In the expansion of \[\left( {1 + x + {x^2}} \right){e^{ - x}}\], the coefficient of \[{x^2}\] is
A. 1
B. \[ - 1\]
C. \[\dfrac{1}{2}\]
D. \[ - \dfrac{1}{2}\]

- Hint: Use the series expansion of the exponential function \[{e^{ - x}}\]. Then multiply it with \[\left( {1 + x + {x^2}} \right)\]. Look at the terms in the product carefully and find out the term having \[{x^2}\] and hence you’’ll get the coefficient of \[{x^2}\].

Formula Used:
The series expansion of \[{e^{ - x}}\] is \[1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + ......\]
The factorial of \[n\] is \[n! = n\left( {n - 1} \right)\left( {n - 2} \right).....2 \cdot 1\]

Complete step-by-step answer:
The given function is \[\left( {1 + x + {x^2}} \right){e^{ - x}}\]
The series expansion of \[{e^{ - x}}\] is \[1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + ......\]
So, the given function becomes \[\left( {1 + x + {x^2}} \right)\left( {1 - \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}} - \dfrac{{{x^3}}}{{3!}} + ......} \right)\]
The factorial of \[n\] is \[n! = n\left( {n - 1} \right)\left( {n - 2} \right).....2 \cdot 1\]
Putting \[n = 1\], we get \[1! = 1\]
Putting \[n = 2\], we get \[2! = 2 \cdot 1 = 2\]
Putting \[n = 3\], we get \[3! = 3 \cdot 2 \cdot 1 = 6\]
Multiply the two expressions using distributive law \[a\left( {b + c} \right) = ab + ac\]
\[\left( {1 + x + {x^2}} \right)\left( 1 \right) - \left( {1 + x + {x^2}} \right)\left( x \right) + \left( {1 + x + {x^2}} \right)\left( {\dfrac{{{x^2}}}{2}} \right) - \left( {1 + x + {x^2}} \right)\left( {\dfrac{{{x^3}}}{6}} \right) + ......\]
\[ = 1 + x + {x^2} - x - {x^2} - {x^3} + \dfrac{1}{2}{x^2} + \dfrac{1}{2}{x^3} + \dfrac{1}{2}{x^4} - ......\]
All other terms are of the degrees greater than \[2\].
Arrange the terms according to the degrees in ascending order.
\[ = 1 + \dfrac{1}{2}{x^2} + ......\]
It is seen that the term having \[{x^2}\] is \[\dfrac{1}{2}{x^2}\]
So, the coefficient of \[{x^2}\] is \[\dfrac{1}{2}\]
Hence, option C is correct.

Note:- After substituting the expansion of \[{e^{ - x}}\] , we get a sum of infinite number of terms. The coefficient of \[{x^2}\] is required. So, in the expansion, the terms including \[{x^2}\] must be written and all other terms of degree more than \[2\] is not necessary.