Question

For $x\in ℝ$ , let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$ .
For $x,y\in ℝ$ , define

$\text{min}\{x,y\}$ $=\{\begin{array}{ll}x& \text{if}x\le y,\\ y& \text{otherwise}\mathrm{.}\end{array}$

Let $f:\left[-2\pi ,2\pi \right]\to ℝ$ be defined by

$f\left(x\right)=sin\left(\text{min}\{x,x-\lfloor x\rfloor \}\right)\text{for}x\in \left[-2\pi ,2\pi \right]\mathrm{.}$

Consider the set $S=\{x\in \left[-2\pi ,2\pi \right]:\text{f is discontinuous at}x\}$ .
Which one of the following statements is TRUE?

1. S has 13 elements
2. S has 7 elements
3. S is an infinite set
4. S has 6 elements

Answer: S has 6 elements

Solution :-

Given that $f\left(x\right)=sin\left(\text{min}\{x,x-\lfloor x\rfloor \}\right)\text{for}x\in \left[-2\pi ,2\pi \right]\mathrm{.}$

We know that $x-\lfloor x\rfloor =\left\{x\right\}$ (fractional part of $x$ ). So, $f\left(x\right)$ becomes $f\left(x\right)=sin\left(\text{min}\{x,\{x\left\}\right\}\right)\text{for}x\in \left[-2\pi ,2\pi \right]$ and we know that $\left\{x\right\}$ is always positive and the value of $\left\{x\right\}$ is in $\left[0,1\right),\forall x\mathrm{.}$

So, if we take $-2\pi \le x\le 0$ , then $\text{min}\{x,\{x\left\}\right\}$ is $x$ .
Therefore, $f\left(x\right)=sin\left(\text{min}\{x,x-\lfloor x\rfloor \}\right)=sin\left(x\right)$ for $x\in \left[-2\pi ,0\right]$ .

If we take $0\le x<1$ , then $\text{min}\{x,\{x\left\}\right\}$ is $x$ , because both $x$ and $\left\{x\right\}$ are the same if $x\in \left[0,1\right)$ .
Therefore, $f\left(x\right)=sin\left(\text{min}\{x,x-\lfloor x\rfloor \}\right)=sin\left(x\right)$ for $x\in \left[0,1\right)$ .

If we take $i\le x<i+1$ , $i\in \{1,2,3,4,5,6\}$ then $\text{min}\{x,\{x\left\}\right\}$ is $\left\{x\right\}$ for each intervals.
Therefore, $f\left(x\right)=sin\left(\text{min}\{x,x-\lfloor x\rfloor \}\right)=sin\left(\left\{x\right\}\right)$ for $x\in \left[i,i+1\right),i\in \{1,2,3,4,5,6\}$ .

SO,

Here, as we can see $f\left(x\right)$ is continuous forall $x\in \left[-2\pi ,1\right)\mathrm{.}$

Now, whenever $x\in \left[i,i+1\right),i\in \{1,2,3,4,5,6\}$ , then $\left\{x\right\}\in \left[0,1\right)\mathrm{.}$

Since, we know that $sinx$ is increasing in $\left[0,\frac{\pi }{2}\approx 1.57\right]$ , so $f\left(x\right)=sinx$ is also increasing in $\left[0,1\right)\mathrm{.}$

Now, let’s analyze all of these intervals $\left[i,i+1\right),i\in \{1,2,3,4,5,6\}$ in these intervals $x$ starts with 1, 2, 3, 4, 5, 6 and ends with approx 2, 3, 4, 5, 6, 7 respectively but in each intervals $\left\{x\right\}$ always starts with 0 and ends with approx 1, so $f\left(x\right)=sin\left(\left\{x\right\}\right)$ attains 0 on each interval’s starting point which is an integer value and increases forward until $x$ gets the end value of each interval.

This means $f\left(x\right)$ is continuous in [1, 2) but discontinuous at 1, again $f\left(x\right)$ is continuous in [2, 3) but discontinuous at 2, and so on. We can observe that $f\left(x\right)$ is discontinuous in interval $\left(0,2\pi \right]$ whenever $x$ attains an integer value and this interval has six integers. So, $f\left(x\right)$ has six point of discontinuity and hence the set S has 6 elements.

So, the correct answer is option-4, $\to$ S has 6 elements