If $$\int\limits_0^2 {\left( {\sqrt {2x} - \sqrt {2x - {x^2}} } \right)dx = \int\limits_0^1 {\left( {1 - \sqrt {1 - {y^2}} - {{{y^2}} \over 2}} \right)dy + \int\limits_1^2 {\left( {2 - {{{y^2}} \over 2}} \right)dy + I} } } $$, then I equals
If y = y(x) is the solution of the differential equation $$\left( {1 + {e^{2x}}} \right){{dy} \over {dx}} + 2\left( {1 + {y^2}} \right){e^x} = 0$$ and y (0) = 0, then $$6\left( {y'(0) + {{\left( {y\left( {{{\log }_e}\sqrt 3 } \right)} \right)}^2}} \right)$$ is equal to
Let P : y2 = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of $${\pi \over 4}$$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is :
Let a triangle ABC be inscribed in the circle $${x^2} - \sqrt 2 (x + y) + {y^2} = 0$$ such that $$\angle BAC = {\pi \over 2}$$. If the length of side AB is $$\sqrt 2 $$, then the area of the $$\Delta$$ABC is equal to :