Question

If $f\left(x\right)=\frac{\left[x\right]}{|x|}$, $x\ne 0$ where $[\cdot ]$ denotes the G.I.F then $f^{\prime }\left(1\right)$ is?

• A. $-1$
• B. $1$
• C. $\infty$
• D. Does not exist

Answer 1

Answer: (A)

$-1$

$\begin{array}{c}f\left(x\right)=\frac{\left[x\right]}{|x|},x\ne 0\\ ifx>0\\ \Rightarrow f\left(x\right)=\frac{x}{x}=1\\ ifx<0\\ \Rightarrow f\left(x\right)=\frac{x-1}{-x}=-1+\frac{1}{x}\\ \Rightarrow \therefore f^{\prime }\left(x\right)=\frac{1}{x^2},\\ f^{\prime }\left(1\right)=\frac{-1}{{\left(1\right)}^2}=-1\end{array}$