F.4 Maths

1.(a)
as (x-3)^2 >=0, when x = 3, f(x) is minimum and minimum value of f(x) = -1.
k - 11 = -1
k = 10

1(b)
let the point on x-intercept be (m,0).
0 = (m-3)^2 - 1
(m-3)^2 = 1
m - 3 = 1 or -1
m = 4 or 2
so, x-intercepts = 2 and 4.

let the point on y-intercept be (0,n).
n = (0-3)^2 - 1
n = 9 - 1 = 8
so, y-intercept = 8.

2(a)
x^2 + 3x - 2 = 0
x = [ -3 ± square root of ( 3^2 - 4(1)(-2) ) ] / [ (2)(1) ]
x = ( -3 ± square root of 17 ) / 2
let α+1 = ( -3 + square root of 17 ) / 2 and β-1 = ( -3 - square root of 17 ) / 2.

α2β2 = (αβ)2
= { [( -3 + square root of 17 ) / 2 - 1][( -3 - square root of 17 ) / 2 + 1] }^2
= [(9 - 17) / 4 + square root of 17 - 1]^2
= (-3 + square root of 17)^2
= 9 - 6*square root of 17 +17
= 26 - 6*square root of 17
= 2(13 - 3*square root of 17)

α^2+β^2
= [( -3 + square root of 17 ) / 2 - 1]^2 + [( -3 - square root of 17 ) / 2 + 1]^2
= [( -3 + square root of 17 ) / 2]^2 - ( -3 + square root of 17 ) + 1
+ [( -3 - square root of 17 ) / 2]^2 + ( -3 - square root of 17 ) + 1
= { [( -3 + square root of 17 ) / 2][( -3 - square root of 17 ) / 2] }^2
- 2*square root of 17 + 2
= [(9 - 17) / 4]^2 - 2*square root of 17 + 2
= 6 - 2*square root of 17
= 2(3 - square root of 17)

2(b)
the required quadratic equation with rootsα^2 and β^2 is
x^2 - (α^2+β^2)x + α2β2 = 0
x^2 + 2[square root of (17) - 3] + 2(13 - 3*square root of 17) = 0

補充, (x-3)² cannot be negative so (x-3)² can only be +1.

WHERE IS THE WRONG?

1. f(x)=(x-3)² + (k-11)

when x = 3, f(x) is at its min. ==> k = 10.

when x = 0; y = f(x) = (0-3)²+(10-11) = 8

when y = 0; (x-3)² -1 = 0; (x-3)² = ±1; x=4 or x=2

2013-12-30 14:58:55 補充：
Sorry! something wrong.

when y = 0; (x-3)² -1 = 0; (x-3) = ±1; x=4 or x=2

2013-12-30 23:31:04 補充：
when y = 0; (x-3)² -1 = 0; (x-3)² = ±1; x=4 or x=2

(x-3)² = ±1 is wrong. It should be (x-3) = ±1