数学Paper 1讨论专区

白羊座aries

原帖由 Ivanlsy 于 2-3-2009 02:27 PM 发表
不告訴你~
去爬貼來看~養成爬貼的好習慣~

你怎样算到(1 - 2i)^3 = (-11 + 2i)

白羊座aries

再来:
When a polynomial f(x) is divided by x-2 the remainder is 7 and when divided by x-1 the remainder is 5.Given that f(x) maybe written in the form (x-1)(x+2)Q(x)+ax+b,where Q(x)is a polynomial and a and b are constants .Find the remainder when f(x) is divided by (x-1)(x-2).

2.Polynomials p(x) and q(x) are given by relationship q(x) = p(x)(x^5-2x+2).
a)If x-2 is a factor of p(x)-5,find the remainder when q(x) is divided by x-2
b)If p(x)is of the form x^2 + ax +b and x-1 is a factor of p(x)-5 , find the values of a and b

Ivanlsy

(1 - 2i)^3 = (1 - 2i)(1 - 2i)(1 - 2i)
                = (1 - 2i - 2i + 4i^2)(1 - 2i)
                = (1 - 4i - 4)(1 - 2i)
                = (-3 - 4i)(1 - 2i)
                = (-3 + 6i - 4i + 8i^2)
                = -3 + 2i - 8
                = -11 + 2i

白羊座aries

原帖由 Ivanlsy 于 2-3-2009 01:00 PM 发表
因爲我們要分子和分母的正負號相反，所以選擇1和4，和2和3。

这个我也明白了,但是我不懂你怎样写那个range,.

原帖由 Ivanlsy 于 2-3-2009 07:56 PM 发表
(1 - 2i)^3 = (1 - 2i)(1 - 2i)(1 - 2i)
                = (1 - 2i - 2i + 4i^2)(1 - 2i)
                = (1 - 4i - 4)(1 - 2i)
                = (-3 - 4i)(1 - 2i)
                = (-3 + 6i - 4i + 8i ...

明了.

Ivanlsy

原帖由 白羊座aries 于 2-3-2009 06:40 PM 发表
When a polynomial f(x) is divided by x-2 the remainder is 7 and when divided by x-1 the remainder is 5.Given that f(x) maybe written in the form (x-1)(x-2)Q(x)+ax+b,where Q(x)is a polynomial and a and b are constants .Find the remainder when f(x) is divided by (x-1)(x-2).

f(x) = (x - 1)(x - 2)Q(x) + ax + b
From Remainder Theorem,
f(2) = 7
f(1) = 5

f(2) = 7
(2 - 1)(2 - 2)Q(x) + 2a + b = 7
2a + b = 7  ---  1

f(1) = 5
(1 - 1)(1 + 2)Q(x) + a + b = 5
a + b = 5  ---  2

2 - 1:
a = 2

From 2,
b = 5 - a
   = 5 - 2
   = 3

f(x) = (x - 1)(x - 2)Q(x) + 2x + 3
Divide both side of the equation with (x - 1)(x - 2),
f(x)/[(x - 1)(x - 2)] = Q(x) + (2x + 3)/[(x - 1)(x - 2)]

Hence (2x + 3) is the remainder when f(x) is divided by (x - 1)(x - 2).

Ivanlsy

原帖由 白羊座aries 于 2-3-2009 06:40 PM 发表
2.Polynomials p(x) and q(x) are given by relationship q(x) = p(x)(x^5-2x+2).
a)If x-2 is a factor of p(x)-5,find the remainder when q(x) is divided by x-2
b)If p(x)is of the form x^2 + ax +b and x-1 is a factor of p(x)-5 , find the values of a and b

a) If (x - 2) is a factor of p(x) - 5, from Factor Theorem,
p(2) - 5 = 0
p(2) = 5

From Remainder Theorem,
When q(x) is divided by x - 2, the remainder is q(2).
Hence
q(2) = p(2)[(2^5) - 2(2) + 2]
        = 5(32 - 4 + 2)
        = 5(30)
        = 150

b) Since (x - 1) is a factor of p(x) - 5, from Factor Theorem,
p(1) - 5 = 0
p(1) = 5
(1)^2 + a(1) + b = 5
1 + a + b = 5
a + b = 4  ---  1

p(2) = 5
(2)^2 + a(2) + b = 5
4 + 2a + b = 5
2a + b = 1  ---  2

2 - 1:
a = -3

From 1,
b = 4 - a
   = 4 - (-3)
   = 7

a = -3, b = 7

白羊座aries

原帖由 Ivanlsy 于 2-3-2009 08:05 PM 发表


f(x) = (x - 1)(x - 2)Q(x) + ax + b
From Remainder Theorem,
f(2) = 7
f(1) = 5

f(2) = 7
(2 - 1)(2 - 2)Q(x) + 2a + b = 7
2a + b = 7  ---  1

f(1) = 5
(1 - 1)(1 + 2)Q(x) + a + b = 5
a + b  ...

原来是这么容易
谢谢你.
日后还需多多指教

Ivanlsy

(x - 4)(x + 2) ≥ 0 when x ≤ -2 or x ≥ 4  --- 1
(x - 4)(x + 2) ≤ 0 when -2 ≤ x ≤ 4  ---  2
(x + 8)(x - 1) > 0 when x < -8 or x > 1  ---  3
(x + 8)(x - 1) < 0 when -8 < x < 1  ---  4

Combining the condition 1 and 4 and 2 and 3,
we get
-8 < x ≤ -2 and 1 < x ≤ 4

我們先考慮一種情況：分子為正，分母為負。

(x - 4)(x + 2) ≥ 0 when x ≤ -2 or x ≥ 4
(x + 8)(x - 1) < 0 when -8 < x < 1

我們考慮使分子為負的x的範圍，也就是 x ≤ -2 or x ≥ 4
注意當x ≥ 4時，分母為正，因爲只有當-8 < x < 1時，分母才是負數。
因此我們捨棄x ≥ 4的範圍。

注意到當x ≤ -2時，分母為負，但是範圍只到-8，且不等於-8。
因爲當 x < -8 時，分母為正數，不符我們所要求。

當 -2 < x ≤ 1時，分母為正數，也不符我們的要求。

所以我們可以肯定 -8 < x ≤ -2 這個範圍符合題意。

以上的説明不容易明白，畫Number line較容易了解。

白羊座aries

1.Show that the quadratic equation with roots 1/(@^3) and 1/(B^3) is b^3x^2 -a^2(a-3b)x+a^3=0
2.The quadratic equation 2x^2 +ax+3 =0 and x^2 +bx+6=0,with a and b as constants,have a common root.Show that 2a^2+2b^2 -5ab+27=0
3.The roots of quadratic equation ax^2 +bx +c =0 with a=\=0 are @and B.If ka is one of the roots of quadratic equation ax^2 +dx +c =0,find the other root in terms of k and B

Ivanlsy

原帖由 白羊座aries 于 4-3-2009 04:09 PM 发表
1.Show that the quadratic equation with roots 1/(@^3) and 1/(B^3) is b^3x^2 -a^2(a-3b)x+a^3=0

這一題的題目不完整~應該註明the equation with roots α and β。
我猜應該是這樣吧~
let the equation ax^2 - ax + b = 0 has roots α and β, in which
α + β = 1
    αβ = b/a

1/(α^3) + 1/(β^3) = (α^3 + β^3) / (αβ)^3
                             = [(α + β)^3 - 3(α^2)β - 3α(β^2)] / (αβ)^3
                             = [(α + β)^3 - 3(αβ)(α + β)] / (αβ)^3
                             = [(1)^3 - 3(b/a)(1)] / (b/a)^3
                             = [1 - 3(b/a)] / (b/a)^3
                             = [a^3 - 3(a^2)b] / b^3
                             = [(a^2)(a - 3b)] / b^3

[1/(α^3)][1/(β^3)] = 1/(αβ)^3
                             = 1/(b/a)^3
                             = (a^3)/(b^3)

x^2 - [1/(α^3) + 1/(β^3)]x + [1/(α^3)][1/(β^3)] = 0
x^2 - {[(a^2)(a - 3b)] / (b^3)}x + (a^3)/(b^3) = 0
(b^3)x^2 +  (a^2)(a - 3b)x + a^3 = 0

Ivanlsy

原帖由 白羊座aries 于 4-3-2009 04:09 PM 发表
2.The quadratic equation 2x^2 +ax+3 =0 and x^2 +bx+6=0,with a and b as constants,have a common root.Show that 2a^2+2b^2 -5ab+27=0

2x^2 + ax + 3 = 0  ---  (1)
x^2 + bx + 6 = 0  ---  (2)

2 X (1) - (2):
3x^2 + (2a - b)x = 0
x[3x + (2a - b)] = 0
x = 0 (No statisfied) or x = -(2a - b)/3
x = -(2a - b)/3

(1) - 2 X (2):
(a - 2b)x - 9 = 0
x = 9/(a - 2b)

-(2a - b)/3 = 9/(a - 2b)
-(2a - b)(a - 2b) = 27
(2a - b)(a - 2b) + 27 = 0
2a^2 - 5ab + 2b^2 + 27 = 0
2a^2 + 2b^2 - 5ab + 27 = 0

Ivanlsy

原帖由 白羊座aries 于 4-3-2009 04:09 PM 发表
3.The roots of quadratic equation ax^2 +bx +c =0 with a=\=0 are @and B.If ka is one of the roots of quadratic equation ax^2 +dx +c =0,find the other root in terms of k and B

應該是kα吧？

白羊座aries

原帖由 Ivanlsy 于 4-3-2009 09:48 PM 发表

應該是kα吧？

对,是k@

白羊座aries

原帖由 Ivanlsy 于 4-3-2009 08:59 PM 发表

這一題的題目不完整~應該註明the equation with roots α and β。
我猜應該是這樣吧~
let the equation ax^2 - ax + b = 0 has roots α and β, in which
α + β = 1
    αβ = b/a

1/(α^3) + 1/(β^ ...

原帖由 Ivanlsy 于 4-3-2009 09:43 PM 发表

2x^2 + ax + 3 = 0  ---  (1)
x^2 + bx + 6 = 0  ---  (2)

2 X (1) - (2):
3x^2 + (2a - b)x = 0
x[3x + (2a - b)] = 0
x = 0 (No statisfied) or x = -(2a - b)/3
x = -(2a - b)/3

(1) - 2 X (2):
(a ...

感激万分

Ivanlsy

原帖由 白羊座aries 于 4-3-2009 04:09 PM 发表
3.The roots of quadratic equation ax^2 +bx +c =0 with a=\=0 are @and B.If ka is one of the roots of quadratic equation ax^2 +dx +c =0,find the other root in terms of k and B

ax^2 + bx + c = 0
αβ = c/a

ax^2 + dx + c = 0
let the other root be Z,
(kα)(Z) = c/a

αβ = kαZ
Z = β/k, where k and α ≠ 0

Ivanlsy

原帖由 白羊座aries 于 4-3-2009 04:09 PM 发表
3.The roots of quadratic equation ax^2 +bx +c =0 with a=\=0 are @and B.If ka is one of the roots of quadratic equation ax^2 +dx +c =0,find the other root in terms of k and B

ax^2 + bx + c = 0
αβ = c/a

ax^2 + dx + c = 0
let the other root be Z,
(kα)(Z) = c/a

αβ = kαZ
Z = β/k, where k and α ≠ 0

白羊座aries

原帖由 Ivanlsy 于 4-3-2009 11:24 PM 发表


ax^2 + bx + c = 0
αβ = c/a

ax^2 + dx + c = 0
let the other root be Z,
(kα)(Z) = c/a

αβ = kαZ
Z = β/k, where k and α ≠ 0

这么容易而已?只是compare constance而已
我看我可以收工回家休息了

白羊座aries

虽然有点容易,但是我算不出答案

express the following in partial fractions

(x^2+x+1)/(x^2+1)(x^2-1)

乖乖小孩

原帖由 白羊座aries 于 6-3-2009 11:47 PM 发表
虽然有点容易,但是我算不出答案

express the following in partial fractions

(x^2+x+1)/(x^2+1)(x^2-1)

(Ax+B)(x^2-1)+(Cx+D)(x^2+1)
哎呀呀，麻烦下，直接乘起来，然后compare！！
Ax^3 - Ax - Bx^2 - b + Cx^3 + Cx +Dx^2 + D
coofficient x^3
A+C=0 ,A=-C
coofficient x
C-A=1
C+C=1
C=1/2
A=-1/2

constant
D-B=1
coofficient x^2
B+D=1
2D=2
D=1
B=0


A,B,C,D 都有了，自己放进去！！

白羊座aries

原帖由 乖乖小孩 于 7-3-2009 12:30 AM 发表
(Ax+B)(x^2-1)+(Cx+D)(x^2+1)
哎呀呀，麻烦下，直接乘起来，然后compare！！
Ax^3 - Ax - Bx^2 - b + Cx^3 + Cx +Dx^2 + D
coofficient x^3
A+C=0 ,A=-C
coofficient x
C-A=1
C+C=1
C=1/2
A=-1/2

consta ...

也是不对..