If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to :

Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$}

and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then

The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to :

For x $$ \in $$ R, x $$ \ne $$ 0, if y(x) is a differentiable function such that

x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt,  then y (x) equals :

(where C is a constant.)