+=
+ ...........
2
)1(1...........
2
)1(1)1()1(
1
1 22x
nnnxx
nnnxxx
x
x nnn
=
+
++
+
+ ...........
21...........
21 2
22
2
xnn
nxxnn
nx
=2
2222
2
22
1 xnn
xnnxxnn
nx
++
++ +.
22221 xnnx + (shown)
Expansion is valid for 1||
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3. Graphing
-ability to obtain equations of asymptotes and turning points of graphs via calculation-ability to do transformation of graphs the forms af(x)+b and/or f(ax+b)
-ability to discern qualitatively the physical meaning of transformations,eg scaling of graph parallel to y axis, translation of graph along x axis etc
-ability to appropriately produce graphs ofparabolas, hyperbolas, ellipsesand circles.
-ability to obtain the graphs of |),(| xf |,)(| xf ),(2
xfy = )(
1
xfy = , y= f 'x
-ability to deduce original graph based on presentation of series of modifiedgraphs
PREDICTED QUESTION STRUCTURES :
a. The diagram shows the graph of y=f(x). On separate diagrams, sketch the graph of
(i) |)(| xfy =
(ii) |)(| xfy =
(iii) )(2
xfy = -1 0 1
(iv))(
1
xfy =
b. The sketches below show the graphs of )(2 xfy = and |)(| xfy = for a certain
function f. Deduce the graph of y=f(x). Draw a sketch of the curves |)(| xfy =
and |)1(| += xfy .
)(2
xfy =
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|)(| xfy =
4. Differentiation (applications of differentiation)
-ability to differentiate directly wrt to the variable contained, functions to bedifferentiated include algebraic, trigonometric, exponential (possibly with acombination of various function types) and techniques such a product/quotientrules, implicit differentiation will be employed.
-ability to generate a stipulated nth order differential equation based on questionrequirements (see predicted question structures b)
-ability to produce steps to find derivatives of basic functions, eg prove that
d
dxtan
1x =
1
1x2
-ability to solve for equations of tangents/normals to a given curve in any fashionstipulated by the question including parametric representations of curves
-appreciate the utilisation of differentiation in the context ofgeometricalminimisation/maximisation problems, eg maximum volume of expanding cylinder
within sphere as well as rates of changes of parameters in geometrical structures,eg rate of change of radius
PREDICTED QUESTION STRUCTURES :
a. Finddx
dyif
* (i) xxy1tantan = (ii) x+y+sin(xy)= 2 (iii) yxyx 13 tancos =+
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b. . If ,ln xeyx= (i) Find
dx
dy.
(ii) Show that xeyxdx
dy
dx
ydx 2)1(
2
2
=++
*c. An inverted cone of base radius 4cm and height 8cm is initially filled with water.
Water drips out from the vertex at a rate of 2 13 scm . Find the rate of decrease in
the depth of the water in the cone 16 seconds after the dripping starts.
d. (i) Given that 42222 =+ yxyx , find an expression for
dx
dyin terms of x and y.
Find the coordinates of each point on the curve 42222 =+ yxyx at which the
tangent is parallel to the x axis.
*(ii) A curve is defined by the parametric equations
32
, tytx == . Show that theequation of the tangent to the curve at the point P ( ),
32 pp is
.032 3 =+ ppxy Show that there is just one point on the curve at which thetangent passes through the point (-3,-5), and determine the coordinates of this
point.
*e. A length of channel of given depth d is to be made from a rectangular sheet ofmetal of width 2a. The metal is to be bent in such a way that the cross sectionABCD is as shown in the figure, with AB+BC+CD=2a and with AB and CDinclined to the line BC at an angle .
A D
d d
B C
Show that BC=2( )cos ecda and that the area of the cross section ABCDis )cos2(cot2
2 ecdad + .
Show that the maximum value of )cos2(cot22
ecdad + , as varies,
is )32( dad .
By considering the length of BC, show that the cross sectional area can only be
made equal to this maximum value if .32 ad
*f . Show that2
1
1
1)(tan
xx
dx
d
+=
SOLUTIONS FOR QUESTIONS MARKED WITH ASTERIX:
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a (i) xxy1tantan = Differentiating both sides wrtx ,
( )
2
12
2
12
1
tan1
1
1tansec
x
xxx
xxx
dx
dyy
+++
=
++=
=dxdy ( )( )( ) ( )( )( ) ( )( ) ( )[ ]212
12
22
12
22
12
tan11
tan1tan11
tan1sec1
tan1
xxx
xxxyxxxx
yxxxx
++++=
++++=
+++
c. hrh
r
2
1
2
1
8
4===
At any time t, volume remaining V=322
12
1)()
2
1(
3
1
3
1hhhhr ==
Differentiating V with respect to ,h 2
4
h
dh
dV =
When t =16,3
12
1h = 128)16(2)8()4(
3
1 32 = h
By the chain rule,dt
dh
dt
dh
dh
dV
dt
dV== 23 )128(
42
1315.0 = cmsdt
dh(shown)
d(ii)
2
3
2
3 2 t
t
t
dx
dy==
At ,P pdx
dy
2
3=
Equation of tangent is ppy2
33 = )( 2px
Rearranging gives 0323 =+ ppxy (shown)
Since it passes through (-3,-5),
2(-5)-3 p (-3)+3p =0 0910
3 =++ pp 0)10)(1(
2 =++ ppp
1=p and point is ( ,12
3
1 ) = ( 1, 1) (shown)(Quadratic polynomial has no real roots since determinant = 03942
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[ ]
ececdd
dycoscot2cos 22 +=
cot2coscoscot2cos0 2 === ecececd
dy
,21
cos = 3
=
Maximum value = =
+
3
22
3
12 2dad )32( dad (shown)
)cos(2 ecdaBC = , putting in3
= ,
= daBC
3
22
Since ,0BC adda 323
2 (shown)
f. Let xy1tan = , then xy =tan
Differentiating both sides wrtx gives
ydx
dy
dx
dyy
2
2
sec
11sec == =
1
1tan2
y=
1
1x2 (shown)
(Similar approaches shall be taken for proving the derivatives of x1sin and )cos1 x
5. Differential equations/Maclaurins Series
-ability to solve first and second order variables separable differential equations
-ability to transform a differential equation into a viable form for resolution by meansof a substitution given in the question
- ability to construct a particular solution for a given differential equationbased on certain given conditional inputs (eg when t=0, v=5m/s)
-ability to sketch a family of solution curves for the differential equation solved
-ability to generate a Maclaurins series of a given function through repeatedprocesses of differentiation to obtain the coefficients of the series itself
-ability to utilise the Maclaurin's series obtained (typically derived in first part ofquestion) for various interpretations, including
(i) deducing the series expansion for other functions/expressions (note that suchmethods could involve integration/differentiation) ,
(ii)to produce suitable approximations (eg use the series expansion to deduce the
value of sin 12 via substituting a suitable value of x within the series)
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PREDICTED QUESTION STRUCTURES :
a. It is given that .tan1ln1
xy += Show that .0)12()1(2
22 =++
dx
dyx
dx
ydx
Find the first three terms of the Maclaurins series for y.Write down the equation of the normal at the point (0,e) on the curve
.tan1ln 1 xy +=
b. Given that ,sinln1
xy = show that .1 2 ydx
dyx = By repeated differentiation of
this result or otherwise, find the Maclaurins expansion for y up to and including
the term in .3x Deduce the approximate value of .6
e
*c. Find the general solution to the differential equation:222 )cos()(sin2)( xyy
dx
dyxy
dx
dy=+
*d. . Determine the Maclaurins expansion for xx tansec , up to and including theterm in 3x . Show that, to this degree of approximation, xx tansec can beexpressed as )1ln( xba ++ where a and b are constants to be determined.
*e. (i) Given that y=tan x , show thatd
2y
dx2=2y
dy
dx . Hence, find the
Maclaurin's series expansion for y , up to and including the term in x3 .(ii) Using the standard series expansion for ln (1+x) and the Maclaurin's series for
y , find the series expansion for ln (1+tanx) in ascending powers of x
up to and including the term in x3.
(iii) Hence show that the first 3 non-zero terms in the expansion ofsec
22x
1tan 2xare 12x8x
2.
*f. Two variables x and y are connected by the differential equationyx
yx
dx
dy
++=
1
1,
use the substitution u= x+y to solve the differential equation. Deduce that
Ayxyx =++ )(2)( 2 , where A is a constant.
SOLUTIONS FOR QUESTIONS MARKED WITH ASTERIX:
c.222 )cos()(sin2)( xyy
dx
dyxy
dx
dy=+
)1(cos)(sin2)(222
= xydxdy
xydx
dy
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)(sin)(sin2)( 222 xydx
dyxy
dx
dy=
0)sin(2
= xydxdy
xdx
dy
yxy
dx
dysin
1sin ==
= xdxdyysin
1
xCx
Aeeycxycoscoscosln + ==+= (shown)
d. . Let xxyxf tansec)( == , then xyxxxxxx
dx
dyxf sec)sec(tansecsectansec)(' 2 ====
)tan)(sec()tansecsec()(''2
2
xydx
dyxxxyx
dx
dy
dx
ydxf +=+==
)tan)(tansec()sectan)(sec()(''' 22
2
3
3
xydx
dyxxxyx
dx
dy
dx
ydx
dx
ydxf ++++==
2)0(''',1)0('',1)0(",1)0( ==== ffff
)1ln(1]32[1321)(3232
xxxxxxxxf +=+=+=By comparison, a=1, b= -1 (shown)
e. (i) y=tanx
Differentiating both sides wrt x givesdy
dx=sec
2x=1tan
2x=1y
2
Differentiating both sides again wrt x gives
d2
y
dx2=2y
dy
dx (shown)
Differentiating both sides a third time wrt x gives
d3y
dx3=2
dy
dx
2
2y d
2y
dx2
f0=0, f '0=1, f' '0=0, f
'' '0=2
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Hence, .......)0(!
..............)0("!2
)0(')0()( )(2
+++++= nn
fn
xf
xxffxf
= x2x
3
3 !=x
x3
3(shown)
(ii) Series expansion for ln 1x =xx
2
2
x3
3........
Hence, ln 1tanx ln [1xx
3
3]
= xx
3
3
1
2x
x3
32
1
3x
x3
33
Collecting all terms up to and including x3 gives
ln 1tanx xx
3
3
x
2
2
1
3
x3=x
x2
2
2x
3
3
(shown)
(iii) ln 1tan2x=2x2x
2
2
2 2x3
3=2x2x2
16 x3
3
d
dx[ln 1tan2x]=
d
dx[2x2x
2
16x3
3]
2sec2
2x
1tan 2x=24x16x
2
Therefore,sec2 2x
1tan 2x=12x8x
2(shown)
f.dx
dy
dx
duyxu +=+= 1
Substituting intoyx
yx
dx
dy
++
=1
1,
uu
u
dx
du
=
+
+=1
2
1
11
( ) = dxduu 21 Cx
uu += 2
2
2
Cxyx
yx +=+
+ 22
)()(
2
Bxyxyx +=++ 4)()(2 2
Byxxy =+ 2)()(2 Ayxyx =++ )(2)( 2 (shown)
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6. Functions
-ability to prove existence/ non existence of inverse function via
(i) horizontal line test(ii) proof by contradiction (eg for fx =x
2, x R , f1=f1=1 2
values of xmap to the same image y ie no inversefunction exists.)
(Sidenote: a popular feature of questions is to demand students toshow that a certain function is strictly increasing/decreasing for a certainset of x values- this can be done by finding f 'x and proving thatf 'x0 when strictly increasing and f 'x0 when strictlydecreasing )
-ability to derive expressions for inverse functions and recognise that
f1x can be obtained graphically via reflecting fx in the line
y=x ; also recognise that ff DR = 1 , ff RD = 1
-ability to solve for fx =f1x (Note that this is achieved easily
by simply solving fx =x )
-ability to discern if composite function eg gfx exists, and to findthe rule and range of the resultant composite function
-ability to resize domains of certain single standing functions such that
the inverse or composite versions exist.( eg fx =x2, x R ,having its domain resized to x0 allows for its inverse to exist).
PREDICTED QUESTION STRUCTURES :
*a. The functions f and g are defined as follows.
,4:2 xxf x
,5: xxg 5x
(i) Explain briefly why the inverse function1f does not exist.
(ii) State the value of A such that }:{ Axx for the inverse function to exist.
(iii) Give a reason why the composite function gf does not exist. Write down the
largest domain of f so that the composite function gfexists. With this
restricted domain of f, write down the rule and domain of gf.
*b. The functions f and g are defined by
,cos: xxf x ,
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-1
y
3
xxxg += 22)(
1 x
-1 0 1 min pt (8
1,
4
1 )
1[: fr 1] ,8
1[: gr )
Since ,gf dr gf exists. (shown)
(ii) xxxgf coscos2)(2 += (shown)
1[= gf dr ,1 ] Using this as the new input domain for )(xg ,
new corresponding range = range of )(xgf = [ ,8
1 3] (shown)
(iii) fg dr , hence fg does not exist. (shown)
Largest possible range of )(xg allowed based on domain of ,8
1
[)( =xf
].
Therefore, the largest possible domain of )(xg based on the above stipulated
domain is [-1, 1.03] (shown)
c. (i) 51)3(2)3(..........)19()23()27( ====== ffff
617)1(.........)37()41()45(2 ===== ffff
1165)45()27( =+=+ ff (shown)
(ii) y
7
3
-7 -6 -4 -2 0 2 4 6 8 10
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(iii)
++=
4
3
2
0
2
3
4
12)73)(2(2
172)( dxxdxxdxxf
=
+
10372
2
0
3x
x [ ]4
32 xx
= 63
242 =
3
236 (shown)