HKCEE 1996

Given x^2-6x+11 ≡ (x+a)^2+b, where x is real.

(a)Find the values of a and b.
Hence,write down the least value of x^2-6x+11.

(b)Using (a),or otherwise, write down the range of possible values of 1/x^2-6x+11.
(a)
x^2-6x+11 ≡ (x+a)^2+b
x^2-6x+11 ≡ (x-3)^2+2
so a=-3, b=2
when x=3, x^2-6x+11 reach the least value 2
(b)
since x^2-6x+11 should be greater than 2
the range of possible values of 1/x^2-6x+11 is
(0,1/2]


2006-12-11 02:02:34 補充：
since x^2-6x 11 should be greater than or equal 2

(a) by using the method of completing the square

x^2-6x+11
= (x^2-2(x)(3)+3^2)-3^2+11
=(x-3)^2-9+11
=(x-3)^2+2

therefore , a = -3 , b=2

the least value of x^2-6x+11 is 2

(b)
(x-3)^2>=0
(x-3)^2+2>2
x^2-6x+11>2
1/(x^2-6x+11)<1/2