Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{R}$. If $f^{\prime}(0)=4 \mathrm{a}$ and $f$ satisfies $f^{\prime \prime}(x)-3 \mathrm{a} f^{\prime}(x)-f(x)=0, \mathrm{a}>0$, then the area of the region $\mathrm{R}=\{(x, y) \mid 0 \leq y \leq f(a x), 0 \leq x \leq 2\}$ is :

Let $$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$$. Then at $$x=2, y^{\prime \prime}+y+1$$ is equal to

The solution of the differential equation $$(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$$, is :

The solution curve, of the differential equation $$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \frac{\mathrm{d} y}{\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :