The number of points in $$\left( { - \infty \,\infty } \right),$$ for which $${x^2} - x\sin x - \cos x = 0,$$ is

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

Let $${S_n} = {\sum\limits_{k = 1}^{4n} {\left( { - 1} \right)} ^{{{k\left( {k + 1} \right)} \over 2}}}{k^2}.$$ Then $${S_n}$$can take value(s)