Rolle’s theorem holds for the function f(x)=x3+bx2+cx, 1≤x≤2  at the point43 , then the value of 100c−500b  must be

Now,   f(1)=f(2) ⇒1+b+c=8+4b+2c⇒3b+c=−7                                   .....(i) Andf'(43)=0⇒3(43)2+2b(43)+c=0⇒16+8b+3c=0                                                                 .....(ii) From Eqs. (i) and (ii), we getb=−5,    c−8⇒  100c−500b=800+2500=3300