Question

Suppose that $f\left(x\right)$ is a differentiable function such that $f^{\prime }\left(x\right)$ is continuous, $f^{\prime }\left(0\right)=1$ and $f^{\prime \prime }\left(0\right)$ does not exist. Let $g\left(x\right)=x{f}^{\prime }\left(x\right)$. Then,

• A. $g^{\prime }\left(0\right)$ does not exist
• B. $g^{\prime }\left(0\right)=0$
• C. $g^{\prime }\left(0\right)=1$
• D. $g^{\prime }\left(0\right)=2$

Answer 1

Answer: (C)

$g^{\prime }\left(0\right)=1$

Given, $f^{\prime }\left(0\right)=1$ and $f^{\prime \prime }\left(0\right)$ does not exist.

Also, given $g\left(x\right)=x{f}^{\prime }\left(x\right)$

$\therefore g^{\prime }\left(x\right)=x{f}^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)$

Put $x=0$, we get

$g^{\prime }\left(0\right)=0f^{\prime \prime }\left(0\right)+f^{\prime }\left(0\right)$

$=0+1=1$