Given $P(x)=x^4+a x^3+b x^2+c x+d$ such that $x=0$ is the only real root of $P^{\prime}(x)=0$. If $P(-1) <…
Given $P(x)=x^4+a x^3+b x^2+c x+d$ such that $x=0$ is the only real root of $P^{\prime}(x)=0$. If $P(-1) < P(1)$, then in the interval $[-1,1]$
$P(-1)$ is the minimum and $P(1)$ is the maximum of $P$
$P(-1)$ is not minimum but $P(1)$ is the maximum of $P$
$P(-1)$ is the minimum and $P(1)$ is not the maximum of $P$
neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of $P$
Solution
$
\begin{aligned}
& P(x)=x^4+a x^3+b x^2+c x+d \\
& P^{\prime}(x)=4 x^3+3 a x^2+2 b x+c \\
& \because x=0 \text { is a solution for } P^{\prime}(x)=0, \Rightarrow c=0 \\
& \therefore P(x)=x^4+a x^3+b x^2+d
\end{aligned}
$
Also, we have $P(-1) < P(1)$
$
\Rightarrow 1-a+b+d < 1+a+b+d \Rightarrow a>0
$
$\because P^{\prime}(x)=0$, 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 $P(-1) < P(1)$
$\therefore$ Max. of $P(x)=\operatorname{Max} .\{P(0), P(1)\}$ \& Min. of $P(x)=\operatorname{Min} .\{P(-1), P(0)\}$
In the interval $[0,1]$,
$
P^{\prime}(x)=4 x^3+3 a x^2+2 b x=x\left(4 x^2+3 a x+2 b\right)
$
$\because P^{\prime}(x)$ has only one root $x=0,4 x^2+3 a x+2 b=0$ has no real roots.
$
\begin{aligned}
& \therefore(3 a)^2-32 b < 0 \Rightarrow \frac{3 a^2}{32} < b \\
& \therefore b>0
\end{aligned}
$
Thus, we have $a>0$ and $b>0$
$
\therefore \mathrm{P}^{\prime}(\mathrm{x})=4 \mathrm{x}^3+3 a \mathrm{x}^2+2 \mathrm{bx}>0, \forall \mathrm{x} \in(0,1)
$
Hence $P(x)$ is increasing in $[0,1]$
$\therefore$ Max. of $P(x)=P(1)$
Similarly, $P(x)$ is decreasing in $[-1,0]$
Therefore Min. $P(x)$ does not occur at $x=-1$