Question

On the real line R, we define two functions f and g as follows:
$f\left(x\right)=min[x-[x],1-x+[x\left]\right]$,
$g\left(x\right)=max[x-[x],1-x+[x\left]\right]$,
where [x] denotes the largest integer not exceeding x.

The positive integer n for which ${\int }_0^n\left(g\right(x)-f(x\left)\right)dx=100$ is?

• A. $100$
• B. $193$
• C. $200$
• D. $202$

Answer 1

The correct option is C $200$

Given, $f\left(x\right)=min[x-[x],1-x+[x\left]\right]=min[f,1-f]$

and $g\left(x\right)=max[x-[x],1-x+[x\left]\right]=max[f,1-f]$

where $f$ is the fractional part of the function.

and $f$ is always $0\le f<1$.

$\therefore$ if $0<f<0.5$, then

$f\left(x\right)=f$
$g\left(x\right)=1-f$

and If $0.5<f<1$, then

$f\left(x\right)=1-f$
$g\left(x\right)=f$

${\int }_0^n\left(g\right(x)-f(x\left)\right)dx=$$n{\int }_0^1\left(g\right(x)-f(x\left)\right)dx$

$=n{\int }_0^{0.5}(1-2x)dx+$$n{\int }_{0.5}^1(2x-1)dx=n(0.5-0.25)+n(0.75-0.5)$

$0.5n=100⟹n=200$