Question

A point $P(x,y)$ moves such that $\left[ x+y+1 \right]=\left[ x \right]$ (where [.] denotes greatest integer function) and $x\in \left( 0,2 \right)$, then the area represented by all the possible positions of P, is
a)$\sqrt{2}$
b) $2\sqrt{2}$
c) $4\sqrt{2}$
d) 2

Answer 1

Hint: We have the following expression: $\left[ x+y+1 \right]=\left[ x \right]$ (where [.] denotes greatest integer function) and $x\in \left( 0,2 \right)$.
As we know that, the greatest integer function returns the nearest integer value. So, [1] = 1. Therefore, the expression can be written as: $\left[ x+y \right]=\left[ x \right]-1$.
If a function lies in the interval $\left( n,n+1 \right)$, the value of greatest integer function is n. Since, we have the interval (0, 2), write it in the form of $\left( n,n+1 \right)$ intervals, and find the value of $\left[ x \right]-1$in those intervals. That is also equal to the value of $\left[ x+y \right]$. Then, plot the point on the graph and find the area covered by the given function.

Complete step-by-step solution:
We are having a greatest integer function as: $\left[ x+y+1 \right]=\left[ x \right].........(1)$
Since, we know that [1] = 1, we can write equation (1) as:
$\begin{align}
  & \Rightarrow \left[ x+y \right]+1=\left[ x \right] \\
 & \Rightarrow \left[ x+y \right]=\left[ x \right]-1......(2) \\
\end{align}$
As we know that, $x\in \left( 0,2 \right)$, so we can say that $0 < x< 1;1\le x< 2$
Now, using the above result, we can find the value of $\left[ x \right]-1$ in the given intervals.
Case-I: When $0< x< 1$
So, we can write: $\left[ x \right]=0$
Therefore,
$\begin{align}
  & \left[ x \right]-1=0-1 \\
 & =-1
\end{align}$
As we have, $\left[ x+y \right]=\left[ x \right]-1$ from equation (2), so we can say that,
$\left[ x+y \right]=-1$
So, we can say that: $-1\le x+y< 0......(3)$
Case-I: When $1\le x< 2$
So, we can write: $\left[ x \right]=1$
Therefore,
$\begin{align}
  & \left[ x \right]-1=1-1 \\
 & =0
\end{align}$
As we have, $\left[ x+y \right]=\left[ x \right]-1$ from equation (2), so we can say that,
$\left[ x+y \right]=0$
So, we can say that: $0\le x+y <1......(4)$
So, from equation (3) and (4), we can plot a graph using the interval $x\in \left( 0,2 \right)$ as shown below.

Hence, we need to find the area of the shaded region in the above graph. This region has four triangles of the same base and the same height.
So, the area of shaded region = 4 $\times $ area of triangle
$\begin{align}
  & =4\times \left( \dfrac{1}{2}\times base\times height \right) \\
 & =4\times \left( \dfrac{1}{2}\times 1\times 1 \right) \\
 & =2\text{ }square\text{ }unit \\
\end{align}$
Hence, option (d) is the correct answer.

Note: The Greatest Integer function is given as: $y=\left[ x \right]$. For all real numbers, x, the greatest integer function returns the largest integer less than or equal to x. In simple terms, it rounds off the given real number to the nearest integer.
For example:
$\begin{align}
  & \left[ 1 \right]=1 \\
 & \left[ 2.5 \right]=2 \\
 & \left[ -3 \right]=-3 \\
 & \left[ -8.6 \right]=-9 \\
\end{align}$