Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

If y = y(x) is the solution curve of the differential equation $${x^2}dy + \left( {y - {1 \over x}} \right)dx = 0$$ ; x > 0 and y(1) = 1, then $$y\left( {{1 \over 2}} \right)$$ is equal to :

If n is the number of solutions of the equation
$$2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1,x \in [0,\pi ]$$ and S is the sum of all these solutions, then the ordered pair (n, S) is :

The function $$f(x) = {x^3} - 6{x^2} + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements :

Statement 1 : there exists x₁, x₂ $$\in$$(2, 4), x₁ < x₂, such that f'(x₁) = $$-$$1 and f'(x₂) = 0.

Statement 2 : there exists x₃, x₄ $$\in$$ (2, 4), x₃ < x₄, such that f is decreasing in (2, x₄), increasing in (x₄, 4) and $$2f'({x_3}) = \sqrt 3 f({x_4})$$.

Then