Question

$f\left(x\right)$ has third continuous derivative and $\underset{x\to 0}{\lim }{\left[1+x+\frac{f\left(x\right)}{x}\right]}^{\frac{1}{x}}=e^3$, then $f\text{'}\text{'}\text{'}\left(x\right)$ is

• A. $x^2$
• B. $3x^2$
• C. $2x^2$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Step 1. Evaluate the given equation:

Given, $\underset{x\to 0}{\lim }{\left[1+x+\frac{f\left(x\right)}{x}\right]}^{\frac{1}{x}}=e^3$

$\Rightarrow$ $\underset{x\to 0}{\lim }{\left[1+x+\frac{f\left(x\right)}{x}\right]}^{\frac{1}{x}}$ $=\underset{x\to 0}{\lim }{\left(1+x\right)}^{\frac{3}{x}}$ $\left[\mathbf{\because }\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}{\left(1+x\right)}^{\frac{\mathbf{n}}{\mathbf{x}}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{e}}^{\mathbf{n}}\right]$

$\Rightarrow$ $\underset{x\to 0}{\lim }{\left[1+x+\frac{f\left(x\right)}{x}\right]}^{\frac{1}{x}}$ $=\underset{x\to 0}{\lim }{\left[{\left(1+x\right)}^3\right]}^{\frac{1}{x}}$

$\Rightarrow$ $\left[1+x+\frac{f\left(x\right)}{x}\right]={\left(1+x\right)}^3$

$=1+x^3+3x^2+3x$ $\left[\mathbf{\because }{\left(a+b\right)}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathbf{3}}\mathbf{+}{\mathit{b}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathit{a}}^{\mathbf{2}}\mathit{b}\mathbf{+}\mathbf{3}\mathit{a}{\mathit{b}}^{\mathbf{2}}\right]$

$\Rightarrow$ $\frac{f\left(x\right)}{x}=1+x^3+3x^2+3x-x-1$

$\Rightarrow$ $\frac{f\left(x\right)}{x}=x^3+3x^2+2x$

$\Rightarrow$ $f\left(x\right)=x\left(x^3+3x^2+2x\right)$

$=x^4+3x^3+2x^2$

Step 2. Find the value of $f\text{'}\left(x\right),f\text{'}\text{'}\left(x\right)$ and $f\text{'}\text{'}\text{'}\left(x\right)$

$f\text{'}\left(x\right)=4x^3+9x^2+4x$

$f\text{'}\text{'}\left(x\right)=12x^2+18x+4$

$f\text{'}\text{'}\text{'}\left(x\right)=\mathbf{24}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{18}$

Hence, Option ‘D’ is Correct.