Ans : (B)

Given, $P(x)= x^4+ax^3+bx^2+cx+d$

$P’(x)=4x^3+3ax^2+2bx+c$

Since, $x=0$ is a solution for $P’(x)=0$,

$C=0$

So, $P(x)= x^4+ax^3+bx^2+d$

Also we have $P(-1)< P(1)$

$1-a+b+d<1+a+b+d$

$A>0$

Since,$ P’(x)=0$, only when $x=0\: V \: and\: P(x)$ is differentiable in

$(-1,1)$, we should have the maximum and minimum at the points

$x= -1,\: 0\: and\: 1$ only

Also, we have $ P(-1)< P(1)$

So,Maximum of $P(x)= Max\{ P(0),P(1)\}$ and

Minimum of $P(x)=Min \{P(-1),P(0)\}$

In the interval $[0,1]$

$P’(x)= 4x^3+3ax^2+2bx=x(4x^2+3ax+2b)$

Since, $P’(x)$ has only one root $x=0$, then $4x^2+3ax+2b=0$ has no real roots.

So, $(3a)^2 – 32b<0$

$ 3a^2/32 > b $

So, $b>0$

Thus, we have $a>0$ and $b>0$

So, $P’(x)= 4x^3+3ax^2+2bx>0, \: for\: x \in (0,1)$

Hence, $P(x)$ is increasing in $[0,1]$ and $P(x)$ is decreasing in $[-1,0]$

Therefore, Maximum of $P(x)=P(1)$ and Minimum $P(x)$ does not occur at

$x=-1$ respectively