Question

Given a function g which has derivative $g^{\prime }\left(x\right)$ for all x satisfying $g^{\prime }\left(0\right)=2$ and $g(x+y)=e^{y}g\left(x\right)+e^{x}g\left(y\right)$ for all $x,yϵR$,$g\left(5\right)=32.$ The value of $\frac{g^{\prime }\left(5\right)-2e^5}{16}$ is

Answer 1

Putting $x=y=0$ in $g(x+y)=e^{y}g\left(x\right)+e^{x}g\left(y\right)$
we get $g\left(0\right)=g\left(0\right)+g\left(0\right)=2g\left(0\right)\Rightarrow g\left(0\right)=0$
So $2=g^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{g\left(h\right)-g\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}$
Also $g^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{g(x+h)-g\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{e^{x}g\left(h\right)+e^{h}g\left(x\right)-g\left(x\right)}{h}$
$=\underset{h\to 0}{lim}(e^x\frac{g\left(h\right)}{h}+\frac{e^h-1}{h}g\left(x\right))$
$=e^x\underset{h\to 0}{lim}\frac{g\left(h\right)}{h}+1.g\left(x\right)$
$=g\left(x\right)+2e^x$
Thus $g^{\prime }\left(5\right)-2e^5=g\left(5\right)=32$