f(x)=(x2−1)2.p(x) where p(x)=a0+a1x+a2x2+a3x3 ∵f(x) has two repeated roots x=1 and x=−1 So f′(x) has atleast 3 roots x=1,x=−1 and x=c Where c∈(−1,1), so mf,=3 Between any two distinct roots of f′(x) there will be atleast one root of f"(x), so mf"=2 mf′+mf"=5