The correct option is B π4<x<3π8
Here, f(x)=sin4x+cos4x
f′(x)=4sin3x⋅cosx+4cos3x(−sinx)
=4sinxcosx(sin2x−cos2x)
=2(sin2x)(−cos2x)⇒f′(x)=−sin4x
Now, f′(x)≥0, if sin4x≤0
⇒(2n−1)π≤4x≤2nπ, n∈Z
⇒(2n−1)π4≤x≤nπ2
For n=1,
π4≤x≤π2
Here, (π4,3π8) is only subset of [π4,π2]
Hence, π4<x<3π8