STPM-学校功课討論区

~HeBe~_@

原帖由 zipp_882000 于 26-8-2006 09:46 AM 发表
9)更离谱我找不到答案，跟完全部指示了

Determine the polynomial P(x) which has the following properties
a)P(x) is of a degree 3
b)(x-1) is a factor of p(x)
c)P(0) =4 and p(-1)=10
d)P(x) has a remiander 16 when divided by (x-2)

Let P(x) = ax^3 + bx^2 + cx + d

P(1 )   = 0         ==>      ? + ? + ? + ?   =0       ---------------(1)
P(0 )   = 4         ==>      ? + ? + ? + ?   =4       ---------------(2)
P(-1)   = 10       ==>      ? + ? + ? + ?   =10      ---------------(3)
P(2 )   = 16       ==>      ? + ? + ? + ?   =16      ---------------(4)


Then, solve the equations......
a=? , b=? , c=? , d=?

Therefore, you already find the P(x) = ..................

zipp_882000

哈哈明白了，谢谢。看来还需要再接再厉。谢谢！！！！

邵逸夫

1) in each of the following expansions,find the terms as stated.
  a) (2a+b)^12 ,5th terms
  b) (x- 1/x)^6, constant terms
2) the coefficient if x^3 is four times the coefficient of x^2 in the expansion of  (1+x)^n. find the value of n.
3) find the coefficient of the terms in x as indicated, in the following expansions.
  a) ( 1+x^2)(2-3x)^7 ,terms in x^3
  b) x(x- x/x^2) ^12 ,terms in x^4

dunwan2tellu

1) in each of the following expansions,find the terms as stated.
  a) (2a+b)^12 ,5th terms
  b) (x- 1/x)^6, constant terms

a) Binomial expansion : 5th term : 12C4 x (2a)^8 (b)^4  
b)(x-1/x)^6 : constant term : 6C3 (x)^3 (-1/x)^3

2) the coefficient if x^3 is four times the coefficient of x^2 in the expansion of  (1+x)^n. find the value of n.

coefficient of x^3 : nC2
coefficient of x^2 : nC1
nC2 = 4nC1 ==> n(n-1)/2 = 4n => n(n-9) = 0 ==> n = 9 (n=/=0)

3) find the coefficient of the terms in x as indicated, in the following expansions.
  a) ( 1+x^2)(2-3x)^7 ,terms in x^3
  b) x(x- x/x^2) ^12 ,terms in x^4

a)
coefficient of x^3 可以是 1 (Ax^3) 或  x^2(Bx)
所以 (2-3x）^7 的 x^3 的 cofficient = 7C3 x (-3x)^3 (2^4)=-15120x^3
x 的 coefficient = 7C1 (-3x)^1 (2)^6 = -1344
加起来 = -1344 + （-15120）= 16464

b) x(x- x/x^2) ^12 ,terms in x^4  <-----很奇怪，你是要表示 x(x-1/x)^12 ?

邵逸夫

不好意思。。type錯了
1) in each of the following expansions,find the terms as stated.
  a) (2a+b)^12 ,10th terms
這題很奇怪。。明明跟著formula走的。。卻得不到答案
b) (x- 1/x)^6, constant terms
可以順便解釋下嗎？

3) find the coefficient of the terms in x as indicated, in the following expansions.
  b) x(x- 2/x^2) ^12 ,terms in x^4

[ 本帖最后由 邵逸夫 于 28-8-2006 07:21 PM 编辑 ]

dunwan2tellu

1) in each of the following expansions,find the terms as stated.
  a) (2a+b)^12 ,10th terms
這題很奇怪。。明明跟著formula走的。。卻得不到答案

你要 10th term , 所以是当 12C9 (2a)^9 (b)^3

3) find the coefficient of the terms in x as indicated, in the following expansions.
  b) x(x- 2/x^2) ^12 ,terms in x^4

要找x^4 的 term 就必须找 (x-2/x^2)^12 里，x^3 的 term .

所以 x^3 term : 12C3 x^9 (2/-x^2)^3 = ...

b) (x- 1/x)^6, constant terms
可以順便解釋下嗎？

constant term 就是说那个 term 里没有 x .所以说， x 的 power 和 1/x 的 power 要一样，那么才会互相 cancel 对方。否则你就 testing 用 binomial expansion 开出来看。

[ 本帖最后由 dunwan2tellu 于 28-8-2006 07:27 PM 编辑 ]

~HeBe~_@

原帖由 zipp_882000 于 26-8-2006 10:12 PM 发表
哈哈明白了，谢谢。看来还需要再接再厉。谢谢！！！！

不用客气。。。。。。。哈哈。。。。

shijim

[ 本帖最后由 shijim 于 5-9-2006 04:35 PM 编辑 ]

dunwan2tellu

直接用 long division 来找他的 remainder

shijim

原帖由 dunwan2tellu 于 5-9-2006 02:49 PM 发表
直接用 long division 来找他的 remainder

我会了
那个power 9的不会

dunwan2tellu

3600 = 2^4 x 3^2 x 5^2

(x+p)^2 是其中一个factor , 所以 p 可能是 3600 的 factor .

一个很有用的 theorem :

given any polynomial P(x) . If P(a) = 0 and P'(a) = 0 . Then (x-a)^2 is a root for P(x)

所以因为 (x+p)^2 是 f(x) 的factor, 那么 f(-p) = 0 .又因为它是 double root , 所以他也一定要是 f'(x) 的 factor , i.e f'(-p) = 0

但是 f'(x) = ...... + 3 , 表示 p 只可能是 + - 3 或 + - 1 .

那么你试试 f(-1),f(1),f(-3),f(3) 看谁 = 0 那么它就是你要找的。

ss_sky87

可是题目说p > 0 那么只试1,3 就可以是吗？

dunwan2tellu

原帖由 ss_sky87 于 5-9-2006 05:21 PM 发表
可是题目说p > 0 那么只试1,3 就可以是吗？

当然你可以取巧。不过别忘了，p=1 时，x = -1 哦！

邵逸夫

Matrix A=  1 2 -3
         ( 3 1  1 )
           0 1 -2
a) find matrix B such that B=A^2 - 100I
b) find (A+I)B and hence find (A+I)^21 B

power 21...要怎樣找？
難道要一個一個去乘21次？
有什麽方法嗎？

[ 本帖最后由 邵逸夫 于 6-9-2006 05:17 PM 编辑 ]

dunwan2tellu

一定是有 pattern 的。你先找
(A+I)B 再找 (A+I)^2 B ...如果还看不出 pattern 就再找多几个 ...

邵逸夫

(A+I)B = -183  -179   275
         -264  -182  -100
         3      -92    85
(A+I)^21 B呢？

shijim

这里又有几题的问题要请问！希望大家多多指教
1．        Find the value of k if the polynomial f(x)= x^10 – 2x^8 – x^3 +2x + k is divisible by x+1
这个k我找到了，是2。 With this value of k, find the remainder when f(x) is divided by (x+1)^2

2.Polynomial-polynomial p(x) and q(x) are defined as p(x)=x^8-1 and q(x)=x^4+4x^3-3x^2-2x+5
Find the remainder when the polynomial 3p(x)+4q(x) is divided by x^2+1
这题我用long division来做。可是怎么也做不到！因为x^2+1中间没有x，不知道怎样除！

2．        When a polynomial is divided by ax-b or bx-a where [a]=/ ([ ] 为modulus，可是这里不明白代表什么意思) the remainder are the same even though the respective quotients q1(x) and q2(x) are not necessarily the same. Show that ax-b is a factor of q2(x) and x-1 is a factor of q1(x) + q2(x).

3．        Given f(x)= x^5-3x^4+2x^3-2x^2+3x+1
Show by long division or otherwise that the remainder when f(x) is divided by x^2+1 is 2x
这样的除数(x^2+1)我每次不会做！（好像第2题）因为除了后比如说乘x^2，可是就是没有x^3，乘了x就是没有x^2。各位可以教教在下吗？然后题目所谓的otherwise的方法，我是这样做
X^2+1=0
所以x= +- i (complex number)
代替 x = + i 就做到了，可是 – i 却做不到。这样的方法可以被接受吗？

4．        Prove the if n is an integer greater than 1, then 26^n – 26 is an odd multiple of 50.
Hence or otherwise, show that the last two digits of 26^n where n>1 is always the same and independent of n.

~HeBe~_@

原帖由 shijim 于 6-9-2006 05:49 PM 发表
2.Polynomial-polynomial p(x) and q(x) are defined as p(x)=x^8-1 and q(x)=x^4+4x^3-3x^2-2x+5
Find the remainder when the polynomial 3p(x)+4q(x) is divided by x^2+1
这题我用long division来做。可是怎么也做不到！因为x^2+1中间没有x，不知道怎样除！

你可以参考这。。。第 32# 帖

http://chinese.cari.com.my/myfor ... xtra=&page=2###

shijim

原帖由 ~HeBe~_@ 于 6-9-2006 06:23 PM 发表


你可以参考这。。。第 32# 帖

http://chinese.cari.com.my/myfor ... xtra=&page=2###

可是里面全部都是除以linear的

dunwan2tellu

4．        Prove the if n is an integer greater than 1, then 26^n – 26 is an odd multiple of 50.
Hence or otherwise, show that the last two digits of 26^n where n>1 is always the same and independent of n.

26^n - 26 = 26(26^(n-1) - 1) = 26(26-1)(26^(n-2) + 26^(n-3) + ... + 1)
= 13 x 50 x (26^(n-2) + 26^(n-3) + ... + 1)

So it is a mutiple of odd 50 since 13 is odd and (26^(n-2) + 26^(n-3) + ... + 1) is also odd .

From above we can write

26^n - 26 = 50(2k+1)  for k=non negative integer

so 26^n = 100k+ 50 + 26 = 100k + 76

So last two digit of 26^n is 76 (因为要直到最后两个 digit 的话，我们就把那个号码除于 100 ,余数就是 last two digit)