If the semi-circular contour D of radius 2 is as shown in the figure, then the value of the integral $$\oint\limits_D {{1 \over {{s^2} - 1}}} ds$$ is

The equation $${x^3} - {x^2} + 4x - 4 = 0\,\,$$ is to be solved using the Newton - Raphson method. If $$x=2$$ taken as the initial approximation of the solution then the next approximation using this method, will be

The value of $$\oint\limits_C {{1 \over {\left( {1 + {z^2}} \right)}}} dz$$ where C is the contour $$\,\left| {z - {i \over 2}} \right| = 1$$ is

The solution of the differential equation $${k^2}{{{d^2}y} \over {d\,{x^2}}} = y - {y_2}\,\,$$ under the boundary conditions (i) $$y = {y_1}$$ at $$x=0$$ and (ii) $$y = {y_2}$$ at $$x = \propto $$ where $$k$$, $${y_1}$$ and $${y_2}$$ are constant is