A problem by Arpita Karkera

Given $f(x).f'(-x)=f(-x).f'(x)$

$f(x).f'(-x)-f(-x).f'(x)=0$

$f(x).f(-x)=c$

$f(0)=3$

$f(0).f(0)=9$

$c=9$

$f(2015).f(-2015)=9$

$f\left( x \right) f^{ ' }\left( -x \right) =f\left( -x \right) f^{ ' }\left( x \right) \\ \frac { f\left( x \right) }{ f\left( -x \right) } =\frac { f^{ ' }\left( x \right) }{ f^{ ' }\left( -x \right) }$ On comparing, we get $f^{ ' }\left( x \right) =kf\left( x \right)$ Let $f\left( x \right)=y$ . So, $ky=y'\\ ky=\frac { dy }{ dx } \\ \int {k dx } =\int { \frac { dy }{ y } } \\ \ln { y } =kx+kC\\ y={ e }^{ kx+kC }\\ y=A{ e }^{ kx }$ $\therefore f\left( x \right)=A{ e }^{ kx }$ Now, f(0)=3 so, A=3 . Hence, $f\left( x \right) =3{ e }^{k x }$ So, $f\left( 2015 \right) f\left( -2015 \right) =3{ e }^{ 2015k }\times 3{ e }^{ -2015k }=9$