discussion of paper
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Discussion of paper. F.6 Pure Mathematics 2009-06-29. JOHN NG. http://johnmayhk.wordpress.com. Q.1 (b) Satisfactory. . y = f( x ). Slope of tangent at x = 0 is f’ (0). a. O. . should be found by definition. Q.1(c) Satisfactory. Prove. is continuous at x = 0,. check:. - PowerPoint PPT PresentationTRANSCRIPT
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
Q.4 (b) Satisfactory
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:
Q.5 (b) Not satisfactory
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
Q.6 (b)(ii) Not satisfactory
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.
Q.7 (b)(ii) Not satisfactory
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
Q.8 (a)(ii) Satisfactory
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.
Q.8 (b) Satisfactory
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.
Q.9 (a)(ii) Satisfactory
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
Q.9 (a)(ii) Satisfactory
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.
Q.9 (b) Not satisfactory
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
Q.10 (b)(ii) Not satisfactory
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