Let fx  be a polynomial of degree 6 in x  in which the coefficient of  x6 is unity and it has extrema at  x=-1and   x=1. If  limx→fxx3=1, then 5f2  is equal to

Suppose that the function fx=x6+ax5+bx4+cx3+dx2+ex+fGiven limx→0fxx3=1it gives c=1,d=e=f=0hence, fx=x6+ax5+bx4+x3differentiate both sides f'x=6x5+5ax4+4bx3+3x2since x=1,x=−1 are stationery points 6+5a+4b+3=0⇒5a+4b=-9 and -6+5a-4b+3=0⇒5a-4b=3solving the above two equations 10a=-6⇒a=-35,b=-32hence, fx=x6−35x5−32x4+x35f2=564-3532-3216+8          =320-96-120+40           =144