A random variable X has the following probability distribution :

X:	1	2	3	4	5
P(X):	K2	2K	K	2K	5K2

Then P(X > 2) is equal to :

Let a, b $$ \in $$ R, a $$ \ne $$ 0 be such that the equation, ax² – 2bx + 5 = 0 has a repeated root $$\alpha $$, which is also a root of the equation, x² – 2bx – 10 = 0. If $$\beta $$ is the other root of this equation, then $$\alpha $$² + $$\beta $$² is equal to :

Given : $$f(x) = \left\{ {\matrix{ {x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr } } \right.$$

and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3 $$, is :

If $$x = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$ and $$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$

for 0 < $$\theta $$ < $${\pi \over 4}$$, then :