Question

Let $f:[0,\infty )\to [0,\infty )$ be defined as$f\left(x\right)=\underset{0}{\overset{x}{\int }}\left[y\right]dy$ where $\left[x\right]$ is the greatest integer less than or equal to $x$. Which of the following is true

• A. $f$ is continuous at every point in $[0,\infty )$ and differentiable except at the integer points
• B. $f$ is both continuous and differentiable except at the integer points in $[0,\infty )$
• C. $f$ is continuous everywhere except at the integer points in $[0,\infty )$
• D. $f$ is differentiable at every point in $[0,\infty )$

Answer 1

The correct option is A $f$ is continuous at every point in $[0,\infty )$ and differentiable except at the integer points

Let $\left[x\right]=n$
$f\left(x\right)=\underset{0}{\overset{x}{\int }}\left[y\right]dy=\underset{0}{\overset{1}{\int }}\left[y\right]dy+\underset{1}{\overset{2}{\int }}\left[y\right]dy+\cdots +{\int }_{n-1}^n\left[y\right]dy+{\int }_n^x\left[y\right]dy$
$=0+1+2+\cdots +(n-1)+n(x-n)$
$=\frac{n(n-1)}{2}+n(x-n)$
$f\left(x\right)=\left[x\right]\left\{x\right\}+\frac{\left[x\right]\left(\right[x]-1)}{2}$
$f\left(x\right)$ is continuous at $x=k,(k\in I)$
$f\left(k^{-}\right)=k-1+\frac{(k-1)(k-2)}{2}=\frac{k^2-k}{2}$
$f\left(k^{+}\right)=k\times 0+\frac{k(k-1)}{2}=\frac{k^2-k}{2}$
$\because f\left(k^{-}\right)=f\left(k^{+}\right)=f\left(k\right)=\frac{k^2-k}{2}$
$LHD=f^{\prime }\left(k^{-}\right)=k-1$
$RHD=f^{\prime }\left(k^{+}\right)=k$
Not differentiable at $x=k$ where $k\in I$