When \[{{x}^{3}}\ \ -\ \ 2{{x}^{2}}\ \ +\ \ ax\ \ -\ \ b\] is divided by \[{{x}^{2}}\ \ -\ \ 2x\ \ -\ \ 3\] , then remainder is \[x\text{ }\text{- }6\] . The values of ‘a’ and ‘b’ is respectively
A. -2 and -6
B. 2 and -6
C. -2 and 6
D. 2 and 6

Hint: To find the values of ‘a’ and ‘b’, first divide \[{{x}^{3}}\ \ -\ \ 2{{x}^{2}}\ \ +\ \ ax\ \ -\ \ b\] by \[{{x}^{2}}\ \ -\ \ 2x\ \ -\ \ 3\] using long division method. The obtained remainder, \[\left( 3x\text{ }+\text{ }ax\text{ }\text{ -}b \right)\] , is equated to the given remainder. Then by comparing the terms in LHS and RHS, the required values are obtained.

Complete step by step answer:
We have to find the values of ‘a’ and ‘b’.
We will have to divide \[{{x}^{3}}-2{{x}^{2}}+\text{ }ax-b\text{ by }{{x}^{2}}-2x-3\] for the solution of this question.
First let us divide ${{x}^{3}}$ by \[{{x}^{2}}-2x-3\] .
The quotient will be $x$ .
Now let us multiply this with the divisor, \[{{x}^{2}}-2x-3\].
So we will get
\[{{x}^{3}}-2{{x}^{2}}-3x\]
Now let us subtract this from the quotient. The remainder thus obtained will be
\[\left( 3x\text{ }+\text{ }ax\text{ }\text{ - }b \right)\]
This is illustrated below:

\[\begin{align}
  & {{x}^{2}}\ -\ 2x\ -\ 3\overset{x}{\overline{\left){{{x}^{3}}\ -\ 2{{x}^{2}}\ +\ ax\ -\ b}\right.}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{\begin{align}
  & \ \ \ {{x}^{3}}\ -\ 2{{x}^{2}}\ -\ 3x \\
 & \left( - \right)\ \left( + \right)\ \ \ \ \ \left( + \right) \\
\end{align}} \\
 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +3x\ +\ ax\ -\ b \\
\end{align}\]
The remainder that is left after this division is \[\left( 3x\text{ }+\text{ }ax\text{ }\text{ - }b \right)\] . But, according to the question, the remainder is \[\left( x\text{ }\text{- }6 \right)\] . So, let us equate \[\left( 3x\text{ }+\text{ }ax\text{ }\text{- }b \right)\] and \[\left( x\text{ }\text{- }6 \right)\] now.
\[\begin{array}{*{35}{l}}
   3x\text{ }+\text{ }ax\text{ }\text{ -}b\text{ }=\text{ }x\text{ }\text{ -}6 \\
   \left( 3\text{ }+\text{ }a \right)\text{ }x\text{ }\text{- }b\text{ }=\text{ }x\text{ }\text{- }6 \\
\end{array}\]
Let us now compare both the sides of the equation.
After comparing, we get:-
\[\begin{array}{*{35}{l}}
   a\text{ }+\text{ }3\text{ }=\text{ }1\text{, }\text{ -}b\text{ }=\text{ }-6 \\
   So,\text{ }a\text{ }=\ -2~~~~~~~~~ \\
   b\text{ }=\text{ }6 \\
\end{array}\]

So, the correct answer is “Option C”.

Note: When performing division, you make errors when multiplying the quotient with the divisor. Note that the whole divisor must be taken for this. The sign must be changed before subtracting. Also when equating the remainder to the given remainder, do not go for solving it, instead, compare them.