Question

If $f(x)$ is discontinuous at only $x=1$ such that ${{f}^{2}}(x)=4$ $\forall x\in R$ , then the number of points $f(x)$ is discontinuous are
A. $4$
B. $6$
C. $8$
D. $10$

Answer 1

Hint:We need to find the number of points in which $f(x)$ is discontinuous. So first find the value of $f(x)$ from ${{f}^{2}}(x)$ by taking the square root. Then find the value of $f(x)$ for different cases, i.e at $(x=1,x\ne 1),(x<1,x\ge 1),(x>1,x\le 1)$ . Each pair of intervals have two possibilities.

Complete step by step answer:
We need to find the number of points in which $f(x)$ is discontinuous.
It is given that ${{f}^{2}}(x)=4$ .
Let us find the root of the above function, i.e
$f(x)=\pm 2$ $\forall x\in R$
It is given that the function is continuous everywhere except at $x=1$ .
So we have the following possibilities:
Firstly, We will look for the points at $x\ne 1$ and $x=1$ ,i.e
\[f(x)=\left\{ \begin{align}
  & \begin{matrix}
   2 & ; & x\ne 1 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & ; & x=1 \\
\end{matrix} \\
\end{align} \right.\]
The next possibility can be obtained by switching the values, i.e
\[f(x)=\left\{ \begin{align}
  & \begin{matrix}
   -2 & ; & x\ne 1 \\
\end{matrix} \\
 & \begin{matrix}
   2 & ; & x=1 \\
\end{matrix} \\
\end{align} \right.\]
The third possibility is obtained for the points at $x<1$ and \[x\ge 1\] , i.e
\[f(x)=\left\{ \begin{align}
  & \begin{matrix}
   2 & ; & x<1 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & ; & x\ge 1 \\
\end{matrix} \\
\end{align} \right.\]
The fourth possibility is obtained by switching the values similar to the second one, i.e
\[f(x)=\left\{ \begin{align}
  & -\begin{matrix}
   2 & ; & x<1 \\
\end{matrix} \\
 & \begin{matrix}
   2 & ; & x\ge 1 \\
\end{matrix} \\
\end{align} \right.\]
The fifth possibility is for $x>1$ and $x\le 1$ ,i.e
\[f(x)=\left\{ \begin{align}
  & \begin{matrix}
   2 & ; & x>1 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & ; & x\le 1 \\
\end{matrix} \\
\end{align} \right.\]
The sixth possibility is obtained similarly as the second and fourth one, i,e
\[f(x)=\left\{ \begin{align}
  & \begin{matrix}
   -2 & ; & x>1 \\
\end{matrix} \\
 & \begin{matrix}
   2 & ; & x\le 1 \\
\end{matrix} \\
\end{align} \right.\]
We have covered all the possible intervals. Thus there are 6 different functions.
Thus the number of points $f(x)$ is discontinuous are 6.
Hence the correct option is B.

Note:
 In questions of these types, take the ranges always in this manner. There is an alternate way to do this.
We know that there are 3 sets of intervals, i.e $(x=1,x\ne 1),(x<1,x\ge 1),(x>1,x\le 1)$ .
So there are $3!$ possibilities, i.e
$3\times 2\times 1=6$
So there are 6 points.