Question

On the real line 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), \\\ \end{aligned}$$ Where [x] denotes the largest integer not exceeding x. The positive integer n for which $$\int\limits_{0}^{n}{\left( g\left( x \right)-f\left( x \right) \right)dx}=100$$ is? (a) 100 (b) 193 (c) 200 (d) 202

Answer 1

Hint: In this question, we first need to write the expression as the sum of the integral terms from 0 to 0.5 and 0.5 to 1 and so on and write the functional terms in those limits. Then on solving we get those values from which we can get a generalisation. So, we can find the value of n.

Complete step-by-step answer:

Let us first write the given functions in the question

& f\left( x \right)=\min \left( x-\left[ x \right],1-x+\left[ x \right] \right)=\min \left( f,1-f \right) \\\ & g\left( x \right)=\max \left( x-\left[ x \right],1-x+\left[ x \right] \right)=\max \left( f,1-f \right) \\\ \end{aligned}$$ Here, f is the fractional part of the function and it is always between 0 and 1 $$0\le f<1$$ Let us consider two cases accordingly If $$0 < f < 0.5$$ then we get, $$\begin{aligned} & f\left( x \right)=f \\\ & g\left( x \right)=1-f \\\ \end{aligned}$$ Now, let us consider the other possible case If $$0.5 < f < 1$$ then we get, $$\begin{aligned} & f\left( x \right)=1-f \\\ & g\left( x \right)=f \\\ \end{aligned}$$ Let us now consider the given equation in the question, $$\Rightarrow \int\limits_{0}^{n}{\left( g\left( x \right)-f\left( x \right) \right)dx}=100$$ Now, this can be also written as $$\Rightarrow n\int\limits_{0}^{1}{\left( g\left( x \right)-f\left( x \right) \right)dx}=100$$ Now, we need to split the equation into two parts from the above conditions considered and write the functional values accordingly. $$\Rightarrow n\left( \int\limits_{0}^{0.5}{\left( g\left( x \right)-f\left( x \right) \right)dx}+\int\limits_{0.5}^{1}{\left( g\left( x \right)-f\left( x \right) \right)dx} \right)=100$$ Now, on substituting the respective functional values we get, $$\Rightarrow n\left( \int\limits_{0}^{0.5}{\left( 1-f-f \right)dx}+\int\limits_{0.5}^{1}{\left( f-\left( 1-f \right) \right)dx} \right)=100$$ Now, this can be further written as $$\Rightarrow n\left( \int\limits_{0}^{0.5}{\left( 1-2x \right)dx}+\int\limits_{0.5}^{1}{\left( 2x-1 \right)dx} \right)=100$$ Now, on integrating and substituting the respective limits we get, $$\Rightarrow n\left( \left( x-{{x}^{2}} \right)_{0}^{0.5}+\left( {{x}^{2}}-x \right)_{0.5}^{1} \right)=100$$ Now, this can be further simplified as $$\Rightarrow n\left( \left( \dfrac{1}{2}-\dfrac{1}{4} \right)+\left( 1-1-\left( \dfrac{1}{4}-\dfrac{1}{2} \right) \right) \right)=100$$ $$\Rightarrow n\times 2\times \dfrac{1}{4}=100$$ Now, on multiplying both sides with 2 we get, $$\therefore n=200$$ Hence, the correct option is (c). Note: It is important to note that the given equation can be written as an integral term from 0 to 1 in the simplified form. Then it is also to be noted that as the given functions contain fractional parts their values change g in between 0 and 1. So, we need to consider the cases accordingly and write the respective functions. If we neglect that between 0 to 1 the functional values become completely opposite before 0.5 and after 0.5 then the complete value of the function in integral on the left hand side becomes 0 which is not possible according to the question. So, we need to check the possible cases.