Question

Given $P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$ such that $x=0$ is the only real root of $P^{\prime }\left(x\right)=0.$ If $P(-1)<P\left(1\right),$ then in the interval $[-1,1],$

• A. $P(-1)$ is the minimum and $P\left(1\right)$ is the maximum value of $P\left(x\right)$
• B. $P(-1)$ is not minimum but $P\left(1\right)$ is the maximum value of $P\left(x\right)$
• C. $P(-1)$ is the minimum and $P\left(1\right)$ is not the maximum value of $P\left(x\right)$
• D. neither $P(-1)$ is the minimum nor $P\left(1\right)$ is the maximum value of $P\left(x\right)$

Answer 1

The correct option is B $P(-1)$ is not minimum but $P\left(1\right)$ is the maximum value of $P\left(x\right)$



$P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$
$P^{\prime }\left(x\right)=4x^3+3a{x}^2+2bx+c$
$P^{\prime }\left(0\right)=0\Rightarrow c=0$

$P^{\prime }\left(x\right)=x(4x^2+3ax+2b)$
As $P\left(x\right)=0$ has no real roots except $x=0,$
discriminant of $4x^2+3ax+2b$ should be less than zero
i.e., ${\left(3a\right)}^2-4\cdot 4\cdot 2b<0$
Then $4x^2+3ax+2b>0$ for all $x\in R$

So, $P^{\prime }\left(x\right)<0$ if $x\in [-1,0)$ i.e., decreasing
and $P^{\prime }\left(x\right)>0$ if $x\in (0,1]$ i.e., increasing
$\therefore$ $x=0$ is the point of local minimum.

Given that $P(-1)<P\left(1\right)$
We conclude that $P\left(0\right)$ is minimum and $P\left(1\right)$ is maximum and $P(-1)$ is not minimum for $x\in [-1,1]$