Question

On the real line $\mathbb{R}$, we define two functions $f$ and $g$ as follows:
$f(x) = \min \\{ x - \left[ x \right],1 - x + \left[ x \right]\\} , \\\ g(x) = \max \\{ x - \left[ x \right],1 - x + \left[ x \right]\\} \\\$
where $\left[ x \right]$ denotes the largest integer not exceeding $x$. The positive integer $n$ for which $\int\limits_0^n {(g(x) - f(x))dx = 100}$ is?
A) $100$
B) $198$
C) $200$
D) $202$

Answer 1

Since $f(x),g(x)$ represent the minimum and maximum of $\\{ x - \left[ x \right],1 - x + \left[ x \right]\\}$, we can use properties of greatest integer function. Then by splitting the integral accordingly we can simplify the expression. Finally by integration we can find the value of $n$.

Useful formula:
For every real number $x$, $\left[ x \right]$ is the greatest integer less than or equal to $x$. So, $0 \leqslant x - \left[ x \right] < 1$.
Integral can be split, that is if $a < b < c,\int\limits_a^c {f(x)dx = \int\limits_a^b {f(x)dx + \int\limits_b^c {f(x)dx} } }$
Also, $\int\limits_0^n {(g(x) - f(x))dx = n\int\limits_0^1 {(g(x) - f(x))dx} }$

Complete step by step solution:
Given, $f(x) = \min \\{ x - \left[ x \right],1 - x + \left[ x \right]\\} ,g(x) = \max \\{ x - \left[ x \right],1 - x + \left[ x \right]\\}$
Let $x - \left[ x \right] = f$.
So, we have, $f(x) = \min \\{ f,1 - f\\} ,g(x) = \max \\{ f,1 - f\\}$
By the definition of greatest integer function that is $\left[ x \right]$,
We have, $0 \leqslant x - \left[ x \right] < 1 \Rightarrow 0 \leqslant f < 1$
So there are two cases. $0 < f < 0.5$ and $0.5 < f < 1$.
If $0 < f < 0.5$, then $f(x) = \min \\{ f,1 - f\\} = f \\\ g(x) = \max \\{ f,1 - f\\} = 1 - f \\\$
and if $0.5 < f < 1$, then $f(x) = \min \\{ f,1 - f\\} = 1 - f \\\ g(x) = \max \\{ f,1 - f\\} = f \\\$
We are asked to find $n$ such that$\int\limits_0^n {(g(x) - f(x))dx = 100}$
Using the results in integration we have,
$\int\limits_0^n {(g(x) - f(x))dx = n\int\limits_0^1 {(g(x) - f(x))dx} }$
$\int\limits_0^1 {dx} = \int\limits_0^{0.5} {dx + \int\limits_{0.5}^1 {dx} }$
So we have,
$\int\limits_0^n {(g(x) - f(x))dx = n(\int\limits_0^{0.5} {(g(x) - f(x))dx} } + \int\limits_{0.5}^1 {(g(x) - f(x))dx} )$
$\int\limits_0^n {(g(x) - f(x))dx = n\int\limits_0^{0.5} {(1 - 2x)dx} } + n\int\limits_{0.5}^1 {(2x - 1)dx}$
Integrating the above equation we get,
$\int\limits_0^n {(g(x) - f(x))dx = n[x - \dfrac{{2{x^2}}}{2}} ]_0^{0.5} + n[\dfrac{{2{x^2}}}{2} - x]_{0.5}^1$
Simplifying we have,
$\int\limits_0^n {(g(x) - f(x))dx = n[x - {x^2}} ]_0^{0.5} + n[{x^2} - x]_{0.5}^1$
Applying the limits,
$\int\limits_0^n {(g(x) - f(x))dx = n[0.5 - 0.25 - (0 - 0)} ] + n[1 - 1 - (0.25 - 0.5)]$
$\Rightarrow \int\limits_0^n {(g(x) - f(x))dx = n[0.5 - 0.25} ] + n[ - (0.25 - 0.5)]$
Simplifying the above equation we have,
$\Rightarrow \int\limits_0^n {(g(x) - f(x))dx = 0.25n} + 0.25n = 0.5n$
We are asked to find $n$ such that$\int\limits_0^n {(g(x) - f(x))dx = 100}$
So by above equation we have,
$0.5n = 100$
Dividing both sides we get,
$\Rightarrow n = \dfrac{{100}}{{0.5}} = 200$
$\therefore n = 200$

Therefore the answer is option C.

Additional information:
Greatest integer function is also called the floor function. There is also another function called ceiling function, which is defined as the lowest integer greater than or equal to the given number.

Note: We should be careful in splitting the integral. Upper limit of the first integral must be equal to the lower limit of the second integral. Otherwise, some portions will be missed from integration or extra portions will be added.