Question

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

• A. $30$
• B. $18$
• C. $20$
• D. $22$

Answer 1

The correct option is D $22$

We know that,
$f^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}\Rightarrow f^{\prime }\left(0\right)=4$

$g\left(x\right)=(x^2+2x+3)f\left(x\right)$
Differentiate w.r.t $x$
$\Rightarrow g^{\prime }\left(x\right)=(2x+2)f\left(x\right)+(x^2+2x+3)f^{\prime }\left(x\right)\Rightarrow g^{\prime }\left(0\right)=2\times f\left(0\right)+3\times f^{\prime }\left(0\right)\Rightarrow g^{\prime }\left(0\right)=2\times 5+3\times 4=22$