Question

Let $P(x)$ be a biquadratic function of $x$ such that ${\mathrm{Lim}}_{x\to 0}{\left(\frac{P(-x)}{2x^3}\right)}^{\frac{1}{x}}=\frac{1}{e^3}$.
Which one of the following statement is correct for $P(x)$ ?

• A. $P(x)$ has minima but no maxima.
• B. $P(x)$ has maxima but no minima.
• C. $P(x)$ increases in $\left(-\infty ,\frac{1}{4}\right)$ and decreases in $\left(\frac{1}{4},\infty \right)$
• D. $P(x)$ decreases in $\left(-\infty ,\frac{-1}{4}\right)$ and increases in $\left(\frac{-1}{4},\infty \right)$.

Answer 1

Answer: (B)

Let $P(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$
Now, $\underset{x\to 0}{\mathrm{Lim}}\left(\frac{P(-x)}{2x^3}-1\right)\frac{1}{x}=-3$
$\therefore \underset{x\to 0}{\mathrm{Lim}}\left(\frac{P(-x)-2x^3}{x^4}\right)=-6\Rightarrow \underset{x\to 0}{\mathrm{Lim}}\left(\frac{a{x}^4-b{x}^3+c{x}^2-dx+e-2x^3}{x^4}\right)=-6$
So, $e=0,d=0,c=0$
$b+2=0\text{ and }a=-6\Rightarrow P(x)=-6x^4-2x^3$

$P(x)=-6x^4-2x^3=-2x^3(3x+1)$
$P^{\prime }(x)=-24x^3-6x^2=-6x^2(4x+1)$

So, $P(x)$ has maxima but no minima.