@budokai....u xpect ppl. 2 laugh at dat







newayz...here goes d proof:







f(0)=f(1)



implies f'(c)=0 ... 0<c<1.....rolle's



assume f(k) not equal to f(k+0.5 ) ...for k belonging to [0,0.5]



=> f'(m) not equal to 0 ...for any m btw. k and k+0.5



...since k itself varies from 0 to 0.5



=> f'(m) not equal to 0 ...for any m btw. 0 and 1



CONTRADICTION.







Hence..proved..



now is dat enlightening enuf..!!