# ISI2015-MMA-59

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In the Taylor expansion of the function $f(x)=e^{x/2}$ about $x=3$, the coefficient of $(x-3)^5$ is

1. $e^{3/2} \frac{1}{5!}$
2. $e^{3/2} \frac{1}{2^5 5!}$
3. $e^{-3/2} \frac{1}{2^5 5!}$
4. none of the above
in Calculus
recategorized

The general expansion of a function using taylor series at any given point  $x=a$  for a function  $f(x)$  is given as :

$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$  (Refer: Taylor Series Definition)

So the coefficient of the term  $(x-3)^{5}$  would be  $\frac{f'''''(3)}{5!}$

Here,  $f(x)=e^{\frac{x}{2}}$    $\Rightarrow$   $f'(x)=e^{\frac{x}{2}}\times (\frac{1}{2})$

$\Rightarrow$   $f''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{2}$

$\Rightarrow$   $f'''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{3}$

So, similarly   $f'''''(x)=e^{\frac{x}{2}}\times (\frac{1}{2})^{5}$

$\therefore$   at  $x=3$,  the value of the coefficient would be :   $e^{\frac{3}{2}}\times \frac{1}{2^{5}5!}$.

Option B is the correct answer.

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