Let $$f(x) = \left| {\matrix{ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

The area enclosed by the closed curve $$\mathrm{C}$$ given by the differential equation

$$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$$ is $$4 \pi$$.

Let $$P$$ and $$Q$$ be the points of intersection of the curve $$\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :

If the orthocentre of the triangle, whose vertices are (1, 2), (2, 3) and (3, 1) is $$(\alpha,\beta)$$, then the quadratic equation whose roots are $$\alpha+4\beta$$ and $$4\alpha+\beta$$, is :

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

$$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to