Question

If $f(x) = f '(x) + f ''(x) + f'''(x) +\dots$ and $ f(0) = 1$, then $ f(x) $ is equal to

• A. e$^{x/2}$
• B. $e^x$
• C. $e^{2x}$
• D. $e^{4x}$

Answer 1

Answer: (A)

Given, $ f(x)=f'(x)+f''(x)+f'''(x)+\ldots$
If $ f(x)=e^{x / 2} $
Then, $ f'(x)=\frac{1}{2} e^{x / 2}, f''(x)=\frac{1}{2^{2}} \cdot e^{x / 2} $
$ f'''(x) =\frac{1}{2^{3}} e^{x / 2} \ldots $ so on
$\therefore f(x)=f'(x)+f''(x)+f'''(x)+\ldots$
$=\frac{1}{2} e^{x / 2}+\frac{1}{2^{2}} \cdot e^{x / 2}+\frac{1}{2^{3}} \cdot e^{x / 2} + ...$ and so on
$=e^{x / 2}\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)$
$=e^{x / 2}\left\{\frac{\frac{1}{2}}{1-\frac{1}{2}}\right\}=e^{x / 2} \cdot\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)=e^{x / 2} \cdot 1$
$\therefore f(x)=e^{x / 2}$
and $f(0)=e^{0}=1$