Question

Let $f\left(x\right)$ be a polynomial of degree $5$ such that $x=\pm 1$ are its critical points. If $\underset{x\to 0}{lim}(2+\frac{f\left(x\right)}{x^3})=4$, then which of the following is not true ?

• A. $f\left(1\right)-4f(-1)=4$.
• B. $x=1$ is a point of maxima and $x=-1$ is a point of minimum of $f$.
• C. $f$ is an odd function.
• D. $x=1$ is a point of minima and $x=-1$ is a point of maxima of $f$.

Answer 1

The correct option is D $x=1$ is a point of minima and $x=-1$ is a point of maxima of $f$.

Given $\underset{x\to 0}{lim}(2+\frac{f\left(x\right)}{x^3})=4$
$\Rightarrow \underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=2$

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}$, Limit exists and it is finite.

$\therefore f\left(x\right)=a{x}^5+b{x}^4+c{x}^3$
$\Rightarrow \underset{x\to 0}{lim}(a{x}^2+bx+c)=2$
$\Rightarrow c=2$

Also, $f^{\prime }\left(x\right)=5a{x}^4+4b{x}^3+6x^2$
$x=\pm 1$ are critical points.
$\Rightarrow f^{\prime }\left(1\right)=5a+4b+6=0$
and $f^{\prime }(-1)=5a-4b+6=0$
$\Rightarrow b=0,a=-\frac{6}{5}$

$\therefore f\left(x\right)=-\frac{6}{5}{x}^5+2x^3$
$\Rightarrow f$ is odd.
$f^{\prime }\left(x\right)=-6x^4+6x^2$
$f^{\prime \prime }\left(x\right)=-24x^3+12x$
$f^{\prime \prime }\left(1\right)<0$ and $f^{\prime \prime }(-1)>0$
So, at $x=-1$, there is local minima and at $x=1$, there is local maxima.

Also, $f\left(1\right)-4f(-1)=4$