The Question Locker Selection Volume 5 Short Preview

7/28/2019 The Question Locker Selection Volume 5 Short Preview

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WHITE GROUP MATHEMATICS 1

A WHITE GROUP MATHEMATICS PUBLICATION

www.whitegroupmaths.com

THE QUESTION LOCKER

SELECTION VOLUME

5

(PREVIEW)


2/15


A FEW WORDSA FEW WORDSA FEW WORDSA FEW WORDS

Never for a moment thought the series would go this far, but it has with your generous support.

In this 5th volume, topics included are: AP/GP, complex numbers, graphing techniques,

applications of integration, applications of differentiation, series/sequences, functions, system of

linear equations, differential equations, vectors, Maclaurins Series, hypothesis testing and

Binomial/Poisson/Normal distributions. It is my sincere hope that you will find my work useful (As

always, solutions were all personally written by me). Peace.

Humbly,

Frederick Koh

BEng (Hons)

BSc (Hons)


3/15


1. The first term of a geometric progression is ,

2

1and its common ratio is .r The sum to n terms

has a value of .256

255On the other hand, if the reciprocal of each term of the original progression is

considered instead, the sum to n terms of this new series would have a value of 510. Find the

value of .r

SOLUTION :

( ) ( ))1(

128

255

1

1

256

255

1

12

1

=

=

r

r

r

r nn

When every term of the original progression is inverted, the new series is still geometric in nature;

however its first term is now equals to 2 and the common ratio is defined as .1

r

Hence, 5101

12

510

11

112

=

=

r

r

r

r

r

r nn

n

5101

12 =

n

n

r

r

r

r

12551

1 =

nn

rr

r

Comparing with (1), it is observed that1

255128

255 = nr or128

11 =nr

Multiplying both sides by r, we have )2(128

=r

rn

Substituting (2) into (1) :


4/15


128

255

1

1281

=

r

r

( )rr

= 1128

255

1281

Rearranging the terms gives

128

127

64

127=r

2

1=r (shown)

2. The sum of the first n terms of a series is given by the expression ( ).31 2n

(i) By finding the nth term of the series, ,nT show that the above series is a geometric progression.

(ii) Find the least value of ksuch that the sum of all terms beginning with the kth term is less than

.3000

1

(iii) Find =

N

n nT1

1in terms of .N

SOLUTIONS :

(i) [ ] [ ] nnnnnnn SST 2)1(2)1(221 333131 ===

( ) ( )nnnn 222222 3813333 + === (shown)

[ ]( ) 9

13

3

3

38

38 22

22

2

)1(2`1 ====

++

n

n

n

n

n

n

T

T

Clearly the above is a geometric progression with common ratio .9

1(shown)


5/15


(ii) ( )kkT 238 =

Sum to infinity with first term as( ) ( ) ( )k

kk

kr

aT 2

22

39

9

8

38

9

11

38

1

==

=

=

For this to be less than ,3000

1 ( )

3000

139 2


6/15


SOLUTION :

[ ] [ ]dadnan

S )1120(2

2

120)1(2

2

+=+=

[ ] )1(119260 += da

Also, )2(148 =+ da

For the first 40 even numbered terms, they constitute an arithmetic progression with common

difference d2 and first term da + (ie second term of the original series is the first even term of the

new series)

Sum of first 40 even numbered terms is

( )[ ] ( ) )3(802202)140()(22

40780

6+=++= dadda

S

From (2), da 814=

Substituting this into (1) gives [ ] dSddS 61801680119162860 +=+=

By (3), ( )daS

802207806

+= becomes

[ ]ddd

80)814(2207806

61801680+=

+

[ ]ddd 801628207801030280 +=+

d1280560+=

Rearranging gives 500250 =d

2=d (shown)

4. From a string, pieces of decreasing lengths are cut. These lengths follow an arithmetic

progression with the 6th piece and 26th piece being 19cm and 15cm respectively.

(i) Find the length of the first piece cut and the common difference of this arithmetic progression.


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(ii) Assuming that the string is sufficiently long, find the number of such pieces that can be cut

from it and hence determine the least possible length of the string.

SOLUTIONS :

(i) )1(195 =+ da

)2(1525 =+ da

(1) :)2( ,5

1420 == dd 20

5

1519 =

=a (shown)

(ii) The length of pieces cut from the string become successively smaller due to a negative

common difference. When the smallest piece is cut, it must still be .0>

0)1( >+ dna

( ) 05

1120 >

+ n

Solving gives 100101 max =< nn (shown)

Least possible length of string is 10105

1

)1100()20(22

100

=

+ cm (shown)

5. The sum of a hundred numbers of an arithmetic progression with first term a and common

difference d is given by S. When every fourth term of this progression is considered, ie

,4T ,8T ,12T . ,100T the sum of these 25 terms is found to be .754+

S

Find the value of

d .

SOLUTION :

[ ] ( ) )1(495010099250)1100(22

100+=+=+= dadadaS


8/15


Also, ( )[ ])4)(125(322

2575

4dda

S++=+

( ) dada 12752510222

25+=+=

Multiplying both sides by 4 gives )2(5100100300 +=+ daS

:)1()2( 2150300 == dd (shown)

6. A function is defined by ,)(2

bx

axxf

= where a and b are both positive constants. Find the

equation of the asymptotes, and hence draw the graph for )(xf when (i) ab >2 (ii) ab


9/15


(i) abab >>2 or ab < :

y

b

b

a

b a 0 a b

bxy += bx =

(ii) abaab


10/15


7. The graph of )(' xfy = is given below:

y

0 2 x

It is further noted that3

1y as ,x and .3)2( =f Sketch the graph of )(xfy = .

SOLUTION :

Certain observations are made:

(i) A point of inflexion exists in each of the two regions 2x and .2


11/15


8. The transformations ,1T 2T and 3T are defined as follows:

:1T A translation of 2 units in the negative x -direction.

:2T A reflection about the x -axis.

3T : A scaling parallel to the y -axis (x -axis invariant) by a factor of 2.

A curve undergoes in succession, the transformations ,1T 2T and 3T and the equation of the

resulting curve is .2

)7(2

+

+=

x

xy Determine the equation of the curve before these tranformations

were effected.

SOLUTION :

Reversing the transformation process,

Before 3T , equation of curve is given by2

)7(

2

)7(2

2

1

+

+=

+

+=

x

x

x

xy

( A scaling parallel to the y -axis by a factor of2

1is carried out)

Before 2T , equation of curve is given by2

7

2

)7(

+

+=

+

+=

x

x

x

xy

( A reflection about the x -axis is carried out)

Before ,1T equation of curve is therefore given byxx

x

x

xy

51

5

2)2(

7)2(+=

+=

+

+= (shown)

(A translation of 2 units in the positive x -direction is carried out)


12/15


9. The sketch below show the graph of ).(xfy = The curve intersects the y axis at the point

2

11,0 , has a maximum point at ),( ba and a minimum point at ).,( dc The equations of the

asymptotes are ,1=x 2=x and 3=y .

y

)(xfy =

3=y

),( dc

0 x

1=x 2

11

),( ba 2=x

On separate diagrams, sketch the following graphs indicating the points corresponding to the

stationary points, asymptotes and axial intercepts where necessary.

(i))(

1xf

y = (ii) )(' xfy =


13/15


SOLUTIONS :

(i)

dc

1,

3

1=y

-1 0 2 x

11

2

)(

1

xfy =

ba

1,

y

(ii)

)(' xfy =

0 a c x

1=x 2=x


14/15


10. By using the fact that 1+x is a factor of ,13 +x express1

23

2

+

+

x

xin partial fractions.

Deduce the coefficient ofnx3 for the ascending series expansion of ( ) 121 + xx

where ,0=n 1, 2..

SOLUTIONS :

)1)(1(

2

1

22

2

3

2

xxx

x

x

x

++

+=

+

+

211 xx

CBx

x

A

+

++

+=

By the cover-up rule, letting ,1=x ( )

( )1

1)1(1

212

2

=+

+=A

Let ,0=x then 1110

20=+=+=

+

+CCCA

Let ,1=x then CBACBA

++=+

++=

+

+

2111211

2123

2

Substituting in ,1=A 1=C gives 012

1

2

3=++= BB

Hence,23

2

1

1

1

1

1

2

xxxx

x

++

+=

+

+(shown)

1

1

1

2

1

13

2

2 +

+

+=

+ xx

x

xx

( ) ( )( ) ( ) 113212 1121 +++=+ xxxxx

( )22 += x ( ) ( )

+

+

+ ..................

!3

)3)(2)(1(

!2

)2)(1(1

33233 xxx

+

+

+ .........

!3

)3)(2)(1(

!2

)2)(1(1 32 xxx


15/15


22 += x ............1............1 32963 ++++ xxxxxx

............1..................2222.............. 32963 ++++++= xxxxxx

Terms of the structurenx3 , ,0=n 1, 2, .

...............1............. 963 +++= xxx

By observation, coefficient ofnx3 is therefore given by ,)1( n ,0=n 1, 2, . (shown)

The Question Locker Selection Volume 5 Short Preview

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