Question

If $f\left(x\right)=f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)+f^{\prime \prime \prime }\left(x\right)+......$ and $f^{\prime }\left(0\right)=1$, then $f\left(x\right)=$

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

Answer 1

The correct option is A $e^{\frac{x}{2}}$

We have, $f\left(x\right)=f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)+f^{\prime \prime \prime }\left(x\right)+......$(1)

Differentiate both sides,

$f^{\prime }\left(x\right)=f^{\prime \prime }\left(x\right)+f^{\prime \prime \prime }\left(x\right)+...............$(2)

Now subtract (2) from (1)

$\Rightarrow f\left(x\right)-f^{\prime }\left(x\right)=f^{\prime }\left(x\right)$

$\Rightarrow f\left(x\right)=2f^{\prime }\left(x\right)$

$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{1}{2}$

Now integrate both sides,

$\Rightarrow \ln f\left(x\right)=\frac{1}{2}x+C$

$\Rightarrow f\left(x\right)=A{e}^{\frac{x}{2}}$

Now since $f\left(0\right)=1\Rightarrow A=1$

Hence $f\left(x\right)=e^{\frac{x}{2}}$