If the equation 22x + 2x · a + 1 = 0 about a has real roots, then the value range of real number a is______ .

If the equation 22x + 2x · a + 1 = 0 about a has real roots, then the value range of real number a is______ .

Let 2x = t > 0, the original equation is T2 + at + 1 = 0. {a = − T2 − 1t = − t − 1t, t > 0 {a ≤ - 2, if and only if t = 1, the equal sign holds. So the value range of real number a is (- ∞, - 2]. So the answer is: (- ∞, - 2]
If the equation 2 ^ 2x-2 ^ x a + A + 1 = 0 about X has two different positive real roots, then the value range of real number a is
If the equation 2 ^ 2x-2 ^ x a + A + 1 = 0 about X has two different positive real roots, then the value range of real number a is (2 + 2 root sign 2, + ∞)
(2^x)^2-a*2^x+a+1=0
t=2^x
t^2-at+a+1=0
Discriminant = a ^ 2-4 (a + 1) = a ^ 2-4a-4 > 0
(a-2)^2>8
a> 2 + 2 root 2 or A0 a > - 1
Take intersection:
a> 2 + 2 2
That is: (2 + 2 radical 2, + ∞)
X satisfying inequality (x + 5) / 2-1 > (AX + 2) / 2 must satisfy inequality 2x-3
2x-3-1
(x+5)/2-1>(ax+2)/2
x+5-2>ax+2
(1-a)x>-1
When 1-A ＞ 0 means a ＜ 1, X ＞ - 1 / (1-A) according to the meaning of the title, then 0 ≤ a ＜ 1
When 1-A ＜ 0, i.e. a ＞ 1, X ＜ - 1 / (1-A), which is inconsistent with the known, so it is not necessary to give up
When 1-A = 0, i.e. a = 1, 0 ＞ - 1 x is all real numbers, which is contradictory to the known ones
So 0 ≤ a < 1
The monotone decreasing interval of the function y = | 3x-5 | is___ .
Because the zero point of function y = | 3x-5 | is 53, combining with the image of function y = | 3x-5 | we can get the monotone decreasing interval of function y = | 3x-5 | is (- ∞, 53), so the answer is: (- ∞, 53]
It is known that the function f (x) = ln (e ^ x + a) (a is a constant, e is the base of natural logarithm) is an odd function on the real number set R, and the function g (x) = λ f (x) + SiNx is a decreasing function on the interval [- 1,1]
(1) Finding the value of a
If f (x) is an odd function, then f (0) = 0, a = 0
F (x) = ln (e ^ x + a) (a is a constant, e is the base of natural logarithm) is an odd function on real number set R
f(0)=ln（e^0+a）=ln（1+a）=ln1=0
So a = 0
F (x) is an odd function,
f(-x)=-f(x)
It satisfies ln (e ^ - x + a) = - ln (e ^ x + a),
Then ln (e ^ - x + a) / ln (e ^ x + a) = - 1
（e^-x+a）/（e^x+a）=e^-1
So a = 0
The decreasing function of interval x 3 is monotone ()
A. （-∞，0）B. （0，+∞）C. （-∞，-1），（1，+∞）D. （-1，1）
Let y ′ = 3x2-3 ＜ 0, we get - 1 ＜ x ＜ 1, and the monotone decreasing interval of the function y = x3-3x is (- 1,1)
For the function f (x), if there is a real number X. such that f (X.) = X. holds, then X. is called the fixed point of F (x)
1. Given the quadratic function f (x) = ax ^ 2 + (B + 1) x + (B-1), when a = 1, B = - 2, find the fixed point of function f (x)
2. If for any real number B, the function f (x) always has two different fixed points, the range of a is obtained
3. Under the condition of 2, if the abscissa of two points a and B on the image y = f (x) is the fixed point of function f (x), and a and B are symmetric about the line y = KX + 1 / (2a ^ 2 + 1), the minimum value of B is obtained
1. A = 1, B = - 2 substitute f (x) f (x) = x & # 178; - x-3 Let f (x) = XX & # 178; - x-3 = XX & # 178; - 2x-3 = 0 (x-3) (x + 1) = 0x = 3 or x = - 1F (x) have fixed points x = 3 and x = - 1.2, ax & # 178; + (B + 1) x + (B-1) = Xax & # 178; + BX + (B-1) = 0f (x) with two different fixed points, and the square discriminant > 0 △ = b &
The monotone decreasing interval of the function y = 32 − 3x2 is______ .
The outer function is y = 3T, the inner function is t = 2-3x2. Because the outer function y = 3T is an increasing function, and the inner function T = x2 + 2x is an increasing function on (- ∞, 0) and a decreasing function on (0, + ∞), the monotonic decreasing of the composite function y = 32-3x2
In the quadratic function f (x) = AX2 + BX + C (a ≠ 0), the number of zeros of the function is ()
A. 0b. 1C. 2D. Uncertainty
∵ a ≠ 0, AC ∫ 0 ∫ a = b2-4ac ∫ 0 ∫ f (x) = 0 has two roots, that is, the function f (x) = AX2 + BX + C has two zeros, so C is selected
Given that the value range of y = 4x-3 · 2x + 3 is [1,7], then the value range of X is ()
A. [2，4]B. （-∞，0）C. （0，1）∪[2，4]D. （-∞，0]∪[1，2]
The range of ∵ y = 4x-3 · 2x + 3 is [1,7], ∵ 1 ≤ 4x-3 · 2x + 3 ≤ 7. ∵ - 1 ≤ 2x ≤ 1 or 2 ≤ 2x ≤ 4. ∵ x ≤ 0 or 1 ≤ x ≤ 2