Q119 / 180 · KCET 2013 · Mathematics
If $f(x)=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots$ and $f(0)=1$, then $f(x)$ is equal to
- Correct answerA$e^{x / 2}$
- B$e^{x}$
- C$e^{2 x}$
- D$e^{4 x}$
Solution
Given, $\quad f(x)=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots$
If $\quad f(x)=e^{x / 2}$
Then, $f^{\prime}(x)=\frac{1}{2} e^{x / 2}, f^{\prime \prime}(x)=\frac{1}{2^{2}} \cdot e^{x / 2}$,
$\begin{aligned}
f^{\prime \prime \prime}(x) &=\frac{1}{2^{3}} e^{x / 2} \ldots \text { so on } \\
\therefore \quad f(x) &=f^{\prime}(x)+f^{\prime \prime}(x)+f^{\prime \prime \prime}(x)+\ldots \\
&=\frac{1}{2} e^{x / 2}+\frac{1}{2^{2}} \cdot e^{x / 2}+\frac{1}{2^{3}} \cdot e^{x / 2}
\end{aligned}$
$=e^{x / 2}\left(\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)$
$=e^{x / 2}\left\{\frac{\frac{1}{2}}{1-\frac{1}{2}}\right\}=e^{x / 2} \cdot\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)=e^{x / 2} \cdot 1$
$\therefore \quad f(x)=e^{x / 2}$
and $f(0)=e^{0}=1$
Hence, the correct answer is option: 'A'.