Function f（ If x + 3) = x2 + 4x-5, then the value range of function f (x) (x ≥ 0) is () A. [−41 4，+∞) B. [-9，+∞） C. [−33 4，+∞) D. [-7，+∞）

order
If x + 3 = t ≥ 0, then x = T2-3
∴f（t）=（t2-3）2+4（t2-3）-5=（t2-1）2-9≥-9
The value range of function f (x) (x ≥ 0) is [- 9, + ∞)
Therefore, B

The value range of function y = | x + 1 | + root sign (x ^ 2-4x + 4)

Y = | x + 1 | + radical (x ^ 2-4x + 4)
=|x+1|+|x-2|
When x ≥ 2; y = x + 1 + X-2 = 2x-1 ≥ 3;
When - 1 ≤ x ≤ 2, y = x + 1 + 2-x = 3;
When x ≤ - 1; y = - X-1 + 2-x = - 2x + 1 ≥ 3;
Y = | x + 1 | + root sign (x ^ 2-4x + 4) value range [3, + ∞)

The domain of SiNx cosx + LG (- xsquare-2x + 3) under the function f (x) = radical

From the meaning of the title
sinx-cosx≥0
√2sin（x-π/4）≥0
2K π + π / 4 ≤ x ≤ 2K π + 5 / 4 π are obtained
-Xsquare - 2x + 3 > 0
Xsquare + 2x-3 < 0
（x-1）（x+3）＜0
-3＜x＜1
So take the intersection
[-3,-3π/4）∪[π/4,1）

It is known that f (x) = - sin2x + SiNx + a (I) when f (x) = 0 has a real number solution, find the value range of real number a; (II) if x ∈ R is always 1 ≤ f (x) ≤ 17 If 4 is true, find the value range of real number a

(1) Because f (x) = 0, that is, a = sin2x − SiNx = (SiNx − 1)
2)2−1
The maximum value of a is equal to (− 1 − 1)
2)2 −1
4=2，
The minimum value of a is equal to - 1
Therefore, a ∈ [− 1
4，2]．
（2）f（x）=-sin2x+sinx+a=−(sinx−1
2)2+1
4+a，∴f(x)∈[−2+a，1
4+a]，
And ∵ 1 ≤ f (x) ≤ 17
4 Heng is established, ν
1≤−2+a
One
4+a≤17
4 ，∴3≤a≤4．
Therefore, the value range of real number a is [3,4]

The known function f (x) = sin (π / 2-x) + SiNx (1) Finding monotone increasing interval of function y = f (x) (2) If f (a - π / 4) = radical 2 / 3, find the value of F (2a + π / 4)

(x) = cosx + sinxf (x) = cosx + sinxf (x) = √ 2Sin (x + π / 4) (1) increasing interval: 2K π - π / 2 ≤ x + π / 4 ≤ 2K π + π / 2: 2K π - 3 / 4 π ≤ x ≤ 2K π + π / 4: 2K π - 3 / 4 π ≤ x ≤ 2K π + π / 4: 2K π - 3 π / 4, 2K π + π / 4], where k ∈ Z (2) f (a - π / 4) = √ 2 / 3, then: Sina = 1 / 3f (3f (1 / 3) if: Sina = 1 / 3f (3f) is: Sina = 1 / 3f (3f) 2A +

The known function f (x) = SiNx + sin (x + π) 2），x∈R． (1) Find the minimum positive period of F (x); (2) Find the maximum and minimum values of F (x); (3) If f (α) = 3 4. Find the value of sin 2 α

（1）∵f(x)=sinx+sin(π
2+x)=sinx+cosx=
2sin(x+π
4) The function f (x) = SiN x + sin (x + π)
2) The minimum positive period of is 2 π
（2）∵x∈R，-1≤sinx≤1
（2）f(x)=sinx+sin(π
2+x)=sinx+cosx=
2sin(x+π
4)
The maximum value of F (x) is
2, the minimum value is-
2… (8 points)
（3）∵f（α）=sinα+sin（α+π
2）=sinα+cosα=3
Four
∴（sinα+cosα）2=sin2α+cos2α+2sinαcosα=1+sin2α=9
Sixteen
∴sin2α=9
16-1=-7
Sixteen

Given the function f (x) = SiNx + sin (x + 3 / π), find the minimum value of the function And find the set of X whose minimum value is f (x)

Solution;
f(x)=sinx+sinxcosπ/3+cosxsinπ/3
=sinx+1/2sinx+√3/2cosx
=3/2sinx+√3/2cosx
=√3sin(x+π/6)
When x + π / 6 = - π / 2 + 2K π
When x = - 2 π / 3 + 2K π
The minimum value of F (x) is: - √ 3
The set of X with minimum value f (x) is: {X / x = 2K π - 2 π / 3, K ∈ Z}

The known function f (x) = SiNx + sin (x + π) 2），x∈R． (1) Find the minimum positive period of F (x); (2) Find the maximum and minimum values of F (x); (3) If f (α) = 3 4. Find the value of sin 2 α

Solve the equation 3x + radical 2 = 0, and solve the inequality 2x radical 3 greater than or equal to 0

There is no process
3x + radical 2 = 0
3 x = 3
2X radical 3 greater than or equal to 0
X is greater than or equal to 2 / radical 3

To solve the equation, 2 (root 2 minus x) minus 3 root sign 2 equals 3x to solve inequality, (1 minus root 2) x is greater than 1 minus root 2

2(√2-x)-3√2=3x
2√2-2x-3√2=3x
-√2-2x=3x
3x+2x=-√2
5x=-√2
x=-√2/5
(1 - √ 2) x > 1 - √ 2 both sides divide by 1 - √ 2, the unequal sign changes direction
X

Function f（ If x + 3) = x2 + 4x-5, then the value range of function f (x) (x ≥ 0) is () A. [−41 4，+∞) B. [-9，+∞） C. [−33 4，+∞) D. [-7，+∞）