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 ${\int }_0^n\left(g\right(x)-f(x)dx=100$ is

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

Answer 1

The correct option is C $200$

I will denote f(x) as f and g(x) as g.

frac(x)= fractional part of x.

$frac\left(x\right)=x-\left[x\right]$

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

$⟹f\left(x\right)=min\left\{frac\right(x),1-frac(x\left)\right\}$

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

$⟹g\left(x\right)=max\left\{frac\right(x),1-frac(x\left)\right\}$

$\text{For}0\le frac\left(x\right)\le 0.5,f=frac\left(x\right);g=1-frac\left(x\right)$

$⟹g-f=1-2frac\left(x\right)$

Similarly,$\text{For}0.5\le frac\left(x\right)\le 1,g=frac\left(x\right);f=1-frac\left(x\right)$

$⟹g-f=2frac\left(x\right)-1$

Given that,

${\int }_0^{n}g\left(x\right)-f\left(x\right)dx=100$

Since, the above function is periodic, i.e. P(k)=P(k+1) where $P\left(K\right)={\int }_k^{k+1}g\left(x\right)-f\left(x\right)dx$

$⟹{\int }_0^1\left(g\right(x)-f(x)dx={\int }_1^{2}g(x)-f(x)dx={\int }_2^3(g\left(x\right)-f\left(x\right)dx={\int }_{n-1}^{n}g\left(x\right)-f\left(x\right)dx$

$⟹{\int }_0^n\left(g\right(x)-f(x)dx=n{\int }_0^{1}g(x)-f(x)dx$ $⟹n{\int }_0^{1}g\left(x\right)-f\left(x\right)dx=n\left({\int }_0^{0.5}g\right(x)-f(x)dx+{\int }_{0.5}^{1}g(x)-f(x\left)dx\right)=100$

$⟹n({\int }_0^{0.5}1-2frac(x)dx+{\int }_{0.5}^{1}2frac(x)-1dx)=100$

In the range from 0 to 1, frac(x)=x

$⟹n({\int }_0^{0.5}1-2xdx+{\int }_{0.5}^{1}2x-1dx)=100$

$⟹n({\int }_{0.5}^{1}2xdx-{\int }_0^{0.5}2xdx)=100$

$⟹n=200$