Question

Find the range of $f\left( x \right)=\ sgn \left( {{x}^{2}}-2x+3 \right)$, is
(A) {-1, 0, 1}
(B) {0, 1}
(C) {1}
(D) {-1, -2, 2, 3}

Answer 1

Hint: We solve this question by first considering the definition of signum function. Then we apply this definition and find the value of the given function in each case. Then we consider the expression ${{x}^{2}}-2x+3$ in the signum function in the given function $f\left( x \right)$. Then we find the range of ${{x}^{2}}-2x+3$ and then use it and the definition we have applied to $f\left( x \right)$ to find the range of $f\left( x \right)$.

Complete step by step answer:
The function we are given is $f\left( x \right)=\ sgn \left( {{x}^{2}}-2x+3 \right)$.
First let us consider the nature of a signum function.
A function $\ sgn \left( x \right)$ is given by
$\ sgn \left( x \right)=\dfrac{\left| x \right|}{x}=\left\{ \begin{matrix}
   1\text{ if }x>0 \\
   -1\text{ if }x<0 \\
   \text{0 if }x=0 \\
\end{matrix} \right.$
So, we can write our given function $f\left( x \right)=\ sgn \left( {{x}^{2}}-2x+3 \right)$ as,
$\ sgn \left( {{x}^{2}}-2x+3 \right)=\dfrac{\left| {{x}^{2}}-2x+3 \right|}{{{x}^{2}}-2x+3}=\left\{ \begin{matrix}
   1\text{ if }{{x}^{2}}-2x+3>0 \\
   -1\text{ if }{{x}^{2}}-2x+3<0 \\
   \text{0 if }{{x}^{2}}-2x+3=0 \\
\end{matrix} \right.\ \ \ \ \ \ ..................\left( 1 \right)$
Now let us consider the value inside the signum function in the given function $f\left( x \right)$, that is ${{x}^{2}}-2x+3$.
We can write it as,
$\begin{align}
  & \Rightarrow {{x}^{2}}-2x+3={{x}^{2}}-2x+1+2 \\
 & \Rightarrow {{x}^{2}}-2x+3=\left( {{x}^{2}}-2x+1 \right)+2 \\
\end{align}$
Now let us consider the formula,
${{x}^{2}}-2ax+{{a}^{2}}={{\left( x-a \right)}^{2}}$
Using this we can write the above equation as,
\[\Rightarrow {{x}^{2}}-2x+3={{\left( x-1 \right)}^{2}}+2\]
Now let us consider the property, square of any real number is positive. So, we have that
\[\begin{align}
  & \Rightarrow {{\left( x-1 \right)}^{2}}>0 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}+2>2 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}+2>0...........\left( 2 \right) \\
\end{align}\]
Now using equations (1) and (2) we get the value of function $f\left( x \right)$ as,
$\ sgn \left( {{x}^{2}}-2x+3 \right)=\dfrac{\left| {{x}^{2}}-2x+3 \right|}{{{x}^{2}}-2x+3}=1$
Hence, we get that the value of $f\left( x \right)$ as 1.
Hence the range of $f\left( x \right)$ is {1}.
Hence answer is Option C.

Note:
The common mistake one does in this question is one might not consider the range of ${{x}^{2}}-2x+3$ in the given function. Then they will just consider the function as,
$f\left( x \right)=\ sgn \left( {{x}^{2}}-2x+3 \right)=\dfrac{\left| {{x}^{2}}-2x+3 \right|}{{{x}^{2}}-2x+3}=\left\{ \begin{matrix}
   1\text{ if }{{x}^{2}}-2x+3>0 \\
   -1\text{ if }{{x}^{2}}-2x+3<0 \\
   \text{0 if }{{x}^{2}}-2x+3=0 \\
\end{matrix} \right.$
and answer the range of $f\left( x \right)$ is {-1, 0, 1} and mark the answer as Option A. So, one need to consider the range of ${{x}^{2}}-2x+3$ and they consider the cases that are in its range and find the range of $f\left( x \right)$.