Question

If Given $f\left(x\right)=\left\{x\right\}+\{x+1\}+\{x+2\}.....\{x+99\}$, then the value of $\left[f\right(\sqrt{2}\left)\right]$ is, where $\{.\}$ denotes fractional part function & $[.]$ denotes the greatest integer fuction

• A. $5050$
• B. $450$
• C. $141$
• D. $14$

Answer 1

The correct option is C $141$

Given $f\left(x\right)=\left\{x\right\}+\{x+1\}+\{x+2\}.....\{x+99\}$

$\Rightarrow f\left(x\right)=x-\left[x\right]+(x+1)-[x+1]+.....(x+99)-[x+99]$

$=x-\left[x\right]+x-\left[x\right]+.......+x-\left[x\right]$

$=100(x-[x\left]\right)=100\left\{x\right\}$

$\therefore \left[f\right(\sqrt{2}\left)\right]=\left[100\sqrt{2}\right]=141$