Application of Derivatives

Given,
P(x)‌=x4+ax3+bx2+cx+d
⇒‌‌P′(x)‌=4x3+3ax2+2bx+c
∵x=0 is the only real root of P′(x) means the cubic equation.
4x3+3ax2+2bx+c=0
∴x=0 must be a triple root, i.e. all other roots are complex or multiplicity is 3 .
Hence, P′(x)=4x3 i.e. 3a=0⇒a=0,2b=0⇒b=0 and c=0
So, P(x)=x4+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 x0.
and is strictly decreasing from ( -1 ) to 0 and then strictly increasing from 0 to 1 .
So, option (a) will be true.