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 $\left[x\right]$ denotes the largest integer not exceeding $x$. The positive integer $n$ for which
$\underset{0}{\overset{n}{\int }}\left(g\right(x)-f(x\left)\right)dx=100$

• A. $100$
• B. $198$
• C. $200$
• D. $202$

Answer 1

The correct option is C $200$

The graph of $f\left(x\right)$

$\underset{0}{\overset{n}{\int }}f\left(x\right)dx=n\underset{0}{\overset{1}{\int }}f\left(x\right)dx=n(\frac{1}{2}\times \frac{1}{2}\times 1)=\frac{n}{4}$

The graph $g\left(x\right)$


$\underset{0}{\overset{n}{\int }}g\left(x\right)dx=n\underset{0}{\overset{1}{\int }}g\left(x\right)dx=n(1-\frac{1}{4})=\frac{3n}{4}$

So
$\underset{0}{\overset{n}{\int }}\left(g\right(x)-f(x\left)\right)dx=100\Rightarrow \frac{3n}{4}-\frac{n}{4}=100\Rightarrow n=200$