Answer:

Solution:

The internal angles a, b and C∵△ABC of ∵△ satisfy sin2a+sin (a-b+c) = sin (c-a-b)+1/2.

∴sin2A+sin2B=-sin2C+ 1/2，

∴sin2A+sin2B+sin2C= 1/2，

∴ 2sina Cosa+2sin (b+c) cos (b-c) =1/2,2sina (cos (b-c)-cos (b+c)) =1/2, and change it to 2sina [-2sinbsin (.

∴sinAsinBsinC= 1/8.

Let the radius of the circumscribed circle be r, which can be obtained from the sine theorem: a/sinA=b/sinB=c/sinC=2R, and S= 1/2absinC, sinasinb sinc = (s/2r 2) = 65438+ 0/8 of the sine theorem, that is, r 2 =.

∫ area s satisfies 1≤S≤2,

∴ 4 ≤ (r 2) ≤ 8, that is, 2≤R≤2√2.

From sinAsinBsinC= 1/8, we can get 8≤abc≤ 16√2. Obviously, options c and d are not necessarily correct.

A.BC (b+c) > ABC ≥ 8, that is, BC (b+c) > 8, correct,

B.ab (a+b) > ABC ≥ 8, that is, ab (a+b) > 8, but ab (a+b) > 16 √ 2 is not necessarily correct.

So choose: a