We have P(x)=x4+ax3+bx2+cx+d ⇒P′(x)=4x3+3ax2+2bx+c But P′(0)=0⇒c=0 ∴P(x)=x4+ax3+bx2+d As given that P(−1)<P(a) ⇒1−a+b+d<1+a+b+d⇒a>0 Now P′(x)=4x3+3ax2+2bx=x(4x2+3ax+2b) As P′(x)=0, there is only one solution x=0, therefore 4x2+3ax+2b=0 should not have any real roots i.e. D<0 ⇒9a2−32b<0 ⇒b>
9a2
32
>0 Hence a,b>0⇒P′(x)=4x3+3ax2+2bx>0 ∀x>0 ∴P(x) is an increasing function on (0,1) ∴P(0)<P(a) Similarly we can prove P(x) is decreasing on (−1,0) ∴P(−1)>P(0) So we can conclude that MaxP(x)=P(1) and MinP(x)=P(0) ⇒P(−1) is not minimum but P(1) is the maximum of P.