Application of Derivatives

Given,$P(x)\;=x^4+a x^3+b x^2+c x+d$$\Rightarrow \;\; P'(x)\;=4 x^3+3 a x^2+2 b x+c$$\because x=0$ is the only real root of $P'(x)$ means the cubic equation.$4 x^3+3 a x^2+2 b x+c=0$$\therefore x=0$ must be a triple root, i.e. all other roots are complex or multiplicity is 3 .Hence, $P'(x)=4 x^3$ i.e. $3a=0 \Rightarrow a=0,\; 2b=0 \Rightarrow b=0$ and $c=0$So, $P(x)=x^4+d$Now, $P(-1)=(-1)^4+d=1+d$$P(l)=(l)^4+d=1+d$Thus, $P(-1)=P(1)$ which contradicting the given condition P(-1)So, our assumption that 0 is a triple root must be false.So, $P'(x)=0$ has exactly one real root and two complex conjugate roots.It means the sign of $P'(x)=0$ does not change around $x=0$ (because there is only one real root)Hence, $P'(x)< 0$ on $(-1,0)$ and $P'(x)> 0$ on $(0,1)$i.e. on the interval $[-1,1]$, the function has a minimum at $x 0$.and is strictly decreasing from ( -1 ) to 0 and then strictly increasing from 0 to 1 .So, option (a) will be true.