Question

Let $P\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$ such that $x=0$ is the only real root of $P\text{'}\left(x\right)=0$. 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 minimum but $P\left(1\right)$ is the maximum of $P$
• C. $P(-1)$ is minimum and is not maximum of $P$
• D. Neither $P(-1)$ is the minimum and$P\left(1\right)$ nor is the maximum of$P$

Answer 1

The correct option is B

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

Explanation for the correct option.

Finding the minimum and maximum of $P$

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

Estimate the derivative of the given function.

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

As $P\prime \left(x\right)=0$ has only root $x=0\Rightarrow c=0$

So the determinant of the equation will be, $D<0$
$\Rightarrow P\left(x\right)=x(4x^2+3ax+2b)\mathbf{\text{[Taking}}\mathit{x}\mathbf{\text{as common]}}$
$\Rightarrow 4x^2+3ax+2b=0$ has non-real roots and $4x^2+3ax+2b>0\forall \in [-1,1].$
At $x=0,f\prime \left(x\right)$ changes sign from negative to positive. Hence, $x=0$ is the point of local minima.

There is no local maxima. Hence, we check at the endpoints.
As $P(-1)<P\left(1\right)\Rightarrow P\left(1\right)$ is the max. of $P\left(x\right)\text{in}[-1,1]$

Therefore, the correct answer is option (B).