Question

求这题完整解题步骤

Answer 1

解：由题意，有
A+k=5/2
-A+k=-1/2
解，得A=3/2,k=1
周期T=2×(7π/12-π/12)=π
则ω=2π/T=2
∴f(x)=(3/2)sin(2x+φ)+1
当x=π/12时，f(π/12)=5/2
即(3/2)sin(π/6+φ)+1=5/2
∴sin(π/6+φ)=1
∴π/6+φ=π/2+2mπ (m∈Z)
而|φ|<π/2
∴此时，m=0,π/6+φ=π/2
∴φ=π/3
故f(x)的解析式为
f(x)=(3/2)sin(2x+π/3)+1
因此，选C项

Answer 2

Fmax = 5/2，Fmin = -1/2
振幅 A = (Fmax-Fmin)/2 = (5/2+1/2)/2 = 3/2
纵向位移 C = Fmax-A = 1
周期 T = 2*（7π/12-π/12）=π
角频率 ω = 2π/T = 2

初相位 φ =π/3

所以f(x) = Asin(ωx+φ) =

Answer 3

啊，好熟悉的题，瞬间让我回想起当年高三～这都是送分题！