The correct option is D (−2,0)∪(2,∞)
1−f(x)−f(x)2>f(1−5x)
Or
f(f(x))>f(1−5x) ...(i)
Now
f′(x)=−1−3x2
Hence
f′(x)<0 for all x. ...(ii)
Hence
f(f(x)>f(1−5x)
But f(x) is a decreasing function (from ii).
Hence
f(x)<1−5x
Or
1−x−x3<1−5x.
Or
4x−x3<0
Or
x3−4x>0
Or
x(x2−4)>0
Or
x>0 and x2−4>0 or
x<0 and x2−4<0
Hence if x>0 x2−4>0 implies xϵ(2,∞)
And
x<0 and x2−4<0 implies (−2,0)
Hence
xϵ(−2,0)∪(2,∞).