Question

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