Question

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

• A. $22$
• B. $14$
• C. $18$
• D. $12$

Answer 1

The correct option is A

$22$

Explanation for the correct option:

Finding the value of $g\text{'}\left(0\right)$:

Given that,

$g\left(x\right)=\left(x^2+2x+3\right)f\left(x\right)...\left(1\right)$

$\begin{array}{rcl}f\text{'}\left(0\right)& =& \underset{x\to 0}{\lim }\frac{\left[f\left(x\right)-5\right]}{x-0}\left[\because f\text{'}\left(x\right)=\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}\right]\\ & =& 4\end{array}$

Now, differentiate the equation $\left(1\right)$ with respect to $x$.

$g\text{'}\left(x\right)=\left(x^2+2x+3\right)f\text{'}\left(x\right)+f\left(x\right)\left(2x+2\right)\mathbf{}\left[\because \frac{d\left(uv\right)}{dx}=vu\text{'}+uv\text{'}\right]$

Substitute $x=0$ in the differentiated equation.

$\begin{array}{rcl}g\text{'}\left(0\right)& =& \left({\left(0\right)}^2+2\left(0\right)+3\right)f\text{'}\left(0\right)+f\left(0\right)\left(2\left(0x\right)+2\right)\\ & =& 3f\text{'}\left(0\right)+2f\left(0\right)\\ & =& 3\times 4+2\times 5\left[\mathrm{Given}\right]\\ & =& 22\end{array}$

Hence, the correct option is A.