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)

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

$\underset{x\to 0}{lim}{\left(\frac{f(-x)}{2x^3}\right)}^{1/x}=e^{-3}$
$\Rightarrow \underset{x\to 0}{lim}{\left(\frac{A{x}^4-B{x}^3+C{x}^2-Dx+E}{2x^3}\right)}^{1/x}=e^{-3}$
For given limit to be finite,
$C=D=E=0$
Also, $\frac{-B}{2}=1\Rightarrow B=-2$
$\therefore \underset{x\to 0}{lim}{\left(\frac{A{x}^4+2x^3}{2x^3}\right)}^{1/x}=e^{-3}$
$\Rightarrow \underset{x\to 0}{lim}{(1+\frac{Ax}{2})}^{1/x}=e^{-3}$
$\therefore \underset{x\to 0}{lim}\frac{Ax}{2}\cdot \frac{1}{x}=-3$
$\Rightarrow \frac{A}{2}=-3\Rightarrow A=-6$

So, $f\left(x\right)=-6x^4-2x^3$