The known function f (x) = x2 + xsinx + cosx (1) Find the minimum value of F (x); (2) If the curve y = f (x) is tangent to the straight line y = B at the point (a, f (a)), find the values of a and B (3) If the curve y = f (x) and the straight line y = B have two different intersections, find the value range of B

(1) By F (x) = x2 + xsinx + cosx,
F '(x) = 2x + SiNx + xcosx - SiNx = x (2 + cosx) (1 point)
Let f '(x) = 0, then x = 0 (2 points)
The list is as follows:
… (4 points)
The function f (x) decreases monotonically on the interval (- ∞, 0),
Monotonically increasing on the interval (0, + ∞),
/ / F (0) = 1 is the minimum value of F (x) (5 points)
(2) ∵ the curve y = f (x) is tangent to the line y = B at the point (a, f (a)),
∴f′（a）=a（2+cosa）=0，b=f（a），… (7 points)
A = 0, B = f (0) = 1 (9 points)
(3) When B ≤ 1, there is only one intersection point between the curve y = f (x) and the straight line y = B at most;
When b > 1, f (- 2b) = f (2b) ≥ 4b-2b-1 > 4b-2b-1 > b, f (0) = 1 < B,
There exists x1 ∈ (- 2b, 0), X2 ∈ (0, 2b), such that f (x1) = f (x2) = B (12 points)
Because the function f (x) is monotone on the interval (- ∞, 0) and (0, + ∞),
When b > 1, curve y=f (x) and straight line y=b have only two different intersections (13 points)
To sum up, if the curve y = f (x) and the straight line y = B have two different intersections, then the value range of B is (1, + ∞) (14 points)

The known function f (x) = x 2 + xsinx + cosx If the curve y = f (x) is tangent to the straight line y = B at (a, f (a)), find the values of a and B

Tangent to the line y = B indicates that the tangent slope is 0
F'(x)=2x+xcosx+sinx-sinx=x(2+cosx)
F '(x) = 0, x = 0 (∵ 2 + cosx > 0)
So, a = 0,
F (0) = 1, so B = 1

If f (x) = xsinx + cosx, then f (1) =? I'm asking what f (1) is equal to, not a formula

f(1)=1sin1+cos1=sin1+cos1=sin57.3°+cos57.3°=0.8415+0.5405=1.382

The known function f (x) = xsinx + cosx-x ^ 2 Let f (x) = xsinx + cosx-x ^ 2. If the curve y = f (x) is tangent to the straight line y = B at the point (a, f (a)), find a and B Is there a problem with this topic? After derivation calculation, we get acosa-2a = 0, and then we get cosa-2 = 0 after removing a a a

0

y =cos²（x-¼π）-1=-sin²（x-¼π）=-(1-cos2(x-¼π))/2=-(1-cos(2x-1/2π))/2=-(1-sin2x)/2=-1/2+sin2x/2
The period is: 2 π / 2 = π

If the minimum value of function f (x) = (A-1) ^ 2-2sin ^ 2x-2cosx (0 ≤ x ≤ π / 2) is - 2, find the value of function a and find the maximum value of F (x)

In the case of 0 ≤ x ≤ π / 2, then 0 ≤ cosx ≤ 1, then, when cosx = 1 / 2, the minimum value of Y is 0-5 / 2 + (A-1) ^ 2 / 2 + (A-1) ^ 2; then, (A-1 ^ 2 ^ 2) ^ 2-5 / 2 + (A-1) ^ 2; because of 0 ≤ x ≤ π / 2, then 0 ≤ cosx ≤ 1, then, when cosx = 1 / 2, y has the minimum value = 0-5 / 2 + (A-1) ^ 2 = - 2; then, (A-1 ^ 2 ^ 2 = - 2; then, (A-1 ^ 2 ^ 2 ^ 2 ^ 2 = - 2; then, (A-1 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 2 ^ 1 ^ 2) 2 = - 2 = - 2= 1 / 1

The maximum value of the function f (x) = 2Sin * 2x + 2cosx-5

Let t = cosx, then | t|

The minimum value of the function y = 2Sin (π / 6-x) + 2cosx (x ∈ R)

y=2sin(π/6-x)+2cosx=2[sin(π/6)×cosx-cos(π/6)×sinx]+2cosx=cosx-√3sinx+2cosx
=3cosx-√3sinx=2√3(√3/2cosx-1/2sinx)=2√3sin(π/3-x)
When π / 3-x = - 1, the minimum value f (x) min = - 2 √ 3

Simplify 1 + sin ^ 2x / SiNx . Please write it down in detail. Thank you. Would you please write down the process? (1 + sin ^ 2x) / SiNx

Because 1 = sin ^ 2x + cos ^ 2x, the original formula = (COS ^ 2x + 2Sin ^ 2x) / SiNx, so the original formula = 1 + SiNx

How to simplify sin ^ 2x - SiNx + 3? Thank you The original problem is to find the maximum value of y = sin ^ 2x - SiNx + 3

The factorization is (sinx-1 / 2) ^ 2 + 11 / 4
If you put the original question out, it will be simple. The definition field of SiNx is between (- 1,1), so the maximum value of (sinx-1 / 2) ^ 2 is 9 / 4, and the minimum value is 0, so the maximum value of Y is 9 / 4 + 11 / 4 = 5