If we have the given condition \[\left| x-1 \right|>5\], then:
A. \[x\in \left( -4,6 \right)\]
B. \[x\in \left[ -4,6 \right]\]
C. \[x\in \left( -\infty ,-4 \right)\cup \left( 6,\infty \right)\]
D. \[x\in \left( -\infty ,-4 \right)\cup \left[ 6,\infty \right)\]

Hint: We know that the modulus of any real number x, denoted by \[\left| x \right|\] is the non-negative value of x without considering its sign. So we will take two cases first for \[x\ge 1\] and the second case for x<1 and we will solve to find the values of x.

Complete step-by-step answer:
We have been given\[\left| x-1 \right|>5\]
Now we will take the first case for \[x\ge 1\].
\[\Rightarrow x\ge 5\]
Since for \[x\ge 1\], the (x-1) is always positive so it will open with a positive sign.
\[\begin{align}
  & \Rightarrow x>5+1 \\
 & \Rightarrow x>6 \\
 & \Rightarrow x\in \left( 6,\infty \right) \\
\end{align}\]
Again, we have the second case for x<1.
\[\Rightarrow -\left( x-1 \right)>5\]
Since for x<1, the (x-1) is always negative, so it will open with a negative sign.
\[\Rightarrow -x+1>5\]
Adding ‘x’ to both the sides of the equality, we get as follows:
\[\begin{align}
  & \Rightarrow -x+1+x>5+x \\
 & \Rightarrow 1>5+x \\
\end{align}\]
Subtracting ‘5’ from both the sides of the equation, we get as follows:
\[\begin{align}
  & \Rightarrow 1-5>5+x-5 \\
 & \Rightarrow -4>x \\
 & \Rightarrow x\in \left( -\infty ,-4 \right) \\
\end{align}\]
Now the complete solution for the given inequality is union of both the cases.
Hence, \[x\in \left( -\infty ,-4 \right)\cup \left( 6,\infty \right)\].
Therefore, the correct answer of the given question is option C.

Note: Be careful while solving the inequality and also take care of the sign while adding or subtracting something to both sides of inequality during calculation. Also, choose the option very carefully as the option C and option D look similar but there is a slight difference between them. We have to take as many possible cases to solve the inequality of the modulus as the number of values obtained by equating the expression inside the modulus equals to zero.