f(x)=(a2=−3a+2)(cos2x/4−sin2x/4)+(a−1)α+sin1 ⇒f(x)=(a−1)(a−2)cos2x+(a−1)x+sin1 ⇒f′(x)=2−1(a−1)(a−2)sin2x+(a−1) ⇒f′(x)=(a−1)[1−2(a−2)sin2x]
If f(x) does not possess critical points, then f′(x)=0 for any x∈R. ⇒(a−1)[1−2(a−2)sin2x]=0 for any x∈R ⇒a=1 and 1−(2a−2)sin2x=0
must not here any solution in R. ⇒a=1 and sin2x=a−22 is not
solvable in R. ⇒a=1 and ∣∣a−22∣∣>1
[for a=2,f(x)⋅x+sin1∴f′(x)=1=0] ⇒a=1 and ∣a−2∣<2 ⇒a=1 and −2<a−2<2 ⇒a=1 and 0<a<4 ⇒a∈(0,1)∪(1,4).