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 $P\left(x\right)=0$. 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 of P
• B. $P(-1)$ is not the minimum but $P\left(1\right)$ is the maximum of P
• C. $P(-1)$ is the minimum and $P\left(1\right)$ is not the maximum of P
• D. neither $P(-1)$ is the minimum nor $P\left(1\right)$ is the maximum of P

Answer 1

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

Given:$P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$

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

Since, $x=0$ is a solution for $P^{\prime }\left(x\right)=0$

$\Rightarrow c=0$

So, $P\left(x\right)=x^4+a{x}^3+b{x}^2+d$

Also we have $P(-1)<P\left(1\right)$

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

$\Rightarrow A>0$

Since $P^{\prime }\left(x\right)=0,$ only when $x=0$

and $P\left(x\right)$ is differentiable in $(-1,1)$, we should have the maximum and minimum at the points

$x=-1,0$ and $1$ only.

Also, we have $P(-1)<P\left(1\right)$

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

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

In the interval $[0,1]$

$P^{\prime }\left(x\right)=4x^3+3a{x}^2+2bx=x(4x^2+3ax+2b)$

Since $P^{\prime }\left(x\right)$ has only one root $x=0$, then $4x^2+3ax+2b=0$ has no real roots.

So,${\left(3a\right)}^2-32b<0$

$\Rightarrow \frac{3a^2}{32}>b$

So,$b>0$

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

So,$P^{\prime }\left(x\right)=4x^3+3a{x}^2+2bx>0,$ for $x\in (0,1)$

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

Therefore, Maximum of $P\left(x\right)=P\left(1\right)$ and Minimum $P\left(x\right)$ does not occur at $x=-1$ respectively.