The correct option is D None of these
h(x)=f(x)−a(f(x))2+a(f(x))3
or h′(x)=f′(x)−2af(x)f′(x)+3a(f(x))2f′(x)
=f′(x)[3a(f(x))2−2af(x)+1]
Now, h(x) increases as f(x) decreases for all real values of x if
3a(f(x))2−2af(x)+1<0 for all x∈R
So,
3a<0 and 4a2−12a≤0
a<0 and a∈[0,3]
So, no such a is possible