f.4 a．maths　help(15 points)

[匿名]

2007-08-27 10:58 pm

1)　Given x^2 - 6x + 11 ≡ (x + a ) ^ 2 + b, where x is real.
a.　Find the values of a and b.
b.　Using (a), or otherwise, write down the range of possible values of 1/x^2 - 6x +11

2)　It is given that (1 + x + ax^2) ^6 =1+ 6x + Ax^2 + Bx^3 + terms involving higher 　　　　　　　　　　　　　　　　　　　　　　　　　　　　　power of x
a.　Express A and B in terms of a.
b.　If 6, A and B are in arithmetic sequence, find the values of a.

更新1:

2(b)∵6,A,B are in A.P. ∴2A=6+B ＜－－－點得出黎嫁．．．？ ∴12a+30=30a+26 ∴a=2/9

回答 (1)

hawk_wing_1999

2007-08-28 12:42 am

✔ 最佳答案

1(a).x^2-6x+11=x^2-6x+9+2
=(x-3)^2+2
∴a=-3,b=2

1(b).For x^2-6x+11
x^2-6x+11=(x-3)^2+2
∴x^2-6x+11>=2 for all real value of x
∴1/(x^2-6x+11)<=1/2 for all real value of x

2(a)(1 + x + ax^2) ^6
=[1+(x+ax^2)]^6
=1+6(x+ax^2)+15(x+ax^2)^2+20(x+ax^2)^3+......
=1+6x+(6a+15)x^2+(30a+20)x^3+......
∴A=6a+15, B=30a+20

2(b)∵6,A,B are in A.P.
∴2A=6+B
∴12a+30=30a+26
∴a=2/9

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f.4 a．maths help(15 points)

回答 (1)

f.4 a．maths　help(15 points)