2sin 2x 6 1 k 0 has a real root in the interval 0, 2, and the value range of k is found

If you have studied derivatives, you can directly use the reciprocal to find the monotonic interval, and then determine the range of k from the values of each extreme point. Otherwise, you can confirm the monotonic interval of -2sin(2x-6) by graphing:

Monotonically decreasing on [0, 3] and increasing monotonically on [3, 2], so that f(x)=-2sin(2x- 6)+1-k, since k is constant, it is clear that his monotonic interval is the same as -2sin(2x- 6).

Therefore f(x) has a minimum value at [0, 2] f( 3) = -1-k

It is also known that there is f(0)=2-k at the endpoint of the defined domain interval, f( 2)==-k, so f(x) has a maximum value of 2-k

If f(x)=0 has a real root in [0, 2], i.e., (-1-k)(2-k)<=0

The solution is -1<=k<=2

Solution: because: x [0, 3].

So: 2x- 6 [- 6, 2].

So: -2sin(2x- 6) [1,2] and the original equation can be reduced to: -2sin(2x- 6)=k-1 So, to make the original equation have a real solution, then: k-1 [-1,2]so: k [0,1].

Solution] 2sin(2x 6) 1 k=0=>k== 2sin(2x 6) 1

0≤x≤π/2

/6≤2x－π/6≤5π/6

1/2≤sin(2x－π/6)≤1

0≤－2sin(2x－π/6)＋1≤3

0 k 3 is the range of k in the range of [0,3].

Let f(x)=7x 2-(k+13)x+k 2-k-2 because 7x 2-(k+13)x+k 2-k-2=0 are divided into (0,1) and (1,2) so f(0)=k 2-k-2 0,f(1)=k 2-2k-8 0,f(2)=k 2-3k 0 so k -1 or k 2,-2 k 4,k 0 or k coarse shirt 3 take the intersection to get -2 k -1 or cavity 3 k 4 i.e. k....

The original proposition is equivalent to:

f(x)=2x 2-3x-2k and the intersection of the x-axis Ling Annihilation is between [-1,1], which is only required.

f(3/4)0

f(1)>0

That is, it can be reburied and solved [-9 16, -1 2).

x∈【0,π/2】

2x+π/3∈[π3,4π/3]

When 2x+3 [ 3, 2) (2,2 3].

Two angles 2x+3 correspond to the same sinusoidal value.

In this case, sin(2x+ sedan rock song3) [closed zheng3 2,1)k=3sin(2x+ zaozi3) [3 3 2,3) that is, the equation has 2 solutions, then the value range of the real number k is [3 3 2,3).

Let f(x)=7x -(k+13)x+k -k-2, then f(0)>0f(1)<=0

f(2)>=0

It would be good to find the intersection after solving the system of equations.

x∈【0,π/2】

2x+π/3∈[π3,4π/3]

When 2x+3 [ 3, 2) (2,2 3].

Only then will the state line appear that two angles 2x+ 3 correspond to the same sinusoidal value.

In this case, sin(2x+ 3) [with resistance3 2,1)k=3sin(2x+ 3) [3 3 2,3), that is, the equation has 2 solutions, then the value range of the real number k is [3 3 2, 3).

Xi Xiaorui has yours in this question.

k 2-1≠0, 4(k+1) 2-4(k 2-1)=8k+8 0k -1, k -1. Bi Pai split.

k 2-1=0, envy.

k=1, there is a solution.

k=-1, no hand closure.

Hence k -1

Substitution x=0 is not enough.

Finally, k = 1 (rounding off).