It is known that when x R, the inequality a cos2x 5 4sinx has a solution, and the value range of the

Because cos2x = 1-2sin x

a+1-2sin^x<5-4sinx

Let sinx=t t t belong to [

a f(t)=2t 2-4t+4 so when t=-1, f(t)max=10

So a is less than 10 (which is the case where there is a solution).

If it is to be constant, then when t=1, f(t)min=2a 2 is looking forward to you, I hope you can see the topic.

cos2x=1-2*(sinx)^2

The inequality can be reduced to 1-2*(sinx) 2+a+4sinx-5<0 ===> 2*(sinx) 2-4sinx-a+4>0

Sort it out into 2*(sinx) 2-4sinx+2>a-2 ===> 2(sinx-1) 2>a-2

The value of six is between -1 and 1, so (sinx-1) 2 is maximum: (-1-1) 2=4, so the maximum on the left is 8, so a-2 as long as it is less than 8.

It's a<10

There is a solution, that is, he only needs to have a solution, and the constant is that the inequality is true when x goes to any value.

For example, sinx<=a has a solution, which means that a>=-1 is sufficient, because when a=-1, x=- 2, so x has a solution.

The meaning of constant is a>=1 because when a=-1, x= 2 yes, the inequality does not hold, so it is not constant, and the constant is true, no matter what the number x is, the inequality must hold forever.

In the case of multiple-choice questions, you can use the special value method to solve the problem:

Let a 0, the inequality be x 0, and the solution set be x≠0, not r, so the answer cannot contain 0, c is wrong.

Let a 1, the inequality is x x 1 0, and the solution set is r, so the answer should contain 1, b is wrong, and d is wrong.

To sum up, choose A.

The general solution is:

Solution 1: Use images.

There is a function f(x) x ax a, which is represented as a parabola with an opening upward.

If the solution set is r, then f(x) 0 is constant, i.e., there is no intersection between the image and the x-axis.

（-a）²－4a＝a（a－4）＜0

Solve a (0,4).

Solution 2: Start directly with the inequality.

The original inequality is 4x 4ax 4a 0

That is, (2x a) a 4a

Considering that (2x a) 0, the original inequality is constant on r and the equivalence is a 4a 0 to solve a (0,4).

Req f(x).

ax²+2x+a

The solution set for ax +2x+a 0 is r, i.e., the image of f(x) is always above the x-axis (with no intersection with the x-axis).

When a=0 and a0, the image of f(x) is not always above the x-axis so a 0, and the discriminant = 2 2-4*a*a=4(1+a)(1-a) 0

i.e. a 0 and a -1 or a 1

So a 1

Let f(x) = ax +2x+a

The solution set for ax +2x+a 0 is r, i.e., the image of f(x) is always above the x-axis (with no intersection with the x-axis).

When a=0 and a0, the image of f(x) is not always above the x-axis so a 0, and the discriminant = 2 2-4*a*a=4(1+a)(1-a) 0

i.e. a 0 and a -1 or a 1

So a 1

ax²+2x+a

a(x²+2x/a+1/a²) a-1/a=a(x+1/a)²+a-1/a

The solution set of the original inequality is r, and must be a>0 and a-1 a>0 to obtain a>1

The solution set of x -ax+1 0 is r, which means that there is x -ax+1>0 regardless of the value of x

Since f(x)=x-ax+1 opens upwards, if f(x)>0 at the lowest point x=a2, there is x-ax+1 evergreen at 0

f(a 2) = a 4-a * a 2 + 1 = 1 - a 4>0, i.e.: -2

It is known that the solution set of the inequality x -ax+1 0 is r

So there is: a 2-4<0 gets: -2

The solution set of the inequality x -ax+1 0 is r

Therefore, the discriminant "0

Discriminant = a 2-4<0

2

The solution set of the implicit ax2+4ax+8>0 is r, so there must be: a 0 0

i.e.: a 04a) 4a 8 0

16a²-32a<0

a²-2a<0

a（a-2）<0

The solution is 0, and the range of values of the real number a mentioned above is 0

Solution: When a=0, the inequality of the banquet beats is 8 0, which is obviously true;

When a≠0, a 0 = 16a2-32a 0 is required, and 0 a 2 is solved

In summary, the value range of the real number a is [0,2) Therefore, c

Draw an image solution. Move the item Chi Lao He.

ax3>=x 2-4x-3 Ax3 and x 2-4x-3 images are made, respectively.

When a>0 is found, the code pie contains a collision when x belongs to -2, and 1 cannot be established constantly.

Solution: -2<=a<=-5