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 :

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

Let $${J_{n,m}} = \int\limits_0^{{1 \over 2}} {{{{x^n}} \over {{x^m} - 1}}dx} $$, $$\forall$$ n > m and n, m $$\in$$ N. Consider a matrix $$A = {[{a_{ij}}]_{3 \times 3}}$$ where $${a_{ij}} = \left\{ {\matrix{ {{j_{6 + i,3}} - {j_{i + 3,3}},} & {i \le j} \cr {0,} & {i > j} \cr } } \right.$$. Then $$\left| {adj{A^{ - 1}}} \right|$$ is :

The area, enclosed by the curves $$y = \sin x + \cos x$$ and $$y = \left| {\cos x - \sin x} \right|$$ and the lines $$x = 0,x = {\pi \over 2}$$, is :