Given the function f (x) = (k ^ 2 + 1) x ^ 2-2kx - (k-1) ^ 2 (K ∈ R), X1 and X2 are the two zeros of F (x), and X1 > x2（ Given the function f (x) = (k ^ 2 + 1) x ^ 2-2kx - (k-1) ^ 2 (K ∈ R), X1 and X2 are the two zeros of F (x), and X1 > x2 (1) (I) prove: X1 = 1; (II) find the value range of x2; (2) Let g (k) be the minimum value of function f (x). When x 2 ∈ [- 2, - 1], find the maximum value of G (k)

Given the function f (x) = (k ^ 2 + 1) x ^ 2-2kx - (k-1) ^ 2 (K ∈ R), X1 and X2 are the two zeros of F (x), and X1 > x2（ Given the function f (x) = (k ^ 2 + 1) x ^ 2-2kx - (k-1) ^ 2 (K ∈ R), X1 and X2 are the two zeros of F (x), and X1 > x2 (1) (I) prove: X1 = 1; (II) find the value range of x2; (2) Let g (k) be the minimum value of function f (x). When x 2 ∈ [- 2, - 1], find the maximum value of G (k)

1) f(1)=k²+1-2k-(k-1)²=(k-1)²-(k-1)²=0
So x = 1 is the zero of F (x)
Because two products = - (k-1) &# 178; / (K & # 178; + 1)

The two different zeros of the function y = x ^ 2 + (M + 1) x + m are X1 and X2, and the sum of the reciprocal squares of X1 and X2 is 2

x1+x2=-(m+1),x1x2=m
1/x1^2+1/x^2=(x1^2+x2^2)/(x1x2)^2=[(x1+x2)^2-2x1x2]/(x1x2)^2=[(m+1)^2-2m]/m^2=(m^2+1)/m^2=2
m^2=1
M = 1 or M = - 1

F (x) = sin2x + acos2x, and π / 4 is the zero point of function y = f (x). 1. Find the value of a and the minimum positive period of function f (x). 2. If ∈ [0, π], find the range of function f (x) and write the value of X when f (x) reaches the maximum
I've got the wrong number of acos2x, which should be ACOS ^ 2 X

(1) ∵ π / 4 is a zero point of the function f (x) = sin2x + ACOS & sup2; X,
∧ f (π / 4) = sin (π / 2) + ACOS & sup2; (π / 4) = 0, i.e. 1 + A / 2 = 0, ∧ a = - 2,
f(x)=sin2x-2cos²x=sin2x-cos2x-1=√2sin(2x-π/4)-1
Period T = π
(2) From (1), f (x) = √ 2Sin (2x - π / 4) - 1
When x ∈ [0, π], 2x - π / 4 ∈ [- π / 4,7 π / 4], sin (2x - π / 4) ∈ [- 1,1],
The range of F (x) is [- √ 2-1, √ 2-1]
When f (x) is the maximum √ 2-1, 2x - π / 4 = π / 2, x = 3 π / 8