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 $\underset{x\to 0}{lim}{\left(\frac{f(-x)}{2x^3}\right)}^{\frac{1}{x}}=e^{-3},$ 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)

Given:

$f\left(x\right)=A{x}^4+B{x}^3+C{x}^2+Dx+E$ and $A,B,C,D,E\in R,A\ne 0$.

$\underset{x\to 0}{lim}{\left(\frac{f(-x)}{2x^3}\right)}^{\frac{1}{x}}=e^{-3}$

$\Rightarrow \underset{x\to 0}{lim}{\left(\frac{A(-x{)}^4+B(-x{)}^3+C(-x{)}^2+D(-x)+E}{2(x{)}^3}\right)}^{\frac{1}{x}}=e^{-3}$
$\Rightarrow \underset{x\to 0}{lim}{\left(\frac{A{x}^4-B{x}^3+C{x}^2-Dx+E}{2x^3}\right)}^{\frac{1}{x}}=e^{-3}$

$\Rightarrow \underset{x\to 0}{lim}(\frac{Ax}{2}-\frac{B}{2}+\frac{C}{2x}-\frac{D}{2x^2}+\frac{E}{2x^3}{)}^{\frac{1}{x}}=e^{-3}$---(i)

Since, $\underset{x\to 0}{lim}(\frac{1}{x^n}{)}^{\frac{1}{x}}=0$

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

Option A,

$A+4B=-6+4(-2)=-6-8=-14\ne 0$

Option B,

$A-3B=-6-3(-2)=0$

Now, substitute the values in $f\left(x\right)$,

$f\left(x\right)=A{x}^4+B{x}^3+C{x}^2+Dx+E$

$\Rightarrow f\left(x\right)=-6x^4-2x^3+0x^2+0x+0$

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

$\Rightarrow f\left(1\right)=-6(1{)}^4-2(1{)}^3=-8$

$\Rightarrow f^{\prime }\left(x\right)=-24x^3-6x$

$\Rightarrow f^{\prime }\left(1\right)=-24(1{)}^3-6(1)=-30$---Option D

Hence, Option B and Option D are correct.