Let a be a constant and a > 1,0 < x ≤ 2 π, then f (x) = cos ^ 2x Let a be a constant, and a > 1, 0 be less than or equal to x, 2 be less than or equal to, and find the maximum value of the function f (x) = cos square x + 2asinx-1

f(x)=cos^2x+2asinx-1
=1-(sinx)^2+2asinx-1
=-(sinx)^2+2asinx
=-(sinx-a)^2+a^2
Let t = SiNx, for this quadratic function, when t = a, find the maximum
But a > 1, so take the maximum value when t = 1, substitute t = 1 into the function, and get the maximum value as - 1 + 2A
That is, when SiNx = 1, the maximum f (x) = 2a-1

RELATED INFORMATIONS

1. Let a > 0 be a constant and f (x) = x ＾ (1 / 2) - ln (x + a) be the maximum of F (x) when a = 3 / 4,
2. The function f (x) = ln (x + 1) - ax + (1-A) / (x + 1) (A & gt; 0.5) (1) is known when the curve y =
3. 1. Given the function f (x) = ln (x + 1) + ax ^ 2-x, a ∈ R (1) When a = 1 / 4, find the extremum of function y = f (x); (2) Is there a real number B ∈ (1,2) such that when x ∈ (- 1, b], the maximum value of function f (x) is f (b)? If there is a range of real number a, if not, please explain the reason I'd like to ask teachers and students to help me solve some problems. We have a holiday these days. First, I'll take a good look at the 20 and 21 questions in my exams. There are more than 70 questions. If I do them one by one, I don't have enough time to take a day off. So I want to take a serious look at the solutions. I hope the process can be more detailed,
4. Let f (x) = (2x + 1) ln (2x + 1); (1) find the minimum of F (x); (2) if x ≥ 0, f (x) ≥ 2aX holds, find the value range of real number a
5. Find the derivative f (x) = ln CTG X / 2 at point x = - π / 6 It's minus one sixth of the point Find the derivative
6. Given that the circle C ': (x-1) ^ 2 + y ^ 2 = a passes through the origin and is symmetric to the circle C with respect to the straight line y = - x, find the equation of the circle C and the length of the chord where the circle C intersects with the circle C'
7. Given that the circle C: x ^ 2 + y ^ 2-10x = 0, the chord length of the line L passing through the origin cut by the circle C is 8, the equation of the line L is obtained
8. The equation of a circle with a line segment length of 10 on the x-axis through the origin and point m (1, - 1) is obtained The equation of a circle with a line segment length of 10 on the x-axis through the origin and point m (1, - 1) is obtained
9. For the circle passing through the origin and a (1.1), the length of the line segment cutting the ground on the x-axis is 3
10. Find the equation of the circle with the length of the line segment of 3 through the origin and a (1,1) on the x-axis
11. The tangent equation y = X-1 at the beginning of point (1, f (1)) for the image with function f (x) = ax + BX + C (a > 0) is known (I) B, C are expressed by A; (II) if f (x) > LNX is constant on [1, ∞], find the value range of A; (III) proof: 1 + 1 / 2 + 1 / 3 +. + 1 / N ≥ ln (n + 1) + n / 2 (n + 1) (n ≥ 1)
12. The function f (x) = - x ^ 3 + ax ^ 2 + BX + C is known. The tangent equation at P (1, - 2) on the image is y = - 3 + 1 If the function f (x) increases monotonically in the interval [- 2,0], the value range of real number B is obtained
13. If the function f (x) = ax ^ 4 + BX ^ 2 + C is known and the tangent equation at x = 1 is y = X-2, then the analytic expression of F (x) is given
14. Given the function f (x) = x & # 178; + 2aX + 2, X ∈ [- 5,5] Use a to represent the maximum value of function f (x) in the interval [- 5,5],
15. (1)2x－3y＝5 x＋2y＝1 (2)u－9.8t－1＝0 u／3－1＝3t (3)x－5y＝16 5y－5x＝4 (4)2x－3y＝5 x＋y＝5／3 （5）21x＋23y＝243 23x＋21y＝241 Find x, Y. quick, I'll make it clear. (1)2x－3y＝5 x＋2y＝1 (2)u－9.8t－1＝0 u／3－1＝3t (3)x－5y＝16 5y－5x＝4 (4)2x－3y＝5 x＋y＝5／3 （5）21x＋23y＝243 23x＋21y＝241
16. To compare the size of two numbers a and B, sometimes we can solve it by the size of A-B and 0: if a = 5x + 2Y, B = 2x = 5Y, and X is less than y, try to compare the size of a and B
17. Given that the tangent equation of curve y = f (x) at point (1, f (1)) is x-2y + 4 = 0, then f (1) + F '(1)=______ ．
18. Given that the tangent equation of the image of the function f (x) = (ax-6) / (x ^ 2 + b) at the point m (- 1, f (- 1)) is x + 2Y + 5 = 0, find the analytic expression graph of the function y = f (x) First, we substitute the point m x = - 1 into the tangent equation and get y = - 2 So the point (- 1, - 2) is the tangent point of F (x) If f (x) is derived, f '(x) = (- ax ^ 2 + 12x + AB) / (x ^ 2 + b) & # 178; So f '(- 1) = (- A-12 + AB) / (1 + b) & # 178; = - 1 / 2 (slope of tangent) ① And f (- 1) = (- a-6) / (B + 1) = - 2 That is, a = 2b-4 From (1) to (2) A = - 6, B = - 1 or a = 2, B = 3 And because (x ^ 2 + b) is the denominator, it is not zero, so B = - 1 is omitted a=2 b=3 Where f '(- 1) = (- A-12 + AB) / (1 + b) &# 178; = - 1 / 2 is based on what? Why is the slope obtained by derivation
19. The parabola x 2 = 4Y P is a tangent line of circle x 2 + (y + 1) 2 = 1 crossed by a moving point on the parabola passing through point P, y = - 2 at two points ab. when Pb is just tangent to the parabola and point P, the area of △ PAB is calculated
20. X ^ 2 = 4Y, the tangent of line L passing through a and B is L1 and L2 (1) Verification of L1 vertical L2 (2) It is proved that the focus of L1 and L2 is on the directrix