Question

Let we have a function \[f:\left[ -1,3 \right]\to \mathbb{R}\] be defined as
\[f=\left\{ \begin{align}
  & \left| x \right|+\left[ x \right],-1\le x<1 \\
 & x+\left| x \right|,1\le x<2 \\
 & x+\left[ x \right],2\le x\le 3 \\
\end{align} \right.\]
Where \[\left[ t \right]\] denotes the greatest integer less than or equal to \['t'\] , then \[f\] is discontinuous at:
(a) Four or more points
(b) Only one point
(c) Only two points
(d) Only three points

Answer 1

Hint: We solve this problem first by dividing the given function at each and every integer because we have step function. Then we check the continuity at each breaking point because the possibility of discontinuity occurs at the division point of function. We check the continuity by using the limits that is if a function \[f\left( x \right)\] is said to be continuous at \[{{x}_{0}}\] is and only if
\[\Rightarrow LHL=RHL\]
Where, \[LHL=\displaystyle \lim_{x \to {{x}_{0}}^{-}}f\left( x \right)\]
\[RHL=\displaystyle \lim_{x \to {{x}_{0}}^{+}}f\left( x \right)\]

Complete step-by-step solution
We are given with a function that is
\[f=\left\{ \begin{align}
  & \left| x \right|+\left[ x \right],-1\le x<1 \\
 & x+\left| x \right|,1\le x<2 \\
 & x+\left[ x \right],2\le x\le 3 \\
\end{align} \right.\]
Now, let us divide the function at each and every integer then we get
\[\Rightarrow f=\left\{ \begin{align}
  & \left| x \right|+\left[ x \right],-1\le x<0 \\
 & \left| x \right|+\left[ x \right],0\le x<1 \\
 & x+\left| x \right|,1\le x<2 \\
 & x+\left[ x \right],2\le x\le 3 \\
\end{align} \right.\]
We know that \[\left[ t \right]\] denotes the greatest integer less than or equal to \['t'\]
This means that \[\left[ t \right]\] gets the value as an greatest integer that is less than \['t'\] and have the same value of \['t'\] if it is an integer.
Now, by using the above definition of step function let us write the values of \[\left[ x \right]\] in the function as
\[\Rightarrow f=\left\{ \begin{align}
  & \left| x \right|+\left( -1 \right),-1\le x<0 \\
 & \left| x \right|+0,0\le x<1 \\
 & x+\left| x \right|,1\le x<2 \\
 & x+2,2\le x\le 3 \\
\end{align} \right.\]
\[\Rightarrow f=\left\{ \begin{align}
  & \left| x \right|-1,-1\le x<0 \\
 & \left| x \right|,0\le x<1 \\
 & x+\left| x \right|,1\le x<2 \\
 & x+2,2\le x\le 3 \\
\end{align} \right.\]
We know that the modulus function is defined as
\[\left| x \right|=\left\{ \begin{align}
  & -x,x\le 0 \\
 & x,x\ge 0 \\
\end{align} \right.\]
Here, we can see that for all negative values of \['x'\] we have
\[\Rightarrow \left| x \right|=-x\]
Also, we can see that for all positive values of \['x'\] we have
\[\Rightarrow \left| x \right|=x\]
By substituting these values in the given function we get
\[\Rightarrow f=\left\{ \begin{align}
  & -x-1,-1\le x<0 \\
 & x,0\le x<1 \\
 & 2x,1\le x<2 \\
 & x+2,2\le x\le 3 \\
\end{align} \right.\]
Now, let us check the continuity at each and every division point.
(i) Let us check the continuity at ‘0’
We know that if a function \[f\left( x \right)\] is said to be continuous at \[{{x}_{0}}\] is and only if
\[\Rightarrow LHL=RHL\]
Where, \[LHL=\displaystyle \lim_{x \to {{x}_{0}}^{-}}f\left( x \right)\]
\[RHL=\displaystyle \lim_{x \to {{x}_{0}}^{+}}f\left( x \right)\]
Now, let us calculate the LHL at ‘0’ then we get
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{0}^{-}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( -x-1 \right)\] when x approaches 0 from left side
By using this value in above equation we get
\[\begin{align}
  & \Rightarrow LHL=\displaystyle \lim_{x \to {{0}^{-}}}\left( -x-1 \right) \\
 & \Rightarrow LHL=-1 \\
\end{align}\]
Now let us find the RHL at ‘0’ then we get
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{0}^{+}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( x \right)\] when x approaches 0 from right side
By using this value in above equation we get
\[\begin{align}
  & \Rightarrow RHL=\displaystyle \lim_{x \to {{0}^{+}}}\left( x \right) \\
 & \Rightarrow RHL=0 \\
\end{align}\]
Here, we can see that
\[\Rightarrow LHL\ne RHL\]
Therefore we can say that the function is not continuous at ‘0’
(ii) Let us check the continuity at ‘1’
Now, let us calculate the LHL at ‘1’ then we get
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{1}^{-}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( x \right)\] when x approaches 1 from left side
By using this value in the above equation we get
\[\begin{align}
  & \Rightarrow LHL=\displaystyle \lim_{x \to {{1}^{-}}}\left( x \right) \\
 & \Rightarrow LHL=1 \\
\end{align}\]
Now let us find the RHL at ‘1’ then we get
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{1}^{+}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( 2x \right)\] when x approaches 0 from right side
By using this value in above equation we get
\[\begin{align}
  & \Rightarrow RHL=\displaystyle \lim_{x \to {{1}^{+}}}\left( 2x \right) \\
 & \Rightarrow RHL=2 \\
\end{align}\]
Here, we can see that
\[\Rightarrow LHL\ne RHL\]
Therefore we can say that the function is not continuous at ‘1’
(iii) Let us check the continuity at ‘2’
Now, let us calculate the LHL at ‘2’ then we get
\[\Rightarrow LHL=\displaystyle \lim_{x \to {{2}^{-}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( 2x \right)\] when x approaches 1 from left side
By using this value in above equation we get
\[\begin{align}
  & \Rightarrow LHL=\displaystyle \lim_{x \to {{2}^{-}}}\left( 2x \right) \\
 & \Rightarrow LHL=4 \\
\end{align}\]
Now let us find the RHL at ‘2’ then we get
\[\Rightarrow RHL=\displaystyle \lim_{x \to {{2}^{+}}}f\left( x \right)\]
From the function we can see that the function\[f\left( x \right)\] have the value of \[\left( x+2 \right)\] when x approaches 0 from right side
By using this value in above equation we get
\[\begin{align}
  & \Rightarrow RHL=\displaystyle \lim_{x \to {{2}^{+}}}\left( x+2 \right) \\
 & \Rightarrow RHL=4 \\
\end{align}\]
Here, we can see that
\[\Rightarrow LHL=RHL\]
Therefore we can say that the function is continuous at ‘2’
Therefore, the given function is discontinuous at \[x=0,1\] which means the function is discontinuous only at two points
So, option (c) is the correct answer.

Note: Students may make mistakes in taking the values in step function.
Here we have a range of \['x'\] as \[-1\le x<0\]
Here, the value of \[\left[ x \right]\] should be ‘-1’ because whatever the values we have in the domain \[-1\le x<0\] the greatest integer less than \['x'\] will be ‘-1’ only.
But students may take the value of \[\left[ x \right]\]in the domain \[-1\le x<0\] as’0’ which will be wrong.
Also, we need to check the continuity at each integer in the domain \[\left( -1,3 \right)\] of the function that means we need to check the continuity at ‘0’ also. But students may miss this point because the function is not divided at ‘0’.