Question

求一道数学题

麻烦带有过程

Answer 1

(I)
f(x)=4√3.sinx.cosx -4(cosx)^2 +m
f(π/6)=7
4√3.(1/2).(√3/2) -4(√3/2)^2 +m =7
3 - 3 +m=7
m=7
(II)
f(x)

=4√3.sinx.cosx -4(cosx)^2 +7
=2√3.sin2x -2(1+cos2x) +7
=(2√3.sin2x -2cos2x) +5
=4sin(2x-π/6) +5
x∈[0,π/4]
max f(x) = f(π/4) = 4sin(π/3) +5 = 2√3 +5
min f(x) = f(0) = 4sin(-π/6) +5 = -2 +5 = 3
c<f(x) < 2c+15
ie
max f(x) =2√3 +5 <2c+15
c> √3 -5 (1) and
c<3 = min f(x) (2)
ie
√3 -5 < c < 3