Let $x=x(y)$ be the solution of the differential equation $y^2 \mathrm{~d} x+\left(x-\frac{1}{y}\right) \mathrm{d} y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

Let $f(x)$ be a real differentiable function such that $f(0)=1$ and $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ for all $x, y \in \mathbf{R}$. Then $\sum_\limits{n=1}^{100} \log _e f(n)$ is equal to :

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