The minimum value of f(x)=sin⁴x +cos⁴x in [0,Π/2 ] is

The minimum value of f(x)=sin⁴x +cos⁴x in [0,Π/2 ] is
(a)1/2√2
(b)1/4
(c)-1/2
(d)1/2

solution : (d)