If f(x + y) = f(x).f(y) $$\forall $$ x, y and f(5) = 2, f'(0) = 3, then
f'(5) is

A triangle with vertices $$\left( {4,0} \right),\left( { - 1, - 1} \right),\left( {3,5} \right)$$ is :

Locus of mid point of the portion between the axes of

$$x$$ $$cos$$ $$\alpha + y\,\sin \alpha = p$$ where $$p$$ is constant is :

If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is