Explanation

(b)
Given, $p(x) = x^{4} + ax^{3} + bx^{2} + cx + d$
$\Rightarrow \ \ \ p^{'}(x) = 4x^{3} + 3ax^{2} + 2bx + c$
$\therefore\ \ x = 0\ is$ a solution for p’(x)=0,
$\Rightarrow \ \ \ c = 0$
$\therefore\ \ p(x) = x^{4} + ax^{3} + bx^{2} + d$ …(i)
$\ \ \$Also , we have p$( - 1$)<p(1)
$\Rightarrow \ 1 - a + b + d < 1 + a + b + d$
$\Rightarrow \ \ a > 0$
$\therefore\ p^{'}(x) = 0,only$ when x=0 $and\ p(x)is\ differentiable\ in\ ( - 1,1),\ we\ should\ have\ the\ maximum\ and\ minimum$
$\ at\ the$
$\ points\ x = - 1,0$ and 1 only
Also ,we have $p( - 1) < p(1)$
$\therefore Maximum\ of\ p(x) = \max\left\{ p(0),p(1) \right\}$
And minimum of P(x)=Min {P(-1), P(0)}
In the interval [0,1]
$\ \ p^{'}(x) = 4x^{3} + 3ax^{2} + 2bx$
$= x\left( 4x^{2} + 3ax + 2b \right)$
$\therefore p^{'}(x)has\ only\ one\ rootx = 0,then\ 4x^{2} + 3ax + 2b = 0\ has$
No real roots.
$\therefore\ \ \ \ (3a)^{2} - 32b < 0$
$\Rightarrow \ \frac{3a^{2}}{32} < b$
$\therefore\ \ \ b > 0$
Thus , we have a>0 and b>0
$\therefore\ \ p^{'}(x) = 4x^{3} + 4ax^{2} + 2bx > 0,\forall x \in (0,1)$
Hence ,$p(x) = p(1)$
Similarly,$p(x)$ is decreasing in [-1,0].
Therefore, Minimum $p(x)$does not occur at $x = - 1$.