Suppose $$\theta \in\left[0, \frac{\pi}{4}\right]$$ is a solution of $$4 \cos \theta-3 \sin \theta=1$$. Then $$\cos \theta$$ is equal to :

For the function

$$f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right],$$

consider the following two statements :

(I) $$f$$ is increasing in $$\left(0, \frac{\pi}{2}\right)$$.

(II) $$f^{\prime}$$ is decreasing in $$\left(0, \frac{\pi}{2}\right)$$.

Between the above two statements,

If $$\mathrm{A}(1,-1,2), \mathrm{B}(5,7,-6), \mathrm{C}(3,4,-10)$$ and $$\mathrm{D}(-1,-4,-2)$$ are the vertices of a quadrilateral ABCD, then its area is :

If $$y=y(x)$$ is the solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$$, then $$y\left(\frac{\pi}{8}\right)$$ is equal to :