If [.] denotes the greatest integer function then the domain of the real-valued function \[{\log _{\left[ {x + \dfrac{1}{2}} \right]}}\left| {{x^2} - x - 2} \right|\] is
A) \[x \in \left[ {\dfrac{3}{2},\infty } \right) - \left\{ 2 \right\}\]
B) \[x \in \left[ {\dfrac{5}{2}, + \infty } \right)\]
C) \[x \in \left( {\dfrac{1}{2},2} \right) \cup \left( {2, + \infty } \right)\]
D) None of these

Hint: We will here use various properties of the logarithm function and the properties of the greatest integer function to get the final answer.
The base of logarithm function is always greater than 0 and is not equal to 1.
Modulus of any value is always greater than zero.
Logarithm function is not defined at 0

Complete step by step solution:
The given function is \[f\left( x \right) = {\log _{\left[ {x + \dfrac{1}{2}} \right]}}\left| {{x^2} - x - 2} \right|\]
Now since we know that logarithm function is not defined at 0 and also, modulus of any value is always greater than zero therefore,
\[
  \left| {{x^2} - x - 2} \right| > 0 \\
   \Rightarrow {x^2} - x - 2 \ne 0 \\
   \Rightarrow {x^2} - 2x + x - 2 \ne 0 \\
   \Rightarrow x\left( {x - 2} \right) + 1\left( {x - 2} \right) \ne 0 \\
   \Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) \ne 0 \\
   \Rightarrow x \ne 2;x \ne - 1 \\
 \]
Now we know that the base of logarithm function is always greater than 0 and is not equal to 1 therefore,
 \[\left[ {x + \dfrac{1}{2}} \right] > 0\] and \[\left[ {x + \dfrac{1}{2}} \right] \ne 1\]
Now according to the property of greatest integer functions,
\[\left[ x \right] = 1{\text{ for }}1 < x < 2\]
Therefore applying this property we get:
\[
  x + \dfrac{1}{2} \notin \left[ {1,2} \right) \\
   \Rightarrow x \notin \left[ {1 - \dfrac{1}{2},2 - \dfrac{1}{2}} \right) \\
   \Rightarrow x \notin \left[ {\dfrac{1}{2},\dfrac{3}{2}} \right) \\
 \]
Also, since \[\left[ {x + \dfrac{1}{2}} \right] > 0\] and the base of the log function is not equal to 1 therefore,
\[
  \left[ {x + \dfrac{1}{2}} \right] \ne 0 \\
  \left[ {x + \dfrac{1}{2}} \right] \ne 1 \\
 \]
This implies:
\[
   \Rightarrow x + \dfrac{1}{2} \geqslant 2 \\
   \Rightarrow x \geqslant 2 - \dfrac{1}{2} \\
   \Rightarrow x \geqslant \dfrac{3}{2} \\
 \]
Therefore considering all the factors we get:
\[
  x \in \left[ {\dfrac{3}{2},2} \right) \cup \left( {2,\infty } \right) \\
   \Rightarrow x \in \left[ {\dfrac{3}{2},\infty } \right) - \left\{ 2 \right\} \\
 \]

Therefore, option A is correct.

Note:
The value of modulus function is always greater than zero.
The base of log function is always greater than 1 and the value of greatest integer function is always :
\[\left[ x \right] = 1{\text{ for }}1 < x < 2\]