In the expansion of (1 + x + x^2)e^{-x},the c | vedclass

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(C) We know that the expansion of $e^{-x}$ is given by $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$Now,the given expression is $(1 + x +

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$0.5$

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(C) We know that the expansion of $e^{-x}$ is given by $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$Now,the given expression is $(1 + x + x^2)e^{-x} = (1 + x + x^2) \left( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots ight)$To find the coefficient of $x^2$,we multiply the terms from the first bracket with the terms of the second bracket such that the product results in $x^2$:$1 	imes (\frac{x^2}{2}) = \frac{1}{2}x^2$$x 	imes (-x) = -x^2$$x^2 	imes (1) = x^2$Summing these coefficients: $\frac{1}{2} - 1 + 1 = \frac{1}{2} = 0.5$Thus,the coefficient of $x^2$ is $0.5$.

In the expansion of $(1 + x + x^2)e^{-x}$,the coefficient of $x^2$ is

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