overall summary for a level maths 1_2

8/6/2019 Overall Summary for a Level Maths 1_2

1/17

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|


2/17

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


3/17

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)


4/17

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 =+


5/17

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||


6/17

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


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 =


7/17

|)(| 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


a. Finddx

dyif

* (i) xxy1tantan = (ii) x+y+sin(xy)= 2 (iii) yxyx 13 tancos =+


8/17

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

+=



9/17

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


10/17

[ ]

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)


11/17


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.


c.222 )cos()(sin2)( xyy

dx

dyxy

dx

dy=+

)1(cos)(sin2)(222

= xydxdy

xydx

dy


12/17

)(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


13/17

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)


14/17

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).


*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 ,


15/17


16/17

-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


17/17

(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)

overall summary for a level maths 1_2

Documents