Given that a is a non-zero constant, the maximum value of the function y = asinx + B is 3 and the minimum value is - 1, the values of a and B are obtained Come on.

Given that a is a non-zero constant, the maximum value of the function y = asinx + B is 3 and the minimum value is - 1, the values of a and B are obtained Come on.

The maximum value of function y = asinx + B is 3 and the minimum value is - 1
(1) When a > 0, there was a significant difference
a+b=3
-a+b=-1
The solution is a = 2, B = 1
(2)a
The minimum positive period of the function y = cos2x-sin2x + 2sinx · cosx is ⊙___ The value range of this function is ⊙___ .
Function y = cos2x-sin2x + 2sinx · cosx = cos2x + sin2x = 2Sin (2x + π 4), so the minimum positive period of function y = cos2x-sin2x + 2sinx · cosx is 2 π 2 = π, and the range of function is: [2,2], so the answer is: π; [2,2]
If the solution set of equation x plus (a + 1) x + B = 0 has only one element a, find a, B
X2+(a+1)X+b=0
We know that the square of (a + 1) - 4B = 0
And a is the solution of the equation, so the square of a + (a + 1) a + B = 0
To solve this system of equations, a = - 1 / 3, B = 1 / 9
If the function f (x) satisfies f (x) + 2F (1 / x) = 3x, then the value of F (2) is
A 1 B-1 C-3/2 D3/2
B) -1
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Analysis:
Let x = 2, have
f(2) + 2f(1/2) = 3*2 = 6 .(1)
Let x = 1 / 2, then we have
f(1/2) + 2f(2) = 3*1/2 = 3/2 .(2)
2 * (2) - (1) get,
3f(2) = -3
Thus, f (2) = - 1
Let x = 2 and 1 / 2, then
f(2)+2f(1/2)=6
f(1/2)+2f(2)=3/2
The solution is f (2) = - 1, f (1 / 2) = 7 / 2
Choose B
B -1
f(2)+2f(1/2)=6
Simultaneous solution of F (1 / 2) + 2F (2) = 3 / 2
B
f(x)+2f(1/x)=3x,
Let 1 / x = t, then f (1 / T) + 2F (T) = 3 / T,
That is 2F (x) + F (1 / x) = 3 / X
f(x)+2f(1/x)=3x,
F (x) = 2 / X - X
Choose B. The solution to this problem is to replace x with 1 / X in the known formula, and get a new sub f (1 / x) + 2F (x) = 3 / x, and then eliminate f (1 / x) by two formulas. We get f (x) = 2 / x-x, so we choose B.
This topic uses the method of constructing equations; we must master that if f (x) - 2F (1 / x) = 3x 2 (1) and x = 1 / x, then f (1 / x) - 2F (x) = 3 / x 2 (2) will (2) * 2
The set a is composed of the real roots of the equation AX & sup2; + 2x + 1 = 0 about X, and there is only one element in the set a
There is only one element in a, that is, the equation has a solution
If a = 0, then it is a linear equation with one variable and satisfies one solution
If a is not equal to 0
Then the quadratic equation has only one solution
So the discriminant = 0
4-4a=0
A=1
So set = {0,1}
{0,1}
Given that the function f (x) satisfies the relation f (x) + 2F (1 / 2) = 3x, find f (x)
Bring 1 / 2 in
f(1/2)+2f(1/2)=3*1/2
f(1/2)=0.5
therefore
f(x)+2*0.5=3x
f(x)=3x-1
When x = 1 / 2, the original formula becomes
f(1/2)+2f(1/2)=3*1/2
3f(1/2)=3*1/2
f(1/2)=1/2
f(x)+2*1/2=3x
f(x)=3x-1
If the equation AX ^ 2 + 2x + 1 = 0 has negative roots, then the value set of a is
If the equation AX ^ 2 + 2x + 1 = 0 has negative roots,
4-4a>=0
A
The value range should be a = 0
A
3f (x + 1) - 2F (x-1) is equal to 2x + 17, f (x) is a function of degree, find f (x)
F (x) is a linear function
f(x)=kx+b
Then f (x + 1) = K (x + 1) + B = KX + (K + b)
f(x-1)=k(x-1)+b=kx+(-k+b)
So 3kx + 3 (K + b) - 2kx-2 (- K + b) = 2x + 17
kx+(5k+b)=2x+17
So k = 2,5k + B = 17
B=7
So f (x) = 2x + 7
Let f (x) = ax + B, then 3A (x + 1) + 3b-2a (x-1) - 2b = 2x + 17
So a = 2, B = 17-5a = 7
f（x）=2x+7
Because f (x) is a linear function, Let f (x) = KX + B, and 3f (x + 1) - 2F (x-1) be equal to 2x + 17, then 3f (x + 1) - 2F (x-1) = 3 ((x + 1) K + b) - 2 ((x-1) K + b) = KX + B + 5K, k = 2, B + 5K = 17, so k = 2B = 7, so f (x) = 2x + 7
Given the set a = {x | AX2 + 2x + 1 = 0, X ∈ r}, a is a real number. (1) if a is an empty set, find the value range of a; (2) if a is a single element set, find the value range of a; (3) if there is at most one element in a, find the value range of A
Solution (1) if a = Φ, then only AX2 + 2x + 1 = 0, no real number solution, obviously a ≠ 0, so only △ = 4-4a < 0, i.e. a > 1. (2) when a = 0, the original equation is transformed into 2x + 1 = 0, and the solution is x = - 12; when a ≠ 0, only △ = 4-4a = 0, i.e. a = 1, so the value of a is 0 or 1; (3) according to (1) (2), when there is at most one element in a, the value of a is 0 or a ≥ 1
It is known that the function f (x) = (1-2 ^ x) / [2 ^ (x + 1) + a] whose domain is R is odd. (1) find the value of a (2) for any real number t
(2) If the inequality f (T ^ 2-2t) + F (2t & sup2; - K) < 0 holds for any real number T ∈ R, the value range of K is obtained
-.
Odd function so f (- x) = - f (x)
f(-1)=-f(1)
f(-1)=(1-1/2)/(1+a)=1/2(1+a)
f(1)=(1-2)/(4+a)=-1/(4+a)
f(-1)+f(1)=1/(2+2a)-1/(4+a)=0
2+2a=4+a a=2
f(x)=(1-2^x)/[2^(x+1)+2]=1/2[(1-2^x)/(1+2^x)]
Derivation f '= [(1-2 ^ x)' (1 + 2 ^ x) - (1-2 ^ x) (1 + 2 ^ x) '] / 2 (1 + 2 ^ x) ^ 2
f'=[-2^xln2(1+2^x)-(1-2^x)2^xln2]/2(1+2^x)^2
=(-2^xln2)(1+2^x+1-2^x)/2(1+2^x)^2=-2^xln2/(1+2^x)^2-1/3
1. F (- x) = [1-2 ^ (- x)] / [2 ^ (- x + 1) + a] (the numerator denominator is multiplied by the x power of 2 at the same time)
=(2^x-1)/(2+a2^x)
-f(x)=(2^x-1)/[a+2^(x+1)]
Because f (- x) = - f (x),
So a = 2
2. It is known from 1 that the denominator of F (x) = (1-2 ^ x) / [2 ^ (x + 1) + 2] is always greater than zero. We just need to discuss the molecules.
(1) Using the definition of odd function f (- x) = - f (x) to calculate a = 2
For example, if f (- 1) = - f (1), a = 2 can be obtained
(2) F (x) = (1-2 ^ x) / [2 ^ (x + 1) + 2] = (1-2 ^ x) / 2 (2 ^ x + 1) using the separation constant method, f (x) = - 1 / 2 + 1 / (2 ^ x + 1) can be known as a decreasing function
Because: for any real number T ∈ R, the inequality f (T ^ 2-2t) + F (2t & sup2... Is expanded
(1) Using the definition of odd function f (- x) = - f (x) to calculate a = 2
For example, if f (- 1) = - f (1), a = 2 can be obtained
(2) F (x) = (1-2 ^ x) / [2 ^ (x + 1) + 2] = (1-2 ^ x) / 2 (2 ^ x + 1) using the separation constant method, f (x) = - 1 / 2 + 1 / (2 ^ x + 1) can be known as a decreasing function
Because: for any real number T ∈ R, the inequality f (T ^ 2-2t) + F (2t & sup2; - K) < 0 holds
So f (T ^ 2-2t) 0 is always true, so the discriminant