Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}$ intersect at two distinct points, is $\mathrm{R}-[\mathrm{a}, \mathrm{b}]$, then the point $(8 \mathrm{a}+12,16 \mathrm{~b}-20)$ lies on the curve :

If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x)$, then $y$ is strictly increasing in :

If the shortest distance between the lines

$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1 , then the sum of all possible values of $\lambda$ is :

If $x=x(t)$ is the solution of the differential equation $(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2$, then, $x(1)$ equals _________.