Let complex numbers $$\alpha \,and\,{1 \over {\overline \alpha }}\,$$ lie on circles $${\left( {x - {x_0}} \right)^2} + \,\,{\left( {y - {y_0}} \right)^2} = {r^2}$$ and $$\,{\left( {x - {x_0}} \right)^2} + \,\,{\left( {y - {y_0}} \right)^2} = 4{r^2}$$ respextively. If $${z_0} = {x_0} + i{y_0}$$ satisfies the equation $$2{\left| {{z_0}} \right|^2}\, = {r^2} + 2,\,then\,\left| a \right| = $$

Let $$f\left( x \right) = x\sin \,\pi x,\,x > 0.$$ Then for all natural numbers $$n,\,f'\left( x \right)$$ vanishes at

The coefficient of three consecutive terms of $${\left( {1 + x} \right)^{n + 5}}$$ are in the ratio $$5:10:14.$$ Then $$n$$ =

Consider the set of eight vectors $$V = \left\{ {a\,\hat i + b\,\hat j + c\hat k:a,\,b,\,c\, \in \left\{ { - 1,\,1} \right\}} \right\}$$. Three non-coplanar vectors can be chosen from v in $${2^p}$$ ways. Then p is