f(x)=(x2−1)2h(x); h(x)=a0+a1x+a2x2+a3x3
Now, f(1)=f(−1)=0
⇒f′(α)=0,α∈(−1,1)
[Rolle's Theorem]
Also, f′(1)=f′(−1)=0
⇒f′(x) has atleast 3 roots −1,α,1 with −1<α<1
f′′(x)=0 will have atleast 2 roots, say β,γ such that −1<β<α<γ<1
[Rolle's Theorem]
So, min(mf′′)=2
and we find mf′+mf′′=5 for f(x)=(x2−1)2h(x)