Question

If a function is given by $f\left( x \right)={{x}^{2}}-2x+4$ then the set of values of x satisfying $f\left( x-1 \right)=f\left( x+1 \right)$ is:
a.$\left\{ -1 \right\}$
b.$\left\{ -1,1 \right\}$
c.$\left\{ 1 \right\}$
d.$\left\{ 1,2 \right\}$

Answer 1

Hint: First of all find $f\left( x-1 \right)$ from $f\left( x \right)={{x}^{2}}-2x+4$ by substituting $\left( x-1 \right)$ instead of x. Similarly, find the value of $f\left( x+1 \right)$ by substituting $\left( x+1 \right)$ instead of x. Now, equate the result of $f\left( x-1 \right)$ & $f\left( x+1 \right)$ and then find the values of x where these two functions are equal.

Complete step-by-step answer:
 is given in the above problem that:
$f\left( x \right)={{x}^{2}}-2x+4$
$f\left( x-1 \right)=f\left( x+1 \right)$
To find the value of $f\left( x-1 \right)$ from $f\left( x \right)={{x}^{2}}-2x+4$ by substituting $\left( x-1 \right)$ instead of x we get,
$f\left( x \right)={{x}^{2}}-2x+4$
$\begin{align}
  & f\left( x-1 \right)={{\left( x-1 \right)}^{2}}-2\left( x-1 \right)+4 \\
 & \Rightarrow f\left( x-1 \right)={{x}^{2}}-2x+1-2x+2+4 \\
 & \Rightarrow f\left( x-1 \right)={{x}^{2}}-4x+7 \\
\end{align}$
To find the value of $f\left( x+1 \right)$ from $f\left( x \right)={{x}^{2}}-2x+4$ by substituting $\left( x+1 \right)$ instead of x we get,
$\begin{align}
  & f\left( x+1 \right)={{\left( x+1 \right)}^{2}}-2\left( x+1 \right)+4 \\
 & \Rightarrow f\left( x+1 \right)={{x}^{2}}+2x+1-2x-2+4 \\
\end{align}$
$\Rightarrow f\left( x+1 \right)={{x}^{2}}+3$
Now, equating the expressions $f\left( x-1 \right)$ & $f\left( x+1 \right)$ we get,
$f\left( x-1 \right)=f\left( x+1 \right)$
${{x}^{2}}-4x+7={{x}^{2}}+3$
As you can see from the above that ${{x}^{2}}$ will be cancelled out from both the sides we get,
$\begin{align}
  & -4x+7=3 \\
 & \Rightarrow -4x=-4 \\
\end{align}$
Dividing -4 on both the sides of the above equation we get,
$x=1$
From the above solution, we have found that only $x=1$ is satisfying $f\left( x-1 \right)=f\left( x+1 \right)$.
Hence, the correct option is (c).

Note: There is a shortcut to solve the above problem.
As you can see that the options have the value of x as -1, 1, 2 so put these values of x in $f\left( x-1 \right)=f\left( x+1 \right)$ and then see which value(s) are satisfying this relation.
Putting $x=-1$ we get,
$f\left( -2 \right)=f\left( 0 \right)$
It is given that $f\left( x \right)={{x}^{2}}-2x+4$. Now, we are going to find:
$\begin{align}
  & f\left( -2 \right)={{\left( -2 \right)}^{2}}-2\left( -2 \right)+4 \\
 & \Rightarrow f\left( -2 \right)=4+4+4=12 \\
\end{align}$
$f\left( 0 \right)=0-0+4=4$
From the above, it is found that $f\left( -2 \right)\ne f\left( 0 \right)$ so $x=-1$ is not the solution.
Putting $x=1$ we get,
$f\left( 0 \right)=f\left( 2 \right)$
It is given that $f\left( x \right)={{x}^{2}}-2x+4$. Now, we are going to find:
We have already found the value of $f\left( 0 \right)$ in the above which is equal to 4.
$\begin{align}
  & f\left( 2 \right)={{\left( 2 \right)}^{2}}-2\left( 2 \right)+4 \\
 & \Rightarrow f\left( 2 \right)=4-4+4=4 \\
\end{align}$
From the above, it is found that $f\left( 0 \right)=f\left( 2 \right)$ so $x=1$ is the solution.
Putting $x=2$ we get,
$f\left( 1 \right)=f\left( 3 \right)$
We have already found the value of $f\left( 1 \right)$ in the above which is equal to 4.
$\begin{align}
  & f\left( 3 \right)={{\left( 3 \right)}^{2}}-2.3+4 \\
 & \Rightarrow f\left( 3 \right)=9-6+4=7 \\
\end{align}$
From the above, it is found that $f\left( 1 \right)\ne f\left( 3 \right)$ so $x=3$ is not the solution.
From the above, we have found that only $x=1$ is satisfying $f\left( x-1 \right)=f\left( x+1 \right)$. Hence, the correct option is (c).