Question

Let $f\left(x\right)=\lambda +\mu \left|x\right|+\nu {\left|x\right|}^2$, where $\lambda$, $\mu$, $\nu$ are real constant. Then $f^{{}^{\prime }}\left(0\right)$ exists if

• A. $\mu =0$
• B. $\nu =0$
• C. $\lambda =0$
• D. $\mu =\nu$

Answer 1

The correct option is A $\mu =0$

$f\left(x\right)=\lambda +\mu \left|x\right|+\nu {\left|x\right|}^{2}f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\lambda -\mu x+\nu x^2& x<0\end{array}\\ \begin{array}{cc}\lambda +\mu x+\nu x^2& x\ge 0\end{array}\end{array}{f}^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-\mu +2\nu x& x<0\end{array}\\ \begin{array}{cc}\mu +2\nu x& x\ge 0\end{array}\end{array}$
Hence $f^{\prime }\left(0\right)$ exists when $-\mu +2\nu .0=\mu +2\nu .0\Rightarrow \mu =0$