If the inequality 3cos3x 2cos2x 1 k holds for any x R, then the minimum value of the real number k

Let y=3cos3x-2cos2x+1

Because 3cos3x-2cos2x+1<=k is true for any x to r.

So for the maximum value of y, there is max(y)<=k, so.

The minimum value of k is max(y).

Now let's find the maximum value of y.

First of all, there is. cos2k=cos^2k-sin^2k=2cos^2k-1cos3k=cos(2k+k)=cos2kcosk-sin2ksink

2cos^2k-1)cosk-2cosk(1-cos^2k)4cos^3k-3cosk

y=4cos 3k-4cos 2k-3cosk+3 t=cosk ,-1<=t<=1

Now the maximum value of y=4t 3-4t 2-3t+3(-1<=t<=1) is required.

y'=12t^2-8t-3

y'< 0, (2-13) 60, t<(2-13) 6 or t>(2+13) 6,y increments.

So max(y)=max(y((2- 13) 6),y(1)) substitute t=(2- 13) 6 into y to y=(46+13 13) 27t=1,y=0

So max(y) = (46 + 13 13) 27 i.e. the minimum value of k is (46 + 13 13) 27

Solution: y=3(cosx) 3-2cos2x+1=3(cosx) 3-4(cosx) 2+3The problem can be reduced to finding the maximum value of the function y=3x3-4x2+3 on [-1,1].

y'=9x^2-8x=9x[x-(8/9)]===>y'(0)=y'(8/9)=0.===>ymax=max=y(0)=3.===>3(cosx)^3-2cos2x+1≤3===>k≥3.

=>kmin=3.

We need to find the range of values of k where the inequality group has only two integer bad solutions.

First, let's analyze the first inequality: 2x - k > 10. This inequality represents a straight line with a slope of positive 2 and an intercept of -10. The solution set of inequalities is all the points above the line.

Next, we analyze the second inequality: 3x - 2 <=0. This inequality represents a straight line with a slope of positive 3 and an intercept of 2 3. The solution set of inequalities is all the points below the line.

Now let's look at the intersection of two straight lines. Let the point where two straight lines intersect with the jujube family is (x0, y0). Then there is the following system of equations:

2x0 - k = 10

3x0 - 2 = 0

Solve this system of equations and get x0 = 4 and k = 2.

Thus, when k = 2, two straight lines intersect at integer points (4, 10). In this case, the inequality group has only two integer solutions.

In order for a group of inequalities to have only two integer solutions, we need to meet the following two conditions:

1. k = 2

2.Straight lines 2x - k = 10 and straight lines 3x - 2 = 0 intersect at x = 4.

Therefore, the range of k is k = 2.

When x=0, 0 0 is constant, and k r is at this time;

When 0cos( x 2) > 0

f'(x)=cos(πx/2)[πx/2-tan(πx/2)]/x^2)

and when 0< x 2< 2, tan( x 2) > x 2, x 2>0

f'(x) < 0, i.e., f(x) decreases monotonically on (0,1);

f(x)>f(1)=1

To make k sin( x 2) x constant true, then k f(1)=1 when x = 1, sin( x 2) kx =>k 1 In summary, to make the inequality sin( x 2) kx hold when 0 x 1, the range of the real number k is (- 1).

When 0 x 1, sin x 2 is the increasing function, the highest point is 1, y=kx is a straight line through the origin, draw the graph, when k>1, it is not satisfied.

3x-k is less than or equal to 0

Solution x k 3

The known positive integer solution is 1,2,3

Then k3 must be greater than or equal to 3

However, if it is greater than or equal to 4, 4 must be included, so it cannot be greater than or equal to 4, but only less than 4

i.e. 3 k 3<4

Solution 9 k<12

Hope it helps you o(o

Because x takes 1 2 3 4 and x k 3 can know 4 k 3<5, the solution gets: 12 k<15, here pay attention to the difference between k and x, k=15, x is not equal to 5, it is recommended that you use the number line to represent the solution set and analyze it carefully, inequality is inseparable from the number line.

x-k<0 (the so-called solution refers to the value of x).

Solution: k > 3x (the range of k is required to be 3x maximum, and it is obvious that the maximum value of 3x is when x takes 3).

k>9

3x-k<=0

x<=k/3

Because the solution of a positive integer with inequality 3x-k less than or equal to 0 is 1,2, 6=

Answer: f (x) = (2) cos( x 2) -k = ( 2) [cos( x 2)-2k ].

Let f (x)=0 have cos( x 2)=2k 0 x 1 0 x 2 2 have 0 cos x 2 know the cover 1 and therefore must: 0 2k 1 0 k 2 easy to know: f(x) has a minimum at x=2k sink- 2k by:

sink-2k 0 has: sink 2k and 0 k 2 0 sink 1

Therefore 2k 1 - root number 2 )2 k (root number 2) 2 synthesis has: 0 k (root number 2) 2

It's been a long time since I've calculated this kind of math problem, so give it a try.

When 0 x 1, f(x)=sin( x 2) is the multiplication, the minimum value of f(x) min(f(x))=f(0)=sin0=0, and the maximum value of f(x) max(f(x))=f(1)=sin(2)=1;

Since sin( x 2) kx at the same time, kx min(f(x))=0;

Since x [0,1], k 0, i.e. the maximum value of k is 0.

I didn't expect to use derivatives to understand this problem. Please refer to it.