收番尋日講過既野
今日Pure Maths Paper 1表現唔太好…
仲係有d唔小心-v-
不過好彩大家都唔識XD
應該會係 80≦x≦86
可以拉番B
Paper 2要繼續加油…拉埋個A-v-
可惜上學期同test dur皮…如果唔係有得爭學科獎…-.-
中化 C
UE C
Pure Maths A/B
Phy B
Chem D
Yeah -v-
番到屋企先做倒9(b)同10(c), 10(b)好多人唔識 post埋上o黎
9(b)
Let S(n) be the
statement “x1α1 x2α2 x3α3…xnαn≦α1x1+α2x2+…+αnxn”
When n = 1,
L.H.S. = x1α1
= x1
R.H.S. = α1 x1
= x1
∴S(1) is true.
Assume S(k) is
true,
i.e. x1α1
x2α2 x3α3…xkαk ≦α1x1+α2x2+…+αk xk
When n = k+1,
There exist a real
number xm such that x1α1 x2α2
= xmα1+α2
è xm = x1α1/(α1+α2)
x2α2/(α1+α2)
Since [α1/(α1+α2)] +[α2/(α1+α2)] = 1,
xm =x1α1/(α1+α2) x2α2/(α1+α2)≦[α1/(α1+α2)] x 1+ [α2/(α1+α2)] x2
L.H.S. =( x1α1
x2α2) x3α3… xkαk xk+1αk+1
= xmα1+α2
x3α3… xkαk
xk+1αk+1
≦(α1+α2)xm+α3x3+α4x4…+αk xk+αk +1xk+1
≦(α1+α2){
[α1/(α1+α2)] x 1+ [α2/(α1+α2)] x2} +α3x3+α4x4+…+αk xk+αk +1xk+1 [ By the result of (a)(ii) that x1αx21-α≦αx1 +
(1-α)x2 ]
=α1x1+α2x2+α3x3+α4x4+…+αk xk+αk +1xk+1
= R.H.S.
Equality holds if
and only if x1=x2=x3=…=xn because in the induction,
each time when n is added by 1, equality holds iff x1 = x2
by the result of (a)(ii).
∴S(k+1) is also true.
By the principle
of mathematical induction, S(n) is true for all natural numbers n.
10(b)
f(r) = (r+1)n
+ (r-1)n = 0
f’(x) = n [(x+1)n-1
+ (x-1)n-1]
= [n/(x+1)] [(x+1)n + (x+1)(x-1)n-1]
= [n/(x+1)] [(x+1)n + (x-1)n
+ 2(x-1)n-1]
= [n/(x+1)] [f(x) + 2(x-1)n-1]
f’(r) = [n/(r+1)]
[f(r) + 2(r-1)n-1]
since f(r)=0, r≠1, then f’(r)≠0.
By the result of
(a), f(r) has no multiple root.
10(c)
g2(r) =
- h2(r) …….(1) （∵r is a root of g2(x) + h2(x)）
2g(r)g’(r) = -2h(r)h’(r)……(2) （∵r is
still a root after differentiation）
(2)÷(1):
2g’(r) /g(r) = 2h’(r)/h(r)
g’(r) = h’(r) [g(r) / h(r)]
[g’(r)]2 =
[h’(r)]2 [g2(r) / h2 (r)]
= [h’(r)]2(-1)
[g’(r)]2 +
[h’(r)]2 = 0
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