Let y = y(x) be the solution of the differential equation,
xy'- y = x²(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to :

Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to :

Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,
$$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$$, then a² + b² is equal to :

The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to
(where C is a constant of integration):