Question

A function $f\left(x\right)$ satisfies the condition, $f\left(x\right)=f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)+f^{\prime \prime \prime }\left(x\right)+...\infty$ where $f\left(x\right)$ is an indefinitely differentiable function and dash denotes the order of derivatives. If $f\left(0\right)=1,$ then $f\left(x\right)$ is

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

Answer 1

Answer: (A)

$e^{x/2}$

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

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

On integrating w.r.t $x$; we get $\ln f\left(x\right)=\frac{1}{2}x+c$

If $x=0;f\left(0\right)=1\Rightarrow c=0$

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