Question

The coefficient of $(x-1{)}^2$ in the Taylor series expansion of $f\left(x\right)=x{e}^x,\left(eϵR\right)$ about the point x = 1 is

• A. $\frac{e}{2}$
  
• B. 2e
• C. $\frac{3e}{2}$
  
• D. 3e

Answer 1

The correct option is C $\frac{3e}{2}$


Taylor series expansion,

$f(a+h)=f\left(a\right)+h{f}^{\prime }\left(a\right)+\frac{h^2}{2!}f"\left(a\right)+......$
In this problem,
h = (x-1)
$a=x{e}^x$
so, the coefficient of $(x-1{)}^2$ will be $\frac{f"\left(x{e}^x\right)}{2!}$
$\because f\left(x\right)=x{e}^x$
First derivative, $f^{\prime }\left(x\right)=x{e}^x+e^x$
Second derivative, $f"\left(x\right)=x{e}^x+e^x$
Hence, coefficient of $(x-1{)}^{2}at$ x = 1 will be $\frac{x{e}^x+2e^x}{2}{]}_{x=1}$
=$\frac{e+2e}{2}$
=$\frac{3e}{2}$