Question

Let $f$ be a biquadratic function of $x$ given by $f\left(x\right)=A{x}^4+B{x}^3+C{x}^2+Dx+E$ where $A,B,C,D,E\in R$ and $A\ne 0.$ If $\underset{x\to 0}{lim}{\left(\frac{f(-x)}{2x^3}\right)}^{1/x}=e^{-3},$ then

• A. $A+4B=0$
• B. $A-3B=0$
• C. $f\left(1\right)=8$
• D. $f^{\prime }\left(1\right)=-30$

Answer 1

Answer: (B), (D)

$f^{\prime }\left(1\right)=-30$

$\underset{x\to 0}{lim}{\left(\frac{f(-x)}{2x^3}\right)}^{1/x}=e^{-3}\Rightarrow \underset{x\to 0}{lim}{\left(\frac{A{x}^4-B{x}^3+C{x}^2-Dx+E}{2x^3}\right)}^{1/x}=e^{-3}\Rightarrow \underset{x\to 0}{lim}{\left(\frac{x^3(Ax-B+\frac{C}{x}-\frac{D}{x^2}+\frac{E}{x^3})}{2x^3}\right)}^{1/x}=e^{-3}$
For the existence of above limit,
$C=D=E=0$
Also, $\frac{-B}{2}=1\Rightarrow B=-2$
$\therefore \underset{x\to 0}{lim}{\left(\frac{Ax+2}{2}\right)}^{1/x}\left(1^{\infty }\text{form}\right)=e^{-3}\Rightarrow \underset{x\to 0}{lim}(\frac{Ax+2}{2}-1)\frac{1}{x}=-3\Rightarrow \frac{A}{2}=-3\Rightarrow A=-6$

So, $f\left(x\right)=-6x^4-2x^3$
$\Rightarrow f\left(1\right)=-6-2=-8$
$f^{\prime }\left(x\right)=-24x^3-6x^2$
$\Rightarrow f^{\prime }\left(1\right)=-24-6=-30$