Consider the function, f(x)=x2m−1(a−x)2n The rolle's theorem is applicable on the above function in the interval [0,a] f′(x)=(2m−1)x2m−2(a−x)−2n(a−x)2n−1x2m−1 When x=c then m>1,n>1 and f′(c)=0 so, (2m−1)c2m−2‌‌=2nc2m−1(a−c)2n−1 ‌