If the solution curve of the differential equation

$$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

If $${{dy} \over {dx}} + {{{2^{x - y}}({2^y} - 1)} \over {{2^x} - 1}} = 0$$, x, y > 0, y(1) = 1, then y(2) is equal to :

If $$y = y(x)$$ is the solution of the differential equation

$$x{{dy} \over {dx}} + 2y = x\,{e^x}$$, $$y(1) = 0$$ then the local maximum value

of the function $$z(x) = {x^2}y(x) - {e^x},\,x \in R$$ is :