Question

If $g\left(x\right)=\left(x^2+2x+3\right)\left[f\left(x\right)\right]$, $f\left(0\right)=5$ and $\underset{x\to 0}{\lim }\frac{f\left(x\right)-5}{x}=4$, then $g\text{'}\left(0\right)=$

• A. $22$
• B. $18$
• C. $20$
• D. $25$

Answer 1

The correct option is A

$22$

Explanation for the correct option

Given, $g\left(x\right)=\left(x^2+2x+3\right)\left[f\left(x\right)\right]$
Thus, $f\left(x\right)=\frac{g\left(x\right)}{\left(x^2+2x+3\right)}$

Also given, $f\left(0\right)=5$

And, $\underset{x\to 0}{\lim }\frac{f\left(x\right)-5}{x}=4$
$\begin{array}{cc}\Rightarrow & \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{g\left(x\right)}{x^2+2x+3}-5}}{x}=4\\ \Rightarrow & \underset{x\to 0}{\lim }\frac{{\displaystyle g\left(x\right)-5\left(x^2+2x+3\right)}}{x\left(x^2+2x+3\right)}=4\\ \Rightarrow & \underset{x\to 0}{\lim }\frac{{\displaystyle g\left(x\right)-5x^2-10x-3}}{x^3+2x^2+3x}=4\end{array}$

Applying L'Hospital's rule i.e.$\underset{x\to 0}{\lim }\left(\frac{h\left(x\right)}{k\left(x\right)}\right)=\underset{x\to 0}{\lim }\left(\frac{h\text{'}\left(x\right)}{k\text{'}\left(x\right)}\right)$

$\begin{array}{cc}\Rightarrow & \underset{x\to 0}{\lim }\frac{{\displaystyle g\text{'}\left(x\right)-10x-10}}{3x^2+4x+3}=4\\ \Rightarrow & \frac{g\text{'}\left(0\right)-10}{3}=4\\ \Rightarrow & g\text{'}\left(0\right)=22\end{array}$

Therefore, $g\text{'}\left(x\right)=22$.

Hence, option(A) is correct.