Let f (x) = g (x) / x.1) find the value of a and B. 2) the inequality f (2 ^ x) - K2 ^ x > = 0 holds on [- 1,1]. 3) the equation f (ABS (2 ^ x-1)) + K [(2 / ABS (2 ^ x-1)) - 3] = 0 has three different real number solutions and the range of K You'd better finish the problem

Let f (x) = g (x) / x.1) find the value of a and B. 2) the inequality f (2 ^ x) - K2 ^ x > = 0 holds on [- 1,1]. 3) the equation f (ABS (2 ^ x-1)) + K [(2 / ABS (2 ^ x-1)) - 3] = 0 has three different real number solutions and the range of K You'd better finish the problem

G '(x) = 2ax-2a, let it be equal to 0, find the stationary point 2ax-2a = 0, x = 1, obviously the stationary point does not belong to the interval given by the title [2,3], then the extreme point must be the end point of the interval G' (x) = 2A, if a > 0, then G '(x) monotonically increases, G' (x) > G '(1) = 0, then G (x) monotonically increases. G (2) = 4A = 4A + 1 + B = 1 + BG (3) = 9a-6a +

F (x) = ax ^ 2-2ax + 2 + B in the interval [2,3], the maximum value is 5, the minimum value is 2, find a, B if B

First of all, a cannot be equal to 0, the symmetry axis X of F (x) is 1, a > 0 monotonically increases in the interval [2,3], A0, f (2) = 2, f (3) = 5, then a = 3 / 4, B = - 3
If a

G (x) = ax ^ 2 - 2aX + 1 + B (a ≠ 0, B 〈 1), in the interval [2,3], the maximum value is 4, the minimum value is 1, Let f (x) = g (x) / x, find the value of a, B
G (x) = ax ^ 2 - 2aX + 1 + B (a ≠ 0, B 〈 1), in the interval [2,3], the maximum value is 4, the minimum value is 1, Let f (x) = g (x) / X
1. Find the value of a and B;
2. If f (2 ^ x) - K · 2 ^ x ≥ 0 is constant in X ∈ [- 1,1], the range of K is obtained

When G (x) = ax & # 178; - 2aX + 1 + B (a ≠ 0, B0), the opening of G (x) is upward,
Increasing in the interval [2,3]
Maximum f (x) max = f (3) = 3A + 1 + B = 4
Minimum f (x) min = f (2) = 1 + B = 1
So,
a=1
b=0
a