Question

Let $f:(0,\infty )\to [1,\infty )$ be defined as $f\left(x\right)=\{\begin{array}{cc}\frac{1}{x},& 0<x<1\\ & \\ \frac{x}{\left[x\right]},& 1\le x<\infty \end{array}$ where $\left[x\right]$ denotes the greatest integer less than or equal to $x.$ Then which of the following statements is (are) CORRECT?

• A. $f\left(x\right)$ is neither injective nor surjective
• B. $f\left(x\right)$ is continuous at $x=1$ but not differentiable at $x=1$
• C. $f\left(x\right)$ has non-removable discontinuity of finite type at $x=2$ with jump equal to $\frac{1}{2}$
• D. $f^{\prime }\left(e\right)=\frac{1}{2}$

Answer 1

Answer: (B), (D)

$f^{\prime }\left(e\right)=\frac{1}{2}$

$f\left(x\right)=\{\begin{array}{c}\frac{1}{x},0<x<1\\ \\ \frac{x}{1},1\le x<2\\ \\ \frac{x}{2},2\le x<3\\ \dots \dots \\ \dots \dots \end{array}$

Clearly, $f\left(\frac{2}{3}\right)=\frac{3}{2}$ and $f\left(\frac{3}{2}\right)=\frac{3}{2}$
So, $f\left(x\right)$ is many-one function.
Also, range of $f\left(x\right)$ is $[1,\infty ),$ so
$f\left(x\right)$ is onto.

$f\left(1^{-}\right)=1=f\left(1^{+}\right)=f\left(1\right)$
But $f^{\prime }\left(1^{-}\right)=-1$ and $f^{\prime }\left(1^{+}\right)=1$
So, $f\left(x\right)$ is continuous but not differentiable at $x=1.$

$f\left(2^{+}\right)=1,f\left(2^{-}\right)=2$
$\therefore \underset{x\to 2}{lim}f\left(x\right)$ does not exist.
Also, jump of discontinuity $=|2-1|=1$

For $x\in (2,3),$
$f\left(x\right)=\frac{x}{2}[\because e\in (2,3\left)\right]$
$\Rightarrow f^{\prime }\left(e\right)=\frac{1}{2}$