Given that the maximum value of function f (x) = - X & sup2; + 2aX + 1-A in the interval [0,1] is 2, find the value of real number a

Given that the maximum value of function f (x) = - X & sup2; + 2aX + 1-A in the interval [0,1] is 2, find the value of real number a

F (x) = - (x-a) ^ 2 + A ^ 2-A + 1, the opening is downward, and the axis of symmetry is x = a
f(0)=1-a,f(1)=a.f(a)=a^2-a+1
The maximum value must be one of the above formulas
A = - 1, consistent with
a> 1, the maximum value is f (1) = a = 2
0=

The definition field of the function f (x) = x / X-1 under the root sign is and the value field is

First, if the denominator is not 0, X-1 ≠ 0, then x ≠ 1;
If x / X-1 ≥ 0, then x ≤ 0 or > = 1
To sum up, the domain x ≤ 0 or x > 1
Analyze the value range according to the definition field
When x is less than or equal to 0, f (x) is [0,1]
(or) when x is greater than 1, f (x) is (1, ∞)
In conclusion, the range f (x) is [0, ∞) and ≠ 1

The definition field of function f (x) = root (2 ^ x-1) is, and the value field is

The domain is
2^x - 1 >=0
=>
2^x > = 1
=>
x >= 0
That is [0, + ∞)
The value range is [0, + ∞)