Let f (x) = 2x 2 - 3x + 1, G (x) = asin (x - π / 6), (a ≠ 0) (1) When 0 ≤ x ≤ π / 2, find the maximum value of y = f (SiNx); (2) If for any x 1 ∈ [0,3], there is always x 2 ∈ [0,3], so that f (x1) = g (x2) holds, find the value range of real number a; (3) The equation f (SiNx) = a-SiNx has two solutions on [0,2 π]

Let f (x) = 2x 2 - 3x + 1, G (x) = asin (x - π / 6), (a ≠ 0) (1) When 0 ≤ x ≤ π / 2, find the maximum value of y = f (SiNx); (2) If for any x 1 ∈ [0,3], there is always x 2 ∈ [0,3], so that f (x1) = g (x2) holds, find the value range of real number a; (3) The equation f (SiNx) = a-SiNx has two solutions on [0,2 π]

（1）
t=sinx∈[0,1]
f(sinx)=2t²-3t+1
The symmetry axis is t = 3 / 4, and the image opening is upward,
When x = 0, f (SiNx) has a maximum value of 1
（2）
That is, the range of F (x) is included in the range of G (x)
f(x)=2x²-3x+1
The symmetry axis is x = 3 / 4, and the image opening is upward,
When x = 3 / 4, the minimum value of F (x) is - 1 / 8
When x = 3, the maximum value of F (x) is 10
That is, the range is [- 1 / 8,10]
x2∈[0,3]
∴ x-π/6∈[-π/6,3-π/6]
∴ sin(x-π/6)∈[-1/2,1]
① A>0
g(x)∈[-A/2,A]
∴ A≥10
②A

If f (x) = asin (x + π / 4) + 3sin (x - π / 4) is an even function, then the value of real number a is

f(-x)=asin(-x+π/4)+3sin(-x-π/4)
=-asin(x-π/4)-3sin(x+π/4)
F (x) = asin (x + π / 4) + 3sin (x - π / 4) is even function
So (x) = F
So asin (x + π / 4) + 3sin (x - π / 4) = - asin (x - π / 4) - 3sin (x + π / 4)
The corresponding coefficients are equal
Obviously a = - 3

If f (x) = asin (x + π) 4)+3sin(x−π 4) If it is an even function, then a=______ .

f(x)＝asin(x+π
4)+3sin(x−π
4)＝a(
Two
2sinx+
Two
2cosx)+3(
Two
2sinx−
Two
2cosx) is even function,
If a = - 3, f (x) = - 3 can be obtained
2cosx is an even function
So the answer is: - 3

If f (x) = asin (x + π) 4)+3sin(x−π 4) If it is an even function, then a=______ .

f(x)＝asin(x+π
4)+3sin(x−π
4)＝a(
Two
2sinx+
Two
2cosx)+3(
Two
2sinx−
Two
2cosx) is even function,
If a = - 3, f (x) = - 3 can be obtained
2cosx is an even function
So the answer is: - 3

If f (x) = asin (x + π) 4)+3sin(x−π 4) If it is an even function, then a=______ .

f(x)＝asin(x+π
4)+3sin(x−π
4)＝a(
Two
2sinx+
Two
2cosx)+3(
Two
2sinx−
Two
2cosx) is even function,
If a = - 3, f (x) = - 3 can be obtained
2cosx is an even function
So the answer is: - 3

If f (x) is an even function with period 2, and when x belongs to (0,1), f (x) = 2 to the power of X-1, then the value of F (log2,10) is?

f(log2,10)=f(log2,10-4)=f(4-log2,10)=2^(4-log2,10)-1
=2^4/2^(log2,10)-1
=16/10-1=3/5

It is known that the function FX is an even function with period 2 when x ∈ (0,1), FX = 2 ^ X-1, f (log2 ^ 10) =? The answer is 5 / 3

log2^8

Let f (x) be an even function defined on R with period 2. If x ∈ (0,1) is known, f (x) = log2 (1-x), then f (x) is on (1,2) Is an increasing function and f (x) > 0

Let m ∈ (- 1,0), then - M ∈ (0,1), so f (- M) = log2 (1 - (- M)) = log2 (1 + m);
Since f (m) is an even function, f (m) = f (- M) = log2 (1 + m) (m ∈ (- 1,0));
Let n ∈ (1,2), then n-2 ∈ (- 1,0), so f (n-2) = log2 (1 + (n-2)) = log2 (n-1);
F (n) is a function with period 2, so f (n) = f (n-2) = log2 (n-1) (n ∈ (1,2))
Therefore, f (x) is a decreasing function on (1,2) and is always less than zero

It is known that the function f (x) whose domain of definition is R is a decreasing function on (8, + ∞) and the function y = f (x + 8) is even, then () A. f（6）＞f（7） B. f（6）＞f（9） C. f（7）＞f（9） D. f（7）＞f（10）

∵ y = f (x + 8) is an even function,
ν f (x + 8) = f (- x + 8), that is, y = f (x) is symmetric about the straight line x = 8
And ∵ f (x) is a minus function on (8, + ∞),
ν f (x) is an increasing function on (- ∞, 8)
From F (8 + 2) = f (8-2), that is, f (10) = f (6),
From 6 < 7 < 8, there is f (6) < f (7), that is, f (7) > F (10)
Therefore, D

Given that the functions f (x) and G (x) are all defined on the real number set R, and satisfy that f (x) is odd function, G (x) is even function, f (x) + G (x) = x2 + X-2, try to find the analytic formula of F (x), G (x)

According to the meaning of the title,
∵ f (x) is an odd function, G (x) is an even function,
And f (x) + G (x) = x2 + X-2 1,
∴f（-x）=-f（x），g（-x）=g（x），
∴f（-x）+g（-x）=（-x）2+（-x）-2，
That is - f (x) + G (x) = x2-x-2 (2);
F (x) = x is obtained from the solution of ① and ②,
g（x）=x2-2．