Given $P(x)=x^4+a x^3+b x^2+c x+d$ such that x = 0 is the only real root of P'(x) = 0. If P(-1) < P(1), then in the interval [-1, 1] 

P(-1) is not minimum but P(1) the maximum of P

We have,

$P(x)=x^4+a x^3+b x^2+c x+d$

$\Rightarrow P'(x)=4 x^3+3 a x^2+2 b x+c$

It is given that x = 0 is a root of P'(x) = 0

∴  $P'(0)=0 \Rightarrow c=$

Putting c = 0, we get

$P'(x)=x\left(4 x^2+3 a x+2 b\right)$

Since x = 0 is the only real root of P'(x) = 0. Therefore,

$4 x^2+3 a x+2 b=0$has no real root

$\Rightarrow 9 a^2-32 b<0$

It is given that

$P(-1)<P(1)$

$\Rightarrow 1-a+b-c+d<1+a+b+c+d$

$\Rightarrow a>0$

But, 9a² - 32b < 0. Therefore, b > 0

∴  $P'(x)=x\left(4 x^2+3 a x+2 b\right)>0$ for all $x \in(0,1]$

$\Rightarrow P(x)$ is increasing in $(0,1]$

$\Rightarrow P(1)$ is the maximum value of $P(x)$

Also, $P'(x)=x\left(4 x^2+3 a x+2 b\right)<0$ for all $x \in[-1,0)$         [∵ $4 x^2+3 a x+2 b>0$ for all x]

$\Rightarrow P(x)$ is decreasing in $[-1,0)$

$\Rightarrow P(-1)$ is not the minimum value of P.