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According to a = {XL- 1 ≤ x ≤ a, a >- 1 and a ∈ r},

We can know that a≥- 1.

Because y=2x- 1 is a monotone function.

therefore

B={yl y=2x- 1,x∈A} = {y| -3 ≤y≤2a- 1} = [-3,2a- 1]

z= x? Is a parabola with an axis of symmetry x=0 and an upward opening.

The discussion about a is as follows:

(1) When a≤0,

For z, the maximum value is obtained when x =-1, and the minimum value is obtained when x=a, where c = {ZL z = the square of x, x ∈ a} = [a? , 1]

At this time, if c is contained in b,

Then 2a- 1≤ 1 and-1≤a≤0 must be satisfied.

Comprehensive-1≤a≤0

(2) When 0≤a≤ 1

For z, x=- 1 takes the maximum value, and x=0 takes the minimum value, and c = {ZL z = the square of x, x ∈ a} = [0, 1].

At this time, if c is contained in b,

Then 2a- 1≤ 1 and 0≤a≤ 1 must be satisfied.

There are 0≤a≤ 1.

(3) When a≥ 1

For z, x=a takes the maximum value, x=0 takes the minimum value, and c = {ZL z = the square of x, x ∈ a} = [0, a? ]

At this time, if c is contained in b,

2a- 1≤ a? And a≥ 1.

2a- 1≤ a? => Answer? - 2a + 1 ≥0 = >(a- 1)? ≥0 holds.

Synthetically get a≥ 1.

Synthesize (1), (2) and (3) to obtain

When a ≥- 1, b always contains C.