Question

Suppose that $f\left(x\right)$ is a differential function such that $f^{\prime }\left(x\right)$ is continuous, $f^{\prime }\left(0\right)=1$ and $f^{\prime }\left(0\right)$ 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

The correct option is C $g^{\prime }\left(0\right)=1$

GIven: $f\left(x\right)$ is differential function and $f^{{}^{\prime }}\left(x\right)$ is continous and $f^{{}^{\prime }}\left(0\right)=1$ and $f^{{}^{\prime }}\left(0\right)$ does not exit

To find: $g^{\prime }\left(0\right)$ if $g\left(x\right)=x{f}^{{}^{\prime }}\left(x\right)$

Sol: $g^{{}^{\prime }}\left(x\right)={lim}_{h\to 0}\frac{g(x+h)-g\left(x\right)}{h}$

$g^{{}^{\prime }}\left(0\right)={lim}_{h\to 0}\frac{g\left(h\right)-g\left(0\right)}{h}$

$g\left(0\right)=0⟹g^{{}^{\prime }}\left(0\right)={lim}_{h\to 0}\frac{g\left(h\right)}{h}$

=${lim}_{h\to 0}\frac{h.f^{{}^{\prime }}\left(x\right)}{h}={lim}_{h\to 0}{f}^{{}^{\prime }}\left(x\right)$

$⟹f^{{}^{\prime }}\left(0\right)=1$

hence, correct answer is $g^{{}^{\prime }}\left(0\right)=1$