Discussion of paper

Discussion of paper

F.6 Pure Mathematics2009-06-29

JOHN NGhttp://johnmayhk.wordpress.com

0for

0forcos1)(f

3

xa

xx

xxx

0for0

0for1cossin2)('f 2

3

x

xx

xxxxx

Q.1 (b) Satisfactory

a)0(f 0)0(f

y = f(x)

a

O

Slope of tangent at x = 0 is f’ (0)

)0(f should be found by definition.

hh

h

)0(f)(flim)0(f0

)(f xQ.1(c) Satisfactory

Prove is continuous at x = 0,

check: )0(f)(flim0

xx

It is not enough to check

)(flim)(flim00

xxxx

We need

)0(f)(flim)(flim00

xxxx

2

1

1

1cosx

xdxd

Q.2 (a) Good

0)12()1( )(2)1()2(2 nnn ynxynyxQ.2 (b) Satisfactory

is valid when n is a POSITIVE INTEGER.Hence, putting x = 0,

)(2)2( nn yny

is valid when n is a POSITIVE INTEGER.Thus, it is WRONG to write

)0(2)2( 0 yy The iterance should stop at

)2(22222 24...)4()2( ynnn

xxdxd sincos

Q.3 (a) Satisfactory

Also, students just used l’hôpital’s rule without usingso-called very important limit

1sinlim0

x

xx

to simplify the work.

Alternative method

20

cos2coslimx

xxx

xxx

xxx cos2cos

1cos2coslim 20

xxx

xxxx cos2cos

1limcos2coslim020

11

12

sin2sin2lim0 x

xxx

xx

xx

x

sin21

22sin2lim

21

0

43

212

21

Q.3 (b) Not satisfactory

xx 11 tanln2

lntan2

ln

)tan2

)(1(limtan

2ln

ln1lim

12

1

xx

xxx xx

0)tan

2(2)

11)(1(

1lim1

22

xxx

xx

Make sure to make indeterminate forms like 0/0, / etc. before using l’hôpital’s rule.

x

xx

xx xx 1

21

tan2

1limtan2

lnln1lim

022

tan2

lim 1

xx

It is a 0/0 form, because


Prove f(x) is bounded on R,we need to find a FIXED number M such that|f(x)| M for ALL x in R.

Hence, it is wrong to say

xx

Mxx)1(g)1(g

)(g)(f f(x) is boundedJust one step further, we can get rid of the x,

10,1)(0for)1(g)1(g)1(g

)(g)(f xxgMMxx xx

Q.5 (a) Not satisfactory

Students may know what [x] is, but notfamiliarized with the operations.Some wrote [2x] = 2[x] or [2x] = 2n - x

x

xx

]2[lim0

Some claimed

0]2[lim

1]2[lim

0

0

x

x

x

xNote:


The graph of y = [2x]/x is strange to students.Students should be taught that, for integer n,

nynyn ][1

xn

xxx ]2[)(f

21

212

nxnnxn

Hence, for

y

x

y = 3/xy =

2/xy = 1/x

Q.5 (b)(ii)

][]2[][)(f)(g xxxxxx

Students tried to prove g(x) is injective.

Just tried some values of x and see that g(x) is NOT.

e.g. For 0 < x < 1/2,

g(x) = 0/x – 0 = 0

showing that is NOT injective.

Q.6 (a)

Mistook that 1)()( )()()( xgxg xhxgxhdxd

Taking logarithm is the trick.

1ln1)(fln1)(f1

xxx nx

xnx

)1(1ln1ln1

)(fln

xdxdnn

dxd

x

xdxd

xx

Q.6 (a) Not satisfactory

Not many students could show that f’(x) > 0.

1

ln1ln1)(f)(f 2

x

xxx

nxnxnnnxx

A negative sign is in the expression, and it is not clear enough that f’(x) > 0. Better

1)1ln(1ln

)(f

1)1ln(ln1ln)(f)(f

2

2

x

xx

xx

x

xxxxx

nx

nn

nnx

nxnnnnnxx

Q.6 (b)(i) Satisfactory

Alternative method

n

k

kkn

k

nnn

n1

1

1

1 111

(By (a))

Q.6 (b)(ii) Not satisfactory

Alternative method

111

1

1

1

n

k

kkn

k

kk nn

Hence

11lim

11lim1

lim1

111

1

1

1

1

1

1

n

k

kk

n

n

k

kk

nn

n

k

kk

n

nn

n

nn

n


There is a misunderstanding.

n

k

kkn1

1

1

is increasing (as n increases) andis bounded from above by 1,does not imply that

11lim1

1

n

k

kk

nn

Q.7 (a)(i) Good

f(x) = 0 has a triple root , some mistook that3)()(f xx 3)()(f xkx

Some weak in basic differentiation rules, e.g.23 ))((3)('f))(()(f xxQxxxQx

Q.7 (a)(ii) Satisfactory

CQxx

Qxx

2)(21)(f

)()(f

Some used integration in this part. They claimed

Also 0)(f Hence, C = 0,

Qxx 2)(21)(f

Thus, is a repeated root of f(x) = 0

Note: the integration is invalid.

Q.7 (b)(i) Good

Students tried to solve the repeated root toprove a2 b.

It may be easier to consider that the quadratic equation g’(x) = 0 has a real root, hence 0.


Not many students could solve the repeated root by elimination:

Many students set up Viète‘s formulas (韋達定理 )and obtained complicated relations.

020363)(g

033)(g22

23

baba

cba

Eliminate 3, 022 cba

Eliminate 2, yield

)(2 2abcab

Q.7 (c)(i)(ii) Good

Many students could solve (c) without using the result in (b)(ii).

Students may find it easy to cope with concrete numerical problems.

Q.8 (a)(i) Satisfactory

11

|1|)(

xxxxxfy

To sketch the graph of

Some students divided cases wrongly like

11 xorx Instead of

1)11( xorxandx


Some students obtained the following wrongly

and

)1(f- )1(f

and cannot draw the conclusion where f is differentiable at x = -1 or not.


If students could obtain the first and second derivatives correctly, it is likely that they could perform very well in this part

Q.8 (c) Satisfactory

Many students ignored that (-1,0) is also a minimum point.

Some said there was no inflection point.


Students should pay attention that g(t) means g is a function of t ONLY, other indeterminate like x and c, are constants with respect to t.

Hence g’(t) is differentiating g with respect to t.

))((f)(f)(f))((f)(f)(f)(g 22 xcccxxtxttxtxct


Also, by applying the mean value theorem, state clearly the range of the value of the d, i.e.

))((g)(g)(g cxdcx

for some d lying between x and c.


Not many students could complete this part.

Put x = ak (k = 1,2,…,n) into result in (a)(ii)

2)(2

)(f))((f)(f)(f ca

dcacca k

kkk

for some dk lying between ak and c.

Summing up and use the f’’(x) < 0 on (a,b), result follows.

Q.9 (c) Not satisfactory

Many students took

instead of

xxx sin)(f

xxx sinln)(f

to obtainn

n

n

cc

aaaaaaaa

sin...

sin...sinsinsin

321

321

Q.9 (c)

Note that

does not satisfy the condition that

xxx sin)(f

),0(0)(f onx

Q.10 (a)(ii) Not satisfactory

Many students tried to prove by M.I. that

but not success in most of the case. Solution:

2141

aan

22

1

)114(21114214

21

1422421

2141

2141

aaa

aa

aaaaa

aa

kk

k

Q.10 (a)(ii)

The easiest way should use the increasing of {an}, that is

and solve the quadratic inequality above.

nn

nnn

aaa

aaaa

2

1

Q.10 (b)(i) Satisfactory

Students may find question in this type is not-so-familiarized, many could not use the fact that

for all n N, to derive

kb

b

n

n 1

NnNN

n

n

N

N

N

N

N

NNn kbkkkb

bb

bb

bb

bbbb

......12

3

1

21


Few could complete this part. Many used the previous result wrongly, they wrote

1122

111 ... n

nn kbkbkbc

Instead of the fact that (b)(i) valid only for n N, i.e.

kbbbb

kkbbbb

kbkbkbbbbb

bbbbbbbc

NN

NN

NnNNNNN

nNNNN

n

1...

...)1(...

......

......

121

2121

2121

21121

Q.10 (c) Not satisfactory

No one could complete this part.No one could show the following key step.

121lim 1

n

n

n bb

Q.10 (c)

...222

......222222222

321

321

aaa

bbb

Now check the condition:

222

21

1

111

1

1lim2

24lim

24

lim

2222

lim2

2lim

lim

nn

nn

n

nnn

n

n

nn

nn

nn

n

n

n

n

n

aaaa

aaa

aaaa

aa

bb

Q.10 (c)

2lim nn

aBy (a), it is easy to have

1211limlim

2

1

n

nn

n

n abb

Hence, we have

Thus, {cn} converges by (b)(ii), i.e. the following sum converges.

...222222222

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