Question

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

Answer 1

Hint: In this question, we have to evaluate the integral in a specific range.
The box function, more commonly known as the greatest integer function, returns the integer just below the value entered, denoted by \[\left[ x \right]\].
If the number is an integer use that integer.
If the number is not an integer use the smaller integer.

Complete step-by-step answer:
It is given that, On the real line \[\mathbb{R}\], we define two functions f and g as follows :
\[f\left( x \right) = \min \left\{ {x - \left[ x \right],1 - x + \left[ x \right]} \right\}\].
\[g\left( x \right) = \max \left\{ {x - \left[ x \right],1 - x + \left[ x \right]} \right\}\].
Where \[\left[ x \right]\] denotes the largest integer not exceeding \[x\].
We need to find out the positive integer $n$ for which
\[ \Rightarrow \int\limits_0^n {\left( {g\left( x \right) - f\left( x \right)} \right)dx = 100} \].
Let us denote,
\[ \Rightarrow f\left( x \right) = x\& g\left( x \right) = g\].
Also, \[m\left( x \right)\]=fractional part of \[x\].
\[ \Rightarrow m\left( x \right) = x - \left[ x \right]\].
Let us consider the term,
\[ \Rightarrow f\left( x \right) = \min \left\{ {x - \left[ x \right],1 - x + \left[ x \right]} \right\}\].
By using \[m\left( x \right) = x - \left[ x \right]\] we get,
\[ \Rightarrow f\left( x \right) = \min \left\{ {m\left( x \right),1 - m\left( x \right)} \right\}\]
Now for \[g\left( x \right)\],
\[ \Rightarrow g\left( x \right) = \max \left\{ {x - \left[ x \right],1 - x + \left[ x \right]} \right\}\]
By using \[m\left( x \right) = x - \left[ x \right]\] we get,
\[ \Rightarrow g\left( x \right) = \max \left\{ {m\left( x \right),1 - m\left( x \right)} \right\}\]
Where \[m\left( x \right)\] is always \[0 \leqslant m\left( x \right) < 1\]
For, \[0 < m\left( x \right) < 0.5\]
\[ \Rightarrow f = m\left( x \right),{\text{ }}g = 1 - m\left( x \right)\]
Thus,
\[ \Rightarrow g - f = 1 - m\left( x \right) - m\left( x \right) = 1 - 2m\left( x \right)\]
\[ \Rightarrow g - f = 1 - 2m\left( x \right)........(1)\]
Similarly for, \[0.5 < m\left( x \right) < 1\]
\[ \Rightarrow f = 1 - m\left( x \right),{\text{ }}g = m\left( x \right)\]
Thus,
\[ \Rightarrow g - f = m\left( x \right) - \left\{ {1 - m\left( x \right)} \right\}\]
Simplifying we get,
\[ \Rightarrow m\left( x \right) - 1 + m\left( x \right) = 2m\left( x \right) - 1\]
\[ \Rightarrow g - f = 2m\left( x \right) - 1.........(2)\]
Given that, \[\int\limits_0^n {\left( {g\left( x \right) - f\left( x \right)} \right)dx = 100} \]
Since the above function is periodic,
That is, \[P(k) = P(k + 1)\] where, \[P(k) = \int\limits_k^{k + 1} {\left\{ {g\left( x \right) - f\left( x \right)} \right\}dx} \]
So we get,
\[ \Rightarrow \int\limits_0^1 {\left\{ {g\left( x \right) - f\left( x \right)} \right\}dx} = \int\limits_1^2 {\left\{ {g\left( x \right) - f\left( x \right)} \right\}dx} = \int\limits_2^3 {\left\{ {g\left( x \right) - f\left( x \right)} \right\}dx = ... = \int\limits_{n - 1}^n {\left\{ {g\left( x \right) - f\left( x \right)} \right\}dx} } \]
Therefore we get, \[\int\limits_0^n {\left( {g\left( x \right) - f\left( x \right)} \right)dx = 100} \]
\[ \Rightarrow n\int\limits_0^1 {\left( {g\left( x \right) - f\left( x \right)} \right)dx = 100} \]
Splitting the limit,
\[ \Rightarrow n\int\limits_0^{0.5} {\left( {g\left( x \right) - f\left( x \right)} \right)dx + n\int\limits_{0.5}^1 {\left( {g\left( x \right) - f\left( x \right)} \right)dx} = 100} \]
By substituting the equations (1) and (2) we get,
\[ \Rightarrow n\int\limits_0^{0.5} {\left\{ {1 - 2m\left( x \right)} \right\}dx + n\int\limits_{0.5}^1 {\left\{ {2m\left( x \right) - 1} \right\}dx} = 100} \]
\[\left[ x \right]\]is the greatest integer function, returning the integer just below the value entered.
\[ \Rightarrow n\int\limits_0^{0.5} {\left( {1 - 2x} \right)dx} + n\int\limits_{0.5}^1 {\left( {2x - 1} \right)dx} = 100\]
Integrating the terms we get,
\[ \Rightarrow n\left[ x \right]\begin{array}{*{20}{c}}
  {0.5} \\
  0
\end{array} - 2n\left[ {\dfrac{{{x^2}}}{2}} \right]\begin{array}{*{20}{c}}
  {0.5} \\
  0
\end{array} + 2n\left[ {\dfrac{{{x^2}}}{2}} \right]\begin{array}{*{20}{c}}
  1 \\
  {0.5}
\end{array} - n\left[ x \right]\begin{array}{*{20}{c}}
  1 \\
  {0.5}
\end{array} = 100\]
Applying the limit values we get,
\[ \Rightarrow n\left( {0.5 - 0} \right) - 0.25n + n\left( {1 - 0.25} \right) - n\left( {1 - 0.5} \right) = 100\]
Simplifying we get,
\[ \Rightarrow 0.5n - 0.25n + 0.75n - 0.5n = 100\]
Add and subtract the terms to simplify,
\[ \Rightarrow 0.5n = 100\]
Solve for \[n\] we get,
\[ \Rightarrow n = 200\]
Hence, \[n = 200\]

$\therefore $(C) is the correct option.

Note: A periodic function is a function that repeats its values at regular intervals, for example, the trigonometric functions, which repeat at intervals of 2π radians. Periodic functions are used throughout science to describe oscillations, waves, and other phenomena that exhibit periodicity. A function \[f\left( x \right)\] is said to be periodic with period p if \[f\left( x \right) = f\left( {x + np} \right)\], for \[n = 1,2,3,.....\]